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Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 1 Numbers Ex 1.4 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 1 Numbers Ex 1.4

Question 1.
Fill in the blanks:
(i) The ones digit in the square of 77 is ________ .
Answer:
9

(ii) The number of non-square numbers between 242 and 252 is ________ .
Answer:
48

(iii) The number of perfect square numbers between 300 and 500 is ________ .
Answer:
5

(iv) If a number has 5 or 6 digits in it, then its square root will have ________ digits.
Answer:
3

(v) The value of Jii lies between integers ______ and ________ .
Answer:
13, 14

Question 2.
Say True or False:
(i) When a square number ends in 6, its square root will have 6 in the unit’s place.
Answer:
True

(ii) A square number will not have odd number of zeros at the end.
Answer:
True

(iii) The number of zeros in the square of 91000 is 9.
Answer:
False

(iv) The square of 75 is 4925.
Answer:
False

(v) The square root of 225 is 15.
Answer:
True

Question 3.
Find the square of the following numbers.
(i) 17
(ii) 203
(iii) 1098
Answer:
(i) 17

(ii) 203

(iii) 1098

Question 4.
Examine if each of the following is a perfect square.
(i) 725
(ii) 190
(iii) 841
(iv) 1089
Answer:
(i) 725
725 = 5 × 5 × 29 = 5² × 29
Here the second prime factor 29 does not have a pair.
Hence 725 is not a perfect square number.

(ii) 190
190 = 2 × 5 × 19
Here the factors 2, 5 and 9 does not have pairs.
Hence 190 is not a perfect square number.

(iii) 841
841 = 29 × 29
Hence 841 is a perfect square

(vi) 1089
1089 = 3 × 3 × 11 × 11 = 33 × 33
Hence 1089 is a perfect square

The factors of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144.

Question 5.
Find the square root by prime factorisation method.
(i) 144
(ii) 256
(iii) 784
(iv) 1156
(v) 4761
(vi) 9025
Answer:
(i) 144
144 = 2 × 2 × 2 × 2 × 3 × 3
√144 = 2 × 2 × 3 = 12

(ii) 256
256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
√256 = 2 × 2 × 2 × 2 = 16

(iii) 784
784 = 2 × 2 × 2 × 2 × 7 × 7
√784 = 2 × 2 × 2 × 2 × 7 × 7 = 28

(iv) 1156
1156 = 2 × 2 × 17 × 17
1156 = 2² × 17²
1156 = (2 × 17)²
∴ \(\sqrt{1156}\) = \(\sqrt{(2 \times 17)^{2}}\) = 2 × 17 = 34
∴ \(\sqrt{1156}\) = 34

(v) 4761
4761 = 3 × 3 × 23 × 23
4761 = 3² × 23²
4761 = (3 × 23)²
√4761 = \(\sqrt{(3 \times 23)^{2}}\)
√4761 = 3 × 23
√4761 = 69

(vi) 9025
9025 = 5 × 5 × 19 × 19
9025 = 5² × 19²
9025 = (5 × 19)²
√925 = \(\sqrt{(5 \times 19)^{2}}\) = 5 × 19 = 95

Question 6.
Find the square root by long division method.
(i) 1764
(ii) 6889
(iii) 11025
(iv) 17956
(v) 418609
Answer:
(i) 1764

√1764 = 42

(ii) 6889

√6889 = 83

(iii) 11025

√11025 = 105

(iv) 17956

√17956 = 134

(v) 418609

√418609 = 647

Roots Calculator is a free online tool that displays the roots of the given quadratic equation.

Question 7.
Estimate the value of the following square roots to the nearest whole number:
(i) √440
(ii) √800
(iii) √1020
Answer:
(i) √440
we have 20² = 400
21²= 441
∴ √440 ≃ 21

(ii) √800
we have 28² = 784
29²= 841
∴ √800 ≃ 28

(iii) √1020
we have 31² = 961
32²= 1024
∴ √1020 ≃ 32

Question 8.
Find the square root of the following decimal numbers and fractions.
(i) 2.89
(ii) 67.24
(iii) 2.0164
(iv) \(\frac{144}{225}\)
(v) \(7 \frac{18}{49}\)
Answer:
(i) 2.89

√2.89 = 1.7

(ii) 67.24

√67.24 = 8.2

(iii) 2.0164

√2.0164 = 1.42

(iv) \(\frac{144}{225}\)

(v) \(7 \frac{18}{49}\)

\(\sqrt{7 \frac{18}{49}}=2 \frac{5}{7}\)

Question 9.
Find the least number that must be subtracted to 6666 so that it becomes a perfect square. Also, find the square root of the perfect square thus obtained.
Let us work out the process of finding the square root of 6666 by long division method.

The remainder in the last step is 105. Is if 105 be subtracted from the given number the
remainder will be zero and the new number will be a perfect square.
∴ The required number is 105. The square number is 6666 – 105 = 6561.

Question 10.
Find the least number by which 1800 should be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square thus obtained.
Answer:
We find 1800 = 2 × 2 × 3 × 3 × 5 × 5 × 2
= 2² × 3² × 5² × 2
Here the last factor 2 has no pair. So if we multiply 1800 by 2, then the number becomes a perfect square.

∴ 1800 × 2 = 3600 is the required perfect square number.
∴ 3600 = 1800 × 2
3600 = 2² × 3² × 5² × 2 × 2
3600 = 2² × 3² × 5² × 2²
= (2 × 3 × 5 × 2)²
\(\sqrt{3600}=\sqrt{(2 \times 3 \times 5 \times 2)^{2}}\)
= 2 × 3 × 5 × 2 = 60
∴ √3600 = 60

Objective Type Questions

Question 11.
The square of 43 ends with the digit .
(A) 9
(B) 6
(C) 4
(D) 3
Answer:
(A) 9
Hint:
Ones digit = 3 × 3 = 9

Question 12.
_______ is added to 24² to get 25².
(A) 4²
(B) 5²
(C) 6²
(D) 7²
Answer:
(D) 7²
Hint:
25² = 25 × 25 = 625
24² = 24 × 24 = 576

Question 13.
√48 is approximately equal to .
(A) 5
(B) 6
(C) 7
(D) 8
Answer:
(C) 7
Hint:
√49 = 7

Question 14.
\(\sqrt{128}-\sqrt{98}+\sqrt{18}\)
(A) √2
(B) √8
(C) √48
(D) √32
Answer:
(D) √32

Question 15.
The number of digits in the square root of 123454321 is ______.
(A) 4
(B) 5
(C) 6
(D) 7
Answer:
(B) 5
Hint:
\(\frac{n+1}{2}=\frac{10}{2}=5\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.6 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.6

To find the value of x, bring the variable to the left side and bring all the remaining values to the right side. Simplify the values to find the result.

Question 1.
Fill in the blanks:
(i) The value of x in the equation x + 5 = 12 is ________ .
Answer:
7
Hint:
Given, x + 5 = 12
x = 12 – 5 = 7 (by transposition method)
Value of x is 7

(ii) The value of y in the equation y – 9 = (-5) + 7 is ________ .
Answer:
11
Hint:
Given, y – 9 = (-5) + 7
y – 9 = 7 – 5 (re-arranging)
y – 9 = 2
∴ y = 2 + 9 = 11 (by transposition method)

(iii) The value of m in the equation 8m = 56 is ________ .
Answer:
7
Hint:
Given, 8m = 56
Divided by 8 on both sides
\(\frac{8 \times m}{8}=\frac{56}{8}\)
∴ m = 7

(iv) The value of p in the equation \(\frac{2 p}{3}\) = 10 is ________ .
Answer:
15
Hint:
Given, \(\frac{2 p}{3}\) = 10
Multiplying by 3 on both sides

Dividing by 2 on both sides

∴ p = 15

(v) The linear equation in one variable has ________ solution.
Answer:
one

Question 2.
Say True or False.
(i) The shifting of a number from one side of an equation to other is called transposition.
Answer:
True

(ii) Linear equation in one variable has only one variable with power 2.
Answer:
False
[Linear equation in one variable has only one variable with power one – correct statement]

Question 3.
Match the following

(A) (i),(ii), (iv) ,(iii),(v)
(B) (iii), (iv), (i) ,(ii), (v)
(C) (iii),(i) ,(iv), (v), (ii)
(D) (iii) , (i) , (v) ,(iv) ,(ii)
Answer:
(C) (iii),(i) ,(iv), (v), (ii)

a. \(\frac{x}{2}\) = 10, multiplying by 2 on both sides, we get
\(\frac{x}{2}\) × 2 = 10 × 2 ⇒ x = 20

b. 20 = 6x – 4 by transposition ⇒ 20 + 4 =6x
6x = 24 dividing by 6 on both sides,
\(\frac{6 x}{6}=\frac{24}{6}\) ⇒ x = 4

c. 2x – 5 = 3 – x
By transposing the variable ‘x’, we get
2x – 5 + x = 3
by transposing – 5 to other side,
2x + x = 3 + 5

∴ x = \(\frac{8}{3}\)

d. 7x – 4 – 8x = 20
by transposing – 4 to other side,
7x – 8x = 20 + 4
– x = 24
∴ x = – 24

e. \(\frac{4}{11}-x=\frac{-7}{11}\)
Transposing \(\frac{4}{11}\) to other side,
– x = \(\frac{-7}{11} \frac{-4}{11}=\frac{-7-4}{11}=\frac{-11}{11}\) = – 1
∴ – x = – 1 ⇒ x = 1

Question 4.
Find x:
(i) \(\frac{2 x}{3}-4=\frac{10}{3}\)
Answer:
Transposing – 4 to other side, it becomes + 4

(ii) \(y+\frac{1}{6}-3 y=\frac{2}{3}\)
Answer:
Transposing \(\) to the other side,

(iii) \(\frac{1}{3}-\frac{x}{3}=\frac{7 x}{12}+\frac{5}{4}\)
Answer:
Transposing \(\frac{-x}{3}\) to the other side, it becomes + \(\frac{x}{3}\)
∴ \(\frac{1}{3}=\frac{7 x}{12}+\frac{5}{4}+\frac{x}{3}\)
Transposing \(\frac{5}{4}\) to the other side, it becomes \(\frac{-5}{4}\)
\(\frac{1}{3}-\frac{5}{4}=\frac{7 x}{12}+\frac{x}{3}\)
Multiply by 12 throughout [we look at the denominators 3, 4, 12, 3 and take the LCM, which is 12]

4 – 15 = 7x + x × 4
– 11 = 7x + 4x
11x = – 11
x = -1

Question 5.
Find x
(i) -3(4x + 9) = 21
Answer:

Expanding the bracket,
-3 × 4 + (-3) × 9 = 21
∴ -12x + (-27) = 21
– 12x – 27 = 21
Transposing – 27 to other side, it becomes +27
– 12 x = 21 + 27 = 48
∴ – 12x = 48 ⇒ 12x = – 48
Dividing by 12 on both sides

(ii) 20 – 2 (5 – p) = 8
Answer:

Expanding the bracket,
20 – 2 × 5 – 2 × (-p) = 8
20 – 10 + 2p = 8
(- 2 × – p = 2p)
10 + 2p = 8 transposing lo to other side
2p = 8 – 10 = – 2
∴ 2p = – 2
∴ p = – 1

(iii) (7x – 5) – 4(2 + 5x) = 10(2 – x)
Answer:

Expanding the brackets,
7x – 5 – 4 × 2 – 4 × 5x = 10 × 2 + 10 × (-x)
7x – 5 – 8 – 20x = 20 – 10x
7x – 13 – 20x = 20 – 10x
Transposing 10x & – 13, we get
7x – 13 – 20x + 10x = 20
7x – 20x + 10x = 20 + 13, Simplifying,
– 3x = 33
3x = – 33
x = \(\frac{-33}{3}\) = – 11
x = – 11

Question 6.
Find x and m:
(i) \(\frac{3 x-2}{4}-\frac{(x-3)}{5}=-1\)
Answer:

(ii) \(\frac{m+9}{3 m+15}=\frac{5}{3}\)
Answer:
Cross multiplying, we get

∴ (m + 9) × 3 = 5 × (3m + 15)
m × 3 + 9 × 3 = 5 × 3m + 5 × 15

27 – 75 = 15m – 3m
– 48 = 12m

⇒ m = – 4

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 7 Information processing Ex 7.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 7 Information processing Ex 7.2

Question 1.
Using repeated division method, find the HCF of the following:
(i) 455 and 26
Answer:

Step 1: The larger number should be dividend 455 & smaller number should be divisor = 26
Step 2: After 1st division, the remainder becomes new divisor & the previous divisor becomes next dividend.
Step 3: This is done till remainder is zero.
Step 4: The last divisor is the HCF L.
∴ Ans: HCF is 13.

(ii) 392 and 256
Answer:
256 is smaller, so it is the 1st divisor

∴ HCF = 8

(iii) 6765 and 610
Answer:

∴ HCF = 5

(iv) 184, 230 and 276
Answer:
First let us take 184 & 230

∴ 46 is the HCF of 184 and 230
Now the HCF of the first two numbers is the dividend for the third number.

∴ Ans: HCF of 184, 230 & 276 is 46

Question 2.
Using repeated subtraction method, find the HCF of the following:
(i) 42 and 70
Answer:
Let number be m & n m > n
We do, m – n & the result of subtraction becomes new ‘m’. if m becomes less than n,
we do n – m and then assign the result as n. We should do this till m n. When m = n then ‘m’ is the HCF.
42 and 70
m = 70 n = 42
70 – 42 = 28

now m = 42, n = 28
42 – 28 = 14.

now m = 28, n = 14
28 – 14 = 14.

now m = 14. n = 14;
we stop here as m = n
∴ HCF of 42 & 70 is 14

(ii) 36 and 80
Answer:

28 – 8 = 20
20 – 8 = 12
12 – 8 = 4
8 = 4 = 4
now m = n = 4
∴ HCF is 4

(iii) 280 and 420
Answer:
Let m = 420, n = 280
m – n = 420 – 280 = 140
now m = 280, n = 140
m – n = 280 – 140 = 140
now m = n = 140
∴ HCF is 140

(iv) 1014 and 654
Answer:
Let m = 1014, n = 654
m – n = 1014 – 654 = 360
now m = 654, n = 360
m – n = 654 – 360 = 294
now m = 360, n = 294
m – n = 360 – 294 = 66
now m = 294 n = 66
m – n = 294 – 66 = 228
now m = 66, n = 228
n – m = 228 – 66 = 162
now m = 162, n = 66
= 162 – 66 = 96
n – m = 96 – 66 = 30
Similarly, 66 – 30 = 36
36 – 30 = 6
30 – 6 = 24
24 – 6 = 18
18 – 6 = 12
12 – 6 = 6 now m = n
∴ HCF of 1014 and 654 is 6

Question 3.
Do the given problems by repeated subtraction method and verify the result.
(i) 56 and 12

Answer:
Let n = 56 & n = 12
m – n = 56 – 12 = 44
now m = 44, n = 12
m – n = 44 – 12 = 32
m – n = 32 – 12 = 20
m – n = 20 – 12 = 8
n – m = 12 – 84
m – n = 8 – 4 = 4. now m = n
∴ HCF of 56 & 12 is 4

(ii) 320, 120 and 95

Answer:
Let us take 320 & 120 first m = 320, n = 120
m – n = 320 – 120 = 200
m = 200, n = 120
∴ m – n = 200 – 120 = 80
120 – 80 = 40
80 – 40 = 40
∴ m = n = 40 → HCF of 320, 120
Now let us find HCF of 40 & 95
m = 95, n = 40
∴ m – n = 95 – 40 = 55
55 – 40 = 15
40 – 15 = 25
25 – 15 = 10
15 – 10 = 5
HCF of 40 & 95 is 5
10 – 5 = 5
∴ HCF of 320 120 & 95 is 5

Question 4.
Kalai wants to cut identical squares as big as she can, from a piece of paper measuring 168mm and by 196mm. What is the length of the side of the biggest square? (To find HCF using repeated subtraction method)
Answer:
Sides are 168 & 196
To find HCF of 168 & 196, we are to use repeated subtraction method.
∴ m = 196, n = 168
m – n = 196 – 168 = 28

now n = 28, m = 168
m – n = 168 – 28 = 140

now m = 140, n = 28
m – n = 140 – 28 = 112

now m = 112, n = 28
m – n = 112 – 28 = 84

now m = 84, n = 28
m – n = 84 – 28 = 56

now m = 56, n = 28
m – n = 56 – 28 = 28
∴ HCF is 28
∴ Length of biggest square is 28

Objective Type Questions

Question 5.
What is the eleventh Fibonacci number?
(a) 55
(b) 77
(c) 89
(d) 144
Answer:
(c) 89
Hint:

∴ 11^th Fibonacci number is 89

Question 6.
If F(n) is a Fibonacci number and n = 8, which of the following is true?
(a) F(8) = F(9) + F( 10)
(b) F(8) = F(7) + F(6)
(c) F(8) = F(10) × F(9)
(d) F(8) = F(7) – F(6)
Answer:
(b) F(8) = F(7) + F(6)
Hint:
Given F(n) is a Fibonacci number & n = 8
∴ F(8) = F(7) + F (6) as any term in Fibonacci series is the sum of preceding 2 terms

Question 7.
Every 3^rd number of the Fibonacci sequence is a multiple of_______
(a) 2
(b) 3
(c) 5
(d) 8
Answer:
(a) 2
Hint:
Every 3^rd number in Fibonacci sequence is a multiple of 2

Question 8.
Every _____ number of the Fibonacci sequence is a multiple of 8
(a) 2^th
(b) 4^th
(c) 6^th
(d) 8^th
Answer:
(c) 6^th

Question 9.
The difference between the 18^th and 17^th Fibonacci number is
(a) 233
(b) 377
(c) 610
(d) 987
Answer:
(d) 987
Hint:
F(18) = F(17) + F(16)
F(18) – F(17) = F(16) = F(15) + F(14)
= 610 + 377 = 987

Factors of 70 are the list of integers that we can split evenly into 70.

Question 10.
Common prime factors of 30 and 250 are
(a) 2 × 5
(b) 3 × 5
(c) 2 × 3 × 5
(d) 5 × 5
Answer:
(a) 2 × 5
Prime factors of 30 are 2 × 3 × 5
Prime factors of 250 are 5 × 5 × 5 × 2
∴ Common prime factors are 2 × 5

Question 11.
Common prime factors of 36,60 and 72 are
(a) 2 × 2
(b) 2 × 3
(c) 3 × 3
(d) 3 × 2 × 2
Answer:
(d) 3 × 2 × 2
Hint:
Prime factors of 36 are 2 × 2 × 3 × 3
Prime factors of 60 are 2 × 2 × 3 × 5
Prime factors of 72 are 2 × 2 × 2 × 3 × 3
∴ Common prime factors are 2 × 2 × 3

Question 12.
Two numbers are said to be co-prime numbers if their HCF is
(a) 2
(b) 3
(c) 0
(d) 1
Answer:
(d) 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Physics Guide Pdf Chapter 2 Kinematics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

11th Physics Guide Kinematics Book Back Questions and Answers

Part – I:
I. Multiple choice questions:

Question 1.
Which one of the following Cartesian coordinate systems is not followed in physics?

Answer:

Question 2.
Identify the unit vector in the following _______.

Answer:

Question 3.
Which one of the following quantities cannot be represented by a scalar?
a) Mass
b) length
c) momentum
d) magnitude of the acceleration
Answer:
c) momentum

Question 4.
Two objects of masses m₁ and m₂ fall from the heights h₁ and h₂ respectively. The ratio of the magnitude of their momenta when they hit the ground is _______. (AIPMT 2012)

Answer:

Question 5.
If a particle has negative velocity and negative acceleration, it speeds _______.
a) increases
b) decreases
c) remains the same
d) zero
Answer:
a) increases

Question 6.
If the velocity is \(\overline{V}\) = \(2 \hat{i}+t^{2} \hat{j}-9 \hat{k}\) then the magnitude of acceleration at t = 0.5s is _______.
a) 1ms^-2
b) 2 ms^-2
c) zero
d) -1ms^-2
Answer:
a) 1ms^-2

Question 7.
If an object is dropped from the top of a building and it reaches the ground at t = 4s, then the height of the building is (ignoring air resistance) (g = 9.8ms^-2)
a) 77.3m
b) 78.4m
c) 80.5
d) 79.2m
Answer:
b) 78.4m

Question 8.
A ball is projected vertically upwards with a velocity v. It comes back to the ground in time t. Which v-t graph shows the motion correctly? (NSEP 00-01)

Answer:

Question 9.
If one object is dropped vertically downward and another object is thrown horizontally from the same height, then the ratio of vertical distance covered by both objects at any instant t is _______.
a) 1
b) 2
c) 4
d) 0.5
Answer:
a) 1

Question 10.
A bail is dropped from some height towards the ground. Which one of the following represents the correct motion of the ball?

Answer:

Question 11.
If a particle executes uniform circular motion in the XY plane in a clockwise direction, then the angular velocity is in _______.
a) +y direction
b) +z direction
c) -z direction
d) -x direction
Answer:
c) -z direction

Question 12.
If a particle executes uniform circular motion, choose the correct statement _______. (NEET 2016)
a) The velocity and speed are constant.
b) The acceleration and speed are constant.
c) The velocity and acceleration are constant.
d) The speed and magnitude of acceleration are constant
Answer:
d) The speed and magnitude of acceleration are constant

Question 13.
If an object is thrown vertically up with the initial speed u from the ground, then the time taken by the object to return back to the ground is _______.
(a) \(\frac{u^{2}}{2 g}\)
(b) \(\frac{u^{2}}{g}\)
(c) \(\frac { u }{ 2g }\)
(d) \(\frac { 2u }{ g }\)
Answer:
(d) \(\frac { 2u }{ g }\)

Question 14.
Two objects are projected at angles 30° and 60° respectively with respect to the horizontal direction. The range of two objects are denoted as R_30° and R_60° Choose the correct relation from the following
a) R_30° = R_60°
b) R_30° = 4R_60°
c) R_30° = R_{\(\frac { 60° }{ 2 }\)}
d) R_30° = 2R_60°
Answer:
a) R_30° = R_60°

Question 15.
An object is dropped in an unknown planet from a height of 50m, it reaches the ground in 2s. The acceleration due to gravity in this unknown planet is _______.
a) g = 20ms^-2
b) g = 25ms^-2
c) g = 15ms^-2
d) g = 30ms^-2
Answer:
b) g = 25ms^-2

II. Short Answer Questions:

Question 1.
Explain what is meant by the Cartesian coordinate system?
Answer:
At any given instant of time, the frame of reference with respect to which the position of the object is described in terms of position coordinates (x,y,z) is called “Cartesian coordinate system”.

If x, y, and z axes are drawn in an anticlockwise direction, then the coordinate system is called a right-handed Cartesian coordinate system.

Question 2.
Define a vector. Give Example.
Answer:
Vector is a quantity which is described by both magnitude and direction. Geometrically a vector is a directed line segment.
Example – force, velocity, displacement.

Question 3.
Define a Scalar. Give Examples.
Answer:
Scalar is a property of a physical quantity which can be described only by magnitude.
Example: Distance, Mass, Temperature, Speed, Energy, etc.

Question 4.
Write short note on the scalar product between two vectors.
Answer:
The Scalar product of two vectors (dot product) is defined as the product of the magnitudes of both the vectors and the cosine of angle between them.
If \(\vec{A}\) and \(\vec{B}\) are two vectors having an angle θ between them, then their scalar or dot product is

Example: W = \(\vec{F}\).\(\vec{dr}\). Work done is a scalar product of force \(\vec{F}\) and \(\vec{r}\)

Question 5.
Write a Short note on vector product between two vectors.
Answer:
The vector product or cross product of two vectors is defined as another vector having a magnitude equal to the product of the magnitudes of two vectors and the sine of the angle between them. The direction of the product vector is perpendicular to the plane containing the two vectors, in accordance with the right hand screw rule or right hand thumb rule. Thus, if\(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are two vectors, then their vector product is written as \(\overrightarrow{\mathrm{A}}\) × \(\overrightarrow{\mathrm{B}}\) which is a vector C defined by \(\overrightarrow{\mathrm{c}}\) = \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) = (AB sin 0) \(\hat{n}\)
The direction \(\hat{n}\) of \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) , i.e., \(\overrightarrow{\mathrm{c}}\) is perpendicular to the plane containing the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\).

Question 6.
How do you deduce that two vectors are perpendicular?
Answer:
If the vector product of the two given vectors is having maximum magnitude.
i.e sinθ = 90°, [ (\(\vec{A}\) x \(\vec{B}\))Max = AB\(\hat{n}\) ] then the two vectors are said to be perpendicular.

In this displacement calculator, we will show you how to find displacement in a matter of seconds.

Question 7.
Define Displacement and distance.
Answer:
Distance is the actual path length traveled by an object in the given interval of time during the motion. It is a positive scalar quantity. Displacement is the difference between the final and initial positions of the object in a given interval of time. It can also be defined as the shortest distance between these two positions of the object. It is a vector quantity.

Question 8.
Define velocity and speed.
Answer:
Velocity – Velocity is defined as the rate of change of position vector with respect to time (or) defined as the rate of change of displacement. It Is a vector quantity.

Speed – Speed is defined as the rate of change of distance. It is a scalar quantity.

Question 9.
Define acceleration.
Answer:
Acceleration is defined as the rate of change of velocity.
Acceleration \(\vec{a}\) = \(\frac{d \vec{v}}{d t}\)
Acceleration is a vector quantity.
Unit – ms^-2
Dimensional formula-[LT^-2]

Question 10.
What is the difference between velocity and average velocity?
Answer:

Question 11.
Define a radian.
Radian is defined as ratio of length of the arc to radians of the arc. One radian is the angle subtended at the center of the circle by an arc that is equal to in length to the radius of the circle.

Question 12.
Define angular displacement and angular velocity.
Answer:

1. Angular displacement: The angle described by the particle about the axis of rotation in a given time is called angular displacement.
2. Angular velocity: The rate of change of angular displacement is called angular velocity.

Question 13.
What is non-uniform circular motion?
Answer:
When an object is moving in a circular path with variable speed, it covers unequal distances in equal intervals of time. Then the motion of the object is said to be a non-uniform circular motion. Here both speed and direction during circular motion change.

Question 14.
Write down the kinematic equations for angular motion.
Answer:
The Kinematic equations for angular motion are ω = ω₀ + αt
θ = ω₀t + \(\frac { 1 }{ 2 }\)αt²
ω² = ω₀² + 2αθ
θ = \(\left(\frac{\omega_{0}+\omega}{2}\right)\) x t
ω₀ → initial angular velocity
ω → final angular velocity
α → angular acceleration
θ → angular displacement
t → time interval

Question 15.
Write down the expression for angle made by resultant acceleration and radius vector in the non-uniform circular motion.
Answer:
In the case of non-uniform circular motion, the particle will have both centripetal and tangential acceleration. The resultant acceleration is obtained as the vector sum of both centripetal and tangential acceleration.

This resultant acceleration makes an angle 6 with a radius vector, which is given by

III. Long Answer Questions:

Question 1.
Explain in detail the triangle law of addition.
Answer:
Let us consider two vector \(\vec{A}\) and \(\vec{B}\) as shown In fig.

Law: To find the resultant of two vectors, the triangular law of addition can be applied as follows.
A and B are represented as the two adjacent sides of a triangle taken in the same order. The resultant is given by the third side of the triangle taken in reverse order.

Magnitude of the resultant vector:
from figure
Let θ be the angle between two vectors.
from ∆ ABN, Sin θ = \(\frac { BN }{ AB }\) ⇒ ∴ BN = B sinθ
Cos θ = \(\frac { AN }{ AB }\) ⇒ ∴ AN = B Cos θ

Which is the magnitude of the resultant \(\vec{A}\) and \(\vec{B}\).

The direction of the resultant vector:
If \(\vec{R}\) makes an angle α with \(\vec{A}\) then

Question 2.
Discuss the properties of scalar and vector products.
Answer:
Properties of scalar product:
formula : \(\vec{A}\).\(\vec{B}\) = ABCosθ

1. The product quantity \(\overline{A}\).\(\overline{B}\) is always a scalar. It is positive if the angle between the vectors is acute (θ< 90°) and negative if angle between them is obtuse (90 < θ < 180)

2. The scalar product is commutative \(\overline{A}\).\(\overline{B}\) = \(\overline{B}\).\(\overline{A}\)

3. The scalar product obey distributive law. \(\overline{A}\).( \(\overline{B}\) + \(\overline{C}\) ) = \(\overline{A}\).\(\overline{B}\) + \(\overline{A}\).\(\overline{C}\)

4. The angle between the vector is θ = Cos^-1\(\frac{\bar{A} \cdot \bar{B}}{A B}\)

5. The scalar product of two vectors will be maximum when cos θ = 1 i.e θ = 0 ie when they are parallel.
[ ( \(\overline{A}\).\(\overline{B}\) ) max = AB.]

6. The scalar product of two vectors will be minimum when cos θ = -1 ie θ = 180°
( \(\overline{A}\).\(\overline{B}\))_mm = – AB [the vector are anti-parallel]

7. If two vector \(\overline{A}\) & \(\overline{B}\) are perpendicular to each other then \(\overline{A}\).\(\overline{B}\) = O. Because cos 90 = 0. Then vectors A & B are mutually orthogonal.

8. The scalar product of a vector with it self is termed as self or dot product and is given by
( \(\overline{A}\) )² = \(\overline{A}\).\(\overline{A}\) = AA cos θ = A²
Here 0=0
The magnitude or norm of the vector \(\overline{A}\) is
|A| = A = \(\sqrt{\bar{A} \cdot \bar{A}}\) = A.

9. Incase of orthogonal unit vectors
\(\hat{n}\).\(\hat{n}\) = 1 x 1cos0 = 1
for eg \(\hat{i}\).\(\hat{i}\) = \(\hat{j}\).\(\hat{j}\) = \(\hat{k}\).\(\hat{k}\) = 1

10. Incase of orthogonal unit vectors \(\hat{i}\), \(\hat{f}\), \(\hat{k}\) then \(\hat{i}\).\(\hat{j}\) = \(\hat{j}\).\(\hat{k}\) = \(\hat{k}\).\(\hat{j}\) = 1.1 cos 90 = 0.

11. In terms of components the scalar product of A and B can be written as

Properties of cross product:
Formula \(\vec{A}\) x \(\vec{B}\) = ABsinθ

1. The vector product of any two vectors is always an another vector whose direction perpendicular to the plane containing these two vectors, ie. Orthogonal to \(\overline{A}\) & \(\overline{B}\) even though \(\overline{A}\) & \(\overline{B}\) may not be mutually orthogonal.

2. Vector product is not commutative
\(\overline{A}\) x \(\overline{B}\) = – \(\overline{B}\).\(\overline{A}\)
\(\overline{A}\) x \(\overline{B}\) ≠ \(\vec{B}\) x \(\vec{A}\)
Here magnitude | \(\overline{A}\) x \(\overline{B}\) | = | \(\overline{B}\).\(\overline{A}\) | are equal but opposite direction.

3. The vector product of two vector is maximum when sine = 1, ie θ = 90°
ie. when \(\overline{A}\) and \(\overline{B}\) are orthogonal to each other.
( \(\overline{A}\) x \(\overline{B}\) ) max = AB \(\hat{n}\).

4. The vector product of two non zero vectors is minimum if |sinθ| = 0. ie. θ = 0 or 180°
( \(\overline{A}\) x \(\overline{B}\) ) m in = 0
Vector product of two non zero vectors is equal to zero if they either parallel or anti parallel

5. The self cross product ie product of a vector with itself is a null vector \(\overline{A}\) x \(\overline{B}\) = AA sinθ = 0

6. The self-vector product of the unit vector is zero
i.e. \(\hat{i}\).\(\hat{j}\) = \(\hat{j}\).\(\hat{j}\) = \(\hat{k}\).\(\hat{k}\) = 0

7. In case of orthogonal unit vectors \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) in accordance with right hand cork screw rule \(\hat{i}\).\(\hat{j}\) =\(\hat{k}\), \(\hat{i}\).\(\hat{k}\) = \(\hat{i}\), \(\hat{k}\).\(\hat{i}\) = \(\hat{j}\) also since cross product is not commutative

9. If two vectors \(\overline{A}\) & \(\overline{B}\) form adjacent sides of a parallelogram then the magnitude of |\(\overline{A}\) x \(\overline{B}\)| will give area 0f parallelogram.

10. Since one can divide a parallelogram into two equal triangles, the area of the triangle is \(\frac { 1 }{ 2 }\) |\(\overline{A}\) x \(\overline{B}\)|.

Question 3.
Derive the kinematic equations of motion for constant acceleration.
Answer:
Consider an object moving in a straight line with uniform or constant acceleration ‘a’. Let u be the initial velocity at t = 0, and v be the final velocity after a time of t seconds

(i) Velocity time relation:
The acceleration of the body at any instant is given by first derivative of the velocity with time
a = \(\frac { dv }{ dt }\)
dv = adt
integrating both sides

Displacement time relation:

(ii) The velocity of the body is given by the first derivative of the displacement with respect to time
But v = ds/dt
∴ dv = v dt
v = u + at
ds = (u + at)dt
ds = udt + atdt
Integrating both sides

Velocity-displacement relation:

also we can derive from the relation v = u + at
v – u = at
Substituting in equation s = ut + \(\frac { 1 }{ 2 }\)at²

Question 4.
Derive the equations of motion for a particle (a) falling vertically (b) projected vertically.
Answer:
For a body falling vertically from a height ‘h’:
Consider an object of mass m falling from height h.
Neglecting air resistance, the downward direction as the positive y-axis.
The object experiences acceleration ‘g’ due to gravity which is constant near the surface of the earth.
In kinematic equations of motion \(\vec{a}\) = g\(\hat{i}\)
By comparing the components a_x = 0, a_g = 0, a_y= g

Case – 1
If the particle is thrown with initial velocity ‘u’ downward then
v = u + gt
y = ut + 1/2gt²
v² – u² = 2gy

Case – 2
Suppose the particle starts from rest,
u = 0
v = gt
y = 1/2gt²
v² = 2gy
For a body projected vertically: Consider an object of mass m thrown vertically upwards with an initial velocity u. Ne-glect air friction. The vertical direction as positive y axis then the acceleration,
a = – g
The kinematic equation of motion are v = u – gt
v = u – gt
s = ut – 1/2 gt²

Projectile motion calculator solving for vertical velocity at time given initial vertical velocity, acceleration of gravity and time.

Question 5.
Derive the equations of motion, range, and maximum height reached by a particle thrown at an oblique angle θ with respect to the horizontal direction.
Answer:
Consider an object thrown with an initial velocity u at an angle θ with horizontal.

Then initial velocity is resolved into two components
u_x = u cos θ horizontally and
u_y = u sin θ vertically
At maximum height u_y = 0 (since acceleration due to gravity is opposite to the direction of the vertical component).
The Horizontal component of velocity
u_x = u cos θ remains constant throughout its motion.
hence after the time t the velocity along the horizontal motion
V_x = U_x + a_xt
= ux = cos θ
The horizontal distance travelled by the projectile in a time ‘t’ is S_x = u_xt + 1/2 a_xt².
Here S_x = x u_x = u cos θ
a_x = 0
∴ x = u cos θt ____ (1)
∴ t = \(\frac { x }{ u cos θ }\) ____ (2)
For vertical motion
V_y = u_y + a_yt
Here v_y = v_y
u_y = u sin θ
a_y = – g
v_y = u sin θ – gt
The vertical distance travelled by the projectile in the same time ‘t’ is
S_y = U_y t + a_y t²
S_y = y, U_y = u sin θ a_y = – g
y = u sin θ t – 1/2 gt² ____ (4)
Substituting the value of t in (4) we get equation:

Which indicates the path followed by the projectile is an inverted parabola.

Expression for Maximum height:
The maximum vertical distance travelled by the projectile during its motion is called maximum height.
We know that
v_y² = u_y² + 2a_ys
Here u_y = u sin 0, a_y = – g, s = h_max
v_y = 0

Expression for horizontal range:
The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range.
Horizontal range = Horizontal component of velocity x time of flight
R = u cos θ x t_f → (1)
Time of flight (t_f) is the time taken by the projectile from point of projection to point the projectile hits the ground again
w.k.t = S_y = u_y t_f + 1/2 a_y t²_f)
Here S_y = 0 u_y = u sin θ, a_y = – g
0 = u sin θ t_f – 1/2g t²_f
1/2 gt t²_f = u sin θ t_f

Question 6.
Derive the expression for centripetal acceleration.
Answer:
In uniform circular motion the velocity vector turns continuously with out changing its magnitude. ie speed remains constant and direction changes. Even though the velocity is tangential to every point is a circle, the acceleration it acting towards the centre of the circle along the radius. This is called centripetal acceleration

Expression:
The centripetal acceleration is derived from a simple geometrical relationship between position and velocity vectors. Let the directions of position and velocity vectors shift through same angle θ in a small time interval ∆t
For uniform circular motion r = \(\left|\bar{r}_{1}\right|\) = \(\left|\bar{r}_{2}\right|\)
and v = \(\left|\bar{v}_{1}\right|\) = \(\left|\bar{v}_{2}\right|\)
If the particle moves from position vector \(\bar{r}_{1}\) to \(\bar{r}_{2}\) the displacement is given by \(\overrightarrow{\Delta r}\) = \(\bar{r}_{2}\) – \(\bar{r}_{1}\)

and change in velocity from \(\bar{v}_{1}\) to \(\bar{v}_{2}\) is given ∆\(\bar { v }\) = \(\bar{v}_{2}\) – \(\bar{v}_{1}\)

The magnitudes of the displacement ∆r and ∆v satisfy the following relation

Here negative sign indicates that ∆v points radially inwards, towards the centre of the circle

For uniform circular motion v=rω where ω is the angular velocity of the particle about the center
The centripetal acceleration a = ω²r.

Question 7.
Derive the expression for total acceleration in the non-uniform circular motion.
Answer:
If the velocity changes both in speed and direction during circular motion, then we get non-uniform circular motion. Whenever the speed is not the same in a circular motion then the particle will have both centripetal and tangential acceleration.

The resultant acceleration is obtained by the vector sum of centripetal and tangential acceleration
Let the tangential acceleration be a_t.
Centripetal acceleration is v²/r.
The magnitude of the resultant acceleration is a_R = \(\sqrt{a_{t}^{2}+\left(\frac{v^{2}}{r}\right)^{2}}\)

IV. Exercises:

Question 1.
The position vector particle has a length of 1m and makes 30° with the x-axis what are the lengths of x and y components of the position vector?
Solution:

Question 2.
A particle has its position moved from \(\left|\bar{r}_{1}\right|\) = 3\(\hat{i}\) + 4\(\hat{j}\) to r\(\left|\bar{r}_{2}\right|\) = \(\hat{i}\)+ 2\(\hat{j}\) calculate the displacement vector ( ∆ \(\vec{r}\) ) and draw the \(\left|\bar{r}_{1}\right|\), \(\left|\bar{r}_{2}\right|\) and ( ∆ \(\vec{r}\) ) vector in a two dimensional Cartesian co-ordinate system.
Solution:

Question 3.
Calculate the average velocity of the particle whose position vector changes from \(\left|\bar{r}_{1}\right|\) = 5\(\hat{i}\) + 6\(\hat{j}\) to \(\left|\bar{r}_{2}\right|\) = 2\(\hat{i}\) + 3\(\hat{j}\) in a time 5 seconds.
Solution:

Question 4.
Convert the vector \(\overline{r}\) = 3\(\hat{i}\) + 2\(\hat{j}\) into a unit vector.
Solution:
A vector divided by its magnitude is a unit vector

Question 5.
What are the resultants of the vector product of two given vector given by
\(\overline{A}\) = 4\(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\) and \(\overline{B}\) = 5\(\hat{i}\) + 3\(\hat{j}\) – 4\(\hat{k}\)?
Solution:

Question 6.
An object at an angle such that the horizontal range is 4 times the maximum height. What is the angle of projection of the object?
Solution:
Incase of obliging projection

Question 7.
The following graphs represent velocity-time graph. Identify what kind of motion a particle undergoes in each graph.

Solution:
(a) When the body starts from rest and moves with uniform acceleration is constant

(b) This graph represents, for a body moving with a uniform velocity or constant velocity. The zero slope of curve indicates zero acceleration.

(c) This v-t graph is a straight line not passing through origin indicates the body has a constant acceleration but greater than fig(i) as slope is more than the first one (more steeper)

(d) Greater changes in velocity (velocity variations are taking place in equal as travels of time. The graph indicates increasing acceleration.

Question 8.
The following velocity-time graph represents a particle moving in the positive x-direction. Analyse its motion from o to 7s calculate the displacement covered and distance traveled by the particle from 0 to 2s.

Solution:
From o to A(o to Is):
At t = os the particle has a zero velocity at t > 0 the particle has a negative velocity and moves in positive x-direction the slope dr/dt is negative. The particle is decelerating. Thus the velocity decreases during this time interval.

From A to B (Is to 2s):
From time Is to 2s the velocity increase and slope dv/dt becomes positive. The particle is accelerating. The velocity increases in this time interval.

From B to C (2s to 5s):
From 2s to 5s the velocity stays constant at 1 m/s. The acceleration is zero.

From C to D (6s to 7s):
From 5s to 6s the velocity decreases. Slope dv/dt is negative. The particle is decelerating. The velocity decreases to zero. The body comes to rest at 6s.

From D to E (6s to 7s)
The particle is at rest during this time interval.

Displacement: in 0 – 2s:
The total area under the curve from 0 to 2s displacement = 1/2bh + 1/2bh
=1/2 x 1.5 x (- 2) + 0.5 x 1
= – 1.5 + 0.25
= – 1.25 m

Distance: is 0 – 2s
The distance covered is = 1.5 + 0.25 = 1.75 m

Question 9.
A particle is projected at an angle of θ with respect to the horizontal direction. Match the following for the above motion.
(a) v_x – decreases and increases
(b) V_y – remains constant
(c) Acceleration – varies
(d) Position vector – remains downwards
Solution:
(a) V_x – remains constant
(b) V_y – decreases and increases
(c) Acceleration (a) – remains downwards
(d) Position vector (r) – varies

Question 10.
A water fountain on the ground sprinkles water all around it. If the speed of the water coming out of the fountains is V. Calculate the total area around the fountain that gets wet.
Solution :
Speed of water = V
(Range)_max = radius = u₂/g = v²/g
This range becomes the radius = (v²/g) of the circle where water sprinkled.
Area covered = Area of circle
= πr² = π\(\left(\frac{v^{2}}{g}\right)\)²
= π \(v^{4} / g^{2}\)

Question 11.
The following table gives the range of the particle when thrown on different planets. All the particles are thrown at the same angle with the horizontal and with the same initial speed. Arrange the planets in ascending order according to their acceleration due to gravity (g value)

Solution:
R = – (sin 2θ)
∵ the initial velocity and angle of projection are constants
R ∝ \(\frac { 1 }{ g }\)
g ∝ \(\frac { 1 }{ R }\)

According to acceleration, due to gravity In ascending order, the solution is. Mercury, Mars, Earth, Jupiter

Question 12.
The resultant of two vectors A and B is perpendicular to vector A and its magnitude is equal to half of the magnitude of vector B. Then the angle between A and B is
(a) 30°
(b) 45°
(c) 150°
(d)120°
Solution:
Let two vectors be A & B

Magnitude of B = B
Magnitude of A = A
∝ = 90°
Given:

Question 13.
Compare the components for the following vector equations.
(a) T\(\hat{j}\) – mg\(\hat{j}\) = ma\(\hat{j}\)
(b) \(\overline{T}\) + \(\overline{F}\) = \(\overline{A}\) + \(\overline{B}\)
(c) \(\overline{T}\) – \(\overline{F}\) = \(\overline{A}\) – \(\overline{B}\)
(d) T\(\hat{j}\) + mg\(\hat{j}\) = ma\(\hat{j}\)
Solution:
We can resolve all vectors in x, y, z components w.r.t. Cartesian co-ordinate system. After resolving the components separately equate x components on both sides y components on both sides and z components on both side we get.
(a) T\(\hat{j}\) – mg\(\hat{j}\) = ma\(\hat{j}\)
T – mg = ma

(b) \(\overline{T}\) + \(\overline{F}\) = \(\overline{A}\) + \(\overline{B}\)
T_x + F_x = A_x + B_x

(c) \(\overline{T}\) – \(\overline{F}\) = \(\overline{A}\) – \(\overline{B}\)
T_x – F_x = A_x – B_x

(d) T\(\hat{j}\) + mg\(\hat{j}\) = ma\(\hat{j}\)
T + mg = ma

Question 14.
Calculate the area of the triangle for which two of its sides are given by the vectors \(\overline{A}\) = 5\(\hat{i}\) – 3\(\hat{j}\) \(\overline{B}\) = 4\(\hat{i}\) + 6\(\hat{j}\).
Solution:

Question 15.
If the earth completes one revolution in 24 hours, what is the angular displacement made by the earth in one hour? Express your answer in both radian and degree.
Solution:
ω = θ/t θ = wt
In 24 hours, angular displacement made
θ = 360° (or) 2π rad
In 1 hours, angular displacement
θ = \(\frac { 360° }{ 24 }\)
θ = 15°
In radian θ = \(\frac { 2π }{ 24 }\) = \(\frac { π }{ 12 }\) radians.
θ = \(\frac { π }{ 12 }\) rad.

Question 16.
An object is thrown with initial speed of 5ms^-1 with an angle of projection of 30°. What is the height and range reached by the particle?
Solution:
u = 5 m/s
θ = 30°
h_max = ?
R = ?
Height reached

Range:

Question 17.
A football player hits the ball with a speed 20m/s with angle 30° with respect to as shown in the figure horizontal directions. The goal post is at a distance of 40 m from him. Find out whether the ball reaches the goal post.

Solution :
In order to find whether the ball is reaching the goal post the range should be equal to 40m so range

= \(\frac { 692.8 }{ 19.6 }\)
= 35.35 m.
Which is less than the distance of the goal post which is 40 m away so the ball won’t reach the goal post.

Question 18.
If an object is thrown horizontally with an initial speed 10 ms^-1 from the top of a building of height 100 m. What is the horizontal distance covered by the particle?
Solution:
u = 10 m/s
h = 100 m
x = ?
x = u x T

x = 45.18 m.

Question 19.
An object is executing uniform circular motion with an angular speed of π/12 radians per second. At t = 0 the object starts at an angle θ = 0. What is the angular displacement of the particle after 4s?
Solution :
ω = π/12 rad/s
ω = θ/t
θ = w x t = π/12 x 4
θ = π/3 radian
θ = \(\frac { 180° }{ 3 }\)
= 60°

Question 20.
Consider the x-axis as representing east, the y-axis as north, and the z-axis as vertically upwards. Give the vector representing each of the following points.

(a) 5m northeast and 2m up.
(b) 4m southeast and 3m up.
(c) 2m northwest and 4m up.
Solution:
5m northeast and 2m up.
(a) The vector representation of 5m N-E and 2m up is (5i + 5j) Cos 45° + 2\(\hat{k}\)

(b) 4m south east and 3m up.

The vector representing 4m south east and 3m up is
(4i – 4j) cos 45 + 3\(\hat{k}\)
\(\frac{4(i-j)}{\sqrt{2}}\) + 3\(\hat{k}\)

(c) 2m north west and 4m up.

The vector representing 2m northwest and 4m up

Question 21.
The moon is orbiting the earth approximately once in 27 days. What is the angle transversed by the moon per day?
Solution :
Angle described in 27 days = 2π rad = 360° days
Angie described in one day = 2π/27 radian
= \(\frac { 360° }{ 27 }\)
θ = 13.3°

Question 22.
An object of mass m has an angular acceleration ∝ = 0.2 rad/s². What is the angular displacement covered by the object after 3 seconds? (Assume that the object started with angle zero with zero angular velocity)
Solution:
∝ = 0.2 rad/s²
θ = ? t = 3s.
w₀ = 0
w.k.T θ = ω₀t + 1/2 ∝ t²
θ = 0 + 1/2 x 0.2 x 9
θ = 0.9 rad
θ = 0 = 0.9 x 57.295° = 51°
The magnitude of the resultant vector R is given by

11th Physics Guide Kinematics Additional Important Questions and Answers

I. Multiple choice questions:

Question 1.
A particle moves in a circle of radius R from A to B as in the figure.

Answer:

Question 2.
The branch of mechanics which deals with the motion of objects without taking force into account is –
(a) kinetics
(b) dynamics
(c) kinematics
(d) statics
Answer:
(c) kinematics

Question 3.
A particle moves in a straight line from A to B with speed v₁ and then from B to A with speed v₂. The average velocity and average speed are _______.

Answer:

Question 4.
A particle is moving in a straight line under constant acceleration. It travels 15m in the 3^rd second and 31m in the 7^th second. The initial velocity and acceleration are _______.
a) 5 m/s, 4 m/s²
b) 4 m/s, 5 m/s²
c) 4 m/s, 4 m/s²
d) 5 m/s, 5 m/s²
Answer:
a) 5 m/s, 4 m/s²

Question 5.
If an object is moving in a straight line then the motion is known as –
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer:
(a) linear motion

Question 6.
A car is moving at a constant speed of 15 m/s. Suddenly the driver sees an obstacle on the road and takes 0.4 s to apply the brake. The brake causes a deceleration of 5 m/s². The distance traveled by car before it stops _______.
a) 6 m
b) 22.5 m
c) 28.5 m
d) 16.2 m
Answer:
c) 28.5 m

Question 7.
A car accelerates from rest at a constant rate for some time after which it decelerates at a constant rate (3 to come to rest. If the total time lapses in ‘t’ seconds, then the maximum velocity reached is _______.

Answer:

Question 8.
Spinning of the earth about its own axis is known as –
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer:
(d) rotational motion

Question 9.
A particle is thrown vertically up with a speed of 40m/s, The velocity at half of the maximum height _______.
a) 20 m/s
b) 20\(\sqrt{2}\)m/s
c) 10 m/s
d) 10\(\sqrt{2}\)m/s
Answer:
b) 20\(\sqrt{2}\)m/s

Question 10.
The ratio of the numerical values of the average velocity and the average speed of the body is always _______.
a) unity
b) unity or less
c) unity or more
d) less than unity
Answer:
b) unity or less

Question 11.
The motion of a satellite around the earth is an example for –
(a) circular motion
(b) rotational motion
(c) elliptical motion
(d) spinning
Answer:
(a) circular motion

Question 12.
One car moving on a straight road covers one-third of the distance with 20 km/h and the rest with 60 km/h. The average speed is _______.
a) 40 km/h
b) 80km/h
c) 46\(\frac { 2 }{ 3 }\) km/hr
d) 36 km/h
Answer:
d) 36 km/h

Question 13.
A 150m long train is moving with a uniform velocity of 45 km/h. The time taken by the train to cross a bridge of length 850m is _______.
a) 56s
b) 68s
c) 80s
d) 92s
Answer:
c) 80s

Question 14.
A particle moves in a straight line with constant acceleration. It changes its velocity from 10 m/s to 20 m/s while passing through a distance of 135 m in ‘t’ seconds. The value of t is _______.
a) 12s
b) 9s
c) 10s
d) 1.8s
Answer:
b) 9s

Question 15.
If a ball is thrown vertically upwards with a speed u the distance covered during the last ‘t’ seconds of its ascent is _______.
a) 1/2 gt²
b) ut – 1/2gt²
c) (u – gt)t
d) ut
Answer:
a) 1/2 gt²

Question 16.
A particle moves along a straight line such that its displacement ‘s’ at any time ‘t’ is given by s = t³ – 6t² + 3t + 4 meters, t being in second. The velocity when acceleration is zero is _______.
a) 3 m/s
b) -12m/s
c) 42 m/s
d) -9 m/s
Answer:
d) – 9 m/s

Question 17.
Which of the following is not a scalar?
(a) Volume
(b) angular momentum
(c) Relative density
(d) time
Answer:
(b) angular momentum

Question 18.
Vector is having –
(a) only magnitude
(b) only direction
(c) bot magnitude and direction
(d) either magnitude or direction
Answer:
(c) both magnitude and direction

Question 19.
The displacement-time graph of a moving particle is shown below.

The instant velocity of the particle is negative at the point
a) D
b) F
c) C
d) E
Answer:
d) E

Question 20.
If two vectors are having equal magnitude and the same direction is known as –
(a) equal vectors
(b) col-linear vectors
(c) parallel vectors
(d) on it vector
Answer:
(a) equal vectors

Question 21.
The velocity-time graph of a body moving in a straight line is shown below _______.

Which are of the following represents its acceleration-time graph?

Answer:

Question 22.
Indicate which of the following graph represents the one-dimensional motion of particle?

Answer:

Question 23.
The variation of velocity of a particle with time moving along a straight line is illustrated in the following figure. The distance travelled by the particle in 4s is _______.

a) 60m
b) 55m
c) 25m
d) 30m
Answer:
b) 55m

Question 24.
An object is moving with a uniform acceleration which is parallel to its instantaneous direction of motion. The displacement (s), velocity (v) graph of this object is _______.

Answer:

Question 25.
A unit vector is used to specify –
(a) only magnitude
(b) only direction
(c) either magnitude (or) direction
(d) absolute value
Answer:
(b) only direction

Question 26.
A vector is not changed if _______.
a) It is rotated through an arbitrary angle
b) It is multiplied by an arbitrary scalar
c) It is cross multiplied by a unit vector
d) It is parallel to itself.
Answer:
d) It is parallel to itself.

Question 27.
Two forces each of magnitude ‘F’ have a resultant of the same magnitude. The angle between two forces
a) 45°
b) 120°
c) 150°
d) 60°
Answer:
b) 120°

Question 28.
The magnitude of a vector can not be-
(a) positive
(b) negative
(e) zero
(cl) 90
Answer:
(b) negative

Question 29.
Six vectors \(\vec{a}\) through \(\vec{f}\) have magnitudes and directions as indicated in figure. Which of the following statement is true?

a) \(\overline{b}\) + \(\overline{e}\) = \(\overline{f}\)
b) 3\(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\) \(\hat{b}\) + \(\hat{c}\) = \(\hat{f}\)
c) \(\hat{d}\) + \(\hat{c}\) = \(\hat{f}\)
d) \(\hat{d}\) + \(\hat{e}\) = \(\hat{f}\)
Answer:
d) \(\hat{d}\) + \(\hat{e}\) = \(\hat{f}\)

Question 30.
A force of 3 N and 4 N are acting perpendicular to an object, the resultant force is-
(a) 9 N
(b) 16 N
(c) 5 N
(d) 7 N
Answer:
(c) 5 N

Question 31.
The figure shows ABCDEF as regular hexagon. What is the value of
\(\overline{AB}\) + \(\overline{AC}\) + \(\overline{AD}\) + \(\overline{AE}\) + \(\overline{AF}\)?

a) \(\overline{A0}\)
b) 2 \(\overline{A0}\)
c) 4 \(\overline{A0}\)
d) 6 \(\overline{A0}\)
Answer:
d) 6 \(\overline{A0}\)

Question 32.
One of the two rectangular components of a force is 20N. And it makes an angle of 30° with the force. The magnitude of the other component is _______.
a) 20/\(\sqrt{3}\)
b) 10/\(\sqrt{3}\)
c) 15/V\(\sqrt{3}\)
d) 40\(\sqrt{3}\)
Answer:
a) 20/\(\sqrt{3}\)

Question 33.
The angle between (\(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)) and (\(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{B}}\)) can be –
(a) only 0°
(b) only 90°
(c) between 0° and 90°
(d) between 0° and 180°
Answer:
(d) between 0° and 180°

Question 34.
If the sum of two unit vectors is a unit vector the magnitude of the difference is _______.
a) \(\sqrt{2}\)
b) \(\sqrt{3}\)
c) 1/\(\sqrt{2}\)
d) \(\sqrt{5}\)
Answer:
b) \(\sqrt{3}\)

Question 35.
If P = mV then the direction of P along-
(a) m
(b) v
(c) both (a) and (b)
(d) neither m nor v
Answer:
(b) v

Question 36.
If \(\overline{A}\) = 2\(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\), \(\overline{B}\) = \(\hat{i}\) + 2\(\hat{j}\) + 3\(\hat{k}\) and \(\overline{C}\) = 6\(\hat{i}\) – 2\(\hat{j}\) – 6\(\hat{k}\) then angle between \(\overline{A}\) + \(\overline{B}\) and \(\overline{C}\) will be _______.
a) 30°
b) 45°
c) 60°
d) 90°
Answer:
d) 90°

Question 37.
The scalar product \(\overrightarrow{\mathrm{A}}\).\(\overrightarrow{\mathrm{B}}\)is equal to-
(a) \(\overrightarrow{\mathrm{A}}\) +\(\overrightarrow{\mathrm{B}}\)
(b) \(\overrightarrow{\mathrm{A}}\). \(\overrightarrow{\mathrm{B}}\)
(c) AB sin θ
(d) (\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)
Answer:
(b) \(\overrightarrow{\mathrm{A}}\). \(\overrightarrow{\mathrm{B}}\)

Question 38.
If \(\overline{A}\) x \(\overline{B}\) = \(\overline{C}\) then which of the following statement is wrong?
a) \(\overline{C}\) ⊥\(\overline{A}\)
b) \(\overline{B}\) ⊥\(\overline{B}\)
c) \(\overline{C}\) ± ( \(\overline{A}\) + \(\overline{B}\) )
d) \(\overline{C}\) ± ( \(\overline{A}\) x \(\overline{B}\) )
Answer:
d) \(\overline{C}\) ± ( \(\overline{A}\) x \(\overline{B}\) )

Question 39.
The scalar product of two vectors will be minimum. When θ is equal to –
(a) 0°
(b) 45°
(c) 180°
(d) 60°
Answer:
(c) 180°

Question 40.
If | \(\overline{A}\) x \(\overline{B}\) |, then value of | \(\overline{A}\) x \(\overline{B}\) | is _______.

Answer:
d) (A² + B² + AB)\(\frac { 1 }{ 2 }\)

Question 41.
The angle between vectors \(\overline{A}\) and \(\overline{B}\) is A. The value of the triple product \(\overline{A}\) ( \(\overline{A}\) x \(\overline{B}\) ) is _______.
a) A² B
b) zero
c) A² B sinθ
d) A² B cos θ
Answer:
d) A² B cos θ

Question 42.
Two adjacent sides of a parallelogram are represented by the two vectors \(\hat{i}\) + 2\(\hat{j}\) + 3\(\hat{k}\) and 3\(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\). The area parallelogram _______.
a) 8
b) 8\(\sqrt{3}\)
c) 3\(\sqrt{8}\)
d) 192
Answer:
b) 8\(\sqrt{3}\)

Question 43.
If \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are two vectors, which are acting along x, y respectively, then \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) lies along-
(a) x
(b) y
(c) z
(d) none
Answer:
(c) z

Question 44.
Galileo writes that for angles of the projectile (45 + θ) and (45 – θ) the horizontal ranges described by the projectile are in the ratio of (if θ ≤ 45)
a) 2:1
b) 1:2
c) 1:1
d) 2:3
Answer:
c) 1:1

Question 45.
A projectile is thrown into the air so as to have the minimum possible range equal to 200. Taking the projection point as the origin the Coordinates of the point where the velocity of the projectile is minimum are _______.
a) 200,50
b) 100,50
c) 100,150
d) 100,100
Answer:
b) 100,50

Question 46.
\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) isequal to –
(a) \(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)
(b) \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)
(c) –\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)
(d) \(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{B}}\)
Answer:
(c) –\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)

Question 47.
The vector product of any two vectors gives a –
(a) vector
(b) scalar
(e) tensor
(d) col-linear
Answer:
(a) vector

Question 48.
A 150 m long train is moving the north at a speed of 10 m/s. A parrot flying towards the south with a speed of 5 m/s crosses the train. The time taken would be _______.
a) 30s
b) 15s
c) 8s
d) 10s
Answer:
d) 10s

Question 49.
A boat is moving with a velocity of 3i+4j with respect to the ground. The water in the river is moving with a velocity of -3i-4j with respect to the ground. The relative velocity of the boat with respect to water _______.
a) 8j
b) -6i -8j
c) 6i + 8j
d) 5\(\sqrt{2}\)
Answer:
c) 6i + 8j

Question 50.
The vector product of two non-zero vectors will be minimum when O is equal to-
(a) 0°
(b) 180°
(e) both (a) and (b)
(d) neither (a) nor (b)
Answer:
(e) both (a) and (b)

II. Long Answer Questions:

Question 1.
What are the different types of motion? State one example for each & explain.
Answer:
The different types of motions are:
a) Linear motion: An object is said to be in linear motion if it moves in a straight line.
Example: An athlete running on a straight tack.

b) Circular motion: It is defined as a motion described by an object traveling a circular path.
Example: The motion of a satellite around the earth

C) Rotational motion: If any object moves in a rotational motion about an axis the motion is rotational motion. During rotation, every point in the object traverses a circular path about an axis.
Example: Spiring of earth about its own axis

D) Vibratory motion: If an object or a particle executes to and fro motion about the fixed point it is said to be in vibratory motion. Sometimes called oscillatory motion.
Example: Vibration of a string on a Guitar.

Question 2.
How will you differentiate motion in one dimension, two dimensions, and in three dimensions?
Answer:
Motion in one dimension: One-dimensional motion is the motion of a particle moving along a straight line.
Example: An object falling freely under gravity close to the earth.

Motion in two dimensions: If a particle is moving along a curved path in-plane, then it is said to be in two-dimensional motion.
Example: Motion of a coin in a carom board.

Motion in three dimensions: A particle moving in usual three-dimensional space has three-dimensional motion.
Example: A bird flying in the sky.

Question 3.
State and define different types of vectors.
Answer:
The different types of vectors are:
1. Equal vectors:
Two vectors \(\vec{A}\) & \(\vec{B}\) are said to be equal when they have equal magnitude and same direction and represent the same physical quantity.

(a) Coilinear vectors: Collinear vectors are those which act along the same line. The angle between them can be 0° or 180°

(i) Parallel vectors – If two vectors \(\vec{A}\) & \(\vec{B}\) act in the same direction along the same line or in parallel lines. Angle between them is equal to zero

(ii) Antiparallel vectors:
Two vectors \(\vec{A}\) & \(\vec{B}\) are said to be antiparallel when they are in opposite direction along the same line or in parallel lines. The angle between them is 180°

2. Unit vector:
A vector divided by its own magnitude is a unit vector.
The unit vector of \(\vec{A}\) is represented as \(\hat{A}\)
Its magnitude is equal to 1 or unity

3. Orthogonal unit vectors:
Let \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) be three unit vectors which specify the direction along positive x-axis, positive y-axis and positive z-axis respectively. These three unit vectors are directly perpendicular to each other
The angle between any two of them is 90°. Then \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) are examples of orthogonal vectors. Two vectors which are perpendicular to each other are called orthogonal vectors.

Question 4.
Explain how two vectors are subtracted when they are inclined to an angle θ.

Let \(\overline{A}\) and \(\overline{B}\) be non zero vectors inclined at an angle θ.
The difference \(\overline{A}\) – \(\overline{B}\) can be obtained as follows.
First obtain – \(\overline{B}\)
The angle between \(\overline{A}\) – \(\overline{B}\)
= 180 – θ.
The difference \(\overline{A}\) – \(\overline{B}\) is the same as the resultant of \(\overline{A}\) – – \(\overline{B}\)

[∵ cos 180 – θ = – cos θ]
(cos 180 – θ = – cos θ)
The gives the resultant magnitude. The resultant is inclined by an angle α₂ to \(\overline{A}\)

This gives the direction of the resultant. \(\vec{A}\) – \(\vec{B}\)

Question 5.
Write short notes on relative velocity.
Answer:
When two objects A and B are moving with uniform velocities then the velocity of one object A with respect to another object B is called the relative velocity of A with respect to B.

Case 1:
Consider two objects A and B moving with uniform velocities \(\overline{V}\)_A and \(\overline{V}\)_B along straight line in same direction with respect to ground.
The relative velocity of object A with respect to object B is \(\vec{V}\)_AB = \(\vec{V}\)_A – \(\vec{V}\)_B
The relative velocity of object B with respect to object A is \(\vec{V}\)_BA = \(\vec{V}\)_B –\(\overline{V}\)_A
Thus, if two objects are moving it’s the same direction the magnitude of the relative velocity of one object with respect to another is equal to the difference in magnitude of the two velocities.

Case 2:
Consider two objects A and B moving with uniform velocities VA and VB along the same track in the opposite direction

The relative velocity of object A with respect to B is
\(\overline{V}\)_AB = \(\overline{V}\)_A – ( – \(\overline{V}\)_B) = \(\overline{V}\)_A + \(\overline{V}\)_B
The relative velocity of object B with respect to A is
\(\vec{V}\)_BA = – \(\vec{V}\)_B – \(\vec{V}\)_A) = – ( \(\vec{V}\)_A + \(\vec{V}\)_B )
Thus if two objects are moving in opposite directions the magnitude of relative velocity of one object with respect to other is equal to the sum of magnitudes of their velocities.

Case 3:
Consider two objects A&B moving with velocities VA and VB at an angle 0 between their directions, then the relative velocity of A with respect to B

tan θ = (β is the angle between \(\overline{V}\)_AB and VB)

Special cases:
(i) When θ = 0, the bodies move along parallel straight lines in the same direction.
V_AB = (V_A – V_B) in the direction of V_A.
V_BA = (V_B – V_A) in the direction of V_B

(ii) When θ = 180° the bodies move along parallel straight lines in opposite direction.
V_AB = V_A – (- V_B) = (V_A + V_B) in the direction of V_A
V_BA = ( V_B + V_A) in the direction of V_B

(iii) If the two bodies are moving at right angles to each other, then θ = 90°
V_AB = \(\sqrt{V_{A}^{2}+V_{B}^{2}}\)

(iv) Consider a person moving horizontally with velocity \(\vec{V}\)_m Let the rain fall vertically with velocity \(\overline{V}\)_R.

An umbrella is held to avoid the rain.
Then relative velocity \(\overline{V}\)_M of rain with respect to man is

Question 6.
Explain Horizontal projection. Derive the equation for its motion, horizontal range & time of flight.
Answer:

Consider an object thrown horizontally with an initial velocity u, from atop of a tower of height h. The horizontal velocity remains constant throughout its motion and the vertical component of velocity go on increases. The constant acceleration acting along the downward direction is g. The horizontal distance travelled is x(t) = x and the vertical distance travelled is y(t)=y. since the motion is two-dimensional the velocity will have both horizontal (u_x) and vertical (u_y) components.

Motion along horizontal direction:
The particle has zero acceleration along the x-direction and so initial velocity ux remains constant throughout its motion.
The distance travelled by projectile in a time’t’ is given by
x = ut+1/2 at²
x = u_xt → (1)
Motion along vertical direction
Here uy =0, a = g, s = y
S = ut + \(\frac { 1 }{ 2 }\) at²
y = \(\frac { 1 }{ 2 }\) gt² → (2)
from (1) t = x/u_x sub in equation (2)
y = \(\frac { 1 }{ 2 }\) g (x/ux)²
y = k x² Where k = \(\frac{g}{2 u_{x}^{2}}\) .x²
This equation resemble the equation of a parabola. Thus the path followed by the projectile is a parabola.

Expression for time of flight:
The time taken for the projectile to complete its trajectory is called the time of flight.
Let h be the height of the tower or the vertical distance traversed.
Let T be the time of flight w.k. S = ut + 1/2 at²
here s = y = h, u = u_y, t = T, a = g

T depends on height of tower or vertical distance & independent of Horizontal velocity.

Expression for Horizontal Range:
The horizontal distance covered by the projectile from the foot of the tower to the point where the projectile hits the ground it called horizontal range.
w. k. t, S = ut + \(\frac { 1 }{ 2 }\) at²
Here,
t = T, a = 0, S = x = R, u = u_x

Hence R ∝ u ∝ & R ∝ \(\frac{1}{\sqrt{g}}\)

Question 7.
Obtain an expression for resultant velocity and the speed of the projectile when it hits the ground in case of a horizontal projection.
Answer:

At any instant t, the projectile has velocity components along both the x and y-axis.
The velocity component at any time t along with horizontal component V_x = u → (1)

Speed of the projectile when it hits the ground:
When the projectile hits the ground after thrown horizontally from top of tower of height h, the time of flight is T = \(\sqrt{\frac{2 h}{g}}\)
The horizontal component of velocity V_x = u
The vertical component of velocity

Conceptual Questions:

Question 1.
Can a body have a constant speed and still have varying velocity?
Answer:
Yes, a particle in uniform circular motion has a constant speed but varying velocity because of the change in its direction of motion at every point.

Question 2.
When an observer is standing on earth appear the trees and houses appear stationary to him. However, when he is sitting in a moving bus or a train all objects appears to move in a backward direction why?
Answer:
For a stationary observer, the relative velocity of trees and houses is zero. For the observer sitting in the moving train, the relative velocity of houses and trees are negative. So these objects appear to move in the backward direction.

Question 3.
Draw position-time graphs for two objects having zero relative velocity?
Answer:

As relative velocity is zero the two bodies A and B have equal velocities. Hence their position-time graphs are parallel straight lines, equally inclined to the time axis.

Question 4.
Can a body be at rest as well as in motion at the same time? Explain.
(OR)
Rest and motion are relative terms. Explain.
Answer:
Yes, the object may be at rest relative to one object and at the same time if maybe in motion relative to another object.

For example, a passenger sitting in a moving train is at rest with respect to his fellow passengers but he is in motion with respect to the objects outside the train. Hence rest and motion are relative terms.

Question 5.
Use integration technique to prove that the distance travelled in-the n^th second of motion is S_th =u + \(\frac { a }{ 2 }\) (2n – 1)
Answer:
By definition of velocity v = \(\frac { ds }{ dt }\)
ds = Vdt = (u + at) dt → (1)
when t = (n – 1) second, let distance travelled = S_n-1
when t = n, second, let distance travelled = S_n

Question 6.
An old lady holding a purse in her hand was crossing the road. She was feeling difficulty in walking. A pickpocket snatched the purse from her and started running away. Can seeing this incident Suresh decided to help that old lady. He informed the police inspector who was standing nearby the inspector chased the pickpocketed and caught hold of him. He collected the purse from the pickpocket and gave back the purse to the old lady.
a) What were the values displayed by Suresh?
b) A police jeep is chasing with a velocity of 45 km/h. A thief in another jeep is moving at 155 km/hr. Police fire a bullet strike the jeep of the thief?
Answer:
The values displayed by Suresh are the presence of mind, helping tendency, and also a sense of social responsibility.
Relative velocity of the bullet with respect to thief’s Jeep = (Vb + Vp)-Vt.
= 180 m/s + 45 km/hr – 155 km/hr
= 180 m/s – 110 x 5/18 m/s
= 180 – 30.5
= 149.5 m/s.

Question 7.
A stone is thrown vertically upwards and then it returns to the thrower. Is it projective?
Answer:
No. It is not a projectile. A projectile should have two-component velocities in two mutually perpendicular directions. But in this case, body has a velocity in only one direction.

Question 8.
Can two non-zero vectors give zero resultant when they multiply with each other?
Answer:
If yes condition for the same. Yes. for example, the cross product of two non-zero vectors will be zero when θ = 0 or θ = 180°.

Question 9.
Justify that a uniform motion is an accelerated motion.
Answer:
In a uniform circular motion, the speed of the body remains the same but the direction of motion changes at every point.

Fig. shows the different velocity vectors at different positions of the particle. At each position, the velocity vector V is perpendicular to the radius vector. Thus the velocity of the body changes continuously due to the continuous change in the direction of motion of the body. As the rate of change is of velocity is acceleration a uniform circular motion is an accelerated motion.

Question 10.
State polygon law of vector addition.
Answer:
If a number of vectors are represented both in magnitude and direction by the sides of an open polygon taken in the same order then their resultant is represented both in magnitude arid direction by the closing side of the polygon taken in the opposite order.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Physics Guide Pdf Chapter 3 Laws of Motion Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

11th Physics Guide Laws of Motion Book Back Questions and Answers

Part – I:
I. Multiple choice questions:

Question 1.
When a car takes a sudden left turn in the curved road, passengers are pushed towards the right due to _______.
a) inertia of direction
b) inertia of motion
c) inertia of rest
d) absence of inertia
Answer:
a) inertia of direction

Question 2.
An object of mass m held against a vertical wall by applying horizontal force F as shown in the figure. The minimum value of the force F is _______ (IIT JEE 1994)

a) Less than mg
b) Equal to mg
c) Greater than mg
d) Cannot determine
Answer:
c) Greater than mg

Question 3.
A vehicle is moving along the positive x-direction, if a sudden brake is applied, then _______.
a) frictional force acting on the vehicle is along negative x-direction
b) frictional force acting on the vehicle is along the positive x-direction
c) no frictional force acts on the vehicle
d) frictional force acts in a downward direction
Answer:
a) frictional force acting on the vehicle is along the negative x-direction

Normal force formula is the support force exerted upon an object that is in contact with another stable object.

Question 4.
A book is at rest on the table which exerts a normal force on the book. If this force is considered as a reaction force, what is the action force according to Newton’s third law?
a) Gravitational force exerted by Earth on the book
b) Gravitational force exerted by the book on Earth
c) Normal force exerted by the book on the table
d) None of the above
Answer:
c) Normal force exerted by the book on the table

Question 5.
Two masses m₁ and m₂ are experiencing the same force where m₁ < m₂. The ration of their acceleration a₁/a₂ is _______.
a) 1
b) less than 1
c) greater than 1
d) all the three cases
Answer:
c) greater than 1

Question 6.
Choose an appropriate free body diagram for the particle experiencing net acceleration along the negative y-direction. (Each arrow mark represents the force acting on the system).

Answer:

Question 7.
A particle of mass m sliding on the smooth double inclined plane (shown in the figure) will experience _______.

a) greater acceleration along the path AB
b) greater acceleration along the path AC
c) same acceleration in both the paths
d) no acceleration in both the paths
Answer:
b) greater acceleration along the path AC

Question 8.
Two blocks of masses m and 2m are placed on a smooth horizontal surface as shown. In the first case, only a force F₁ is applied from the left. Later only a force F₂ is applied from the right. If the force acting at the interface of the two blocks in the two cases is the same, then F₁ : F₂ is _______. (Physics Olympiad 2016)

a) 1 : 1
b) 1 : 2
c) 2 : 1
d) 1 : 3
Answer:
c) 2 : 1

Question 9.
Force acting on the particle moving with constant speed is _______.
a) always zero
b) need not be zero
c) always non zero
d) cannot be concluded
Answer:
b) need not be zero

Question 10.
An object of mass m begins to move on the plane inclined at an angle θ. The coefficient of static friction of inclined surface is us. The maximum static friction experienced by the mass is _______.
a) mg
b) μ_s mg
c) μ_s mg sin θ
d) μ_s mg cos θ
Answer:
d) μ_s mg cos θ

Question 11.
When the object is moving at a constant velocity on the rough surface _______.
a) net force on the object is zero
b) no force acts on the object
c) only external force acts on the object
d) only kinetic friction acts on the object
Answer:
a) net force on the object is zero

Question 12.
When an object is at rest on the inclined rough surface _______
a) static and kinetic frictions acting on the object is zero
b) static friction is zero but kinetic friction is not zero.
c) static friction is not zero and kinetic friction is zero.
d) static and kinetic frictions are not zero.
Answer:
c) static friction is not zero and kinetic friction is zero.

Question 13.
The centrifugal force appears to exist _______.
a) only in inertial frames
b) only in rotating frames
c) in an accelerated frame
d) both in inertial and non-inertial frames
Answer:
b) only in rotating frames

Question 14.
Choose the correct statement from the following.
a) Centrifugal and centripetal forces are action-reaction pairs.
b) Centripetal forces is a natural force.
c) Centrifugal force arises from the gravitational force.
d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion.
Answer:
d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion.

Question 15.
If a person moving from pole to equator, the centrifugal force acting on him
a) increase
b) decreases
c) remains the same
d) increase and then decreases
Answer:
a) increase

II. Short Answer Questions:

Question 1.
Explain the concept of Inertia. Write two examples each for Inertia of motion, inertia of rest and inertia of direction.
Answer:
The inability of objects to move on its own or change its state of motion is called inertia. Inertia means resistance to change its state. There are three types of inertia:
1. Inertia of rest:
The inability of an object to change its state of rest is called inertia of rest.
Example:

• When a stationary bus starts to move, the passengers experience a sudden backward push.
• A book lying on the table will remain at rest until it is moved by some external agencies.

2. Inertia of motion:
The inability of an object to change its state of uniform speed (constant speed) on its own is called inertia of motion.
Example:

• When the bus is in motion, and if the brake is applied suddenly, passengers move forward and hit against the front seat.
• An athlete running is a race will continue to run even after reaching the finishing point.

3. Inertia of direction:
The inability of an object to change its direction of motion on its own is called inertia of direction.
Example:

• When a stone attached to a string is in whirling motion, and if the string is cut suddenly, the stone will not continue to move in circular motion but moves tangential to the circle.
• When a bus moving along a straight line takes a turn to the right. The passengers are thrown towards left.

Question 2.
State Newton’s second law:
Answer:
Newton second law state that” The force acting on an object is equal to the rate of change of its momentum:
F = \(\frac { dp }{ dt }\)
If m is the mass_of the object, and v its velocity of motion then \(\overline{p}\) = m\(\overline{v}\):
The above equation can be written as
F = \(\frac { dp }{ dt }\)(m\(\overline{v}\) ) = m\(\frac { dv }{ dt }\)
∴ F = ma

Question 3.
Define one newton:
Answer:
One newton is defined as the force which acts on 1 kg of mass to give an acceleration 1 ms^-2 in the direction of the force.

Question 4.
Show that impulse is the change of momentum.
Answer:
If a very large force acts on an object for a very short time, then the force is impulsive force or impulse. If a Force (F) acts on an object in a very short time (∆t), then from Newton’s second law dp = f.dt
Integrating \(\int_{i}^{f} d p=\int_{t_{1}}^{t_{2}} f d t\)
P_f – P_i =F[t₁ – t₂]
P_i – initial momentum of the object at time t₁
P_f – Final momentum of the object at time t₂
P_f – P_i = ∆P = > Change is momentum
(t₂ – t₁) = ∆t = > Time interval
The above equation can be written as
∴ ∆P = F ∆ t.
ie \(\int_{t_{1}}^{t_{2}} F d t\) = J is called the impulse and t₁ it is equal to change in momentum of the object is ∆P = F ∆ t. When F is kept constant

Question 5.
Using free body diagram, show that it is easy to pull an object than to push it.
Answer:
It is easier to pull on object than to push it.
An object pushed at a angle θ.

Case 1:
When a body is pushed at an arbitrary angle θ (0 to π/2) the applied force. F is resolved into two components
F sin θ = Horizontally – parallel to surface
F cos θ = Vertically – perpendicular to surface
The total downward force = mg + F cos 0 This is equal to normal force (reaction)
N_Push = mg + F cos θ … (1)
The static friction is equal to
P^s_(max) = μS N_push = μ_s (mg + F cos θ) … (2)

Case 2:
An object pulled at an angle θ

When an object is pulled at an angle 0 the applied force is resolved into two components.
F sin θ – Horizontally – parallel to the surface
F cos θ – Vertically – perpendicular to the surface
The total downward force = mg – F cos θ
This is equal to the normal force (reaction)
N_pull = mg – F cos θ … (3)
The static friction is equal to
F_s^(max) = μ_s N_pull = μ_s (mg – F cos θ) … (4)
Conclusion:
From equations (1) & (3) (or) from (2) and (4) it is clear that normal force or reaction due to pulling is less than that of push. So it is easier to pull an object than push it out.

Question 6.
Explain the various types of friction suggest a few methods to reduce friction.
Answer:
There are two types of Friction:
(1) Static Friction:
Static friction is the force which opposes the initiation of motion of an object on the surface. The magnitude of static frictional force f_s lies between
\(0 \leq f_{s} \leq \mu_{s} \mathrm{N}\)
where, µ_s – coefficient of static friction
N – Normal force

(2) Kinetic friction:
The frictional force exerted by the surface when an object slides is called as kinetic friction. Also called as sliding friction or dynamic friction,
f_k – µ_kN
where µ_k – the coefficient of kinetic friction
N – Normal force exerted by the surface on the object

Methods to reduce friction:
Friction can be reduced

• By using lubricants
• By using Ball bearings
• By polishing
• By streamlining

Question 7.
What is the meaning of “Pseudo force”?
Answer:
“A force which does not actually act on particles but appears due to acceleration of the frame is called – Pseudo force”. In order to use Newton I & II law in the rotational frame of reference, one needs to include a “Pseudo force” called “Centrifugal force”. A pseudo force has no origin, It arises due to the non-inertial nature of the frame considered. In order to solve circular motion problems from a rotating frame of reference, pseudo force is necessary.

Question 8.
State of the empirical laws of static and kinetic friction.
Answer:

Question 9.
State Newton’s third law.
Answer:
Newton’s third law states that for every action there is an equal and opposite reaction.

Question 10.
What are inertial frames?
Answer:
A frame of reference in which Newton’s I law of motion holds good is called an inertial frame of reference. In such a frame if no force acts on a body it continues to be at rest or in uniform motion. So it is called as an inertial frame.

Question 11.
Under what condition will a car skid on a leveled circular road?
Answer:
On a leveled circular road, if the static friction is not able to provide enough centripetal ’force to turn, the vehicle will start to skid \(\mu_{s}<\frac{v^{2}}{r g}\)

III. Long Answer Questions:

Question 1.
Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is fired from it.
Answer:
Law: If there are no external forces acting on the system, then the total linear momentum of the system ( ^–_tot) is always a constant vector. In other words, the total linear momentum of the system is conserved in time.
Proof:
By combing Newton’s second and third laws the law of conservation of total linear momentum can be proved. When two particles interact with each other, they exert equal and opposite forces on each other
Consider two particles 1 & 2.
Let F₂₁ be the force exerted on 2 by 1
Let F₁₂ be the force exerted on 1 by 2
According to Newtons III law
\(\bar { F }\)₂₁ = – \(\bar { F }\)₁₂ → (1)
In terms of momentum, according to Newtons II Law,
F₁₂ = \(\frac { d }{ dt }\)\(\bar { P }\)₁ → (2)
F₂₁ = \(\frac { d }{ dt }\)\(\bar { P }\)₂ → (3)
Where is the momentum of p₁ particle P₂ is the momentum of II particle Sub (2) & (3) in (1)

This implies that (P₁ + P₂) = constant vector
\(\bar { P }\)₁ + \(\bar { P }\)₂ = \(\bar { P }\)_tot– is the total linear momentum of the two particle system.
F₁₂ & F₂₁ are internal force and no external force acting on the system form outside. So total linear momentum is conserved.

Question 2.
What are concurrent forces? State Lamis theorem.
Answer:
Concurrent forces:
A collection of forces is said to be concurrent if the lines of forces act at a common point. If the concurrent forces are in same plane they are coplanar also, in additional to concurrent forces.

Lamis is a theorem:
If a system of three concurrent and coplanar forces is in equilibrium, then Lamis theorem states that ” the magnitude of each force of the system is proportional to sine of the angle between other two forces”.

Proof:
Let F1, F2 and F3 be three coplanar and concurrent forces act at a common point 0 as in figure.

If the point 0 is in equilibrium then according to Lamis theorem.

Question 3.
Explain the motion of blocks connected by a string in 1) vertical motion 2) horizontal motion.
Answer:
When blocks are connected by strings and force F is applied vertically or horizontally, it produces Tension (T) in the string which affects acceleration to some extent. Let us discuss vertical and horizontal motion here

Case 1:
1) Vertical motion of connected bodies:

Consider two blocks to masses m₁ and m₂ (m₁ > m₂) connected by light and in an extensible string that passes over the pulley.

Let T be the tension in the string and a be the acceleration. When the system is released m₂ move vertically up and m₁ move vertically down with acceleration a. The gravitational force m₁g on m₁ is used to lift m₂. The upward direction is chosen as y. The free body diagram of both masses can be drawn as

Applying Newton II law for mass m₂
T\(\hat{j}\) – m₂g\(\hat{j}\) = m₂a\(\hat{j}\) → (1)
By comparing the components on both sides We get
T – m₂g = m₂a
Similarly for mass m₁

If m₁ = m₂ ie both masses are equal, a = 0 This shows that if masses are equal, there is no acceleration, and the system as a whole will be at rest.

Tension on the string:
Substitute the value of ‘a’ from (4) in (1)

Equation (4) gives the magnitude of acceleration for m₁ the acceleration vector is
For m₁ the acceleration vector is

Case 2:
2) Horizontal motion of connected masses:
Let mass m₂ kept on the horizontal surface or table and m₁ is hanging through a small pulley.

Assume there is no friction on the surface. As both the blocks are connected to the unstretchable string, m₁ moves with an acceleration downward then m₂ also moves with the same acceleration ‘a’ horizontally.

The forces acting on m₂ are

1. Downward gravitational force (m₂ g)
2. Upward normal force (N) exerted by the surface.
3. Horizontal Tension (T) exerted by the string.

Free body diagram

i) Downward gravitational force m₁g
ii) Tension ‘T’ – upwards.
Applying II Law for m₁.
T\(\hat{j}\) – m₁g\(\hat{j}\) = – m₁a\(\hat{j}\)
m₁g\(\hat{j}\) – T\(\hat{j}\) = m₁a\(\hat{j}\)
By comparing components
m₁g – T = m₁a → (1)
Apply Newton’s II Law for m₂
T\(\hat{i}\) = m₂a\(\hat{i}\) (along x)
Comparing components
N = m₂ → (2)
There is no acceleration along y direction
N = m₂a → (3)
Substituting (2) in (1)

Tension can be got by substituting (4) in (2)

Conclusion: By comparing motion in both cases it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings.

Question 4.
Briefly explain the origin of friction show that in an inclined plane, angle of friction is equal to angle of repose.
Answer:
Frictional is an opposing force exerted by the surface on the object which resists it motion. This force is frictional force. Friction force always opposes the relative motion between two surfaces which are in contact.

“when a force parallel to the surface is applied on the object, the force tries to move the object with respect to the surface. This relative motion is opposed by the surface by exerting a frictional force on the objects in a direction opposite to the applied force. Frictional force always act on the object parallel to surface on which the object is placed”.

To show angle of repose = angle of friction in an inclined plane.

Angle of friction = tan θ = F_s/N
Consider an inclined plane on which an object is placed.
Let θ be the angle of inclination of the plane with horizontal.
“Angle of repose is the angle of inclined plane with horizontal such that an object placed on it begins to slide”.
The gravitational force mg is resolved into two components.
mg sin θ – parallel to inclined plane.
mg cos θ – perpendicular to inclined plane.
The component mg sin θ parallel to inclined plane tries to move the object down.
The component mg cos θ perpendicular to inclined plane is balanced by normal reaction N.
N = mg cos θ … (1)
When the object just begins to slide static friction attains its maximum value

tan θ = angle of repose = angle of friction.
Which is equal to tan θ = μ_s
Hence proved.

Question 5.
State Newton’s three laws and discuss their significance.
Answer:

1. I law: Every object continues to be in the state of rest or off uniform motion unless there is external force acting on it.
2. II law: The force acting on an object is equal to the rate of change of its momentum F = \(\frac{d(\bar{p})}{d t}\)
3. III law: For every action there in an equal and an opposite reaction.

Discussion:
1) Newton’s laws are vector laws. The equation \(\bar { F }\) = \(\bar { ma }\) is a vector equation and it is essentially equal to three scalar equations. In Cartesian co-ordinate this equation can be written as
F_x\(\hat{i}\) + F_y\(\hat{i}\) + F_y\(\hat{k}\) = ma_x\(\hat{i}\) + ma_y\(\hat{j}\) + ma_z\(\hat{k}\) From this we can infer Fz cannot affect a_y and a_z and vice versa.

2) The acceleration experienced at a time ‘t’ depends on the force and the body at that instant does not depend on the force which acted on the body before F(t) = ma(t) In general direction of force may be different form direction of motion.

Case 1: Force and motion in same direction:
Example: when an apple falls towards earth the direction of motion and the force are in same downward direction.

Case 2: Force and motion are not in same direction
Example: The moon experiences a force towards the earth. But if actually moves in an elliptical orbit. In this case direction of motion and force are different.

Case 3: Force and motion in opposite direction:
If an object is thrown vertically upwards direction of motion is upward, but a gravitational force is downward.

Case 4: Zero net force, but there is motion:
Example: When a raindrop gets detached from the cloud it experiences both downward gravitational force and upward air drag force. As it descends the upward drag force increases and cancels the downward force. So the raindrop moves with a constant velocity called terminal velocity till it touches the surface of the earth, Hence the raindrop comes with zero net force and therefore with zero acceleration but with non zero terminal velocity.

Case 5: if multiple forces
\(\bar { F }\)₁, \(\bar { F }\)₂, \(\bar { F }\)₃, …… \(\bar { F }\)_n act on the same body, then the total force \(\bar { F }\) net is equal to vector sum of individual forces, which provides the acceleration.

\(\bar { F }\)_net = \(\bar { F }\)₁, \(\bar { F }\)₂, \(\bar { F }\)₃, …… \(\bar { F }\)_n
\(\bar { F }\)net = m\(\bar { a }\)
Example: Bow & arrow

Case 6: Newtons II Law can be written as
F = m\(\frac { dv }{ dt }\) = m.\(\frac{d^{2} r}{d t^{2}}\)
Newtons II law is basically a second order derivative of position vector, which is not zero, there must be a force acting on it.

Case 7: if no force acts an a body then
m. \(\frac{d \bar{v}}{d t}\) = 0
Which implied \(\bar { V }\) = constant. If is essentially a I law. So Newtons II law is consistent with I law, but cannot be derived form each other. Newton II law is a cause and effect relation. Force is the cause and effect is the acceleration,
(effect) ma = F(cause)
\(\frac{d \bar{p}}{d t}\) = F.

Question 6.
Captain the similarities and differences of between centripetal and centrifugal forces.
Answer:

Question 7.
Briefly explain centrifugal force with suitable examples.
Answer:
Circular motion can be analyzed from two different frames of reference. One is the Inertial frame where Newton’s laws are obeyed. The other is the rotating frame of reference which is noninertial as it is accelerating. To use Newton’s I and II law in the rotational frame of reference the pseudo force called as centrifugal force is needed.

The centrifugal forces appear to act on objects with respect to rotating frames. To explain consider an example, In the case of a whirling motion of a stone tied to a string, assume the stone has angular velocity ω in an inertial frame. If the motion of the stone is observed from a frame which is also rotating along with the stone with the same angular velocity ω then the stone appears to be at rest.

This implies that in addition to the inward centripetal force (-mrω²) there must be an equal and opposite force that acts on outwards equal to (+ mrω₂). So the total force acting on the stone in the rotating frame is equal to zero (- mrω² + mrω² = 0) This outward force acting on the stone + mrω² is called centrifugal force.

Question 8.
Briefly explain ‘rolling friction’.
Answer:
When an object moves on a surface essentially it is sliding on it. But the wheels move on the surface through rolling motion. In case of rolling motion when a wheel moves on a surface the point of contact is always at rest. Since the point of contact is at rest, there is no relative motion between the wheel and the surface. Hence the frictional force is very less.

At the same time if an object moves without a wheel, there is a relative motion between the object and the surface. As a result frictional force is larger. This makes it difficult to move the object.

Ideally in pure rolling, motion of point of contact with the surface should be at rest, but in practice it is not so. Due to elastic nature of the surface at the point of contact there will be same deformation on the wheel or on the surface. Due to this deformation there will be minimal friction between wheel and surface. It is called rolling friction. In fact rolling friction if much smaller than kinetic friction.

Question 9.
Describe the method of measuring angle of repose.
Answer:
Angle of repose is the minimum angle that an inclined plane makes with horizontal when a body placed on it just begins to slide. Consider a body of mass m placed on an inclined plane. The angle of inclination 0 of the inclined plane is adjusted that the body on the plane begins to slide down. Thus 0 is the angle of repose.

Various forces acting on the body are:
(a) Weight mg of the body acting vertically downwards

(b) The limiting friction fs (max) in upward direction along the inclined plane. It balances the component mg sin θ of the weight mg along the inclined plane.
∴ mg sin θ = f_s(max) …. (1)

(c) The normal reaction ‘N’ perpendicular to the inclined plane. It is balanced by the component mg cos θ, perpendicular to the inclined plane.

Where θ is the angle of repose.

Question 10.
Explain the need for banking of tracks.
Answer:
In a leveled circular road skidding mainly depends on the co-efficient of static friction n_s. The coefficient of static friction depends on the nature of surface which has a maximum limiting value. To avoid this usually “the outer edge of the road is slightly raised compared to inner edge”. This is called banking of roads or tracks. The angle of inclination called banking angle.

Let the surface of the road make angle θ with horizontal surface. Then the normal force makes an angle θ with vertical. When the car takes a turn, two forces are acting on the car.
(a) Gravitational force mg (downwards)
(b) Normal force N (Perpendicular to surface).
Normal force ‘N’ can be resolved into two components N cos θ and N sin θ and balances downward gravitational force.
N sin θ provides necessary centripetal acceleration, According to II law
N cos θ = mg
N sin θ = \(\frac{m v^{2}}{r}\)
Dividing the above equations,
tan θ = \(\frac{v^{2}}{r g}\)
V = \(\sqrt{r g \tan \theta}\)
∴ The banking angle θ and radius of curvature of the road or track determines the safe speed of car at the turning.
If the speed exceeds this safe limit, then it starts to skid outward but the frictional force comes into effect and provides an additional centripetal force to prevent outward skidding. But at the same time if the speed is less than the safe limit it starts to skid inward and again frictional force come into effect which reduces centripetal force to prevent inward skidding

However if the speed of the vehicle is sufficiently greater than the correct speed the frictional force cannot stop the car from skidding. So to avoid skidding in circular road or tracks they are banked.

Question 11.
Calculate the centripetal acceleration of moon towards the earth.
Answer:
The moon orbits the earth once in 27.3 days in an almost circular orbit.
Radius of earth = 6.4 x 10⁶ m.

Solution:
The centripetal accelerations given by a = \(\frac{ v^{2}}{r}\)
This can be related with moon a_m = R_mω²
ω → angular velocity
a_m → centripetal acceleration of the moon due to earths gravity to angular velocity
R_m → Distance between the center of earth to moon. Which is 60 times the radius of the earth.

IV. Conceptual Questions:

Question 1.
Why it is not possible to push a car from inside?
Answer:
While trying to push a car from outside, he pushes the ground backward at an angle. The ground offers an equal reaction in the opposite direction, so car can be moved. But the person sits inside means car and the person becomes a single system, and the force given will be a internal force. According to Newton’s third law, total internal force acting on the system is zero and it cannot accelerate the system.

Question 2.
There is a limit beyond which the polishing of a surface increases frictional resistance rather than decreasing it. Why?
Answer:
When surfaces are highly polished the area of contact between them increases. As result of this a large number of atoms and molecules lying on both surfaces start exerting strong attractive forces on each other. Therefore the frictional force increases.

Question 3.
Can a single isolated force exist in nature? Explain your answer?
Answer:
No. According to Newton’s third law, for every action, there is an equal and opposite reaction. So, whatever case we consider, if there is an action there is always a reaction. So it is impossible.

Question 4.
Why does a parachute descend slowly?
Answer:
When a parachute is descending down in the atmosphere an equal and opposite force to motion of parachute is acting due to air resistance. As the area of the parachute is large the air resistance resists the motion and so it descends slowly.

Question 5.
When walking on ice one should take short steps. Why?
Answer:
When a person is walking on ice he presses the ice downward with his feet and in turn the ice pushes the person with an equal force. Since ice slippery the person is not able to press it hardly. So the action legs and so the reaction which is also less. So by making small steps, with larger normal force one can walk without slipping.

Question 6.
When a person walks on a surface the frictional force exerted by the surface on the person is opposite to the direction of motion. True or false.
Answer:
False. In frictional force exerted by the surface on the person is in the direction of his motion. The frictional force acts as an external force to move the person. When the person trying to move, he gives a push to ground on the backward direction and by Newton’s third law he is pushed by the ground in the forward direction. Hence frictional force acts along the direction of motion.

Question 7.
Can the coefficient of friction be more than one?
Answer:
No, it cannot be more than 1 for normal plane surfaces. But when surfaces are so irregular that they have sharp minute projections and cavities on them. Then the coefficient of friction may be more than one.

Question 8.
Can we predict the direction of motion of a body from the direction of force on it?
Answer:
Yes. The direction of motion is always opposite to the force of kinetic friction. By using the principle of equilibrium, the direction of force of static friction can be determined. When the object is in equilibrium, the frictional force must point in the direction which results as a net force is zero.

Question 9.
The momentum of a system of particles is always conserved. True or false.
Answer:
True: The Linear momentum of the system is always a constant vector, as long as no external force are acting on it.

V. Numerical Problems:

Question 1.
A force of 50N acts on the object of mass 20kg shown in the figure. Calculate the acceleration of the object in x and y-direction.

Solution:
m = 20kg, F = 50N
Force can be resolved into 2 components.
m = 20kg
F_x = F cos θ = 50 cos 30 = \(\frac{50 \sqrt{3}}{2}\)\(\frac { 1 }{ 2 }\)
F_y = F sin θ = 50 sin 30 = 50 x \(\frac { 1 }{ 2 }\) = 25N

Question 2.
A spider of mass 50 g is hanging on a string of a cobweb as shown in the figure. What is the tension in the string?
Solution:

Question 3.
What is the reading shown in spring balance?
Solution:

∵ Forces on both sides are equal the reading in the spring balance is zero.

(ii)

The spring is pulled by a force along the inclined plane so.
F = mg sin θ
= 2 x 9.8 x sin 30
= 9.8 x 2 x 1/2
F = 9.8 N

Question 4.
The physics books are stacked on each other in the sequence +1 volumes 1 and 2, +2 volumes 1 and 2 on a table, a) identify the forces acting on each book and draw the free body diagram, b) Identify the forces exerted by each books on the other.
Solution:

Force on book A
1) Downward gravitational force exerted by earth (m_A g)
2) Upward normal force (N_B) exerted by book B (N_B).

Force on book B
i) Downward gravitational force exerted by earth (m_B g)
ii) Downward force exerted by books A (N_A)
iii) Upward normal force exerted by book C (Nc).

Force on book C
i) Downward gravitational force exerted by earth (m_C g)
ii) Downward force exerted by books B (N_B)
iii) Upward normal force exerted by book D (N_D).

Force on book D

i) Downward gravitational force exerted by earth (m_D g)
ii) Downward force exerted by books C (N_c)
iii) Upward force exerted by the N_Table.

Question 5.
A bob attached to the string oscillates back and forth. Resolve the forces acting on the bob into components. What is the acceleration experience by the bob at an angle 0?
Solution:

(i) Gravitational force(mg) acting downwards.
(ii) Tension (T) exerted by the string on the bob whose position determines the direction of T.
The bob is moving is a circular arc and so it has centripetal acceleration. At points A and C bob comes to rest momentarily and then its velocity increases when move towards B.
Hence tangential acceleration is along the arc.

The gravitational force can be resolved into two components.
mg cos θ along the string
mg sin θ perpendicular to the string
At point A & C T=mg cos θ and at all other points T is greater than mg cos θ.
∴ Centripetal force = T – mg cos θ
∴ ma_c = T- mg cos θ
Centripetal acceleration
a_c = \(\frac { T – mg cos θ}{ 2 }\)m

Question 6.
Two masses m₁ and m₂ are connected with a string passing over a frictionless pulley fixed at the corner of the table as shown in the figure. The coefficient of static friction of mass m₁ with the table is µ_s. Calculate the minimum mass m₃ that may be placed on m₁ to prevent it from sliding. Check if m₁ = 15kg, m₂ = 10kg, m₃ = 25kg and µ_s = 0.2.

Solution:
(1) The mass m₁ and m₂ provides maximum static friction on the table
fS_(max) = µ_sN
= µ_s(m₁ + m₃)g
fS_(max) = 0.2(m₁ + m₃)g …. (1)
Tension T in the string provided by
= T = m₂g …. (2)
T < fs_(max) then m₁ and m₂ won’t move.
If m₃ is not present
T = fs (for just sliding)
When m₃ is placed on m₁ it won’t slide

(2) If m₁ = 15kg
m₂ = 10kg
m₃ = 25kg
µ_s = 0.2.
Then µ(m₁+m₃)g = 0.2(15±10)g
= 2.5 x 0.2g = 5 g =5g N
= 5g N
But m₂ g = 10g N
∵ m₂g = 10g N = T >µ_s(m₁ + m₃)g
∵ T > µs (m₁ + m₃)
m₁ + m₃ will slide on the table.

Question 7.
Calculate the acceleration of the bicycle of mass 25 kg as shown in fig 1 & 2.

Solution:
Case 1:
Resultant force =
forward force – Frictional force
= 500 – 400
F_R = 100N
F_R = ma = 100N
a = \(\frac { 100 }{ 25 }\)
a = 4 ms^-2

Case 2:
Resultant force = forward force – Frictional force = 400 – 400 = 0 N
a = 0.

Question 8.
Apply Lamis theorem on sling shot and calculate the Tension in each string?

Solution:

Question 9.
A foot ball player kicks a 0.8 kg ball and imparts it a velocity 12 ms^-1. The contact between foot and ball is only for one sixtieth of a second find the average kicking force.
Solution:
m = 0.8 kg
v = 12 ms^-1
t = \(\frac { 1 }{ 60 }\)S
F = ?
Change in momentum = impulse
P_f – P_i = F_t
mv – 0 = Ft

Question 10.
A stone of mass 2kg is attached to a string of length 1m. The string can withstand a maximum tension of 200N, What is the maximum speed that stone can have during the whirling motion.
Solution:
m = 2 kg
I = 1 m = r
T = F = 200 N
V = ?
During whirling motion, the force acting on the stone is a centripetal force which provides the necessary Tension in the string.
T = F = \(\frac{m v^{2}}{r}\)
200 = \(\frac{2 \times v^{2}}{1}\)
V_max = 10 ms^-1

Question 11.
Imagine that the gravitational force between Earth and Moon is provided by an invisible string that exists between the Moon and Earth. What is the tension that, exists in this invisible string due to earth’s centripetal force?
Solution:
Mass of moon = 7.34 x 10²² kg
Distance between the moon and earth = 3.84 x 10⁸ m

Question 12.
Two bodies of masses 15kg and 10kg are connected with light string kept on a smooth surface. A horizontal force F = 500N is applied to a 15kg as shown in the figure. Calculate the tension acting in the string.
Answer:

Let m₁ = 15kg
m₂ = 10kg
F = 500N
All the two blocks move with a common acceleration (a) under the force F = 500 N.
F = (m₁+m₂)

Question 13.
People often say “for every action there is an equivalent opposite reaction”. Here they meant ‘action of a human’. Is it correct to apply Newton’s third law to human actions? What is mean by ‘action’ in Newton’s third law? Give your arguments based on Newton’s laws.
Solution:
Newton’s third law is applicable to only humans actions that involve physical force. Third law is not applicable to humans’ psychological actions or thoughts.

Question 14.
A car takes a turn with velocity 50ms^-1 on the circular road of the radius of curvature 10m. Calculate the centrifugal force experienced by a person of mass 60kg inside the car?
Solution :
V = 50 ms^-1
r = 10m
centrifugal reaction = ?
m = 60kg
F = \(\frac{m v^{2}}{r}\)
= \(\frac { 60×50×50 }{ 10 }\)
= 3 x 5 x 10³
F = 15,000 N

Question 15.
A long stick rests on the surface. A person standing 10m away from the stick with what minimum speed an object of mass 0.5 kg should he threw so that it hits the stick. (Coefficient of kinetic friction is 0.7)

Solution:
Force on the mass = μ_kN = μ_k mg
F = 0.7 x 0.5 x 9.8
F = 3.43 N
But F = ma
a = \(\frac { 3.43 }{ 0.5 }\)
= 6.86ms^-2
m = 0.5kg
W.K.T
v² – u² = 2as
|u²| = 2as
u = \(\sqrt{2 \times 6.86 \times 10}\)
= 11.71 ms^-2
Velocity with which the mass is thrown = 11.71 ms^-1

11th Physics Guide Laws of Motion Additional Important Questions and Answers

I. Multiple choice questions:

Question 1.
The concept “force causes motion” was given by –
(a) Galileo
(b) Aristotle
(c) Newton
(d) Joule
Answer:
(b) Aristotle

Question 2.
A body of 5 kg is moving with a velocity of 20m/s. If a force of 100 N is applied on it for 10s in the same direction as its velocity, what will now be the velocity of the body _______.
a) 2000 m/s
b) 220 m/s
c) 240 m/s
d) 260 m/s
Answer:
b) 220 m/s

Question 3.
The inability of objects to move on their own or change their state of motion is called as –
(a) force
(b) momentum
(c) inertia
(d) impulse
Answer:
(c) inertia

Question 4.
A force vector applied on a mass is represented as \(\vec { f }\) = 6\(\vec { i }\) – 8\(\vec { j }\) + 10\(\vec { k }\) and accelerates with 1ms^-2, what will be the mass of the body in kg _______.
a) 10\(\sqrt{2}\)
b) 20
c) 2\(\sqrt{10}\)
d) 10
Answer:
a) 10\(\sqrt{2}\)

Question 5.
A particle of mass m is at rest at the origin at time t = 0. It is subjected to force F(t)= F₀e^-bt in the x direction. Its speed v(t) is depicted by which of the following curves?

Answer:

Question 6.
If the brake is applied in the moving bus suddenly, passengers move forward is an example for –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(a) Inertia of motion

Question 7.
Two masses m₁ and m₂ are connected by a light string passing over a smooth pulley. When set free m₁ moves down by 2m in 2s the ratio of m₁/m₂ is _______.
a) 9/7
b) 11/ 9
c) 13/11
d) 15/13
Answer:
b) 11/ 9

Question 8.
Two blocks are connected by a string as shown in the figure. The upper block is hung by another string. A force ‘F’ applied on the upper string produces an acceleration of 2ms^-2. in the upward direction in both the blocks. If T and T¹ in both parts of the string then _______.

a) T = 84N; T¹ = 50N
b) T = 70N; T¹ = 70N
c) T = 84N; T¹ = 60N
d) T = 70N; T¹ = 60N
Answer:
c) T = 84N; T¹ = 60N

Question 9.
Four identical blocks each of mass m linked by threads as shown. IF the system moves with constant acceleration under the influence of force F, the Tension T² _______.

a) F
b) F/2
c) 2F
d) F/4
Answer:
b) F/2

Question 10.
Rate of change of momentum of an object is equal to –
(a) acceleration
(b) work done
(c) force
(d) impulse
Answer:
(c) force

Question 11.
A block of mass 10 kg is pushed on a smooth inclined plane of inclination 30°, so that it has an acceleration 2ms^-2. The applied force is _______.
a) 50 N
b) 60 N
c) 70 N
d) 80 N
Answer:
c) 70 N

Question 12.
Two blocks of masses 6 kg and 3 kg are connected by the string as shown over a frictionless pulley. The acceleration of the system is _______.

a) 4 ms^-2
b) 2 ms^-2
c) Zero
d) 6 ms^-2
Answer:
c) Zero

Question 13.
The acceleration of the systems shown in the figure is _______.

a) 1 ms^-2
b) 2 ms^-2
c) 3 ms^-2
d) 4 ms^-2
Answer:
a) 1 ms^-2

Question 14.
A uniform rope of length L is pulled by constant force P as shown, The Tension in the rope at a distance x from the end where the force is applied _______.

a) P
b) p[a – \(\frac { x }{ L }\) ]
c) px/L
d) p(1 + \(\frac { x }{ L }\) )
Answer:
b) p[a – \(\frac { x }{ L }\) ]

Question 15.
The law which is valid in both inertial and non-inertial frame is –
(a) Newton’s first law
(b) Newton’s second law
(c) Newton’s third law
(d) none
Answer:
(c) Newton’s third law

Question 16.
One end of a string of length / is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards centre) is _______.
a) T
b) T – \(\frac{m v^{2}}{r}\)
c) T + \(\frac{m v^{2}}{r}\)
d) 0
Answer:
a) T

Question 17.
A block is kept on a frictionless inclined surface with the angle of inclination α. The incline is given an acceleration ‘a’ to keep the block stationary. Then a is equal to _______.

a) g
b) g tan α
c) g / tan α
d) g cosec α
Answer:
b) g tan α

Question 18.
The action and reaction forces acting on –
(a) same body
(b) different bodies
(c) either same or different bodies
(d) none of the above
Answer:
(b) different bodies

Question 19.
If an elevator moving vertically up with an acceleration g, the force entered on the floor by a passenger of mass M is _______.
a) mg
b) 1 / 2 mg
c) zero
d) 2 mg
Answer:
d) 2 mg

Question 20.
A man of weight 80 kg, stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5 ms^-2. What would be the reading on the scale? [g = 10 ms^-2].
a) 400 N
b) 1200 N
c) 800 N
d) zero
Answer:
b) 1200 N

Question 21.
A person is standing in an elevator in which situation he find his weight more than the actual weight _______.
a) the elevator moves upwards with constant acceleration
b) the elevator moves downwards with constant acceleration
c) the elevator moves upwards with uniform velocity
d) the elevator moves downwards with uniform velocity
Answer:
b) the elevator moves downwards with constant acceleration

Question 22.
Newton’s second law gives –
(a) \(\overrightarrow{\mathrm{F}} \propto \frac{d \overrightarrow{\mathrm{P}}}{\mathrm{dt}}\)
(b) \(\overrightarrow{\mathrm{F}}=\frac{d \overrightarrow{\mathrm{P}}}{\mathrm{dt}}\)
(c) \(\overrightarrow{\mathrm{F}}=m \vec{a}\)
(d) all the above
Answer:
(d) all the above

Question 23.
A mass of 1 kg is suspended by a string. Another string C is connected to its lower end (as in the figure). If a sudden jerk is given to c, then _______.

a) The portion AB of the string break
b) The portion BC of the string will break
c) The mass will be rotating
d) none of the above
Answer:
a) The portion AB of the string break

Question 24.
In the above questions, if the string c is stretched slowly then _______.
a) The portion AB of the string break
b) The portion BC of the string will break
c) The mass will be rotating
d) none of the above
Answer:
a) The portion AB of the string break

Question 25.
As shown in the figure, the tension in the horizontal card is 30N. The weight w and the tension in the string OA in Newton are _______.

a) 30, \(\sqrt{3}\), 30
b) 30\(\sqrt{3}\), 60
c) 60\(\sqrt{3}\), 30
d) none of the above
Answer:
b) 30\(\sqrt{3}\), 60

Question 26.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-1. How long does the body take to stop?
(a) 0.75 s
(b) 1.33 s
(c) 6 s
(d) 35 s
Answer:
Acceleration a = \(\frac{-F}{m}\) = \(\frac{50}{20}\) = – 2.5 ms^-2
u = l5 ms^-1
v = 0
t = ?
v = u + at
0 = 15 – 2.5t
t = \(\frac{15}{2.5}\) = 6s

Question 27.
A block of mass 10 kg is suspended through two light spring balances as in figure.

a) both scales will read 10 kg
b) both scales will read 5 kg
c) the upper scale will read 10 kg and the lower zero
d) the reading may be anything but their sum will be 10 kg
Answer:
a) both scales will read 10 kg

Question 28.
Two blocks A and B of masses 2m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as in figure. The magnitude of the acceleration of A and B after the string is cut, are respectively.

a) g, g/2
b) g/2, g
c) g, g
d) g/2, g/2
Answer:
b) g/2, g

Question 29.
Choose the correct statement _______.
a) The frictional forces are dependent on the roughness of the surface.
b) The kinetic friction is proportional to normal reaction
c) The friction is independent of area of contact
d) All statements are correct
Answer:
d) All statements are correct

Question 30.
When an object of mass m slides on a frictionless surface inclined at an angle θ, then the normal force exerted by the surface is-
(a) g cos θ
(b) mg cos θ
(c) g sin θ
(d) mg tan θ
Answer:
(b) mg cos θ

Question 31.
Two cars of unequal masses are similar types. If they are moving at the same initial speed, the minimum stopping distance _______.
a) is smaller for the heavier car
b) is smaller for the lighter car
c) is same for both car
d) depends on volume of the car
Answer:
c) is same for both car

Question 32.
An ice block is kept on an inclined plane of angle of 30°. The coefficient of kinetic friction between the block and the inclined plane is \(\frac{1}{\sqrt{3}}\). The acceleration of the block is _______.
a) zero
b) 2 ms^-2
c) 1.5 ms^-2
d) 5 ms^-2
Answer:
d) 5 ms^-2

Question 33.
Starting from rest a body slides down at 45° inclined plane in twice the time, it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is __________.
a) 0.33
b) 0.25
c) 0.75
d) 0.80
Answer:
c) 0.75

Question 34.
A uniform metal chain if placed on a rough table such that the one end of the chain hangs down over the edge of the table. When one-third of its length hangs over the edge, the chain starts sliding. Then the coefficient of static friction is _______.
a) 3/4
b) 1/4
c) 2/3
d) 1/2
Answer:
d) 1/2

Question 35.
If two masses m₁ and m₂ (m₁ > m₂) tied to string moving over a frictionless pulley, then the acceleration of masses –
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g
(b) \(\frac{m_{1}+m_{2}}{\left(m_{1}-m_{2}\right)}\) g
(c) \(\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\) g
(d) \(\frac{m_{1} m_{2}}{2 m_{1} m_{2}}\) g
Answer:
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g

Question 36.
While walking on ice one should take small steps to avoid slipping. This is because smaller steps ensure _______.
a) large friction
b) smaller friction
c) larger normal force
d) smaller normal force
Answer:
c) larger normal force

Question 37.
A box is placed on inclined plane and has to be pushed down. The angle of inclination is _______.
a) equal to angle of friction
b) more than angle of friction
c) equal to angle repose
d) less than angle of repose
Answer:
d) less than angle of repose

Question 38.
Three masses in contact is as shown above. If force F is applied to mass m₁ then the contact force acting on mass m₂ is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)

Question 39.
A block B is placed on block A. The mass block B is less than the mass of block friction that exists between the blocks whereas the ground on which block A is placed is taken to be smooth. A horizontal force ‘F’ increasing linearly with time beings to act on B. The acceleration a_A and a_B of blocks A and B respectively are plotted against Y. The correctly plotted graph is _______.

Answer:

Question 40.
Two blocks of mass m₁ 6 kg and m₂ = 3 kg as in figure coefficient of friction between m₁, and m₂ and between m₁ and surface is 0.5 and 0.4 respectively. The maximum horizontal force to can be applied to the mass m₁ so that they move without separation is _______.

a) 41 N
b) 61 N
c) 81 N
d) 101 N
Answer:
c) 81 N

II. Additional Questions:

Question 1.
Identify the internal and external forces acting on the following systems.
a) Earth alone as a system
b) Earth and sun as a system
c) Our body as a system white walking
d) our body and earth as a system
Solution:
(a) Earth alone as a system:

Earth orbits the sun due to the gravitational attraction of the sun. If we consider earth as a system, then the sun’s gravitational force is an external force. If the moon is also taken into account, it also exerts an external force on earth.

(b) (Earth + sun) as a system:

In this case, there are two internal forces which form action and reaction pair, i.e gravitational force on the sun by earth & vice versa.

(c) Our body as a system:
While walking, we exert a force on earth and earth exerts an equal and opposite force on our body. If our body is considered as a system then the force exerted by the earth on our body is external.

(d) Our body + earth as a system.
In this case, there are two internal forces present in the system one is the force exerted by our body on earth and the other is the equal and opposite force exerted by the earth on our body.

Question 2.
When a cricket player catches the ball he pulls his hands gradually is the directions of the ball’s motion. Why?
Answer:
If the man stops his hands soon after catching the ball the ball comes to rest quickly. It means that the momentum of the ball is brought to rest very quickly. So the average force acting on the body will be very large. Due to the large average force, the hands will get hurt. To avoid this the player brings the ball to rest slowly.

Question 3.
An impulse is applied to a moving object with a force at an angle of 20° w.r.t. velocity vector, what is the angle between the impulse vector and change in momentum vector?
Answer:
Impulse and change in momentum are in the same direction. Therefore the angle between these two vectors is zero.

Question 4.
Why a high jumper after crossing the bar made to fall on a spongy floor instead of the cemented floor.
Answer:
After high jumping, landing on a cemented floor is more dangerous than jumping on a spongy surface. A spongy surface brings the body to rest slowly than the cemented floor So that the average force experienced by the body will be lesser.

Question 5.
Obtain an expression for centrifugal force acting on a man on the surface of the earth.
Answer:

Even though earth is treated as an inertial frame, it is actually not so. Earth spins about its own axis with an angular velocity co Any object at the surface of the earth (rotational frame) experiences a centrifugal force. The centrifugal force appears to act exactly in opposite direction from the axis of rotation.

The centrifugal force on a man standing on the surface of earth is = mrω² where r is the perpendicular distance of man from the axis of rotation. From fig. r = R cos θ where R= radius of earth.
θ = latitude of the earth where man is standing

Question 6.
A stone when thrown on a glass window smashes the window pane to pieces, but a bullet from the gun passes through, by making a clean hole. Why?
Answer:
Due to small speed the stone remains in contact with the window pane for a long time. It transfer its motion to the pane and break them into pieces. But the particles of window pane near a hole are unable to share the fast motion of the bullet and so remain undisturbed.

Question 7.
China wares are wrapped in a straw paper before packing why?
Answer:
The straw paper between the china-wares increases the time of experiencing the jerk during transportation. Hence they strike against each other with less force and are less likely to get damaged.

Question 8.
Why it is necessary to bend knees while jumping from greater height?
Answer:
During the jump, our feet comes to rest at once and for this smaller time a larger force a acts on feet. If we bend the knees slowly, the value of time impact increases and less force acts on our feet. So we get less hurt.

Question 9.
Why is it difficult to drive a nail into a wooden block without supporting it?
Answer:
When we hit the nail with the hammer the nail and unsupported block together move forward as a single system. There is no force of reaction. When the block is rested against support, the reaction of the support holds the 14. block in position and nail is driven into the block.

Question 10.
Why is static friction called a self-adjusting force?
Answer:
As applied force increases, the static friction also increases and becomes equal to the applied force. That is why static friction is called a self-adjusting force.

Question 11.
Carts with rubber tyres are easier to fly than those with iron wheels. Why?
Answer:
The coefficient of friction between rubber tyres and road is much smaller than that between iron wheels and road.

Question 12.
Why ball bearings are used in machinery?
Answer:
By using ball bearings between the moving parts of the machinery the sliding friction gets converted into rolling friction. The rolling friction is much lesser than the sliding friction. This reduces power dissipation.

Question 13.
A bird is sitting on the floor of a closed glass cage and the cage is in the hand of a girl. Will the girl experience any change in weight of the cage when the bird –
(i) starts flying in the cage with a constant velocity
(ii) flies upwards with acceleration
(iii) fies downwards with acceleration.
Solution:
In a closed cage, the inside air is bound with a cage.
(i) As the acceleration is zero, there is no change in the weight of the cage.
(ii) while flying upwards R – Mg = Ma
R = M (a+g)
The cage will appear heavier

(iii) while flying downwards
Mg – R – Ma
R = M (g – a)
The cage will appear lighter.

Question 14.
A long rope is hanging, passing over a pulley. Two monkeys of equal weight climb up from the opposite ends of the rope. One of them climbs up more rapidly relative to rope. Which monkey will reach the top first? Pulley is frictionless and the rope is mass less and inextensible.
Answer:
There is no external force which may provide momentum to any monkey. The monkeys themselves give equal momenta to each other. Therefore two monkeys will climb up the rope at the same rate relative to earth. As their masses are equal they will reach the top simultaneously.

Question 15.
A light string passing over a smooth pulley connects two blocks of masses m1 and m2 (vertically). If the acceleration of the system is g/8. Find the ratio of the two masses.
Answer:

Question 16.
Briefly explains how a horse is able to pull a cart?
Answer:

Consider a cart connected to a horse by a string. The horse while pulling the cart produces a Tension T in the string in their forward direction (action). The cart in turn pulls the horse by an equal force T in opposite direction.

Initially, the horse presses the ground with a force ‘F’ in an inclined direction. The reaction ‘R’ of the ground acts on the horse in the opposite direction.

The reaction ‘R’ can be resolved into two perpendicular components.
1) The vertical components ‘V balances the weight of the horse.
2) The horizontal components ‘H’ helps the horse to move forward.
Let F be the force of friction.
The horse moves forward if H > T.
In that case net force acting on horse = H – T
If m is mass of horse and ‘a’ be its acceleration.
H – T = ma → (1)
The cart moves forward if T > F.
In this case, net force acting on the cart = T – F.
The weight of the cart is balanced by the reaction of the ground acting on it.
If M is the man of cart and a is its acceleration T-F = Ma -»(2).
Adding (1) & (2)
H – F = (M+m)a
a = \(\frac { H-F }{ M+m }\)
obviously a is positive of H-F is positive or H > F. Thus the system moves.

Question 17.
A man of mass m is standing on the floor of a lift. Find his apparent weight when the lift is (i) moving upwards with uniform acceleration ‘a’ (ii) moving downwards with uniform acceleration a (iii) at rest or moving with uniform velocity (v) (iv) falling freely. Take acceleration due to gravity as g.
Answer:
Consider a man of mass ‘m’ standing on a weighing machine placed in a lift. The actual weight of the man is mg. It acts vertically downwards through center of gravity ‘G’ of the man, which acts in a weighting machine which offers a normal reaction ‘R’ read by a machine.
So R is the apparent weight of the man.

Case 1:
When the lift moves upwards with an acceleration ‘a’
The net upward force on the man = R – mg = ma
Apparent weight R = m(g+a)
So when the lift accelerates upwards, the apparent weight of man increases

Case 2:
When lift moves downwards with acceleration ‘a’ Net downward force: mg – R = ma R = m(g – a)
So when lift accelerates down, the apparent weight of man decreases.

Case 3:
When lifting at rest or moving with uniform velocity.
So acceleration ‘a’ = 0
Net force on man is R – mg = mx 0
R – mg = 0
R = mg
Apparent weight = actual weight

(iv) when lift falls freely a = g
the net downward force mg – R = ma
mg – R = mg
R = 0
The apparent weight of the man equal to zero.
Hence the person develops a feeling of weightlessness when he falls freely under gravity.

Question 18.
Derive the law of conservation of linear momentum from Newton’s third law of motion.
Answer:
Consider two bodies A and B of masses m₁ and m₂ moving in the same directions along the same line with velocities u₁ and u₂ respectively (U₁>u₂).
They collide for a time At. After the collision, their velocities be v₁ and v₂

During collision the body A exerts a force F^BA on B and the body B exerts a force F^AB on A such that
F^BA = – F^AB
Impulse of F^AB = \(\overline{F}\)_AB ∆t = change in momentum is A
\(\overline{F}\)_BA ∆t = m₁\(\overline{v}\)₁ – m₁\(\overline{u}\)₁
Impulse of F_BA = \(\overline{F}\)_BA ∆t = change in momentum

Total linear momentum after collision = Total linear momentum before collision
This proves the law of conservation of linear momentum.

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 1 Numbers Ex 1.5 Textbook Questions and Answers, Notes.

Simplifying Rational Expressions Calculator.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 1 Numbers Ex 1.5

Question 1.
Fill in the blanks
(i) The ones digits in the cube of 73 is __________ .
Answer:
7

(ii) The maximum number of digits in the cube of a two digit number is __________ .
Answer:
6

(iii) The smallest number to be added to 3333 to make it a perfect cube is __________ .
Answer:
42

(iv) The cube root of 540×50 is __________ .
Answer:
30

(v) The cube root of 0.000004913 is __________ .
Answer:
0.017

Question 2.
Say True or False.
(i) The cube of 24 ends with the digit 4.
Answer:
True

(ii) Subtracting 103 from 1729 gives 93.
Answer:
True

(iii) The cube of 0.0012 is 0.000001728.
Answer:
False

(iv) 79570 is not a perfect cube.
Answer:
True

(v) The cube root of 250047 is 63.
Answer:
True

Question 3.
Show that 1944 is not a perfect cube.
Answer:

1944 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3

= 2³ × 3³ × 3 × 3
There are two triplets to make further triplets we need one more 3.
∴ 1944 is not a perfect cube.

Question 4.
Find the smallest number by which 10985 should be divided so that the quotient is a perfect cube.
Answer:

We have 10985 = 5 × 13 ×13 × 13
= 5 × 13 ×13 × 13
Here we have a triplet of 13 and we are left over with 5.
If we divide 10985 by 5, the new number will be a perfect cube.
∴ The required number is 5.

Question 5.
Find the smallest number by which 200 should be multiplied to make it a perfect cube.
Answer:


Grouping the prime factors of 200 as triplets, we are left with 5 × 5
We need one more 5 to make it a perfect cube.
So to make 200 a perfect cube multiply both sides by 5.

1000 = 2 × 2 × 2 × 5 × 5 × 5 × 5 × 5
Now 1000 is a perfect cube.
∴ The required number is 5.

Question 6.
Find the cube root 24 × 36 × 80 × 25.
Answer:

Question 7.
Find the cube root of 729 and 6859 prime factorisation.
Answer:
(i)


= 3 × 3
\(\sqrt[3]{729}\) = 9

(ii) \(\sqrt[3]{6859}\) = \(\sqrt[3]{19 \times 19 \times 19}\)
\(\sqrt[3]{6859}\) = 19

Question 8.
What is the square root of cube root of 46656?
Answer:

We have to find out \(\sqrt{(\sqrt[3]{46656})}\)
First we will find \(\sqrt[3]{46656}\)

∴ The required number is 6.

Question 9.
If the cube of a squared number is 729, find the square root of that number.
Answer:

(729)^1/3 = 3 × 3 = 9
∴ The cube of 9 is 729.
9 = 3 × 3 [ie 3 is squared to get 9]

We have to find out √3,
√3 = 1.732

Question 10.
Find two smallest perfect square numbers which when multiplied together gives a perfect cube number.
Answer:
Consider the numbers 2² and 4²
The numbers are 4 and 16.
Their procluct 4 × 16 = 64
64 = 4 × 4 × 4
∴ The required square numbers are 4 and 16

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.5 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5

Question 1.
Solve 2x² + x – 15 ≤ 0
Answer:
The given inequality is
2x² + x – 15 ≤ 0 ——— (1)
2x² + x – 15 = 2x² + 6x – 5x – 15
= 2x (x + 3) – 5 (x + 3)
= (2x – 5)(x + 3)
2x² + x – 15 = 2\(\left(x-\frac{5}{2}\right)\))(x + 3) ——— (2)
The critical numbers are x – \(\frac{5}{2}\) = 0 or x + 3 = 0
The critical numbers are x = \(\frac{5}{2}\) or x = – 3
Divide the number line into three intervals

(i) (- ∞, – 3)
When x < – 3 say x = – 4
The factor x – \(\frac{5}{2}\) = – 4 – \(\frac{5}{2}\) < 0 and
x + 3 = – 4 + 3 = – 1 < 0
x – \(\frac{5}{2}\) < 0 and x + 3 < 0
⇒ \(\left(x-\frac{5}{2}\right)\) (x + 3) > 0
Using equation (2) 2x² + x – 15 > 0
∴ 2x² + x – 15 ≤ 0 is not true in (- ∞, – 3)

(ii) \(\left(-3, \frac{5}{2}\right)\)
When – 3 < x < \(\frac{5}{2}\) say x = 0
The factor x – \(\frac{5}{2}\) = 0 – \(\frac{5}{2}\) = – \(\frac{5}{2}\) < 0 and
x + 3 = 0 + 3 = 3 > 0
x – \(\frac{5}{2}\) < 0 and x + 3 > 0
⇒ \(\left(x-\frac{5}{2}\right)\) (x + 3) < 0
using equation (2) 2x² + x – 15 < 0
∴ 2x² + x – 15 ≤ 0 is true in \(\left(-3, \frac{5}{2}\right)\)

(iii) \(\left(\frac{5}{2}, \infty\right)\)
When x > \(\frac{5}{2}\) say x = 3
The factor x – \(\frac{5}{2}\) = 3 – \(\frac{5}{2}\) > 0 and
x + 3 = 3 + 3 > 0
x – \(\frac{5}{2}\) > 0 and x + 3 > 0
= \(\left(x-\frac{5}{2}\right)\) (x + 3) > 0
Using equation (2) 2x² + x – 15 > 0
∴ 2x² + x – 15 ≤ 0 is not true in \(\left(\frac{5}{2}, \infty\right)\)

We have proved the inequality 2x² + x – 15 ≤ 0 is true in the interval \(\left(-3, \frac{5}{2}\right)\)
But it is not true in the interval

Inequality solver that solves an inequality with the details of the calculation: linear inequality, quadratic inequality.

Question 2.
Solve x² + 3x – 2 ≥ 0
Answer:
The given inequality is
– x² + 3x – 2 ≥ 0
x² – 3x + 2 < 0 ——– (1)
x² – 3x + 2 = x² – 2x – x + 2
= x(x – 2) – 1(x – 2)
x² – 3x + 2 = (x – 1) (x – 2) ——— (2)
The critical numbers are
x – 1 = 0 or x – 2 = 0
The critical numbers are
x = 1 or x = 2
Divide the number line into three intervals
(- ∞, 1), (1, 2) and (2, ∞).

(i) (- ∞, 1)
When x < 1 say x = 0
The factor x – 1 = 0 – 1 = – 1 < 0 and
x – 2 = 0 – 2 = – 2 < 0
x – 1 < 0 and x – 2 < 0
⇒ (x – 1)(x – 2) > 0
Using equation (2) x² – 3x + 2 > 0
∴ The inequality x² – 3x + 2 ≤ 0 is not true in the interval (- ∞, 1 )

(ii) (1, 2)
When x lies between 1 and 2 say x = \(\frac{3}{2}\)
The factor x – 1 = \(\frac{3}{2}\) – 1 = \(\frac{1}{2}\) > 0 and
x – 2 = \(\frac{3}{2}\) – 2 = – \(\frac{1}{2}\) – < 0
x – 1 > 0 and x – 2 < 0
⇒ (x – 1)(x – 2) < 0
Using equation (2) x² – 3x + 2 < 0
∴ The inequality x² – 3x + 2 ≤ 0 is true in the interval (1, 2 )

(iii) (2, ∞)
When x > 2 say x = 3
The factor x – 1 = 3 – 1 = 2 > 0 and
x – 2 = 3 – 2 = 1 > 0
x – 1 > 0 and x – 2 > 0
= (x – 1)(x – 2) > 0
Using equation (2) x² – 3x + 2 > 0
∴ The inequality x² – 3x + 2 ≤ 0 is not true in the interval (2, ∞)
We have proved the inequality x² – 3x + 2 ≤ 0 is true in the interval [ 1, 2 ].
But it is not true in the interval
(- ∞, 1) and (2, ∞)
∴ The solution set is [1, 2]

Students can download Maths Chapter 8 Statistics and Probability Ex 8.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.1

Question 1.
Find the range and coefficient of range of the following data.
(i) 63, 89, 98, 125, 79, 108, 117, 68
(ii) 43.5, 13.6, 18.9, 38.4, 61.4, 29.8
Answer:
(i) Here the largest value (L) = 125
The smallest value (S) = 63
Range = L – S = 125 – 63 = 62

Question 2.
If the range and the smallest value of a set of data are 36.8 and 13.4 respectively, then find the largest value.
Solution:
If the range = 36.8 and
the smallest value = 13.4 then
the largest value = L = R + S
= 36.8 + 13.4 = 50.2

Question 3.
Calculate the range of the following data.

Answer:
Smallest value (S) = 400
Largest value (L) = 650
Range = L – S = 650 – 400 = 250

Question 4.
A teacher asked the students to complete 60 pages of a record notebook. Eight students have completed only 32, 35, 37, 30, 33, 36, 35 and 37 pages. Find the standard deviation of the pages yet to be completed by them.
Answer:
The remaining number of pages to be completed is 60 – 32; 60 – 35; 60 – 37; 60 – 30; 60 – 33; 60 – 36; 60 – 35 and 60 – 37
The pages to be completed are, 28, 25, 23, 30, 27, 24, 25, and 23
Arrange in ascending order we get, 23, 23, 24, 25, 25, 27, 28 and 30

Free online sample standard deviation calculator and variance calculator with steps.

Question 5.
Find the variance and standard deviation of the wages of 9 workers given below:
₹ 310, ₹ 290, ₹ 320, ₹ 280, ₹ 300, ₹ 290, ₹ 320, ₹ 310, ₹ 280.
Answer:
Arrange in ascending order we get,
280, 280, 290, 290, 300, 310, 310, 320 and 320
Assumed mean = 300


Variance = 222.222
Standard deviation = √Variance = √222.222 = 14.907 = 14.91
Variance = 222.22
Standard deviation = 14.91

Question 6.
A wall clock strikes the bell once at 1 o’clock, 2 times at 2 o’clock, 3 times at 3 o’clock and so on. How many times will it strike in a particular day? Find the standard deviation of the number of strikes the bell make a day.
Answer:
Wall clock strikes the bell in 12 hours
1, 2, 3, 4, 5,… ,12
Wall clock strikes in a day (24 hours)
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.
Assumed mean = 14



The standard deviation of bell strike in a day is 6.9

Question 7.
Find the standard deviation of the first 21 natural numbers.
Answer:
Here n = 21
The standard deviation of the first ‘n’ natural numbers,

The standard deviation of the first 21 natural numbers = 6.06

Question 8.
If the standard deviation of a data is 4.5 and if each value of the data is decreased by 5, then find the new standard deviation.
Solution:
If the standard deviation of a data is 4.5 and each value of the data decreased by 5, the new standard deviation does not change and it is also 4.5.

Question 9.
If the standard deviation of a data is 3.6 and each value of the data is divided by 3, then find the new variance and new standard deviation.
Answer:
The standard deviation of the data = 3.6
Each value of the data is divided by 3
New standard deviation = \(\frac{3.6}{3}\) = 1.2
New Variance = (1.2)² = 1.44 [∴ Variance = (S.D)²]
New standard Deviation = 1.2
New variance = 1.44

Question 10.
The rainfall recorded in various places of five districts in a week are given below.

Find its standard deviation.
Answer:
Assumed mean = 60

Question 11.
In a study about viral fever, the number of people affected in a town were noted as

Find its standard deviation.
Answer:
Assumed mean = 35

Question 12.
The measurements of the diameters (in cms) of the plates prepared in a factory are given below. Find the standard deviation.

Answer:
Assumed mean = 34.5

Question 13.
The time taken by 50 students to complete a 100 meter race are given below. Find its standard deviation.

Answer:
Assumed mean = 11


Standard deviation (σ) = 1.24

Question 14.
For a group of 100 candidates, the mean and standard deviation of their marks were found to be 60 and 15 respectively. Later on, it was found that the scores 45 and 72 were wrongly entered as 40 and 27. Find the correct mean and standard deviation.
Answer:
Number of candidates = 100
n = 100
Mean (\(\bar{x}\)) = 60
standard deviation (σ) = 15
Mean (\(\bar{x}\)) = \(\frac{\Sigma x}{n} \Rightarrow 60=\frac{\Sigma x}{100}\)
Σx = 6000
Correct total = 6000 + (45 – 40) + ( 72 – 27) = 6000 + 5 + 45 = 6050
Correct mean (\(\bar{x}\)) = \(\frac{6050}{100}\) = 60.5
Given standard deviation = 15


Correct mean = 60.5
Correct standard deviation (σ) = 14. 61

Question 15.
The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4, 10, 12 and 14, then find the remaining two observations.
Answer:
Let the missing two observations be ‘a’ and ‘b’
Arithmetic mean = 8

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Computer Science Guide Pdf Chapter 2 Number Systems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 2 Number Systems

11th Computer Science Guide Number Systems Text Book Questions and Answers

Part – I

I. Choose The Correct Answer:

Question 1.
Which refers to the number of bits processed by a computer’s CPU? .
a) Byte
b) Nibble
c) Word length
d) Bit
Answer:
c) Word length

Question 2.
How many bytes does 1 KiloByte contain?
a) 1000
b) 8
c) 4
d) 1024
Answer:
d) 1024

Question 3.
Expansion for ASCII
a) American School Code for Information Interchange
b) American Standard Code for Information Interchange
c) All Standard Code for Information Interchange
d) American Society Code for Information Interchange
Answer:
b) American Standard Code for Information Interchange

Question 4.
2^50 is referred as
a) Kilo
b) Tera
c) Peta
d) Zetta
Answer:
c) Peta

Question 5.
How many characters can be handled in Binary Coded Decimal System?.
a) 64
b) 255
c) 256
d) 128
Answer:
a) 64

Question 6.
For 11012 whai is the Hexadecimal equivalent?
a) F
b) E
c) D
d) B
Answer:
c) D

Hex calculator for performing addition, subtraction, multiplication and division of hexadecimal numbers.

Question 7.
What is the 1’s complement of 00100110?
a) 00100110
b) 11011001
c) 11010001
d) 00101001
Answer:
b) 11011001

Question 8.
Which amongst this is not an Octal number?
a) 645
b) 234
c) 876
d) 123
Answer:
c) 876

Part II

Very Short Answers.

Question 1.
What is data?
Answer:
The term data comes from the word datum which means a raw fact. The data is a fact about people, places, or some objects.
Example: Rajesh, 16, XI.

Question 2.
Write the l’s complement procedure.
Answer:
The steps to be followed to find l’s complement of a number:
Step 1: Convert given Decimal number into Binary
Step 2: Check if the binary number contains 8 bits, if less add O at the left most bit, to make it as 8 bits.
Step 3: Invert all bits (i.e. Change 1 as 0 and 0 as 1)

Example: Find 1’s complement for (-24)₁₀

Question 3.
Convert (46)₁₀ into a Binary number.
Answer:

Question 4.
We cannot find l’s complement for (28)₁₀ State reason.
Answer:
Since it is a positive number. 1’s complement will come only for negative numbers.

Question 5.
List the encoding systems for characters in memory.
Answer:
There are several encoding systems used for computers. They are

• BCD – Binary Coded Decimal
• EBCDIC – Extended Binary Coded Decimal Interchange Code
• ASCII – American Standard Code for Information Interchange
• Unicode
• ISCII – Indian Standard Code for Information Interchange

Part III

III. Very Short Answers

Question 1.
What is radix of a number system? Give example.
Answer:
Each number system Is uniquely Identified by Its base value or radix. Radix or base Is the count of number of digits In each number system. Radix or base is the general Idea behind positional numbering system. Ex.

Number system	Base / Radix
Binary	2
Octal	8
Decimal	10
Hexadecimal	16

Question 2.
Write a note on the binary number system.
Answer:
There are only two digits in the Binary system, namely, 0 and 1. The numbers in the binary system are represented to the base 2 and the positional multipliers are the powers of 2.

The leftmost bit in the binary number is called the Most Significant Bit (MSB) and it has the largest positional weight. The rightmost bit is the Least Significant Bit (LSB) and has the smallest positional weight.

Question 3.
Convert (150)₁₀ into Binary, then convert that Binary number to Octal.
Answer:
Decimal to Binary conversion 150

Binary to octal conversion
LSB to MSB divide the number into three-digit binary and write the equivalent octal digit

Question 4.
Write a short note on ISCII.
Answer:
ISCII – Indian Standard Code for Information Interchange (ISCII) is the system of handling the character of Indian local languages. This is an 8 – bit coding system. Therefore it can handle 256 (2⁸) characters. It is recognized by the Bureau of Indian Standards (BIS). It is integrated with Unicode.

This system is formulated by the Department of Electronics in India in the year 1986-88 and recognized by the Bureau of Indian Standards (BIS). Now, this coding system is integrated with Unicode.

Question 5.
Add : a) -22₁₀ + 15₁₀ b) 20₁₀ + 25₁₀.
Answer:
a) -22₁₀ + 15₁₀

Answer in 2’s complement form . 11111001 is 2’s complement of 7 which is the answer.

Part IV

IV. Detail Answers.

Question 1.
a) Write the procedure to convert fractional Decimal to Binary.
Answer:
Conversion of fractional Decimal to Binary
The method of repeated multiplication by 2 has to be used to convert such kinds of decimal fractions.

The steps involved in the method of repeated multiplication by 2:

Step 1. : Multiply the decimal fraction by 2 and note the integer part. The integer part is either 0 or 1.
Step 2: Discard the integer part of the previous product. Multiply the fractional part of the previous product by 2. Repeat Step 1 until the same fraction repeats or terminates (0).
Step 3: The resulting integer part forms a sequence of 0’s and 1’s that becomes the binary equivalent of a decimal fraction.
Step 4: The final answer is to be written from the first integer part obtained till the last integer part obtained.
Integer part (last integer part obtained)

Write the integer parts from top to bottom to obtain the equivalent fractional binary number.
Hence
(0.2)₁₀ = (0.00110011.,.)₂ = (0.00110011)₂.

b) Convert (98.46)₁₀ to Binary.
Convert (98.46)₁₀ to Binary
Procedure: Conversion of an integral part:

(98)₁₀ = (1100010)₂ Conversion of fractional part:

Question 2.
Find l’s Complement and 2’s Complement for the following Decimal number.
a) -98
b) -135
Answer:
a) Conversion of (98)₁₀ into binary

Conversion of (135)₁₀ into binary

Question 3.
a) Add 1101010₂ + 101101)₂
Answer:
a) Add 1101010₂ + 101101₂

b) Subtract 1101011₂ – 111010₂.
Subtract 1101011₂ – 111010₂

11th Computer Science Guide Number Systems Additional Questions and Answers

Part I

Choose The Correct Answer.

Question 1.
The simplest method to represent a negative binary number is called ………………..
(a) signed magnitude
(b) sign bit or parity bit
(c) binary
(d) decimal
Answer:
(a) signed magnitude

Question 2.
Computer understand ________________language.
a) High level
b) Assembly
c) Machine
d) All the above
Answer:
c) Machine

Question 3.
Expansion for BCD ………………..
(a) Binary coded decimal
(b) binary complement decimal
(c) binary computer decimal
(d) binary convert decimal
Answer:
(a) Binary coded decimal

Question 4.
__________is the basic unit of data in computer.
a) BIT
b) BYTE
c) NIBBLE
d) WORD
Answer:
a) BIT

Question 5.
The ……………….. operator is defined in boolean algebra by the use of the dot (.) operator.
(a) AND
(b) OR
(c) NOT
(d) NAND
Answer:
(a) AND

Question 6.
Binary digit means __________
a) 0
b) 1
c) either 0 or 1
d) None of these
Answer:
c) either 0 or 1

Question 7.
The convert (65)₁₀ into its equivalent octal number ………………..
(a) (101)₈
(b) (101)₁₀
(c) (101)₁₂
(d) (101)₄
Answer:
(a) (101)₈

Question 8.
A collection of 8 bits is called __________
a) BIT
b) BYTE
C) NIBBLE
d) WORD
Answer:
b) BYTE

Question 9.
……………….. is the general idea behind the positional numbering system.
(a) Radix
(b) Computer memory
(c) Binary number
(d) Decimal number
Answer:
(a) Radix

Question 10.
__________refers to the number of bits processed by a computer’s CPU.
a) Word length
b) Nibble
c) Word size
d) None of these
Answer:
a) Word length

Question 11.
Bit means ………………..
(a) nibble
(b) byte
(c) word length
(d) binary digit
Answer:
(d) binary digit

Question 12.
__________is a valid word length of a computer.
a) 64
b) 32
c) 16
d) All the above
Answer:
d) All the above

Question 13.
The computer can understand ……………….. languages.
(a) computer
(b) machine
(c) post
(d) pre
Answer:
(b) machine

Question 14.
1 KiloByte equals to __________bytes.
a) 1024
b) 256
c) 1000
d) 128
Answer:
a) 1024

Question 15.
How many bytes does 1 zettabyte contain?
(a) 2⁹⁰
(b) 2⁸⁰
(c) 2⁷⁰
(d) 2⁶⁰
Answer:
(c) 2⁷⁰

Question 16.
1024 MegaBytes equals to _________
a) 1 GigaByte
b) 1 TeraByte
c) 1 YottaByte
d) None of these
Answer:
a) 1 GigaByte

Question 18.
1-kilo byte represents ……………….. bytes.
(a) 512
(b) 256
(c) 1024
(d) 64
Answer:
(c) 1024

Question 18.
1Kb equals to _________bytes.
a) 2¹⁰
b) 2²⁰
c) 2³⁰
d) 2⁴⁰
Answer:
a) 2¹⁰

Question 19.
How many megabytes does 1 GB contain?
(a) 2²⁰
(b) 2¹⁰
(c) 2³⁰
(d) 2⁴⁰
Answer:
(b) 2¹⁰

Question 20.
1 GB equals to ________ bytes.
a) 2¹⁰
b) 2²⁰
c) 2³⁰
d) 2⁴⁰
Answer:
c) 2³⁰

Question 21.
What is the 1’ s complement of 11001?
(a) 11100110
(b) 01010101
(c) 11110000
(d) 100100111
Answer:
(a) 11100110

Question 22.
1 PetaByte(PB) equals to _________bytes.
a) 2⁵⁰
b) 2⁶⁰
c) 2⁷⁰
d) 2⁸⁰
Answer:
a) 2⁵⁰

Question 23.
The hexadecimal equivalent of 15 is ………………..
(a) A
(b) B
(c) E
(d) F
Answer:
(d) F

Question 24.
1 ZettaByte (1ZB) equals to _______ bytes.
a) 2⁵⁰
b) 2⁶⁰
c) 2⁷⁰
d) 2⁸⁰
Answer:
c) 2⁷⁰

Question 25.
The radix of a hexadecimal number is ………………..
(a) 2
(b) 8
(c) 16
(d) 10
Answer:
(c) 16

Question 26.
Computer memory is normally represented in terms of ________ bytes.
a) Kilo
b) Mega
c) Kilo or Mega
d) None of these
Answer:
c) Kilo or Mega

Question 27.
The most commonly used number system is ………………..
(a) binary
(b) decimal
(c) octal
(d) hexadecimal
Answer:
(b) decimal

Question 28.
The most commonly used coding scheme to represent character set and the number is ________
a) BCD
b) ASCII
c) EBCDIC
d) All the above
Answer:
b) ASCII

Question 29.
What does MSB mean?
(a) Major sign bit
(b) Most sign bit
(c) Minor sign bit
(d) Most significant bit
Answer:
(d) Most significant bit

Question 30.
The ASCII value for blank space is _________
a) 43
b) 42
c) 32
d) 62
Answer:
c) 32

Question 31.
The binary equivalent of hexadecimal number B is ………………..
(a) 1011
(b) 1100
(c) 1001
(d) 1010
Answer:
(a) 1011

Question 32.
The most commonly used numbering system in real life is the _________number system.
a) Hexadecimal
b) Octal
c) Binary
d) Decimal
Answer:
d) Decimal

Question 33.
What is the range of ASCII values for lower case alphabets?
(a) 65 to 90
(b) 65 to 122
(c) 97 to 122
(d) 98 to 122
Answer:
(c) 97 to 122

Question 34.
_________is the count of number of digits in each number system.
a) base
b) radix
c) base or radix
d) symbols
Answer:
c) base or radix

Question 35.
What is the ASCII value for blank space?
(a) 8
(b) 2
(c) 18
(d) 32
Answer:
(d) 32

Question 36.
Identify the true statement from the following.
a) In the positional number system, each decimal digit is weighted relative to its position in the number.
b) A numbering system is a way of representing numbers.
c) The speed of a computer depends on the number of bits it can process at once.
d) All the above
Answer:
d) All the above

Question 37.
Which one of the following bits has the smallest positional weight?
(a) MSB
(b) LSB
(c) UPS
(d) USB
Answer:
(b) LSB

Question 38.
The rightmost bit in the binary number is called as the __________
a) MSB
b) LSB
c) FSB
d) None of these
Answer:
b) LSB

Question 39.
Name the person who proposed the basic principles of Boolean Algebra?
(a) Wiliam Boole
(b) George Boole
(c) James Boole
(d) Boolean George
Answer:
(b) George Boole

Question 40.
_______ numbers are used as a shorthand form of a binary sequence.
a) Hexadecimal
b) Octal
c) Decimal
d) None of these
Answer:
a) Hexadecimal

Question 41.
What is the other name for a logical statement?
(a) Truth values
(b) Truth functions
(c) Truth table
(d) Truth variables
Answer:
(b) Truth functions

Question 42.
In hexadecimal number system letter ‘E’ represents _______
a) 12
b) 13
c) 14
d) 15
Answer:
c) 14

Question 43.
The NOT operator is represented by the symbol.
(a) over bar
(b) single apostrophe
(c) a and b
(d) plus
Answer:
(c) a and b

Question 44.
_______is a method to convert decimal number to binary number.
a) Repeated division by 2
b) Sum of powers of 2
c) Repeated addition by 2
d) Either A or B
Answer:
d) Either A or B

Question 45.
The output for the AND operator is ………………..
(a) A + B
(b) –
(c) A.B
(d) AB + C
Answer:
(c) A.B

Question 46.
Computer can handle _______ numbers.
a) signed
b) unsigned
c) signed and unsigned
d) None of these
Answer:
c) signed and unsigned

Question 47.
Which gate takes only one input?
(a) OR
(b) AND
(c) NOT
(d) XOR
Answer:
(c) NOT

Question 48.
In the signed magnitude method, the leftmost bit is called _______bit.
a) sign
b) parity
c) sign or parity
d) None of these
Answer:
c) sign or parity

Question 49.
Which is not a derived date?
(a) AND
(b) NAND
(c) NOR
(d) XOR
Answer:
(a) AND

Question 50.
The numbers are represented in computers in _______method.
a) Signed magnitude representation
b) 1’s complement
c) 2’s complement
d) All the above
Answer:
d) All the above

Question 51.
The statement “C equal the complement of A or B” means
(a) C = A + B
(b) C = \(\overline{\mathrm{A}}+\overline{\mathrm{B}}\)
(c) C = \(\overline{\mathrm{A}}\) + \(\overline{\mathrm{B}}\)
(d) C = \(\overline{\mathrm{A}\mathrm{B}}\)
Answer:
(a) C = A + B

Question 52.
If the number has_______sign, it will be considered as negative in signed magnitude representation.
a) +
b) no
c) –
d) A or B
Answer:
c) –

Question 54.
What is the output of the XOR gate?
(a) C = A% B
(b) C = A \(\otimes\) A
(c) C = A \(\odot\) B
(d) C = A \(\oplus\) B
Answer:
(d) C = A \(\oplus\) B

Question 54.
2’s complement of (0001i000)2 is_______
a) 11100111
b) 00011001
c) 11101000
d) None of these
Answer:
c) 11101000

Question 55.
Find A + \(\overline{\mathrm{A}}\) .B = ………………..
(a) A + B
(b) A.B
(c) \(\overline{\mathrm{A}}\).B
(d) A.\(\overline{\mathrm{B}}\)
Answer:
(d) A.\(\overline{\mathrm{B}}\)

Question 56.
When two binary numbers are added _______will be the output.
a) sum
b) carry
c) sum and carry
d) None of these
Answer:
c) sum and carry

Question 57.
When subtracting 1 from 0, borrow 1 from the next _______
a) LSB
b) MSB
c) either A or B
d) None of these
Answer:
b) MSB

Question 58.
Find the wrong pair from the following:
(a) Null element : A + 1 = 1
(b) Involution : \(\overset { = }{ A }\) = A
(c) Demorgan’s : \(\overline{\mathrm{A+B}}\) =\(\overline{\mathrm{A}}\) . \(\overline{\mathrm{A}}\)
(d) Commutative : A + B = B . A
Answer:
(d) Commutative : A + B = B . A

Question 59.
_______ is the character encoding system.
a) BCD and ISCII
b) EBCDIC
c) ASCII and Unicode
d) All the above
Answer:
d) All the above

Question 60.
With 2 inputs in the truth table, how many sets of values will be obtained.
(a) 4
(b) 8
(c) 2
(d) 1
Answer:
(a) 4

Question 61.
EBCDIC stands for _______
a) Extensive Binary Coded Decimal Interchange Code
b) Extended Binary Coded Decimal Interchange Code
c) Extended Binary Coded Digit Interchange Code
d) Extended Bit Coded Decimal Interchange Code.
Answer:
b) Extended Binary Coded Decimal Interchange Code

Question 62.
ASCII stands for _______
a) Arithmetic Standard Code for Information Interchange
b) American Structured Code for Information Interchange
c) American Standard Code for Information Interchange
d) American Standard Code for Instant Interchange
Answer:
c) American Standard Code for Information Interchange

Question 63.
ISCII stands for_______
a) International Standard Code for Information Interchange
b) Indian Structured Code for Information Interchange
c) India’s Standard Code for Information Interchange
d) Indian Standard Code for Information Interchange
Answer:
d) Indian Standard Code for Information Interchange

Question 64.
BCD is _______bit code.
a) 6
b) 7
c) 8
d) None of these
Answer:
a) 6

Question 65.
EBCDIC is_______ bit code
a) 6
b) 7
c) 8
d) None of these
Answer:
c) 8

Question 66.
ASCII is________ bit code
a) 6
b) 7
c) 8
d) None of these
Answer:
b) 7

Question 67.
Unicode is _______ bit code
a) 16
b) 7
c) 8
d) None of these
Answer:
a) 16

Question 68.
ISCII is_______ bit code
a) 16
b) 7
c) 8
d) None of these
Answer:
c) 8

Question 69.
_______coding system is formulated by IBM.
a) BCD
b) EBCDIC
c) ISCII
d) None of these
Answer:
b) EBCDIC

Question 70.
IBM stands for_______
a) Indian Business Machine
b) International Basic Machine
c) International Business Method
d) International Business Machine
Answer:
d) International Business Machine

Question 71.
_______is the system of handling the characters of Indian local languages.
a) ASCII
b) Unicode
c) BCD
d) ISCII
Answer:
d) ISCII

Question 72.
ISCII system is formulated fay the _______ in India.
a) Department of Electronics
b) Department of Electricity
c) Department of E-commerce
d) Department of Economics
Answer:
a) Department of Electronics

Question 73.
SCO system can handle___________characterscharacters.
a) 64
b) 128
c) 256
d) 65536
Answer:
a) 64

Question 74.
EBCDIC system can handle _______ characters.
a) 64
b) 128
c) 256
d) None of these
Answer:
c) 256

Question 75.
ASCII system can handle _______characters.
a) 64
b) 128
c) 256
d) None of these
Answer:
c) 256

Question 76.
Unicode system can handle _______ characters.
a) 64
b) 128
c) 256
d) 65536
Answer:
d) 65536

Question 77.
ISCII system can handle _______ characters.
a) 64
b) 128
c) 256
d) 65535
Answer:
c) 256

Question 78.
__________ language characters are not represented by ASCII.
a) Tamil
b) Malayalam
c) Telugu and Kannada
d) All the above
Answer:
d) All the above

Question 79.
Tamil, Malayalam, Telugu, and Kannada language characters are represented by _______ code.
a) ASCII
b) Unicode
c) BCD
d) ISCII
Answer:
b) Unicode

Question 80.
_______scheme is denoted by hexadecimal numbers
a) ASCII
b) Unicode
c) BCD
d) ISCII
Answer:
b) Unicode

Question 81.
ISCII code was formulated in the year_______
a) 1986 – 88
b) 1984 – 86
c) 1988
d) 1987
Answer:
a) 1986 – 88

Question 82.
_______coding system is integrated with Unicode.
a) ASCII
b) EBCDIC
c) BCD
d) ISCII
Answer:
d) ISCII

Question 83.
_______was generated to handle all the coding system of Universal languages.
a) ASCII
b) Unicode
c) BCD
d) ISCII
Answer:
b) Unicode

Question 84.
The popular coding scheme after ASCII is_______
a) EBCDIC
b) Unicode
c) BCD
d) ISCII
Answer:
b) Unicode

Question 85.
BCD system is_______bit encoding system.
a) 2⁸
b) 2¹⁶
c) 2⁶
d) 2⁴
Answer:
c) 2⁶

Question 86.
EBCDIC system is _______bit encoding system.
a) 2⁸
b) 2¹⁶
c) 2⁶
d) 2⁷
Answer:
a) 2⁸

Question 87.
ASCII system is a bit encoding system.
a) 2⁸
b) 2¹⁶
c) 2⁶
d) 2⁷
Answer:
d) 2⁷

Question 88.
Unicode system is _________bit encoding system.
a) 2⁸
b) 2¹⁶
c) 2⁶
d) 2⁷
Answer:
b) 2¹⁶

Question 89.
ISCII svstem is _________bit encoding system.
a) 2⁸
b) 2¹⁶
c) 2⁶
d) 2⁷
Answer:
a) 2⁸

Question 90.
The input code in ASCII can be converted into _________system.
a) EBCDIC
b) Unicode
c) BCD
d) ISCII
Answer:
a) EBCDIC

Question 91.
What is ASCII value for ‘A’ in a decimal number,
a) 97
b) 65
c) 98
d) 32
Answer:
b) 65

Question 92.
What is the ASCII value for ‘A’ in a binary number?
a) 01100001
b) 01000001
c) 01100010
d) 00100000
Answer:
b) 01000001

Question 93.
What is the ASCII value for ‘A’ in an octal number?
a) 141
b) 101
c) 142
d) 40
Answer:
b) 101

Question 94.
What is the ASCII value for ‘A’ in hexadecimal numbers?
a) 61
b) 41
c) 62
d) 20
Answer:
b) 41

Question 95.
Find the false statement in the following.
a) Computers can handle positive and negative numbers.
b) MSB is called a sign bit
c) LSB is called a parity bit
d) All the above
Answer:
c) LSB is called a parity bit

Question 96.
Match the following.
a) 78 – (1) Binary number
b) linn – (2) Octal number
c) CAFE – (3) Decimal number
d) 71 – (4) Hexadecimal number

a) 3, 1, 4, 2
b) 4, 3, 2, 1
c) 1, 3, 2, 4
d) 3, 1, 2, 4
Answer:
a) 78 – (1) Binary number

Question 97.
In signed magnitude representation,_________ in the sign bit represents negative number.
a) 0
b) 1
c) No symbol
d) None of these
Answer:
b) 1

Question 98.
In signed magnitude representation, __________in the sign bit represents positive number.
a) 0
b) 1
c) No symbol
d) None of these
Answer:
a) 0

Question 99.
The term data comes from the word __________
a) datum
b) date
c) fact
d) None of these
Answer:
a) datum

Part II

Very Short Answers.

Question 1.
What is nibble?
Answer:
Nibble is a collection of 4 bits. A nibble is half a byte.

Question 2.
Define information.
Answer:
Information is a processed fact and obtained from the computer as output. It conveys meaning.

Question 3.
What is radix?
Answer:
The base value of a number is also known as the radix.

Question 4.
Define Bit and Byte.
Answer:
Bit: A bit is the short form of a Binary digit which can be ‘0’ or ‘1’. It is the basic unit of data in computers.
Byte: A collection of 8 bits is called Byte. It is the basic unit of measuring the memory size in the computer.

Question 5.
Expand: BCD, EBCDIC, ASCII
Answer:
BCD – Binary Coded Decimal; EBCDIC – Extended Binary Coded Decimal Interchange Code; ASCII – American Standard Code for Information Interchange.

Question 6.
What are the different types of coding schemes to represent the character sets?
Answer:
The different coding schemes are

• BCD – Binary Coded Decimal
• EBCDIC – Extended Binary Coded Decimal Interchange Code
• ASCII – American Standard Code for Information Interchange
• Unicode
• ISCII – Indian Standard Code for Information Interchange.

Question 7.
What are the methods of converting a number from decimal to binary?
Answer:

1. Repeated division by two.
2. Sum of powers of 2.

Question 8.
What does base or radix mean?
Answer:
Radix or base is the count of a number of digits in each number system. Radix or base is the general idea behind the positional numbering system.

Question 9.
What are the various ways for Binary representation of signed numbers?
Answer:

1. Signed magnitude representation
2. 1’s complement
3. 2’s complement

Question 10.
Write a note on the decimal number system.
Answer:
It consists of 0,1,2,3,4,5,6,7,8,9(10 digits). It is the oldest and most popular number system used in our day-to-day life. In the positional number system, each decimal digit is weighted relative to its position in the number.
Its base or radix is 10.

Question 11.
Write about the octal number system.
Answer:
Octal number system uses digits 0,1,2,3,4,5,6 and 7 (8 digits). Each octal digit has its own positional value or weight as a power of 8. Its base or radix is 8.

Question 12.
How will you convert decimal to hexadecimal?
Answer:
To convert Decimal to Hexadecimal, “Repeated division by 16” method can be used) In this method, we have to divide the given number by 16.
Example: Convert (31)10 into its equivalent hexadecimal number.

Question 13.
Give the procedure to Octal to Binary.
Answer:
Procedure: For each octal digit in the given number write its 3 digits binary equivalent using positional notation.
Example: Convert (6213)₈ to equivalent Binary number.

Question 14.
How will you convert Hexadecimal to Binary?
Answer:
Procedure: Write 4 bits Binary equivalent for each Hexadecimal digit for the given number using the positional notation method.
Example:
Convert (8BC)₁₆ into an equivalent Binary number.

Question 15.
Write short note on Binary Coded Decimal (BCD).
Answer:
This is 2⁶ bit encoding system. This can handle 2⁶ = 64 characters only. This encoding system is not
in the practice right now.

Question 16.
Write note on EBCDIC encoding system.
Answer:
Extended Binary Coded Decimal Interchange Code (EBCDIC) is similar to ASCII Code with 8 bit representation. This coding system is formulated by International Business Machine (IBM). The coding system can handle 256 characters. The input code in ASCII can be converted to EBCDIC system and vice – versa.

Question 17.
Write a note on the ISCII encoding system.
Answer:
ISCII is the system of handling the character of Indian local languages. This is an 8-bit coding system. Therefore it can handle 256 (28) characters. This system is formulated by the Department of Electronics in India in the year 1986-88 and recognized by the Bureau of Indian Standards (BIS). Now, this coding system is integrated with Unicode.

Part III

III. Very Short Answers

Question 1.
Write about the binary number system.
Answer:
There are only two digits in the Binary system, namely, 0 and 1. The numbers in the binary system are represented to the base 2 and the positional multipliers are the powers of 2.
The leftmost bit in the binary number is called the Most Significant Bit (MSB) and it has the largest positional weight. The rightmost bit is the Least Significant Bit (LSB) and has the smallest positional weight.

Question 2.
What is the octal number system?
Answer:
The octal number system uses digits 0, 1, 2, 3, 4, 5, 6, and 7 (8 digits): Each octal digit has its own positional value or weight as a power of 8.
Example: The Octal sequence (547)₈ has the decimal equivalent:

Question 3.
Give the procedure to convert decimal to octal.
Answer:
To convert Decimal to Octal, “Repeated Division by 8” method can be used) In this method, we have to divide the given number by 8.
Example:
Convert (65)₁₀ into its equivalent Octal number.

Question 4.
Give the procedure to convert Octal to Decimal
Answer:
To convert Octal to Decimal, we can use positional notation method)

• Write down the Octal digits and list the powers of 8 from right to left (Positional Notation).
• For each positional notation of the digit write the equivalent weight.
• Multiply each digit with its corresponding weight.
• Add all the values.

Example:
Convert (1265)₈ to equivalent Decimal number.

(1265)₈ = 512 x 1 + 64 x 2 + 8 x 6 + 1 x 5
= 512 + 128 + 48 + 5
(1265)₈ = (693)₁₀

Question 5.
How will you convert Hexadecimal to Decimal?
Answer:
To convert Hexadecimal to Decimal we can use the positional notation method.

• Write down the Hexadecimal digits and list the powers of 16 from right to left (Positional Notation)
• For each positional notation written for the digit, now write the equivalent weight.
• Multiply each digit with its corresponding weight
• Add all the values to get one final value.

Example:
Convert (25F)₁₆ into its equivalent Decimal number.

(25F)₁₆ = 2 x 256 + 5 x 16 + 15 x 1
= 512 + 80 + 15 (25F)₁₆
= (607)₁₀

Question 6.
Write about binary representation for signed numbers.
Answer:
Computers can handle both positive (unsigned) and negative (signed) numbers. The simplest method to represent negative binary numbers is called Signed Magnitude. In signed magnitude method, the leftmost bit is the Most Significant Bit (MSB), which is called the sign bit or parity bit.
The numbers are represented in computers in different ways:

• Signed Magnitude representation
• 1’s Complement
• 2’s Complement

Question 7.
Explain ASCII code in detail.
Answer:
This is the most popular encoding system recognized by the United States. Most of the computers use this system. Remember this encoding system can handle English characters only. This can handle 27 bit which means 128 characters.

In this system, each character has an individual number. The new edition ASCII -8, has 28 bits and can handle 256 characters are represented from 0 to 255 unique numbers.

The ASCII code equivalent to the uppercase letter ‘A’ is 65. The binary representation of the ASCII (7 bit) value is 1000001. Also 01000001 in ASCII-8 bit.

Question 8.
Explain Unicode in detail.
Answer:
This coding system is used in most modern computers. The popular coding scheme after ASCII is Unicode. ASCII can represent only 256 characters. Therefore English and European Languages alone can be handled by ASCII. Particularly there was a situation when the languages like Tamil, Malayalam, Kannada, and Telugu could not be represented by ASCII.

Hence, Unicode was generated to handle all the coding system of Universal languages. This is a 16-bit code and can handle 65536 characters. The unicode scheme is denoted by hexadecimal numbers.

Part IV

IV. Detail Answers.

Question 1.
Explain decimal to binary conversion using Repeated Division by 2 methods.
Answer:
To convert Decimal to Binary “Repeated Division by 2” method can be used. Any Decimal number divided by 2 will leave a remainder of 0 or 1. Repeated division by 2 will leave a sequence of 0s and Is that become the binary equivalent of the decimal number.

Suppose it is required to convert the decimal number N into binary form, dividing N by 2 in the decimal system, we will obtain a quotient N1 and a remainder Rl, where R1 can have a value of either 0 or 1. The process is repeated until the quotient becomes 0 or 1. When the quotient is ‘0’ or ‘1’, it is the final remainder value. Write the final answer starting from the final remainder value obtained to the first remainder value obtained.

Example:
Convert (65)₁₀ into its equivalent binary number

Question 2.
Explain decimal to binary conversion using Sum of powers of 2 methods.
Answer:
A decimal number can be converted into a binary number by adding up the powers of 2 and then adding bits as needed to obtain the total value of the number.
a) Find the largest power of 2 that is smaller than or equal to 65.
6510 > 6410

b) Set the 64’s bit to 1 and subtract 64 from the original number
65 – 64 = 1

c) 32 is greater than the remaining total.
Therefore, set the 32’s bit to 0.

d) 16 is greater than the remaining total.
Therefore, set the 16’s bit to 0
.
e) 8 is greater than the remaining total.
Therefore, set the 8’s bit to 0.

f) 4 is greater than the remaining total.
Therefore, set the 4’s bit to 0.

g) 2 is greater than the remaining total.
Therefore, set the 2’s bit to 0.

h) As the remaining value is equivalent to l’s bit, set it to 1.
1 – 1 = 0
Conversion is complete 65₁₀ = (1000001)₂

Example:
The conversion steps can be given as follows:
Given Number: 65
Equivalent or value less than the power of 2 is: 64
(1) 65 – 64 = 1
(2) 1 – 1= 0

65₁₀ = (1000001)₂.

Question 3.
Explain the procedure to convert fractional decimal to Binary.
Answer:
The method of repeated multiplication by 2
has to be used to convert such kind of decimal fractions.
The steps involved in the method of repeated multiplication by 2:

Step 1: Multiply the decimal fraction by 2 and note the integer part. The integer part is either 0 or 1.
Step 2: Discard the integer part of the previous product. Multiply the fractional part of the previous product by 2. Repeat Step 1 until the same fraction repeats or terminates (0).
Step 3: The resulting integer part forms a sequence of Os and Is that becomes the binary equivalent of a decimal fraction.
Step 4: The final answer is to be written from the first integer part obtained till the last integer part obtained.

Write the integer parts from top to bottom to obtain the equivalent fractional binary number. Hence (0.2)₁₀= (0.00110011…)₂ = 0.00110011)₂

Question 4.
How will you convert Binary to Decimal?
Answer:
To convert Binary to Decimal we can use the positional notation method.
Step 1: Write down the Binary digits and list the powers of 2 from right to left (Positional Notation)
Step 2: For each positional notation written for the digit, now write the equivalent weight.
Step 3; Multiply each digit with its corresponding weight
Step 4: Add all the values.

Example:
Convert (111011)2 into its equivalent decimal number.

32 + 16 + 8 + 0 + 2 + 1 = (59)₁₀
(111011)₂ = (59)₁₀

Question 5.
How will you convert Binary to Octal?
Answer:
Step 1.: Group the given binary number into 3 bits from right to left.
Step 2: You can add preceding O to make a group of 3 bits if the leftmost group has less than 3 bits.
Step 3: Convert equivalent octal value using “2’s power positional weight method”

Example
Convert (11010110)₂ into an octal equivalent number

Step 1: Group the given number into 3 bits from right to left.
011 010 110
The left-most groups have less than 3 bits, so 0 is added to its left to make a group of 3 bits.

Step-2: Find the Octal equivalent of each group.

Question 6.
Give the procedure to convert Binary to Hexadecimal.
Answer:
Step 1: Group the given number into 4 bits from right to left.
Step 2: You can add preceding 0’s to make a group of 4 bits if the leftmost group has less than 4 bits.
Step 3: Convert equivalent Hexadecimal value using “2’s power positional weight method”.

Example
Convert (1111010110)₂ into Hexadecimal number
Step 1: Group the given number into 4 bits from right to left. 1
0011 1101 0110
0’s are added to the leftmost group to make it a group of 4 bits.

Question 7.
Give the procedure to convert fractional Binary to Decimal equivalent.
Answer:
The steps to convert fractional Binary number to its decimal equivalent:
Step 1 : Convert an integral part of Binary to Decimal equivalent using positional notation method.
Step 2 : To convert the fractional part of binary to its decimal equivalent.
Step 2,1 : Write down the Binary digits in the fractional part.
Step 2,2 : For all the digits write powers of 2 from left to right starting
from 2^-1, 2^-2, 2^-3 2^-n, now write the equivalent weight.
Step 2.3 : Multiply each digit with its corresponding weight.
Step 2.4 : Add all the values which you obtained in Step 2.3.

Step 3 : To get final answer write the integral
part (after conversion), followed by a decimal point(.) and the answer arrived at Step 2.4

Example:
Convert the given Binary number (11.011)₂ into its decimal equivalent Integer part (11)₂ = 3

Question 8.
Explain the method of representing signed binary numbers in the Signed Magnitude representation.
Answer:
The value of the whole numbers can be determined by the sign used before it. If the number has a ‘+’ sign or no sign it will be considered as positive. If the number has signed it will be considered negative.

Example:
+ 43 or 43 is a positive number
– 43 is a negative number

In signed binary representation, the leftmost bit is considered as a sign bit. If this bit is 0, it is a positive number and if it 1, it is a negative number. Therefore a signed binary number has 8 bits, only 7 bits used for storing values (magnitude), and 1 bit is used for signs.

Question 9.
Explain the method of representing signed binary numbers in l’s complement representation.
Answer:
This is an easier approach to represent signed numbers. This is for negative numbers only i.e. the number whose MSB is 1.

The steps to be followed to find l’s complement of a number:

Step 1: Convert given Decimal number into Binary
Step 2: Check if the binary number contains 8 bits, if less add 0 at the left most bit, to make it as 8 bits.
Step 3: Invert all bits (i.e. Change 1 as 0 and 0 as 1)

Example: Find 1’s complement for (-24)₁₀

Question 10.
Explain the method of representing signed binary numbers in 2’s complement representation.
Answer:
The 2’s-complement method for the negative number is as follows:
a) Invert all the bits in the binary sequence (i.e., change every 0 to 1 and every 1 to 0 ie.,l’s complement)
b) Add 1 to the result to the Least Significant Bit (LSB).
Example: 2’s Complement represent of (-24)₁₀

Question 11.
Explain binary addition with a suitable example.
Answer:
The following table is useful when adding two binary numbers.

In 1 + 1 = 10, is considered as sum 0 and the 1 as carry bit. This carry bit is added with the previous position of the bit pattern.
Example: Add: 1011₂ + 1001₂

Example: Perform Binary addition for the following:
23₁₀ + 12₁₀
Step 1: Convert 23 and 12 into binary form

Step 2: Binary addition of 23 and 12:

Question 11.
Explain binary subtraction with a suitable example.
Answer:
The table for Binary Subtraction is as follows:

When subtracting 1 from 0, borrow 1 from the next Most Significant Bit, when borrowing from the next Most Significant Bit, if it is 1, replace it with 0. If the next Most Significant Bit is 0, you must borrow from a more significant bit that contains 1 and replace it with 0 and 0s upto that point become Is.
Example : Subtract 1001010₂ — 10100₂.

Example: Perform Binary addition for the
following:
(-21)₁₀ + (5)₁₀
Step 1: Change -21 and 5 into binary form

Workshop

Question 1.
Identify the number system for the following numbers.
Answer:

Question 2.
State whether the following numbers are valid or not. If invalid, give a reason.
Answer:

Question 3.
Convert the following Decimal numbers to their equivalent Binary, Octal, Hexadecimal.
i) 1920
ii) 255
iii) 126
Answer:
i) 1920

ii) 255

iii) 126

Question 4.
Convert the given Binary number into its equivalent Decimal, Octal, and Hexadecimal numbers.
i)101110101
ii) 1011010
iii) 101011111
Answer:
i) 101110101
Binary to Decimal (Multiply by positional value and then add)

ii) 1011010
Binary to decimal

iii) 101011111
Binary to decimal

Question 5.
Convert the following Octal numbers into Binary numbers.
a) 472
b) 145
c) 347
d) 6247
e) 645
Answer:
Procedure: Write three digits binary number for every octal digit that will give the equivalent binary number.

Ans.
(472)₈ = (100111010)₂

Answer:
(145)₈ = (001100101)₂

Answer:
(347)₈ = (011100111)₂

Answer:
(6247)₈ = (110010100111)₂

Answer:
(645)₈ = (110100101)₂.

Question 6.
Convert the following Hexadecimal numbers to Binary numbers
a) A6
b) BE
c) 9BC8
d) BC9
Answer:
Procedure: Write four digits binary number for every Hexadecimal digit that will give the equivalent binary number.

Answer:
(A6)₁₆ = (10100110)₂.

Answer:
(BE)₁₆ = (1011 1110)₂

Answer:
(9BC8)₁₆ = (1001101111001000)₂

Answer:
(BC9)₁₆ = (101111001001)₂

Question 7.
Write the l’s complement number and 2’s complement number for the following decimal numbers:
Perform the following binary computations:
a) -22
b) -13
c) -65
d) -46
Answer:
a) -22

Question 8.
a) 10₁₀ + 15₁₀
b) – 12₁₀ + 5₁₀
c) 14₁₀ – 12₁₀
d) (-2)₁₀ – (-6)₁₀
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 5th Science Guide Pdf Term 1 Chapter 3 Energy Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 5th Science Solutions Term 1 Chapter 3 Energy

Samacheer Kalvi 5th Science Guide Energy Text Book Back Questions and Answers

Evaluation
I. Choose the correct Answer:

Question 1.
When diesel is burnt chemical energy is converted into _______.
a) wind energy
b) heat energy
c) solar energy
d) sound energy
Answer:
b) heat energy

Kinetic Energy calculator – online mechanical engineering tool to calculate the observed energy of a body due to its motion, in both US customary & metric (SI)

Question 2.
Running water possesses _______.
a) potential energy
b) chemical energy
c) kinetic energy
d) sound energy
Answer:
c) kinetic energy

Question 3.
Unit of energy is _______.
a) Kilogram
b) Newton
c) Kelvin
d) Joule
Answer:
d) Joule

Question 4.
Which one of the following requires wind energy?
a) Bicycle
b) Photosynthesis
c) Parachute
d) Automobiles
Answer:
c) Parachute

Question 5.
Cow dung possesses _______ .
a) kinetic energy
b) chemical energy
c) solar energy
d) heat energy
Answer:
b) chemical energy

II. Find out the energy conversion that takes place in the following:

Question 1.
Iron box: _______.
Answer:
Chemical energy to Heat energy

Question 2.
Electric Iron box: _______.
Answer:
Electric energy to Heat energy

Question 3.
Electric fan:
Answer:
Electric energy to Mechanical energy

Question 4.
Speaker:
Answer:
Electric energy to Sound energy

Question 5.
Generator: _______.
Answer:
Mechanical energy to Electrical energy

III. Find out form of energy possessed by the following things:

Question 1.
A rock on the top of a hill.
Answer:
Pontential Energy

Question 2.
A rolling ball.
Answer:
Kinetic Energy

Question 3.
Charcoal.
Answer:
Heat Energy

Question 4.
Water falls.
Answer:
Kinetic Energy

Question 5.
Battery.
Answer:
Chemical Energy

IV. Match the following:

Answer:

1. e
2. d
3. a
4. c
5. b

V. Say True or False:

Question 1.
An apple falling from a tree is an example for kinetic energy.
Answer:
True

Question 2.
Electrical energy is used to run electric trains.
Answer:
True

Question 3.
Heat energy cannot be produced by friction.
Answer:
False

Potential Energy Calculator is a free online tool that displays the potential energy of the body.

Question 4.
Potential energy and heat energy are the two forms of mechanical energy.
Answer:
False

Question 5.
The unit of energy is joule.
Answer:
True

VI. Answer in brief:

Question 1.
What is energy?
Answer:
Energy is defined as capacity to do work. S.I unit of work is Joule.

Question 2.
What are the different forms of energy?
Answer:
There are different forms of energy like mechanical energy, heat energy, light energy, wind energy, solar energy, electrical energy, and chemical energy.

Question 3.
What are the uses of mechanical energy?
Answer:

• In hydroelectric plants, kinetic energy of water is converted into electrical energy.
• Windmills convert kinetic energy of winds into electrical energy.
• Mechanical energy of the hammer is used to apply a force on a nail.
• Mechanical energy can bring a moving body to rest and make a body at rest to move.

Question 4.
State the Law of conservation of energy.
Answer:
Law of conservation of energy states that energy Can neither be created nor be destroyed. One form of energy is converted into another form of energy.

Question 5.
Give the uses of Light energy.
Answer:

• We are able to see objects with the help of light energy.
• Plants use light energy to synthesis their food.
• With the help of light energy, our skin is able to synthesis Vitamin D.
• Electricity can be produced with the help of light energy.

VII. Answer in detail:

Question 1.
Explain the types of Mechanical energy.
Answer:
Energy possessed by an object due its position is called mechanical energy.
Mechanical energy can be classified into two;

• Kinetic energy
• Potential energy

a) Kinetic energy: Energy possessed by a moving objectis known as kinetic energy. It is also known as energy of motion.
Eg: Moving car, cricket ball bowled by a player, bullet coming out of a gun.

b) Potential energy: Energy possessed by an object which is at rest is known as potential energy. It is also known as stores energy of position.
Eg: Object lifted above, stone in the stretched rubber, water in the dam.

Question 2.
Explain Conservation of energy.
Answer:
Energy cannot be created cannot be destroyed also. It is changed from one form to another form or transferred from one object to another object. Examples for conservation of energy in our daily life are :

1. Water dam :

• Water stored in water dams posseses potential energy.
• When water falls down, potential energy of water is converted, into kinetic energy.
• Kinetic energy of water rotates the turbines and electric energy is generated.

2. Electrical appliances:

• Electrical energy is used in many domestic appliances such as electric stove, iron box and fan.
• Electric energy flows into the coil in the devices.
• As current flows, it heats up the coil.
• With the help of this heat energy, we do many useful works.
• Thus, electrical energy is converted into heat energy.
• Electical energy is converted to mechanical energy in fan, light enrgy in bulb and sound energy in computer.

3. Driving a car:

• We use fuel in the form of petrol or diesel or gas to run vehicles.
• When this fuel burns in the engine, chemical energy is converted into heat energy.
• Burning fuel produces hot gases which pushes the piston in the engine to move the vehicle.
• Thus heat energy is converted into mechanical energy.

Samacheer Kalvi 5th Science Guide Energy Additional Questions and Answers

I. Choose the correct Answer:

Question 1.
___________ is the only form of energy visible to human eye.
a) Heat energy
b) Light energy
c) Electrical energy
d) Chemical energy
Answer:
b) Light energy

Question 2.
Light travels at a speed of _________.
a) 3,00,000 km/s
b) 4,00,000 km/s
c) 5,00,000 km/s
d) 6,00,000 km/s
Answer:
a) 3,00,000 km/s

Question 3.
Sunlight takes _________ to reach earth.
a) 5 minutes
b) 6 minutes
c) 7 minutes
d) 8 minutes
Answer:
a) 5 minutes

Question 4.
Which of the following device converts chemical energy into electrical energy?
a) Battery
b) Loud speaker
c) Solar cell
d) Electrical motor
Answer:
a) Battery

Question 5.
___________ stands first in generating electricity from wind mill.
a) Kerala
b) Tamil Nadu
c) Karnataka
d) Andhra pradesh
Answer:
b) Tamil Nadu

Question 6.
The energy, possessed by a object due to its position is called _______.
a) Kinetic energy
b) Potential energy
c) Mechanical energy
d) Electrical energy
Answer:
b) Potential energy

Question 7.
The energy possessed by a cricket ball bowled by a player is _______.
a) Potential energy
b) Kinetic energy
c) Mechanical energy
d) Electrical energy.
Answer:
b) Mechanical energy

Question 8.
Electric eel is a _______.
a) fish
b) snake
c) frog
d) none of the above
Answer:
a) fish

Question 9.
Which of the following is the example potential energy?
a) Object lifted above
b) Bullet coming out of a sun
c) Moving car
d) Cricket ball bowled by a player
Answer:
c) Moving car

Question 10.
What energy is possessed by stretched rubber.
a) Nuclear
b) Potential
c) Kinetic
d) Thermal
Answer:
b) Potential

II. Fill in the blanks:

Question 1.
_______ is a measure of heat in a body.
Answer:
Temperature

Question 2.
________ use wind energy to generate electricity.
Answer:
Windmills

Question 3.
The form of energy that produces feeling of hotness is called as _______.
Answer:
heat

Question 4.
Light energy helps our skin to synthesis _________.
Answer:
vitamin – b

Question 5.
When stored water falls down, potential energy of water is converted into _______.
Answer:
Kinetic energy

Question 6.
Electric eel can generate _______.
Answer:
electricity

Question 7.
Law of conversation of energy given by _________.
Answer:
Julius Robert Mayar

III. Say True or False:

Question 1.
Light does not require any medium to travel.
Answer:
True

Question 2.
Mechanical energy can be used for pumping water.
Answer:
False

Question 3.
Heat energy is used to run vehicles.
Answer:
True

Question 4.
Heat is measured in Joule.
Answer:
True

Question 5.
Water stored in water dams possesses kinetic energy.
Answer:
False

Question 6.
Photosynthesis changes solar energy into chemical energy.
Answer:
True

Question 7.
Wind mills convert kinetic energy of winds into electrical energy.
Answer:
True

IV. Circle the odd one:

Question 1.
a) Proton
b) Dentron
c) Electron
d) Neutron
Answer:
b) Dentron

Question 2.
a) Nuclear power plants
b) Hydroelectric plants
c) Water heaters
d) Windmills
Answer:
c) Water heaters

V. Match the following:

Question 1:

Answer:

1. d
2. a
3. b
4. c

Question 2.

Answer:

1. d
2. c
3. b
4. a

VI. Answer briefly:

Question 1.
What is mechanical energy?
Answer:

• Energy possessed by an object due to its position is called mechanical energy.
• Mechanical energy can be classified into two.
  • • Kinetic energy
    • Potential energy

Question 2.
What is energy of motion? Give examples.
Answer:
Energy possessed by a moving object is known as kinetic energy. It is also known as energy of motion.
Examples: Moving car, Cricket ball bowled by a player, Bullet coming out of a gun.

Question 3.
Write any two uses to mechanical energy.
Answer:

• Windmills convert kinetic energy of winds into electrical energy.
• Mechanical energy of the hammer is used to apply a force on a Class nail.

Question 4.
What is wind energy and write its uses?
Answer:
Energy possessed by the wind is known as wind energy.
Uses of wind energy

• Wind mills use wind energy to generate electricity.
• Ships sail by the power of wind.
• Sports like wind surfing, sailing, kite surfing use wind energy.
• Wind energy can be used for pumping water.

Question 5.
Name the particles of atom?
Answer:
Atoms possess particles like protons, electrons, and neutrons.

Question 6.
What is electric energy?
Answer:
Movement of electron in the objects causes energy. This energy is called electric energy.

Question 7.
What are the different ways electricity can be generated?
Answer:
Electric energy is also generated from nuclear power plants, hydroelectric plants, and windmills. It is also generated from solar energy.

Question 8.
Write the uses of Electric energy.
Answer:

• Electric energy is needed for the working of fan, light, television, washing machine, refrigerator, etc.
• Electric iron box, electric stove, and electric water heater work by electrical energy.
• It is used to run cars and trains.
• It is used in factories to produce materials.

Question 9.
What is law of conservation of energy?
Answer:
Law of conservation of energy states that energy can neither be created nor be destroyed. One form of energy is converted into another form of energy. This law was given by Julius Robert Mayor.

Question 10.
What energy transfer happen in a car engine?
Answer:
When Petrol or diesel or gas burns in the engine, chemical energy is converted into heat energy. Burning fuel produces hot gases which pushes the piston in the engine to move the vehicle. Thus heat energy is converted into mechanical energy.

VII. Answer in detail:

Question 1.
Write the uses of heat energy and light energy.
Answer:
Uses of heat energy:

• Heat energy obtained from power stations is used to generate electricity.
• Heat energy obtained from petrol and diesel is used to run vehicles.
• We cook food with the help of heat. Heat energy renders the food material soft and easy to digest.
• Hard substances like iron are heated to mold them into different shapes.
• Heat is used to dry clothes and other wet substances.

Uses of light energy:

• We are able to see objects with the help of light energy.
• Plants use light energy to synthesis their food.
• With the help of light energy, our skin is able to synthesis Vitamin-D.
• Electricity can be produced with the help of light energy.

Question 2.
Explain about chemical energy and its uses.
Answer:
Chemical energy:
Chemical energy is stored in substances when atoms join together to form chemical compounds. When two or more chemical substances react with each other, this energy is released.

Uses of chemical energy:

• The food we eat contains chemical energy.
• Chemical energy in wood provides heat energy which helps us to cook food.
• Chemical energy in coal is used to generate electricity.
• Batteries we use in our daily life contain chemical energy.
• Fuels like petrol and diesel possess chemical energy which is used to run vehicles.

Question 3.
Draw the diagram to show how the electric energy generated from water.
Answer: