Category: Class 11

Students can download 11th Business Maths Chapter 5 Differential Calculus Ex 5.1 Questions and Answers, Notes, Samcheer Kalvi 11th Business 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 11th Business Maths Solutions Chapter 5 Differential Calculus Ex 5.1

Samacheer Kalvi 11th Business Maths Differential Calculus Ex 5.1 Text Book Back Questions and Answers

Question 1.
Determine whether the following functions are odd or even?
(i) f(x) = \(\left(\frac{a^{x}-1}{a^{x}+1}\right)\)
(ii) f(x) = log(x² + \(\sqrt{x^{2}+1}\))
(iii) f(x) = sin x + cos x
(iv) f(x) = x² – |x|
(v) f(x) = x + x²
Solution:
(i) f(x) = \(\left(\frac{a^{x}-1}{a^{x}+1}\right)\)

Thus f(-x) = -f(x)
∴ f(x) is an odd function.

(ii) f(x) = log(x² + \(\sqrt{x^{2}+1}\))
f(-x) = log((-x)² + \(\sqrt{(-x)^{2}+1}\))
= log(x² + \(\sqrt{x^{2}+1}\))
Thus f(-x) = f(x)
∴ f(x) is an even function.

(iii) f(x) = sin x + cos x
f(-x) = sin(-x) + cos(-x)
= -sin x + cos x
= -[sin x – cos x]
Since f(-x) ≠ -f(x) (or) f(x) ≠ -f(x)
∴ f(x) is neither odd nor even function.

(iv) Given f(x) = x² – |x|
f(-x) = (-x)² – |-x|
= x² – |x|
= f(x)
∴ f(x) is an even function.

(v) f(x) = x + x²
f(-x) = (-x) + (-x)² = -x + x²
Since f(-x) ≠ f(x), f(-x) ≠ -f(x).
∴ f(x) is neither odd nor even function.

Question 2.
Let f be defined by f(x) = x³ – kx² + 2x, x ∈ R. Find k, if ‘f’ is an odd function.
Solution:
For a polynomial function to be an odd function each term should have odd powers pf x. Therefore there should not be an even power of x term.
∴ k = 0.

Question 3.
If f(x) = \(x^{3}-\frac{1}{x^{3}}\), then show that f(x) + f(\(\frac{1}{x}\)) = 0
Solution:
f(x) = \(x^{3}-\frac{1}{x^{3}}\) …….. (1)
\(f\left(\frac{1}{x}\right)=\left(\frac{1}{x}\right)^{3}-\frac{1}{\left(\frac{1}{x}\right)^{3}}\) = \(\frac{1}{x^{3}}-x^{3}\) …….. (2)
(1) + (2) gives \(f(x)+f\left(\frac{1}{x}\right)=x^{3}-\frac{1}{x^{3}}+\frac{1}{x^{3}}-x^{3}=0\)
Hence Proved.

Question 4.
If f(x) = \(\frac{x+1}{x-1}\), then prove that f(f(x)) = x.
Solution:

Hence proved.

Question 5.
For f(x) = \(\frac{x-1}{3 x+1}\), write the expressions of f(\(\frac{1}{x}\)) and \(\frac{1}{f(x)}\).
Solution:

Question 6.
If f(x) = e^x and g(x) = log_e x then find
(i) (f + g) (1)
(ii) (fg) (1)
(iii) (3f) (1)
(iv) (5g) (1)
Solution:
(i) (f+g) (1) = e¹ + log_e 1 = e + 0 = e
(ii) (fg) (1) = f(1) g(1) = e¹ \(\log _{e}^{1}\) = e × 0 = 0
(iii) (3f) (1) = 3 f(1) = 3 e¹ = 3e
(iv) (5g) (1) = 5 (g) (1) = 5 \(\log _{e}^{1}\) = 5 × 0 = 0

Question 7.
Draw the graph of the following functions:
(i) f(x) = 16 – x²
(ii) f(x) = |x – 2|
(iii) f(x) = x|x|
(iv) f(x) = e^2x
(v) f(x) = e^-2x
(vi) f(x) = \(\frac{|x|}{x}\)
Solution:
(i) f(x) = 16 – x²
Let y = f(x) = 16 – x²
Choose suitable values for x and determine y. Thus we get the following table.

Plot the points (-4, 0), (-3, 7), (-2, 12), (-1, 15), (0, 16), (1, 15), (2, 12), (3, 7), (4, 0).
The graph is as shown in the figure.

(ii) Let y = f(x) = |x – 2|


Plot the points (2, 0), (3, 1) (4, 2), (5, 3), (0, 2), (-1, 3), (-2, 4), (-3, 5) and draw a line.
The graph is as shown in the figure.

(iii) Let y = f(x) = x|x|


Plot the points (0, 0), (1, 1) (2, 4), (3, 9), (-1, -1), (-2, -4), (-3, -9) and draw a smooth curve.
The graph is as shown in the figure.

(iv) For x = 0, f(x) becomes 1 i.e., the curve cuts the y-axis at y = 1.
For no real value of x, f(x) equals to 0. Thus it does not meet the x-axis for real values of x.

(v) For x = 0, f(x) becomes 1 i.e., the curve cuts the y-axis at y = 1.
For no real value of x, f(x) equal to 0. Thus it does not meet the x-axis for real values of x.

(vi) If f: R → R is defined by


The domain of the function is R and the range is {-1, 0, 1}.

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 5 Differential Calculus Ex 5.3

Samacheer Kalvi 11th Business Maths Differential Calculus Ex 5.3 Text Book Back Questions and Answers

Question 1.
Examine the following functions for continuity at indicated points.

Solution:
(a) f(x) = \(\frac{x^{2}-4}{x-2}\), also given that f(2) = 0

[∵ x = 2 – h, where h → 0, x → 2]


∴ The given function is not continuous at x = 2.

(b) Given that f(x) = \(\frac{x^{2}-9}{x-3}\) and f(3) = 6

[∵ x = 3 – h, where h → 0, x → 3]


[∵ x = 3 + h, where x → 3, h → 0]


= 0 + 6
= 6
Also given that f(3) = 6

∴ The given function f(x) is continuous at x = 3.

Question 2.
Show that f(x) = |x| is continuous at x = 0.
Solution:

[∵ x = 0 – h]



[∵ |x| = x if x > 0]

∴ f(x) is continuous at x = 0.

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 4 Trigonometry Ex 4.4

Samacheer Kalvi 11th Business Maths Trigonometry Ex 4.4 Text Book Back Questions and Answers

Question 1.
Find the principal value of the following:
(i) sin^-1 (\(-\frac{1}{2}\))
(ii) tan^-1 (-1)
(iii) cosec^-1 (2)
(iv) sec^-1 (-√2)
Solution:
(i) sin^-1 (\(-\frac{1}{2}\))
Let sin^-1 (\(-\frac{1}{2}\)) = y
[where \(\frac{-\pi}{2} \leq y \leq \frac{\pi}{2}\)]
\(-\frac{1}{2}\) = sin y
sin y = \(-\frac{1}{2}\) (∵ \(\sin \frac{\pi}{6}=\frac{1}{2}\))
sin y = sin(\(-\frac{\pi}{6}\)) [∵ \(\sin \left(-\frac{\pi}{6}\right)=-\sin \left(\frac{\pi}{6}\right)\)]
∴ y = \(-\frac{\pi}{6}\)
∴ The principal value of sin-1 (\(-\frac{1}{2}\)) is \(-\frac{\pi}{6}\)

(ii) tan^-1 (-1) = y
(-1) = tan y where \(\frac{-\pi}{2} \leq y \leq \frac{\pi}{2}\)
(or) tan y = – 1
tan y = tan(\(-\frac{\pi}{4}\)) (∵ \(\tan \frac{\pi}{4}\) = 1)
∴ y = \(-\frac{\pi}{4}\) [∵ \(\tan \left(-\frac{\pi}{4}\right)=-\tan \left(\frac{\pi}{4}\right)=-1\)]
∴ The principal value of tan^-1 (-1) is \(-\frac{\pi}{4}\).

(iii) Let cosec^-1 (2) = y
2 = cosec y
(or) cosec y = 2
⇒ \(\frac{1}{\sin y}\) = 2
⇒ sin y = \(\frac{1}{2}\) (Take reciprocal)
⇒ sin y = \(\sin \left(\frac{\pi}{6}\right)\)
⇒ y = \(\frac{\pi}{6}\)
The principal value of cosec^-1 (-1) is \(\frac{\pi}{6}\).

(iv) Let sec^-1 (-√2 ) = y
-√2 = sec y
sec y = -√2
\(\frac{1}{\cos y}\) = -√2
Taking reciprocal cos y = \(\frac{-1}{\sqrt{2}}\) [where 0 ≤ y ≤ π]

∴ The principal value of sec^-1 (-√2) is \(\frac{3 \pi}{4}\)

Question 2.
Prove that
(i) 2 tan^-1 (x) = \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\)
(ii) \(\tan ^{-1}\left(\frac{4}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)=\frac{\pi}{4}\)
Solution:
(i) Let tan^-1 x = θ
x = tan θ

(ii) \(\tan ^{-1}\left(\frac{4}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)=\frac{\pi}{4}\)

Question 3.
Show that \(\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{2}{11}\right)=\tan ^{-1}\left(\frac{3}{4}\right)\)
Solution:
We know that tan^-1 x + tan^-1 y = \(\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\)
Now LHS = \(\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{2}{11}\right)\)

Question 4.
Solve: tan^-1 2x + tan^-1 3x = \(\frac{\pi}{4}\).
Solution:
Given tan^-1 (2x) + tan^-1 (3x) = \(\frac{\pi}{4}\)

⇒ 5x = 1(1 – 6x²)
⇒ 6x² + 5x – 1 = 0
⇒ (x + 1) (6x – 1) = 0
⇒ x + 1 = 0 (or) 6x – 1 = 0
⇒ x = -1 (or) x = \(\frac{1}{6}\)
x = -1 is rejected. It doesn’t satisfies the question.
Note: Put x = -1 in the given question.

So the question changes.

Question 5.
Solve: tan^-1 (x + 1) + tan^-1 (x – 1) = \(\tan ^{-1}\left(\frac{4}{7}\right)\)
Solution:


⇒ 7x = 2(2 – x²)
⇒ 7x = 4 – 2x²
⇒ 2x² + 7x – 4 = 0
⇒ (x + 4) (2x – 1) = 0
⇒ x + 4 = 0 (or) 2x – 1 = 0
⇒ x = -4 (or) x = \(\frac{1}{2}\)
x = -4 is rejected, since does not satisfies the question.
∴ x = \(\frac{1}{2}\)

Question 6.
Evaluate
(i) cos[tan^-1(\(\frac{3}{4}\))]
(ii) \(\sin \left[\frac{1}{2} \cos ^{-1}\left(\frac{4}{5}\right)\right]\)
Solution:
(i) Let \(\tan ^{-1}\left(\frac{3}{4}\right)\) = θ
\(\frac{3}{4}\) = tan θ
tan θ = \(\frac{3}{4}\)
∴ cos θ = \(\frac{4}{5}\)

Now \(\cos \left(\tan ^{-1} \frac{3}{4}\right)\) = cos θ = \(\frac{4}{5}\)

(ii) Let \(\cos ^{-1}\left(\frac{4}{5}\right)\) = A
Then \(\frac{4}{5}\) = cos A
cos A = \(\frac{4}{5}\)

We know that

Question 7.
Evaluate: \(\cos \left(\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{12}{13}\right)\right)\)
Solution:

Let \(\sin ^{-1}\left(\frac{4}{5}\right)\) = A
sin A = \(\frac{4}{5}\)
∴ cos A = \(\frac{3}{5}\)

Let \(\sin ^{-1}\left(\frac{12}{13}\right)\) = B
\(\frac{12}{13}\) = sin B
sin B = \(\frac{12}{13}\)
∴ cos B = \(\frac{5}{13}\)
Now \(\cos \left(\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{12}{13}\right)\right)\) = cos (A + B)
= cos A cos B – sin A sin B
= \(\frac{3}{5} \times \frac{5}{13}-\frac{4}{5} \times \frac{12}{13}\)
= \(\frac{15}{65}-\frac{48}{65}\)
= \(-\frac{33}{65}\)

Question 8.
Prove that \(\tan ^{-1}\left(\frac{m}{n}\right)-\tan ^{-1}\left(\frac{m-n}{m+n}\right)=\frac{\pi}{4}\)
Solution:

Question 9.
Show that \(\sin ^{-1}\left(-\frac{3}{5}\right)-\sin ^{-1}\left(-\frac{8}{17}\right)=\cos ^{-1}\left(\frac{84}{85}\right)\)
Solution:
\(\sin ^{-1}\left(-\frac{3}{5}\right)-\sin ^{-1}\left(-\frac{8}{17}\right)=-\sin ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1}\left(\frac{8}{17}\right)\)
= \(\sin ^{-1}\left(\frac{8}{17}\right)-\sin ^{-1}\left(\frac{3}{5}\right)\)

Let \(\sin ^{-1}\left(\frac{8}{17}\right)\) = A
\(\frac{8}{17}\) = sin A
sin A = \(\frac{8}{17}\)
∴ cos A = \(\frac{15}{17}\)

Let \(\sin ^{-1}\left(\frac{3}{5}\right)\) = B
sin B = \(\frac{3}{5}\)
∴ cos B = \(\frac{4}{5}\)
Consider cos(A – B) = cos A cos B + sin A sin B
= \(\frac{15}{17} \times \frac{4}{5}+\frac{8}{17} \times \frac{3}{5}\)
= \(\frac{60}{85}+\frac{24}{85}\)
cos (A – B) = \(\frac{84}{85}\)
∴ A – B = \(\cos ^{-1}\left(\frac{84}{85}\right)\)

Question 10.
Express \(\tan ^{-1}\left[\frac{\cos x}{1-\sin x}\right]\), \(-\frac{\pi}{2}<x<\frac{3 \pi}{2}\) in the simplest form.
Solution:
\(\tan ^{-1}\left[\frac{\cos x}{1-\sin x}\right]\)


[∵ a² – b² = (a + b) (a – b)]

[∵ Divide each term by cos \(\frac{x}{2}\)]

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 4 Trigonometry Ex 4.3

Samacheer Kalvi 11th Business Maths Trigonometry Ex 4.3 Text Book Back Questions and Answers

Question 1.
Express each of the following as the sum or difference of sine or cosine:
(i) sin\(\frac{A}{8}\) sin\(\frac{3A}{8}\)
(ii) cos(60° + A) sin(120° + A)
(iii) cos\(\frac{7 A}{3}\) sin\(\frac{5 A}{3}\)
(iv) cos 7θ sin 3θ
Solution:
(i) sin\(\frac{A}{8}\) sin\(\frac{3A}{8}\)

[∵ 2 sin A sin B = cos(A – B) – cos(A + B)

[∵ cos(-θ) = cos θ]

(ii) cos(60° + A) sin(120° + A) = \(\frac{1}{2}\) [2 cos(60° + A) sin(120° + A)] [Multiply and divide by 2]
= \(\frac{1}{2}\) [sin((60° + A) + (120° + A))] – sin((60° + A) – (120° + A))]
[∵ 2 cos A sin B = sin(A + B) – sin(A – B)]
= \(\frac{1}{2}\) [sin(180° + 2A) – sin(60° + A – 120° – A)]
= \(\frac{1}{2}\) [(-sin 2A) – sin(-60°)]
= \(\frac{1}{2}\) [-sin 2A + sin 60°]
= \(\frac{1}{2}\) [-sin 2A + \(\frac{\sqrt{3}}{2}\)]

(iii) cos\(\frac{7 A}{3}\) sin\(\frac{5 A}{3}\)

(iv) cos 7θ sin 3θ = \(\frac{1}{2}\) [sin(7θ + 3θ) – sin(7θ – 3θ)]
= \(\frac{1}{2}\) (sin 10θ – sin 4θ)

Question 2.
Express each of the following as the product of sine and cosine
(i) sin A + sin 2A
(ii) cos 2A + cos 4A
(iii) sin 6θ – sin 2θ
(iv) cos 2θ – cos θ
Solution:
(i) sin A + sin 2A = 2 sin(\(\frac{A+2 A}{2}\)) cos(\(\frac{A-2 A}{2}\))
[∵ sin C + sin D = sin(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))]
= 2 sin \(\frac{3A}{2}\) cos \(\frac{A}{2}\) [∵ cos(-θ) = cos θ]

(ii) cos 2A + cos 4A = 2 cos(\(\frac{2 \mathrm{A}+4 \mathrm{A}}{2}\)) cos(\(\frac{2 \mathrm{A}-4 \mathrm{A}}{2}\))
[∵ cos C + cos D = 2 cos(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))
= 2 cos(\(\frac{6 \mathrm{A}}{2}\)) cos(\(\frac{6 \mathrm{-2A}}{2}\))
= 2 cos(3A) cos (-A) [∵ cos(-θ) = cos θ]
= 2 cos 3A cos A

(iii) sin 6θ – sin 2θ = 2 cos(\(\frac{6 \theta+2 \theta}{2}\)) cos(\(\frac{6 \theta-2 \theta}{2}\))
[∵ sin C – sin D = 2 cos(\(\frac{C+D}{2}\)) sin(\(\frac{C-D}{2}\))
= 2 cos(\(\frac{8 \theta}{2}\)) sin(\(\frac{4 \theta}{2}\))
= 2 cos 4θ sin 2θ

(iv) cos 2θ – cos θ = -2 sin(\(\frac{2 \theta+\theta}{2}\)) sin(\(\frac{2 \theta-\theta}{2}\))
[∵ cos C – cos D = -2 sin(\(\frac{C+D}{2}\)) sin(\(\frac{C-D}{2}\))
= -2 sin(\(\frac{3 \theta}{2}\)) sin(\(\frac{\theta}{2}\))

Question 3.
Prove that
(i) cos 20° cos 40° cos 80° = \(\frac{1}{8}\)
(ii) tan 20° tan 40° tan 80° = √3.
Solution:
(i) cos 20° cos 40° cos 80° = \(\left(\frac{2 \sin 20^{\circ}}{2 \sin 20^{\circ}}\right)\) cos 20° cos 40° cos 80°
[multiply and divide by 2 sin 20°]

(Multiply and divide by 2)


[∵ sin(180° – θ) = sin θ]

(ii) tan 20° tan 40° tan 80°

Consider sin 20° × sin 40° sin 80°
= sin 20° sin (60° – 20°) sin (60° + 20°)
= sin 20° [sin² 60° – sin² 20°]

cos 20° × cos 40° cos 80° = \(\frac{1}{8}\) [∵ from (i)] …… (2)
divide (1) by (2) we get, tan 20° tan 40° tan 80° = \(\frac{\frac{\sqrt{3}}{8}}{\frac{1}{8}}\) = √3

Question 4.
Prove that
(i) (cos α – cos β)² + (sin α – sin β)² = 4 \(\sin ^{2}\left(\frac{\alpha-\beta}{2}\right)\)
(ii) sin A sin(60° + A) sin(60° – A) = sin 3A
Solution:
(i) LHS = (cos α – cos β)² + (sin α – sin β)²

(ii) LHS = 4 sin A sin (60° + A) . sin (60° – A)
= 4 sin A {sin (60° + A) . sin (60° – A)}
= 4 sin A {sin² 60° – sin² A}
= 4 sin A {\(\frac{3}{4}\) – sin² A}
= 3 sin A – 4 sin³ A
= sin 3A
= RHS

Question 5.
Prove that
(i) sin (A – B) sin C + sin (B – C) sin A + sin(C – A) sin B = 0
(ii) \(2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0\)
Solution:
Consider sin (A – B) sin C
= (sin A cos B – cos A sin B) sin C
= sin A cos B sin C – cos A sin B sin C …….. (1)
Similarly sin(B – C) sin A = sin B cos C sin A – cos B sin C sin A …….. (2)
[Replace A by B, B by C, C by A in (1)]
and sin(C – A) sin B [Replace A by B, B by C, C by A in (2)]
= sin C cos A sin B – cos C sin A sin B …….. (3)
Adding (1), (2) and (3) we get
sin (A – B) sin C + sin (B – C) sin A + sin(C – A) sin B = 0

(ii) \(2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0\)

[∵ cos C + cos D = 2 cos(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))]

[∵ cos(-θ) = cos θ]

[take 2 cos \(\frac{\pi}{3}\) as commom)

Hence proved.

Question 6.
Prove that
(i) \(\frac{\cos 2 A-\cos 3 A}{\sin 2 A+\sin 3 A}=\tan \frac{A}{2}\)
(ii) \(\frac{\cos 7 \mathbf{A}+\cos 5 \mathbf{A}}{\sin 7 \mathbf{A}-\sin 5 \mathbf{A}}=\cot \mathbf{A}\)
Solution:
(i) \(\frac{\cos 2 A-\cos 3 A}{\sin 2 A+\sin 3 A}=\tan \frac{A}{2}\)

(ii) \(\frac{\cos 7 \mathbf{A}+\cos 5 \mathbf{A}}{\sin 7 \mathbf{A}-\sin 5 \mathbf{A}}=\cot \mathbf{A}\)



Hence proved.

Question 7.
Prove that cos 20° cos 40° cos 60° cos 80° = \(\frac{3}{16}\).
Solution:
LHS = cos 20° cos 40° cos 60° cos 80°
= cos 20° cos 40° (\(\frac{1}{2}\)) cos 80° [∵ cos 60° = \(\frac{1}{2}\)]
= \(\frac{1}{2}\) (cos 20° cos 40° cos 80°)
= \(\frac{1}{2}\left(\frac{2 \sin 20^{\circ}}{2 \sin 20^{\circ}}\right)\) (cos 20° cos 40° cos 80°)
[multiply and divide by 2 sin 20°]

[multiply and divide by 2]

Hence Proved.

Question 8.
Evaluate:
(i) cos 20° + cos 100° + cos 140°
(ii) sin 50° – sin 70° + sin 10°
Solution:
(i) LHS = (cos 20° + cos 100°) + cos 140°
= 2 \(\cos \left(\frac{20^{\circ}+100^{\circ}}{2}\right) \cos \left(\frac{20^{\circ}-100^{\circ}}{2}\right)\) + cos 140°
[∵ cos C + cos D = 2 cos(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))]
= 2 cos 60° cos(-40°) + cos 140°
= 2 × \(\frac{1}{2}\) × cos(-40°) + cos(180° – 140°)
[∵ cos(-θ) = cos θ, cos 60° = \(\frac{1}{2}\)
= cos 40° – cos 40°
= 0
Hence Proved.

(ii) LHS = (sin 50° – sin 70°) + sin 10°
= 2 \(\cos \left(\frac{50^{\circ}+70^{\circ}}{2}\right) \sin \left(\frac{50^{\circ}-70^{\circ}}{2}\right)\) + sin 10°
[∵ sin C – sin D = 2 cos(\(\frac{C+D}{2}\)) sin(\(\frac{C-D}{2}\))]
= 2 cos 60° sin(-10°) + sin 10°
= 2 × \(\frac{1}{2}\) (-sin 10°) + sin 10° [∵ sin(-θ) = -sin θ]
= -sin 10° + sin 10°
= 0
= RHS

Question 9.
If cos A + cos B = \(\frac{1}{2}\) and sin A + sin B = \(\frac{1}{4}\), prove that \(\tan \left(\frac{\mathbf{A}+\mathbf{B}}{2}\right)=\frac{\mathbf{1}}{2}\)
Solution:
Given that cos A + cos B = \(\frac{1}{2}\)
\(2 \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=\frac{1}{2}\) …….. (1)
Also given that sin A + sin B = \(\frac{1}{4}\)
\(2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=\frac{1}{4}\) ……. (2)
\(\frac{(2)}{(1)}\) gives

Question 10.
If sin(y + z – x), sin(z + x – y), sin(x + y – z) are in A.P, then prove that tan x, tan y and tan z are in A.P.
Solution:
In A.P. commom difference are equal, namely t₂ – t₁ = t₃ – t₂
sin(z + x – y) – sin(y + z – x) = sin(x + y – z) – sin(z + x – y)


cos z sin (x – y) = cos x sin (y – z)
cos z (sin x cos y – cos x sin y) = cos x (sin y cos z – cos y sin z)
Divide bothsides by cos x cos y cos z we get

tan x – tan y = tan y – tan z
Multiply both sides by (-1) we get,
tan y – tan x = tan z – tan y
This means tan x, tan y, and tan z are in A.P.
Hence proved.

Question 11.
If cosec A + sec A = cosec B + sec B prove that cot(\(\frac{A+B}{2}\)) = tan A tan B.
Solution:
Given that cosec A + sec A = cosec B + sec B

Arrange T-ratios of the sine and cosine in the separate side

[∵ sin C – sin D = 2 cos(\(\frac{C+D}{2}\)) sin(\(\frac{C-D}{2}\))]

Students can download 11th Business Maths Chapter 4 Trigonometry Ex 4.2 Questions and Answers, Notes, Samcheer Kalvi 11th Business 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 11th Business Maths Solutions Chapter 4 Trigonometry Ex 4.2

Samacheer Kalvi 11th Business Maths Trigonometry Ex 4.2 Text Book Back Questions and Answers

Question 1.
Find the values of the following:
(i) cosec 15°
(ii) sin (-105°)
(iii) cot 75°
Solution:
(i) cosec 15° = \(\frac{1}{\sin 15^{\circ}}\)
Consider sin 15° = sin(45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°

cosec 15° = \(\frac{1}{\sin 15^{\circ}}\) = \(\frac{2 \sqrt{2}}{\sqrt{3}-1}\)

(ii) sin (-105°) = -sin (105°) (∵ sin (-θ) = – sin θ)
= -[sin(60° + 45°)]
= -[sin 60° cos 45° + cos 60° sin 45°]

(iii) cot 75° = \(\frac{1}{\tan 75^{\circ}}\)
Consider tan 75° = tan (30° + 45°)

cot 75° = \(\frac{1}{\tan 75^{\circ}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\)

Question 2.
Find the values of the following:
(i) sin 76° cos 16° – cos 76° sin 16°
(ii) \(\sin \frac{\pi}{4} \cos \frac{\pi}{12}+\cos \frac{\pi}{4} \sin \frac{\pi}{12}\)
(iii) cos 70° cos 10° – sin 70° sin 10°
(iv) cos² 15° – sin² 15°
Solution:
(i) Given that, sin 76° cos 16° – cos 76° sin 16° (∴ This is of the form sin(A – B))
= sin(76° – 16°)
= sin 60°
= \(\frac{\sqrt{3}}{2}\)

(ii) This is of the form sin(A + B) = \(\sin \left(\frac{\pi}{4}+\frac{\pi}{12}\right)\)
= \(\sin \left(\frac{3 \pi+\pi}{12}\right)\)
= \(\sin \frac{4 \pi}{12}\)
= \(\sin \frac{\pi}{3}\)
= \(\frac{\sqrt{3}}{2}\) (∵ sin 60° = \(\frac{\sqrt{3}}{2}\))

(iii) Given that cos 70° cos 10° – sin 70° sin 10°
(This is of the form of cos (A + B), A = 70°, B = 10°)
= cos (70° + 10°)
= cos 80°

(iv) cos² 15° – sin² 15°
[∵ cos 2A = cos² A – sin² A, Here A = 15°]
= cos (2 × 15°)
= cos 30°
= \(\frac{\sqrt{3}}{2}\)

Question 3.
If sin A = \(\frac{3}{5}\), 0 < A < \(\frac{\pi}{2}\) and cos B = \(\frac{-12}{13}\), π < B < \(\frac{3 \pi}{2}\), find the values of the following:
(i) cos(A + B)
(ii) sin(A – B)
(iii) tan(A – B)
Solution:
Given that sin A = \(\frac{3}{5}\), 0 < A < \(\frac{\pi}{2}\) (i.e., A lies in first quadrant)
Since A lies in first quadrant cos A is positive.

cos A = \(\frac{\text { Adjacent side }}{\text { Hypotenuse }}=\frac{4}{5}\)
tan A = \(\frac{3}{4}\)
AB = \(\sqrt{5^{2}-3^{2}}\) = 4
Also given that cos B = \(\frac{-12}{13}\), π < B < \(\frac{3 \pi}{2}\) (i.e., B lies in third quadrant)
Now sin B lies in third quadrant. sin B is negative.

CA = \(\sqrt{13^{2}-12^{2}}\) = 5
sin B = \(\frac{-\text { Opposite side }}{\text { Hypotenuse }}=\frac{-5}{13}\)
tan B = \(\frac{-\text { Opposite side }}{\text { Adjacent }}=\frac{5}{12}\) [B lies in 3rd quadrant. tan B is positive.]
(i) cos(A + B) = cos A cos B – sin A sin B

(ii) sin(A – B) = sin A cos B – cos A sin B

(iii) tan(A – B)

Question 4.
If cos A = \(\frac{13}{14}\) and cos B = \(\frac{1}{7}\) where A, B are acute angles prove that A – B = \(\frac{\pi}{3}\)
Solution:
cos A = \(\frac{13}{14}\), cos B = \(\frac{1}{7}\)

cos(A – B) = cos A cos B + sin A sin B

cos(A – B) = cos 60°
A – B = 60° = \(\frac{\pi}{3}\)

Question 5.
Prove that 2 tan 80° = tan 85° – tan 5°.
Solution:
Consider tan 80° = tan(85° – 5°)


∴ 2 tan 80° = tan 85° – tan 5°
Hence Proved.

Question 6.
If cot α = \(\frac{1}{2}\), sec β = \(\frac{-5}{3}\), where π < α < \(\frac{3 \pi}{2}\) and \(\frac{\pi}{2}\) < β < π, find the value of tan(α + β). State the quadrant in which α + β terminates.
Solution:
Given that cot α = \(\frac{1}{2}\) where π < α < \(\frac{3 \pi}{2}\) (i.e,. α lies in third quadrant)
tan α = \(\frac{1}{\frac{1}{2}}\) = 2 [∵ In 3rd quadrant tan α is positive]
Also given that sec β = \(\frac{-5}{3}\) where \(\frac{\pi}{2}\) < β < π (i.e., β lies in second quadrant cos β and tan β are negative)


tan (α + β) = \(\frac{2}{11}\) which is positive.
α + β terminates in first quandrant.

Question 7.
If A + B = 45°, prove that (1 + tan A) (1 + tan B) = 2 and hence deduce the value of tan 22\(\frac{1}{2}\).
Solution:
Given A + B = 45°
tan (A + B) = tan 45°
\(\frac{\tan A+\tan B}{1-\tan A \tan B}=1\)
tan A + tan B = 1 – tan A . tan B
tan A + tan B + tan A tan B = 1
Add 1 on both sides we get,
(1 + tan A) + tan B + tan A tan B = 2
1(1+ tan A) + tan B (1 + tan A) = 2
(1 + tan A) (1 + tan B) = 2 ……. (1)
Put A = B = 22\(\frac{1}{2}\) in (1) we get
(1 + tan 22\(\frac{1}{2}\)) (1 + tan 22\(\frac{1}{2}\)) = 2
⇒ (1 + tan22\(\frac{1}{2}\))² = 2
⇒ 1 + tan 22\(\frac{1}{2}\) = ±√2
⇒ tan 22\(\frac{1}{2}\) = ±√2 – 1
Since 22\(\frac{1}{2}\) is acute, tan 22\(\frac{1}{2}\) is positive and therefore tan 22\(\frac{1}{2}\) = √2 – 1

Question 8.
Prove that
(i) sin(A + 60°) + sin(A – 60°) = sin A.
(ii) tan 4A tan 3A tan A + tan 3A + tan A – tan 4A = 0
Solution:
(i) LHS = sin (A + 60°) + sin (A – 60°)
= sin A cos 60° + cos A sin 60° + sin A cos 60° – cos A sin 60°
= 2 sin A cos 60°
= 2 sin A \(\left(\frac{1}{2}\right)\)
= sin A
= RHS

(ii) 4A = 3A + A
tan 4A = tan (3A + A)
tan 4A = \(\frac{\tan 3 \mathrm{A}+\tan \mathrm{A}}{1-\tan 3 \mathrm{A} \tan \mathrm{A}}\)
on cross multiplication we get,
tan 3A + tan A = tan 4A (1 – tan 3A tan A) = tan 4A – tan 4A tan 3A tanA
i.e., tan 4A tan 3A tan A + tan 3A + tan A = tan 4A
(or) tan 4A tan 3A tan A + tan 3A + tan A – tan 4A = 0

Question 9.
(i) If tan θ = 3 find tan 3θ
(ii) If sin A = \(\frac{12}{13}\), find sin 3A.
Solution:
(i) tan θ = 3

(ii) If sin A = \(\frac{12}{13}\)
We know that sin 3A = 3 sin A – 4 sin³ A

Question 10.
If sin A = \(\frac{3}{5}\), find the values of cos 3A and tan 3A.
Solution:

Given sin A = \(\frac{3}{5}\)
cos A = \(\frac{\text { Adjacent side }}{\text { Hypotenuse }}=\frac{4}{5}\)
and tan A = \(\frac{\text { Opposite side }}{\text { Adjacent side }}=\frac{3}{4}\)
We know that cos 3A = 4 cos³ A – 3 cos A

Question 11.
Prove that \(\frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin (C-A)}{\cos C \cos A}+\frac{\sin (A-B)}{\cos A \cos B}=0\)
Solution:
Consider \(\frac{\sin (B-C)}{\cos B \cos C}\)
= \(\frac{\sin \mathrm{B} \cos \mathrm{C}-\cos \mathrm{B} \sin \mathrm{C}}{\cos \mathrm{B} \cos \mathrm{C}}\)
= \(\frac{\sin B \cos C}{\cos B \cos C}-\frac{\cos B \sin C}{\cos B \cos C}\)
= tan B – tan C ……… (1)
Similarly we can prove \(\frac{\sin (C-A)}{\cos C \cos A}\) = tan C – tan A …….(2)
and \(\frac{\sin (A-B)}{\cos A \cos B}\) = tan A – tan B …….. (3)
Add (1), (2) and (3) we get
\(\frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin (C-A)}{\cos C \cos A}+\frac{\sin (A-B)}{\cos A \cos B}=0\)

Question 12.
If tan A – tan B = x and cot B – cot A = y prove that cot(A – B) = \(\frac{1}{x}+\frac{1}{y}\).
Solution:


Hence proved.

Question 13.
If sin α + sin β = a and cos α + cos β = b, then prove that cos(α – β) = \(\frac{a^{2}+b^{2}-2}{2}\)
Solution:
Consider a² + b² = sin²α + sin²β + 2 sin α sin β + cos²α + cos²β + 2 cos α cos β
a² + b² = (sin²α + cos²α) + (sin²β + cos²β) + 2[cos α cos β + sin α sin β]
a² + b² = 1 + 1 + 2 cos(α – β)
∴ cos(α – β) = \(\frac{a^{2}+b^{2}-2}{2}\)

Question 14.
Find the value of tan\(\frac{\pi}{8}\).
Solution:
Method 1:
\(\frac{\pi}{8}=\frac{180^{\circ}}{8}=\frac{45^{\circ}}{2}=22 \frac{1}{2}\)
We know that tan 2A = \(\frac{2 \tan A}{1-\tan ^{2} A}\)
Put A = 22\(\frac{1}{2}\) in the above formula

On cross multiplication we get


Here a = 1, b = 2, c = -1

Since 22\(\frac{1}{2}\) is acute tan 22\(\frac{1}{2}\) is positive tan 22\(\frac{1}{2}\) = tan \(\frac{\pi}{8}\)
= -1 + √2
= √2 – 1

Method 2:



∴ \(\tan ^{2} 22 \frac{1}{2}=(\sqrt{2}-1)^{2}\)
Taking square root, \(\tan ^{2} 22 \frac{1}{2}\) = ±(√2 – 1)
But \(22 \frac{1}{2}\) lies in first quadrant, tan \(22 \frac{1}{2}\) is positive.
∴ tan 22\(\frac{1}{2}\) = √2 – 1

Method 3:
consider tan A = \(\frac{\sin 2 A}{1+\cos 2 A}\)
Put A = \(22 \frac{1}{2}\)



tan 22\(\frac{1}{2}\) = √2 – 1

Question 15.
If tan α = \(\frac{1}{7}\), sin β = \(\frac{1}{\sqrt{10}}\). Prove that α + 2β = \(\frac{\pi}{4}\) where 0 < α < \(\frac{\pi}{2}\) and 0 < β < \(\frac{\pi}{2}\).
Solution:
Given that tan α = \(\frac{1}{7}\)
We wish to find tan(α + 2β)

Students can download 11th Business Maths Chapter 4 Trigonometry Ex 4.1 Questions and Answers, Notes, Samcheer Kalvi 11th Business 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 11th Business Maths Solutions Chapter 4 Trigonometry Ex 4.1

Samacheer Kalvi 11th Business Maths Trigonometry Ex 4.1 Text Book Back Questions and Answers

Question 1.
Convert the following degree measure into radian measure
(i) 60°
(ii) 150°
(iii) 240°
(iv) -320°
Solutions:
(i) 1°= \(\frac{\pi}{180}\) radians
∴ 60° = \(\frac{\pi}{180}\) × 60 radians = \(\frac{\pi}{3}\) radians.

(ii) 150° = \(\frac{\pi}{180}\) × 150 radians = \(\frac{5 \pi}{6}\) radians.

(iii) 240° = \(\frac{\pi}{180}\) × 240 radians = \(\frac{4 \pi}{3}\) radians.

(iv) -320° = \(\frac{\pi}{180}\) × -320 = \(\frac{-16 \pi}{9}\) radians

Question 2.
Find the degree measure corresponding to the following radian measure.
(i) \(\frac{\pi}{8}\)
(ii) \(\frac{9 \pi}{5}\)
(iii) -3
(iv) \(\frac{11 \pi}{18}\)
Solution:
We know that, one radian = \(\frac{180^{\circ}}{\pi}\)
(i) \(\frac{\pi}{8}\)
\(\frac{\pi}{8}=\frac{180^{\circ}}{\pi} \times \frac{\pi}{8}\) degrees
= \(\frac{45}{2}\)
= 22.5°
= 22°30′ [∵ 0.5° = (0.5 × 60)’ = 30′]

(ii) \(\frac{9 \pi}{5}\)
\(\frac{9 \pi}{5}=\frac{180^{\circ}}{\pi} \times \frac{9 \pi}{5}\) degrees
= 36 × 9 degrees
= 324°

(iii) -3

= -171.81°
= -171°48′ (∵ 0.8° = (0.8 × 60)’ = 48′)

(iv) \(\frac{11 \pi}{18}\)
\(\frac{11 \pi}{18}=\frac{180}{\pi} \times \frac{11 \pi}{18}\)
= 10 × 11°
= 110°

Question 3.
Determine the quadrants in which the following degree lie.
(i) 380°
(ii) -140°
(iii) 1195°
Solution:
(i) 380° = 360°+ 20°
This is of the form 360° + θ
∴ After one completion of the round, the angle is 20°, 380° lies in the I quadrant.

(ii) -140° = -90° + (-50°)
The angle is negative it moves in the anti-clockwise direction.
-140° lies in the III quadrants.

(iii) 1195° = (3 × 360°) + 90° + 25°
∴ After three completion round, the angle will lie in the II quadrant.
1195° lies in the II quadrant.

Question 4.
Find the values of each of the following trigonometric ratios.
(i) sin 300°
(ii) cos(-210°)
(iii) sec 390°
(iv) tan(-855°)
(v) cosec 1125°
Solution:
(i) sin 300° = sin(360° – 60°)
[For 360° – 60°. No change in T-ratio. 300° lies in 4th quadrant ‘sin’ is negative]
= -sin 60°
= \(-\frac{\sqrt{3}}{2}\)

(ii) cos(-210°) = cos 210° (∵ cos(-θ) = cos θ)
[∵ 180 + 30°. No change in T-ratio. 210° lies 3rd quadrant ‘cos’ is negative]
= cos(180° + 30°)
= -cos 30°
= \(-\frac{\sqrt{3}}{2}\)

(iii) sec 390° = sec(360° + 30°)
= sec 30°
= \(\frac{1}{\cos 30^{\circ}}\)
= \(\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}\)
= \(\frac{2}{\sqrt{3}}\)

(iv) tan(-855°) = -tan 855° (∵ tan(-θ) = – tan θ)
[∵ Multiplies of 360° are dropped out. For 180° – 45°. No change in T-ratio. 180° – 45° lies in 2nd quadrant ‘tan’ is negative]
= -tan(2 × 360° + 135°)
= -tan 135°
= -tan(180° – 45°)
= -(-tan 45°)
= -(-1)
= 1

(v) cosec 1125° = cosec(3 × 360°+ 45°)
= cosec 45°
= \(\frac{1}{\sin 45^{\circ}}\)
= \(\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}\)
= √2

Question 5.
Prove that:
(i) tan(-225°) cot(-405°) – tan(-765°) cot(675°) = 0.
(ii) 2 sin² \(\frac{\pi}{6}\) + cosec² \(\frac{7 \pi}{6}\) cos² \(\frac{\pi}{3}\) = \(\frac{3}{2}\)
(iii) \(\sec \left(\frac{3 \pi}{2}-\theta\right) \sec \left(\theta-\frac{5 \pi}{2}\right)+\tan \left(\frac{5 \pi}{2}+\theta\right) \tan \left(\theta-\frac{5 \pi}{2}\right)=-1\)
Solution:
(i) tan(-225°) = -(tan 225°)
= -(tan(180° + 45°))
= – tan 45°
= – 1
cot(-405°) = -(cot 405°)
= – cot(360° + 45°) [∵ For 360° + 45° no change in T-ratio.]
= -cot 45°
= -1
tan(-765°) = -tan 765°
= -tan(2 × 360° + 45°)
= -tan 45°
= -1
cot 675° = cot (360°+ 315°)
= cot 315°
= cot(360° – 45°)
= -cot 45°
= -1
LHS = tan(-225°) cot(-405°) – tan(-765°) cot(675°)
= (-1) (-1) – (-1) (-1)
= 1 – 1
= 0
= RHS.
Hence proved.

(ii) 2 sin² \(\frac{\pi}{6}\) + cosec² \(\frac{7 \pi}{6}\) cos² \(\frac{\pi}{3}\) = \(\frac{3}{2}\)
LHS = 2 sin² \(\frac{\pi}{6}\) + cosec² \(\frac{7 \pi}{6}\) cos² \(\frac{\pi}{3}\)
[∵ \(\frac{7 \pi}{6}\) = 210°, 210° = 180° + 30°. For 180° + 30° no change in T-ratio.
210° lies in 3rd quadrant, cosec θ is negative.]
= 2\(\left(\sin \frac{\pi}{6}\right)^{2}\) + (cosec (180° + 30°))² \(\left(\cos \frac{\pi}{3}\right)^{2}\)
= 2 \(\left(\frac{1}{2}\right)^{2}\) + (-cosec 30°)² . \(\left(\frac{1}{2}\right)^{2}\)
= \(2 \times \frac{1}{4}+(-2)^{2} \frac{1}{4}\)
= \(\frac{2}{4}+\frac{4}{4}=\frac{6}{4}\)
= \(\frac{6}{4}\)
= \(\frac{3}{2}\)
= RHS

(iii) sec(\(\frac{3 \pi}{2}\) – θ) = sec (270° – θ) = -cosec θ
[∵ For 270° – θ change T-ratio. So add ‘co’ infront ‘sec’, it becomes ‘cosec’]
sec(θ – \(\frac{5 \pi}{2}\)) = \(\sec \left(-\left(\frac{5 \pi}{2}-\theta\right)\right)\)
= sec(\(\frac{5 \pi}{2}\) – θ) [∵ sec(-θ) = θ]
= sec(450° – θ)
= sec (360° + (90° – θ))
= sec (90° – θ)
= cosec θ
[∵ For 90° – θ change in T-ratio. So add ‘co’ in front of ‘sec’ it becomes ‘cosec’]
tan(\(\frac{5 \pi}{2}\) + θ) = tan(450° + θ)
[∵ For 90° + θ, change in T-ratio. So add ‘co’ in front of ‘tan’ it becomes ‘cot’]
= tan (360° + (90° + θ))
= tan (90° + θ)
= -cot θ
\(\tan \left(\theta-\frac{5 \pi}{2}\right)=\tan \left(-\left(\frac{5 \pi}{2}-\theta\right)\right)\)
= \(-\tan \left(\frac{5 \pi}{2}-\theta\right)\) [∵ tan(-θ) = -tan θ]
= -tan(450° – θ)
= -tan(360° + (90° – θ))
= -tan(90° – θ)
= -cot θ
LHS = \(\sec \left(\frac{3 \pi}{2}-\theta\right) \sec \left(\theta-\frac{5 \pi}{2}\right)+\tan \left(\frac{5 \pi}{2}+\theta\right) \tan \left(\theta-\frac{5 \pi}{2}\right)\)
= -cosec θ (cosec θ) + (-cot θ) (-cot θ)
= -cosec² θ + cot² θ
= -(1 + cot² θ) + cot² θ [∵ 1 + cot² θ = cosec² θ]
= -1
= RHS

Question 6.
If A, B, C, D are angles of a cyclic quadrilateral, prove that: cos A + cos B + cos C + cos D = 0.
Solution:
Note: If the vertices of a quadrilateral lie on the circle then the quadrilateral is called a cyclic quadrilateral.
In a cyclic quadrilateral sum of opposite angles are 180°.

Since A, B, C, D are angles of cyclic quadrilateral
A + C = 180° and B + D = 180°
LHS = cos A + cos B + cos C + cos D
= cos A + cos B + cos(180° – A) + cos(180° – B)
= cos A + cos B – cos A – cos B
= 0
= RHS

Question 7.
Prove that

(ii) sin θ . cos{sin(\(\frac{\pi}{2}\) – θ) . cosec θ + cos(\(\frac{\pi}{2}\) – θ) . sec θ} = 1
Solution:

(ii) sin θ . cos{sin(\(\frac{\pi}{2}\) – θ) . cosec θ + cos(\(\frac{\pi}{2}\) – θ) . sec θ} = 1

Question 8.
Prove that: cos 510° cos 330° + sin 390° cos 120° = -1.
Solution:
LHS = cos 510° cos 330° + sin 390° cos 120°
= cos(360° + 150°) cos(360° – 30°) + sin(360° + 30°) × cos(180° – 60°)
= cos 150° cos 30° + sin 30° (-cos 60°)
= cos(180° – 30°) cos 30° + sin 30° cos 60°
= -cos 30° cos 30° + \(\frac{1}{2} \times\left(\frac{-1}{2}\right)\)
= \(-\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}-\frac{1}{2} \times \frac{1}{2}\)
= \(-\frac{3}{4}-\frac{1}{4}\)
= \(\frac{-3-1}{4}\)
= -1

Question 9.
Prove that:
(i) tan(π + x) cot(x – π) – cos(2π – x) cos(2π + x) = sin² x.

Solution:
(i) tan(π + x) cot(x – π) – cos(2π – x) cos(2π + x) = (tan x) (-cot(π – x) – cos x cos x
[∵ cot(x – π) = cot(-(π – x)) = -cot(π – x) = cot x]
= tan x cot x – cos² x
= 1 – cos² x
= sin² x [∵ sin² x + cos² x = 1 ⇒ sin² x = (1 – cos² x)]

Question 10.
If sin θ = \(\frac{3}{5}\), tan φ = \(\frac{1}{2}\) and \(\frac{\pi}{2}\) < θ < π < φ < \(\frac{3 \pi}{2}\), then find the value of 8 tan θ – √5 sec φ.
Solution:

Given that sin θ = \(\frac{3}{5}=\frac{\text { Opposite side }}{\text { Hypotenuse }}\)
∵ AB = \(\sqrt{5^{2}-3^{2}}=\sqrt{25-9}=\sqrt{16}\) = 4
Here θ lies in second quadrant [∵ \(\frac{\pi}{2}\) < θ < π]
∵ tan θ is negative.
tan θ = \(-\frac{3}{4}\)
Also given that tan Φ = \(\frac{1}{2}=\frac{\text { Opposite side }}{\text { Adjacent side }}\)
∴ PR = \(\sqrt{\mathrm{PQ}^{2}+\mathrm{QP}^{2}}=\sqrt{4+1}=\sqrt{5}\)
Here Φ lies in third quadrant (∵ π < Φ < \(\frac{3 \pi}{2}\))
∴ sec Φ is negative.
\(\sec \phi=\frac{1}{\cos \phi}=-\frac{1}{\left(\frac{2}{\sqrt{5}}\right)}=-\frac{\sqrt{5}}{2}\)
Now 8 tan θ – √5 sec Φ = \(8\left(-\frac{3}{4}\right)-\sqrt{5}\left(\frac{-\sqrt{5}}{2}\right)\)
= 2 × (-3) + \(\frac{5}{2}\)
= -6 + \(\frac{5}{2}\)
= \(\frac{-12+5}{2}\)
= \(\frac{-7}{2}\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Physics Guide Pdf Chapter 1 Nature of Physical World and Measurement Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

11th Physics Guide Nature of Physical World and Measurement Book Back Questions and Answers

Part – I:
I. Multiple choice questions:

Question 1.
One of the combinations from the fundamental physical constants is \(\frac { hG }{ G }\), The unit of this expression is.
(a) Kg²
(b) m³
(c) S^-1
(d) m
Answer:
(a) Kg²

Question 2.
If the error in the measurement of radius is 2%, then the error in the determination of volume of the sphere will be:
(a) 8%
(b) 2%
(c) 4%
(d) 6%
Answer:
(d) 6%

Question 3.
If the length and time period of an oscillating pendulum have errors of 1% and 3% respectively then the error in measurement of acceleration due to gravity is: [Related to AMPMT 2008]
(a) 4%
(b) 5%
(c) 6%
(d) 7%
Answer:
(d) 7%

Question 4.
The length of a body is measured as 3.15m, if the accuracy is 0.01m, then the percentage error in the measurement is:
(a) 351%
(b) 1%
(c) 0.28%
(d) 0.035%
Answer:
(c) 0.28%

Question 5.
Which of the following has the highest number of significant figure?
(a) 0.007 m²
(b) 2.64 x 1024 Kg
(c) 0.0006032 m²
(d) 6.3200 J
Answer:
(d) 6.3200 J

Question 6.
If π = 3.14, then the value of π² is:
(a) 9.8596
(b) 9.860
(c) 9.86
(d) 9.9
Answer:
(c) 9.86

Question 7.
Which of the following pairs of physical quantities have same dimension?
(a) force and power
(b) torque and energy
(c) torque and power
(d) force and torque
Answer:
(b) torque and energy

Question 8.
The dimensional formula of planck’s constant h is _____________. [AMU, Main, JEE, NEET]
(a) [ML² T^-1]
(b) [ML²T^-3]
(c) [MLT^-1]
(d) [ML³T^-3]
Answer:
(a) [ML² T^-1]

Question 9.
The velocity of a particle v at an instant t is given by v = at + bt² the dimension of b is _____________
(a) [L]
(b) [LT^-1]
(c) [LT^-2]
(d) [LT^-3]
Answer:
(d) [LT^-3]

Question 10.
The dimensional formula for gravitational constant G is _____________. [Related to AIPMT 2004]
(a) [ML³T^-2]
(b) [M^-1L³T^-2]
(c) [M^-1 L^-3T^-2]
(d) [ML^-3T²]
Answer:
(b) [M^-1L³T^-2]

Question 11.
The density of a material is CGS system of units Is 4 g cm^-3. In a system of units in which unit of length is 10cm and unit of mass is 100g, then the value of density of material will be:
(a) 0.04
(b) 0.4
(c) 40
(d) 400
Answer:
(c) 40

Question 12.
If the force is proportional to square of velocity, then the dimension of proportionality constant is _____________. [JEE 2000]
(a) [ML T⁰]
(b) [ML T^-1]
(c) [ML^-2T]
(d) [ML^-1T⁰]
Answer:
(d) [ML^-1T⁰]

Question 13.
The dimension of (µ⁰ε⁰)^{\(\frac { -1 }{ 2 }\)} is _____________. [Main AIPMT 2011]
(a) length
(b) time
(c) velocity
(d) force
Answer:
(c) velocity

Question 14.
Planck’s constant (h), speed of light in vacuum (c) and Newton’s gravitational constant (G) are takers as three fundamental constants. Which of the following combinations of these has the dimension of length? [NEET 2016 (phase II)]

Answer:

Question 15.
A length-scale (l) depends on the permittivity (c) of a dielectric material Boltzmann constant (Kb), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression for L is dimensionally correct? [JEE (advance(d) 2016 ]

Answer:

II. Short Answer Questions:

Question 1.
Briefly explain the types of physical quantities?
Answer:
Physical quantities are classified into two types. There are fundamental and derived quantities. Fundamental or base quantities are quantities which cannot be expressed in terms of any other physical quantities. These are length, mass, time, electric current, temperature, luminous intensity, and amount of substance.
Quantities that can be expressed in terms of fundamental quantities are called derived quantities. For example, area, volume, velocity, acceleration, force.

Question 2.
How will you measure the diameter of the moon using the parallax method?
Answer:
In order to determine the diameter of the moon, initially, a distance of the moon is calculated using the parallax method. Let D be the distance of the moon from the earth. Let d be the diameter of the moon. Let ∝ be the angular size of the angular diameter of the moon (ie) the angle subtended by d at the earth.

We have ∝ = d/D
d = ∝ D
The angle ∝ can be measured from the same location on the earth. When two diametrically opposite points of the moon are viewed through a telescope, the angle between the two directions gives the angular size or angular diameter. Since D is the known to size or diameter d of the moon can be determined.

Question 3.
Write the rules for determining significant figures.
Answer:
(1) All non zero digits are significant
Example: 1342 has 4 significant figures

(2) All zeros between two non-zero digits are significant
Example: 2008 has four significant figures

(3) All zeros to the right of non-zero digit but to the left of the decimal point are significant.
Example: 3070.00 has 4 significant figures.

(4) The trailing zeros are not significant, ie in the number without a decimal point. All zeros are significant if they come from the measurement.
Example: 4000 has one significant figure.

(5) If a number is less than 1, the zero (s) on the right of the decimal point but to the left of the first non-zero digit are not significant.
Example: 0.0034 has 2 significant figures.

(6) All zeros to the right of the decimal point and to the right of non zero digits are significant
Example: 40.00 has four significant figures.

(7) The number of significant figures does not depend on the system of units used.
Example: 1.53cm, 0.0150cm, 0.0000153 Km all have three significant figures.

(8) The power of 10 is irrelevant to the determination of significant figures
Example: 5.7 x 10² cm has two significant figures.

Question 4.
What are the limitations of dimensional analysis?
Answer:
(1) This method gives no information about the dimensionless constants in the formula. Like 1, 2,7i, e etc. ie they can not be determined using this analysis.

(2) This method can not decide whether the given quantity is a scalar or a vector.

(3) Using this method one cannot derive relations involving trigonometric, exponential and logarithmic functions.

(4) It cannot be applied to an equation involving more than three physical quantities.

(5) It can be used to check whether a given physical relation is dimensionally correct or not. The physical correctness can not be checked using this
For example:
s = ut + 1/3 at² is dimensionally correct were as physically not correct, as the correct equation is s = ut + 1/2at².

Question 5.
Define precision and accuracy. Explain with one example.
Answer:
The accuracy of a measurement is a measure of how close the measured value is to the true value of the quantity. The precision of measurement is the closeness of two or more measured values to each other.

The true value of a certain length is near 5.678 cm. In one experiment, using a measuring instrument of resolution 0.1 cm, the measured value is found to be 5.5 cm. In another experiment using a measuring instrument of greater resolution, say 0.01 cm, the length is found to be 5.38 cm. We find that the first measurement is more accurate as it is closer to the true value, but it has lesser precision. On the contrary, the second measurement is less accurate, but it is more precise.

III. Long Answer Questions:

Question 1.
State the principle of homogeneity and explain with an example.
Answer:
The principle of homogeneity of dimensions states that the dimensions of all the term in the physical expression should be same. This principle is used to check the correctness of the equation
For example
V² = U² + 2as
Writing dimensions on both sides
[LT^-1]² = [LT^-1]² + [LT^-1]²
[L² T²] = [L² T²] + [L²T^-2]
Here the dimensions of all the terms in the expression are same and equal to[L²T^-2]
So the equation is dimensionally correct.

Question 2.
Write short notes on the following.
(a) unit
(b) rounding off
(c) dimensionless quantity
Answer:
(a) Unit:
Unit of a physical quantity is defined as an arbitrarily chosen standard of measurement of a quantity which is accepted internationally.

The units in which the fundamental quantities are measured are called fundamental or base units and the units of measurement of all other physical quantity which can be obtained by a suitable multiplication or division of powers of fundamental units are called as derived units, example area, volume.

(b) Rounding off:
While doing calculations, the result got should not has too many figures. If no case the result have more significant figures than the figures involved in the data used for calculating. The result of calculation with numbers containing more than one uncertain digits should be rounded off.

Example :
18.35 when rounded off to 3 digits 18.4
19.45 when rounded off to 3 digits 19.4
101.55 x 10⁶ when rounded off to four digits 101.6 x 10⁶.

(c) Dimensionless quantity:
There are two types of dimensionless quantities – (i) dimensionless variable and (ii) dimensionless constant.

• Dimensionless variables – Physical quantities which have no dimensions but have variable values are called dimensionless variables.
  Examples: specific gravity, strain, refractive index, etc.
• Dimensionless constants – Quantities which have constant values and also have no dimension are called dimensionless constants.
  Example: π, e, numbers, etc.

Question 3.
What do you mean by the propagation of errors? Explain the propagation of errors in addition and multiplication.
Answer:
A number of measured quantities may be involved in the final calculation of an experiment. Different types of instruments might have been used for observation.

So the errors in the final result depends on –
(i) The error in individual measurements.
(ii) On the nature of mathematical operations.

The various possibilities of the propagation or combination of errors in different arithmetical operations are called propagation of errors.

Error in addition (or) sum of two quantities:
Let ∆A and ∆B be the absolute errors in measuring two quantities A and B respectively. Then
Measured value of A = A ± AA
Measured value of B = B ± AB
Consider sum A + B = Z
The error ∆Z in Z is given by
Z + ∆Z = (A ± ∆A) + (B ± ∆B).
= (A + B ) ± (∆A + ∆B)
Z + ∆Z = Z ± (∆A + ∆B)
∆Z = ∆A + ∆B
The maximum possible error in the sum of two quantities is equal to the sum of the absolute errors in the individual quantities.

Errors in multiplication (or) product of two quantities:
Let ∆A and ∆B be the absolute errors in the two quantities A and B respectively. Consider the product Z = AB
The error ∆Z in z given by
Z ± ∆Z = (A ± ∆A) (B ± ∆B)
= (AB) ± (A ∆ B) ± (B ∆ A) ± (∆A . ∆B)
Z ± ∆ Z = Z ± (A ∆ B) (B ∆ A) + (∆A . ∆B)
LHS by Z & RHS by AB = Z.
1± \(\frac { ∆ Z }{ Z }\) = 1 ± \(\frac { ∆ B }{ B }\) ± \(\frac { ∆ A }{ A }\) ± \(\frac { ∆A.∆B }{ AB }\)
Here \(\frac { ∆A.∆B }{ AB }\) can be neglected as \(\frac { ∆A}{ A }\) & \(\frac { ∆B }{ B }\) are small
∴ The maximum fractional error in Z is
\(\frac { ∆ Z }{ Z }\) = ±\(\frac { ∆A }{ A }\) x \(\frac { ∆B }{ B }\)
∴ The maximum fractional error in the product of two quantities is equal to the sum of fractional errors is the individual quantities.

Question 4.
Explain in detail various types of errors.
Answer:
The uncertainty in a measurement is called an error.
There are 3 types of errors namely –

1. Random error
2. Systematic error
3. Gross error.

1. Systematic errors – These are reproducible inaccuracies that are consistently in the same direction. These occur often due to problem that persists throughout the experiment. Systematic errors are further classified as

• instrumental error
• imperfection in experimental technique or procedure
• personal errors
• errors due to external causes
• least count error.

(a) Instrumental error: When an instrument is not calibrated properly at the time of manufacture instrumental errors may occur.

Example: If the measurement is made with a meter scale whose end is worn out the result obtained will have errors.

Correction – These errors can be corrected by choosing the instrument carefully.

(b) Imperfections in experimental technique or procedure: These errors arise due to limitations in the experimental arrangement.

Example: While performing experiments with a calorie meter, if there is no proper insulation, there will be radiation losses. This results in an error.

Correction – Necessary steps and corrections should be applied and followed while performing experiments.

(c) personal errors: These errors are due to individuals performing the experiments., maybe due to incorrect initial setting up of the experiment or carelessness of the individuals making the observation due to improper precautions.

(d) Errors due to external causes: The change in external conditions during experiments can cause error in measurement.

Example: Changes in temperature, humidity or pressure during measurement may affect the result of the measurement.

(e) Least count error: Least count is the smallest value that can be measured by the measuring instrument and the error due to this measurement is the least count error. The instrument’s resolution is the cause of the error. The error is half of the least value measured by the device.

Correction – Least count error can be reduced by using a high precision instrument for measurement.

(2) Random errors – Random errors may arise due to random and unpredictable variations is experimental conditions like pressure, temperature voltage supply etc., Errors may also due to persona! errors by the observer. Random errors are sometimes called “Chance errors”.

Example: While measuring the thickness of a wire using a screw gauge, different readings are taken in different trails.

Correction – By taking the arithmetic mean of all readings observed may reduce the random error and the mean value is taken as best possible true value.

(3) Gross errors – The error caused due to sheer carelessness of an observer is called gross error.
Examples: Improper setting of the instrument Making wrong observations without bothering about the sources of errors and precautions. Using wrong values in calculation Recording wrong observations

Correction – This error can be minimized only when the observer is careful and mentally alert.

Question 5.
(i) Explain the use of screw gauge and vernier calipers in measuring smaller distances:
(ii) Write a note a triangular method and radar method to measure large distances.
Answer:
Measurement of small distances by screw gauge and vernier calipers screw gauge : The screw gauge is an instrument used for measuring accurately the dimensions of objects up to a maximum of about 50 mm. The principle of the instrument is the magnification of linear motion using the circular motion of a screw. The least count of screw gauge is 0.01 mm.

Vernier calipers: A vernier caliper is a versatile instrument for measuring the dimensions of an object namely diameter of a hole or the depth of a hole. The least count of vernier caliper is a 0.01 cm.

Measurement of larger distances: For measuring larger distance such as height of a tree, distance of a moon or a planet from earth, the triangulation method, parallax method and radar method are used.

(a) Triangulation method for the height of an accessible object:

Let AB = h, be the height of a tree, to be measured. Let C be the point of observation at a distance X from B. Using a range finder placed
at C, ∠ACB = θ is measured.
∴ Considering ∆ ABC,
tan θ = \(\frac { AB }{ BC }\)
h = x tan θ
By knowing x, h can be calculated.

(b) Radar method:
RADAR is an acronym of Radio detection and ranging. A RADAR can be used to measure the distance of near planet, moon, enemy planes, moving as well as stationary targets etc. In this process, Radio signals are transmitted from the transmitter and after reflection from target, the radio signals are received by the receiver.

The time interval is recorded between the two instants i.e from time of transmission to time of reception. By knowing velocity and time, distance can be measured.

IV. Numerical Problems:

Question 1.
In a submarine equipped with sonar the time delay between the generation of a pulse and its echo after reflection form an enemy submarine is observed to be 80s. If the speed of sound in water is 1460 ms^-1. What is the distance of enemy submarine?
Solution:
Time taken = 80s
Velocity of sound = V = 1460 m/s
Distance of enemy submarine d = ?
V = \(\frac{2d}{t}\)
d = \(\frac{Vt}{2}\)
= \(\frac{1460×80}{2}\)
= 1460 x 40
= 58400 m
d = 58.4k m

Question 2.
The radius of the circle is 3.12 m calculate the area of the circle with regard to significant figures.
Solution:
Given: radius: 3.12 m (Three significant figures)
Solution:
Area of the circle = πr² = 3.14 × (3.12 m)² = 30.566
If the result is rounded off into three significant figure, the area of the circle = 30.6 m²

Question 3.
Assuming that the frequency γ of the vibrating string may depend up on
(i) applied force (F)
(ii) Length (l)
(iii) mass per unit length (m) prove that γ ∝ \(\frac { 1 }{ l }\)\(\sqrt{\frac{F}{m}}\) using dimensional analysis.
Solution:
γ ∝ F^a l^bm^c
Writing dimension on both sides

Hence Proved.

Question 4.
Jupiter is at a distance of 824.7 million Km from the earth. Its angular diameter is measured to be 35.72″ calculate the diameter of Jupiter
Solution:

Given Distance of Jupiter = 824.7 × 10⁶ km = 8.247 × 10¹¹ m
angular diameter = 35.72 × 4.85 × 10^-6rad = 173.242 × 10^-6 rad = 1.73 × 10^-4 rad
∴ Diameter of Jupiter D = D × d = 1.73 × 10^-4 rad × 8.247 × 10¹¹ m
= 14.267 × 1o⁷ m = 1.427 × 10⁸ m (or) 1.427 × 10^{5 Km}

Question 5.
The measurement value of length of a simple pendulum is 20 cm known with 2mm accuracy. The time for 50 oscillations was measured to be 40s with in Is resolution. Calculate the percentage accuracy in the determination of acceleration due to gravity g from the above statement.
Solution:
l = 20 x 10^-2m = 20 cm
∆l = 2mm = 0.2 cm
Time for 50 oscillations = 40s
Time for 1 oscillation = T = \(\frac { 40 }{ 50 }\)
= \(\frac { 4 }{ 5 }\)s
∆T = 1s.
∆T = \(\frac { 1 }{ 50 }\)S

= 1% + 5%
= 6%

11th Physics Guide Nature of Physical World and Measurement Additional Important Questions and Answers

I. Multiple choice questions:

Question 1.
The unit of surface tension ……
(a) MT^-2
(b) Nm^-2
(c) Nm
(d) Nm^-1
Answer:
(d) Nm^-1

Question 2.
One astronomical unit is equal to _______.
(a) 1.510 x 10¹²m
(b) 1.5 x 10¹²Km
(c) 1.5 x 10¹¹m
(d) 1.5 x 10¹²Cm
Answer:
(c) 1.5 x 10¹¹m

Question 3.
One light-year is ……
(a) 3.153 × 10⁷ m
(b) 1.496 × 10⁷ m
(c) 9.46 × 10¹² km
(d) 3.26 × 10¹⁵ km
Answer:
(c) 9.46 × 10¹² km

Question 4.
The dimensional formula for the coefficient of viscosity is _______.
(a) M⁰ L^-1T^-1
(b) M¹ L¹T¹
(c) M¹ L^-1T¹
(d) M² L²T⁰
Answer:
(c) M¹ L^-1T¹

Question 5.
One parsec is …..
(a) 3.153 × 10⁷ m
(b) 3.26 × 10¹⁵ m
(c) 30.84 × 10¹⁵ m
(d) 9.46 × 10¹⁵ m
Answer:
(c) 30.84 × 10¹⁵ m

Question 6.
The dimensional formula for (ε₀) permittivity of free space _______.
(a) M^-1 L³ T⁴ A²
(b) M^-1 L³ A²
(c) M^-1L^-3T⁴A²
(d) M¹L³T^-4A^-2
Answer:
(c) M^-1L^-3T⁴A²

Question 7.
One Angstrom is ………
(a) 10^-9 m
(b) 10^-10m
(c) 10^-12 m
(d) 10^-15 m
Answer:
(b) 10^-10 m

Question 8.
In the formula x = 3yz², x and y have dimensions of capacitance and magnetic induction respectively what are the dimensions of y?
(a) M³ L² T² Q⁴
(b) M³ L^-2T⁴Q⁴
(c) M³ L² T⁴ Q⁴
(d) M³L^-2T⁴Q⁴
Answer:
(d) M³L^-2T⁴Q⁴

Question 9.
\(\frac{1}{12}\) of the mass of carbon 12 atom is …..
(a) 1 TMC
(b) mass of neutron
(c) 1 amu
(d) mass of hydrogen
Answer:
(d) mass of hydrogen

Question 10.
The resistance of a conductor R = V/I where v = (50±2)v and I = (9 ± 0.3) A find the percentage error in R.
(a) 8.5%
(b) 3.7%
(c) 7.8%
(d) 7.3%
Answer:
(d) 7.3%

Question 11.
The study of forces acting on bodies whether at rest or in motion is …..
(a) classical mechanics
(b) quantum mechanics
(c) thermodynamics
(d) condensed matter physics
Answer:
(a) classical mechanics

Question 12.
In the measurement of pressure if maximum errors in the measurement of force and length of a square plate are 3% and 2% respectively. The maximum error is _______.
(a) 7%
(b) 8%
(c) 4%
(d) 5%
Answer:
(a) 7%

Question 13.
Which of the following is not a dimensionless physical quantity?
(a) Mechanical equivalent of heat
(b) volumetric strain
(c) atomic mass unit
(d) Avogadro’s number
Answer:
(c) atomic mass unit

Question 14.
The study of production and propagation of sound waves …..
(a) Astrophysics
(b) Acoustics
(c) Relativity
(d) Atomic physics
Answer:
(b) Acoustics

Question 15.
The physical quantities not having the same dimensions are _______.
(a) torque and work
(b) Linear momentum and planks constant
(c) stress and youngs modulus
(d) speed and (ε₀μ₀)^-1/2
Answer:
(a) torque and work

Question 16.
Which two of the following five physical parameters have the same dimension?
(1) energy density
(2) refractive index
(3) dielectric content
(4) youngs modulus
(5) magnetic field

(a) 1 and 4
(b) 1 and 5
(c) 2 and 4
(d) 3 and 5
Answer:
(a) 1 and 4

Question 17.
The astronomers used to observe distant points of the universe by …….
(a) Electron telescope
(b) Astronomical telescope
(c) Radio telescope
(d) Radar
Answer:
(c) Radio telescope

Question 18.
The youngs modulus of a material of the wire is 12.6 x 10¹¹ dyne/cm². Its value is MKS system is _______.
(a) 12.6 x 10¹² N/M²
(b) 12.6 x 10¹⁰ N/M²
(c) 12.6 x 10⁶ N/M²
(d) 12.6 x 10⁸ N/M²
Answer:
(a) 12.6 x 10¹² N/M²

Question 19.
The dimensionless quantity _______.
(a) never has a unit
(b) always has a unit
(c) may has a unit
(d) does not exist
Answer:
(c) may has a unit

Question 20.
Which one of the following is not a fundamental quantity?
(a) length
(b) luminous intensity
(c) temperature
(d) water current
Answer:
(d) water current

Question 21.
The time dependence of a physical quantity p is givers by \(p_{0} e^{-\alpha t^{2}}\) = P, where α is a constant and t is time. The constant α is _______
(a) a dimensionless
(b) has the dimension of T²
(c) has the dimension as that of P
(d) has the dimension equal to dimensions of PT^-2
Answer:
(a) a dimensionless

Question 22.
A force F is given F = at + bt² Where t is time. What are the dimensions of a & b?
(a) ML T^-3 and ML²T^-4
(b) MLT^-3 and MLT^-4
(c) MLT^-1 and MLT⁰
(d) MLT^-4and MLT^-1
Answer:
(a) ML T^-3 and ML²T^-4

Question 23.
The triple point temperature of the water is ……
(a) -273.16 K
(b) 0K
(c) 273.16 K
(d) 100 K
Answer:
(d) 100 K

Question 24.
The force F on a sphere of radius ‘a’ moving in a medium with a velocity v is given by F = 6 π av. The dimension of η are
(a) ML^-1T^-2
(b) MT^-1
(c) MLT^-2
(d) ML^-3
Answer:
(d) ML^-3

Question 25.
The unit of moment of force ……
(a) Nm²
(b) Nm
(c) N
(d) NJ rad
Answer:
(b) Nm

Question 26.
If the orbital velocity of a planet is given by v = G^aM^bR^c then _______.
(a) a = 1/3, b = 1/3, c = – 1/3
(b) a = 1/2, b =1/2, c = – 1/2
(c) a = 1/3, b = – 1/2, c = 1/2
(d) a = 1/2, b = – 1/2, c = – 1/2
Answer:
(a) a = 1/3, b = 1/3, c = – 1/3

Question 27.
The period T of a soap bubble under SHM given by T = P^aD^bS^c where P is the pressure, d is the density of water and E is the total energy of the explosion, then the value of a, b and c are:
(a) -3/2, 1/2, 1
(b) -5/6, 1/2, 1/3
(c) 5/6, 1/2, 1/3
(d) -5/6, -1/2, 1/3
Answer:
(a) -3/2, 1/2, 1

Question 28.
One degree of arc is equal to …….
(a) 1.457 × 10² rad
(b) 1.457 × 10^-2 rad
(c) 1.745 × 10² rad
(d) 1.745 × 10^-2 rad
Answer:
(b) 1.457 × 10^-2 rad

Question 29.
Frequency is the functions of density p length ‘l’ and tension T. The period of oscillation is proportional to _______.
(a) ρ^1/2 λ² T^-1/2
(b) ρ^1/2 λ^3/2T^-1/2
(c) ρ^1/2 λ^3/2 T^-3/4
(d) ρ^1/2 λ^1/2 T^3/2
Answer:
(a) ρ^1/2 λ² T^-1/2

Question 30.
The frequency of vibration of string is given by γ^l = \(\frac{p}{2 \ell} \sqrt{\frac{T}{m}}\) Here l is the length, P is the number of segments in the string. T is tension is the string, the dimensional formula for ‘m’ will be.
(a) M⁰ L T^-1
(b) M L⁰ T^-1
(c) M L^-1 T⁰
(d) M⁰ L⁰ T⁰
Answer:
(c) M L^-1 T⁰

Question 31.
1 second of arc is equal to ………………..
(a) 0.00027°
(b) 1.745 × 10^-2 rad
(c) 2.91 × 10^-4 rad
(d) 4.85 × 10^-6 rad
Answer:
(a) 0.00027°

Question 32.
If force (f), length (L) and time (T) are assumed to be fundamental units then the dimensional formula of the mass will be _______.
(a) FL^-1T²
(b) FL^-1T²
(c) FL^-1 T^-1
(d) FL²T²
Answer:
(a) FL^-1T²

Question 33.
If pressure ‘p’ velocity v and time T are taken as fundamental physical quantities the dimensional formula for force is _______.
(a) [PV² T²]
(b) [P^-1V²T^-2]
(c) [PV T²]
(d) [P^-1V T²]
Answer:
(a) [PV² T²]

Question 34.
The range of distance can be measured by using direct methods is …..
(a) 10^-2 to 10^-5 m
(b) 10^-2 to 10² m
(c) 10² to 1(T5 m {d) 10″2 to 105 m
Answer:
(b) 10^-2 to 10² m

Question 35.
The speed of light (c) gravitational constant G and planks constant h are taken as fundamental units. The dimension of time in the new system will be _______.
(a) G^1/2 h^1/2 C^-5/2
(b) g^1/2 h^1/2 C^1/2
(c) G^1/2 h^1/2 C^-3/2
(d) G^1/2 h^1/2 C^1/2
Answer:
(a) G^1/2 h^1/2 C^-5/2

Question 36.
The rate of flow(Q) (volume of liquid flowing per unit volume through a pipe depends on radius r, Length f of pipe, pressure difference P across the ends of pipe and coefficient of viscosity of liquid η as Q α r^aρ^b η^cL^dthen _______
(a) a = 4, b = 1, c = – 1, d = – 1
(b) a = 4, b = – 1, c = 1, d = – 1
(c) a = 4, b = 1, c = 1, d = – 1
(d) values of a,b,c and d cannot be determined
Answer:
(d) values of a,b,c and d cannot be determined

Question 37.
The dimensions of universal gas constant is _______
(a) ML²T^-2θ^-1
(b) ML²T^-2θ
(c) ML³ T^-1 θ^-1
(d) none of these
Answer:
(a) ML²T^-2θ^-1

Question 38.
Find odd one out.
(a) Newton
(b) metre
(c) candela
(d) Kelvin
Answer:
(a) Newton

Question 39.
Which of the following combinations have the dimensions of time? L, C, R represent inductance, capacitance, and resistance respectively.
(a) RC
(b) \(\sqrt{LC}\)
(c) L/R
(d) C/L
Answer:
(b) \(\sqrt{LC}\)

Question 40.
The dimensions of mobility are _______
(a) M^-1 LA T^-2
(b) ML A^-1T^-2
(c) MA^-1T^-2
(d) M^-1A T²
Answer:
(d) M^-1A T²

Question 41.
The smallest physical unit of time is
(a) second
(b) minute
(c) microsecond
(d) shake
Answer:
(d) shake

Question 42.
The length, breadth and thickness of strip are given by l = (10.0±0.1)cm, h=(1.00±0.01)cm t = (0.100±0.001)cm the most probable error in volume will be _______
(a) 0.03 cm³
(b) 0.111 cm³
(c) 0.012 cm³
(d) 0.12 cm³
Answer:
(a) 0.03 cm³

Question 43.
The measured mass and volume of a body are 22.42g and 4.7cm³ respectively with possible errors of 0.01 g and 0.1cm³. The maximum error in density is _______.
(a) 0.2%
(b) 2%
(c) 5%
(d) 10%
Answer:
(b) 2%

Question 44.
Half the lifetime of a free neutron is in the order of ……
(a) 10°
(b) 10¹ s
(c) 10² s
(d) 10³ s
Answer:
(d) 10³ s

Question 45.
An experiment measures quantities a, b, c and y is calculated from a formula. y = \(\frac{a b^{2}}{c^{3}}\) If the percentage errors in a,b and c are ± 1%, ± 3%, ± 2% respectively, the percentage error in calculating y is _______
(a) ± 13%
(b) ± 7%
(c) ±4%
(d) ± 1%
Answer:
(a) ± 13%

Question 46.
A student measures the distance traversed in a free fail of a body, initially at rest in a given time. He uses this data to estimate g, the acceleration due to gravity. If maximum percentage error in measurement of the distance and the time are e₁ and e₂ respectively the percentage error in the estimation of g is _______
(a) e₂ – e₁
(b) e₁+ 2e₂
(c) e₁ + e₂
(d) e₁ – 2e₂
Answer:
(b) e₁+ 2e₂

Question 47.
The heat generated in a circuit is given by Q = I²Rt. Where I is current, R is the resistance and t is the time. If an error in measuring current, resistance, and time are 2%, 1%, and 1% respectively. The maximum error in measuring heat will be _______
(a) 2%
(b) 4%
(c) 6%
(d) 8%
Answer:
(c) 6%

Question 48.
Imperfections in experimental procedure give ……………….. error.
(a) random
(b) gross
(c) systematic
(d) personal
Answer:
(c) Systematic

Question 49.
A student performs an experiment for determination of g = \(\frac{4 \pi^{2} l}{T^{2}}\) an error of ∆l. For that, he takes the time of n oscillations with the stopwatch of least count ∆T and he commits a human error of 0.1s. For which of the following data, the measurement of g will be most accurate? ∆l, ∆T, n
(a) 5m, 0.2s, 10
(b) 5mm, 0.2s, 20
(c) 5mm, 0.1s, 20
(d) 1mm, 0.1s, 50
Answer:
(d) 1mm, 0.1s, 50

Question 50.
A Screw gauge gives the following reading when used to measure the diameter of the wire.
Main scale reading = 0.
Circular scale reading = 52 divisions
Given that 1 mm on the main scale corresponds to 100 division on a circular scale. The diameter of the wire is _______.
(a) 0.52 cm
(b) 0.052 cm
(c) 0.0026 cm
(d) 0.005 cm
Answer:
(b) 0.052 cm

Question 51.
The error caused due to the sheer carelessness of an observer is called as ……………. error.
(a) Systematise
(b) Gross
(c) Random
(d) Personal
Answer:
(b) Gross

Question 52.
The force (F) velocity(v) and time “T” are taken as fundamental units then the divisions of mass are _______.
(a) [FVT²]
(b) [FV^-1T^-1]
(c) [FV T^-1]
(d) [FV^-1T]
Answer:
(d) [FV^-1T]

Question 53.
Attempting to explain diverse physical phenomenon with few concepts and law is _______.
(a) unification (or) reductionism
(b) neither unification nor reductionism
(c) unification
(d) reductionism
Answer:
(c) unification

Question 54.
An attempt to explain a microscopic system in terms of its microscopic constituents _______.
(a) unification
(b) reductionism
(c) neither unification or reductionism
(d) neither unification nor reductionism
Answer:
(b) reductionism

Question 55.
The study of nature of particles is _______ a branch of physics.
(a) nuclear physics
(b) quantum mechanics
(c) condensed another physics
(d) high energy physics
Answer:
(d) high energy physics

Question 56.
The ratio of the mean absolute error to the mean value is called …………….
(a) absolute error
(b) random error
(c) relative error
(d) percentage error
Answer:
(c) Relative error

Question 57.
1″ is equal to _______ radian.
(a) 1.745 x 10^-2 rad
(b) 1.78 x 10^-3 rad
(c) 2.91 x 10^-4 rad
(d) 4.847 x 10^-6 rad
Answer:
(d) 4.847 x 10^-6 rad

Question 58.
From a point on the ground, the top of a tree is seen to have an angle of elevation of 60°. The distance between the tree and a point is 50 m. The height of the tree is _______.
(a) 86.6 m
(b) 90.6 m
(c) 92.8 m
(d) 80.6 m
Answer:
(a) 86.6 m

Question 59.
The maximum possible error in the sum of two quantities is equal to …….
(a) Z = A + B
(b) ∆Z = ∆A + ∆B
(c) ∆Z = ∆A/∆B
(d) ∆Z = ∆A – ∆B
Answer:
(b) ∆Z = ∆A + ∆B

Question 60.
Number of a significant digit in 0.030400
(a) 6
(b) 5
(c) 4
(d) 3
Answer:
(b) 5

II. Short Answer Type Questions:

Question 1.
A new unit of length is chosen such that the speed of light in a vacuum is unity. What is the distance between the sun and the earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?
Answer:
Speed of light in vacuum, c = 1 new unit of length s^-1
t = 8 min. 20 sec, = 500 s
x = ct= 1 new unit of length s^-1 × 500s
x = 500 new unit of length

Conceptual Questions:

Question 1.
Why it is convenient to express the distance of stars in terms of light-year (or) parsec rather than in Km?
Solution:
One light-year = 9.46 x 10¹⁵ m = 9.46 x 10¹² Km. As the distance of stars is extraordinarily large, so it is convenient to express them in light-year rather than in meters or in kilometers.

Question 2.
Show that a screw gauge of pitch 1 mm and 100 divisions is more precise than a vernier caliper with 20 divisions on the sliding scale.
Solution:
The device that has a minimum least count is said to be more precise.
In case of screw gauge: Least count

= 0.05 mm
Out of this screw, gauges is having the minimum least count. So screw gauge is more precise.

Question 3.
What is the difference between mN, Nm, and nm?
Answer:
mN means milli newton, 1 mN = 10^-3 N, Nm means Newton meter, nm means nanometer.

Question 4.
Having all units in atomic standards is more useful. Explain.
Answer:
It became necessary to redefine all units in atomic standards because the prototype offers the following difficulties.

1. It is difficult to preserve prototype models.
2. It is difficult to produce replicas of prototypes for their use in different countries.
3. The techniques used for producing replicas are not of very high accuracy.
4. Atomic standard units can be reproduced anywhere and at any time.
5. It is variant in time and space.
6. It is unaffected by environmental conditions like temperature, pressure, etc.
7. It has an accuracy of 1 part in 10⁹.

Question 5.
Why dimensional methods are applicable only up to three quantities?
Answer:
The dimensional analysis is not applicable on more than 3 physical quantities because the equating powers of M, L & T, we get three unknowns. Similar constraints are present for electrical or other non-mechanical quantities also

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Commerce Guide Pdf Chapter 3 Classification of Business Activities Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Commerce Solutions Chapter 3 Classification of Business Activities

11th Commerce Guide Classification of Business Activities Text Book Back Questions and Answers

I. Choose the Correct Answer.

Question 1.
The industries engaged in extraction of iron ore are known as ………………
a. Construction Industries
b. Manufacturing Industries
c. Extraction Industries
d. Genetic Industries
Answer:
c. Extraction Industries

Question 2.
Auxiliaries to trade is also called as ……………
a. Trade
b. Advertisement
c. Warehousing
d. Aids to Trade
Answer:
d. Aids to Trade

Question 3.
Production which involves several stages for manufacturing finished products is known as …………………..
a. Analytical Industry
b. Synthetic Industry
c. Processing Industry
d. None of the above
Answer:
c. Processing Industry

Question 4.
Normally high level risk involved in ……………….
a. Industry
b. Commerce
c. Trade
d. All of the above
Answer:
a. Industry

Question 5.
Commerce is mainly concerned with ………………….
a. Connecting producer and consumer
b. Pricing of Goods
c. Buying and Selling of goods
d. Manufacturing of goods
Answer:
a. Connecting producer and consumer

II. Very Short Answer Questions

Question 1.
Define commerce:
Answer:
According to Evelyn Thomas, “Commercial operations deal with the buying and selling of goods, the exchange of commodities and the contribution of finished products”.

Question 2.
What do mean by industry?
Answer:
Industry refers to economic activities which are connected with conversion of resources into useful goods. The term is used for activities in which mechanical appliances and technical skills are involved. E.g Electronic industry would include all firms producing electronic goods, and so on.

Question 3.
What is trade?
Answer:
The term ‘trade’ is used to denote buying and selling. It is an essential part of commerce.

Question 4.
Write a short note on transportation.
Answer:
Transport or transportation is the medium which helps the movement of humans, animals and goods from one location to another. Since all the goods produced cannot be consumed in the place of production, it should be move to the places where they are demanded. The process of moving goods is known an transportation.

III. Short Answer Questions

Question 1.
Distinguish between Extractive industries and genetic industries.
Answer:
Extractive industries:

• These industries extract or draw out products from natural sources.
• Extractive industries supply some basic raw materials that are mostly products of the geographical or natural environment.

Genetic industries:

• These industries remain engaged in breeding plants and animals for their use in further reproduction.
• The seeds, nursery companies poultry, dairy, piggery, hatcheries, nursery, fisheries, apiary etc.

Question 2.
What do you mean by tertiary industries?
Answer:
The industries which produces utility services and sell them at the profit. These industries help trade, commerce and industry. The auxiliaries to trade like banking, insurance, warehouse, advertisement etc. are included in this.

Question 3.
Write any three characteristics of commerce.
Answer:
1. Economic Activity: Commerce is an economic activity because it consists of activities which are undertaken for earning profits. A trader buys goods with the aim of selling them at a profit.

2. Exchange of Goods and Services: Commerce involves the exchange and distribution of goods and services. Goods may be purchased or produced for sale. Commerce comprises both trade and aids to trade.

3. Profit Motive: The motive of commercial activities is to earn profits. Any activity which does not have the aim of profit will not be a part of commerce.

Question 4.
Narrate commerce with an example.
Answer:
Commerce includes all the activities which help in bringing goods from the producer to the’ ultimate consumer. According to Evelyn Thomas, “Commercial operations deal with the buying and selling of goods, the exchange of commodities and the contribution of finished products”. Commerce includes services such as transport, warehousing, packaging, insurance, banking and sales promotion which are incidental or auxiliaries to trade.

IV. Long Answer Questions

Question 1.
Explain the various kinds of industries on the basis of size.
Answer:
1. Micro Units: A unit wherein investment in plant and machinery is upto Rs. 25 lakhs in case of manufacturing and upto Rs. 10 lakhs in case of service enterprises.

2. Small Units: A manufacturing unit wherein investment in plant and machinery is more than Rs. 25 lakhs but does not exceed Rs.5 crore. In the case of service enterprises, these limits are Rs. 10 lakhs and Rs. 2 crores respectively.

3. Medium Units: A manufacturing unit wherein investment in plant and machinery is more than Rs.5 crore but does not exceed Rs. 10 crore. In the case of service enterprises, these limits are Rs.2 crore and Rs.5 crore respectively.

4. Large Units: A manufacturing unit wherein, investment in plant and machinery exceeds Rs.10 crore. In the case of a service unit investment in equipment exceeds Rs.5 crore.

Question 2.
Compare industry, commerce, and trade.
Answer:

SI. No	Variables	Industry	Commerce	Trade
1.	Meaning	Extraction,reproduction,conversion, processing and construction of useful products	Activities involving the distribution of goods and services	Purchase and sales of goods and services
2.	Scope	Consists of all activities involving conversion of material and semi-finished into finished goods.	Comprises trade auxiliaries to trade	Comprises exchange of good. and services
3.	Capital	a large amount of capital is required	Need for capital is comparatively less	Small capital is needed to maintain stock and to grant credit
4.	Risk	High risk is involved	Relatively less risk is involved	Relatively less risk is involved
5.	Side	It represents the supply side of goods and services	It represents the demand side of goods and services	It represents both supply and demand
6.	Utility creation	It creates form utility by changing the form or shape of materials	It creates place utility by moving goods from producers to consumers	It creates possession utility through exchange.

Question 3.
What are the characteristics of commerce?
Answer:
1. Economic Activity:
Commerce is an economic activity because it consists of activities which are undertaken for earning profits. A trader buys goods with the aim of selling them at a profit.

2. Exchange of Goods and Services:
Commerce involves the exchange and distribution of goods and services. Goods may be purchased or produced for sale. Commerce comprises both trade and aids to trade.

3. Profit Motive:
The motive of commercial activities is to earn profits. Any activity which does not have the aim of profit will not be a part of commerce.

4. Regularity of Transaction:
An isolated transaction does not imply commerce.

5. Creation of Utilities:
Commerce creates several types of utilities. It creates place utility by carrying goods to the place where they are needed. It makes goods available as and when demanded thereby creating time utility. By creating these utilities commerce helps to increase the volume of trade.

Question 4.
Write short notes on:

1. Analytical industry
2. Genetic industry and
3. Construction industry.

Answer:
1. Analytical industry:
These industries are called Secondary industries. These industries analyses and separates different elements from the same material, as in the case of oil refinery. In these industries the raw material is broken down into several useful materials.
E.g an oil industry separates crude oil into keroseñe, gasóline, diesel oil and petrol etc.

2. Genetic Industry:
The word ‘Genetic’ means parentage or hereditary. Genetic industries are concerned mainly with producing  breeding or multiplying of certain species of plants or animals with the object of earning profits from their sale. Examples of these types are nurseries, forestry, cattle-breeding, and commercial kennels. Animal husbandry is one type of Genetic Industry.

3. Construction Industry:
These industries are involved in the construction of building, dams, bridges, roads as well as tunnels and canals.

Question 5.
Briefly explain the auxiliaries to trade.
Answer:
Meaning:

Auxiliaries of trade may be classified into five categories:
1. Transportation:
Selling all the goods produced at or near the production place is not possible. Hence, goods are to be sent to different places where they are demanded.

2. Banking and Finance:
Nowadays we cannot think of a business without a bank. A bank is an organization which accepts deposits of money from the public, withdrawals on demand or otherwise, and lends the same to those who need it. Necessary funds can be obtained by businessmen from a bank. Thus, banking helps business activities to overcome the v problem of finance.

3. Insurance:
Business involves various types of risks. Materials and goods held in stock or in transit are subject to the risk of loss or damage. Insurance provides protection in all such cases. On payment of a nominal premium, the amount of loss or damage and compensation for an injury, if any, can be recovered from the insurance company.

4. Warehousing:
Goods are held in stock to make them available as and when required. Special arrangements must be made for the storage of goods to prevent loss or damage. Warehousing helps business firms to overcome the problem of storage and facilities the availability of goods when needed.

5. Advertising:
Advertising is one of the most important methods of promoting the sale of products, particularly, consumer goods like electronic goods, automobiles, soaps, detergents, etc. Advertising helps in providing information about available goods and services and inducing customers to buy particular items.

11th Commerce Guide Classification of Business Activities Additional Important Questions and Answers

I. Choose the Correct Answer:

Question 1.
All business activities can be classified into ……………. broad categories.
(a) Two
(b) Three
(c) Four
(d) Five
Answer:
(a) Two

Question 2.
Manufacturing industries may be categorized into ………………
a) Two
b) Three
c) Four
d) Ten
Answer:
c) Four

Question 3.
Horticulture is an example for …………….
(a) Primary industry
(b) Secondary industry
(c) Tertiary industry
(d) Local industry
Answer:
(a) Primary industry

Question 4.
…………………. industries produce utility services and sell them at a profit.
a) Construction Industries
b) Tertiary Industries
c) Analytical Industries
d) Manufacturing Industries
Answer:
b) Tertiary Industries

Question 5.
Professional or specialized skills and high technology are used to provide ……………. type of services.
(a) Personalised
(b) Public
(c) Distributive
(d) Quaternary
Answer:
(d) Quaternary

Question 6.
Expand MSME:
a) Major Small Medium Enterprises
b) Micro Scale. Middle Enterprises
c) Micro Small Medium Enterprises
d) Medium Small Micro Enterprises
Answer:
c) Micro Small Medium Enterprises

Question 7.
The service Enterprises with a maximum investment of 5 crores Is known as …………
a) Micro Unit
b) Small Unit
c) Large Unit
d) Medium Unit
Answer:
d) Medium Unit

Question 8.
Special arrangements must be made for goods to prevent loss or damage.
(a) Transportation
(b) Pricing
(c) Storage
(d) Advertising
Answer:
(c) Storage

Question 9.
Which of the following is not categorized as commerce?
a) Buy goods with ‘the aim of selling at profit.
b Exchange and Distribution of goods and services.
c) An individual sells his asset.
d) None of the above.
Answer:
c) An individual sells his asset.

Question 10.
Auxiliaries of trade are also called……………………
a) Trade
b) Advertisement
c) Warehousing
d) Aids to trade
Answer:
d) Aids to trade

Question 11.
The production which Involves several stages for manufacturing finished products Is known as………….
a) Analytical Industry
b) Synthetic Industry
c) Processing Industry
d) None of the above
Answer:
c) Processing Industry

Question 12.
Normally high-level risk involved in ………………….
a) Industry
b) Commerce
c) Trade
d) All the above
Answer:
a) Industry

Question 13.
Commerce is mainly concerned with
a) Distribution of Goods
b) Pricing of Goods
c) Buying and Selling of Goods
d) Manufacturing of Goods
Answer:
b) Pricing of Goods

II. Very Short Answer Questions

Question 1.
What is the Extractive industry?
Answer:
Extractive industries extract or draw out products from natural sources. Extractive industries supply some basic raw materials that are mostly products of the geographical or natural environment.

Question 2.
What do you mean by Tertiary Industries?
Answer:
The industry which produces utility services and sells them at a profit is known as tertiary industries:
They do not produce goods. They help in trade, industry, and commerce. It is also known as the Service industry.
E.g: The tourism and Hospitality industry, banking industries, etc.

Question 3.
What is the Secondary industry?
Answer:
Secondary Industries are concerned with using the materials which have already been extracted at the primary stage. These industries process such materials to produce goods for final consumption or for further processing by other industrial units.

III. Short Answer Questions.

Question 1.
Write any three categories of Manufacturing industries.
Answer:

1. Analytical Industry analyses and separates different elements from the same materials, as in the case of an oil refinery.
2. The synthetic Industry combines various ingredients into a new product, as in the case of cement.
3. The processing Industry involves successive stages for manufacturing finished products, as in the case of sugar and paper.

IV. Long Answer Questions

Question 1.
Write short notes on Primary Industries.
Answer:
These industries concerned with the production of goods with the help of nature. It is a nature-oriented industry, which requires very little human effort, for example, Agriculture, farming, forestry, fishing, horticulture, etc.
These industries are subdivided as follows:

Extractive Industries:
These industries extract or draw out products from natural resources. These industries supply some basic raw materials that are mostly products of the geographical or natural environment. These products are transformed into other manufacturing industries. These industries include farming, mining, lumbering, hunting and fishing operations.

Genetic Industries:
These industries engaged in breeding plants and animals for their use in further reproduction. The seeds, nursery companies, poultry, dairy, piggery, hatcheries, nursery, fisheries, apiary, etc are some of the examples of genetic industries.

Question 2.
Write a short note on Secondary Industries.
Answer:
These industries produced goods for final, consumption or for further processing by other industrial units. For example, the mining of iron ore is a primary industry, but the manufacturing of steel is a secondary industry.

It can be categorized as follows:

• Manufacturing Industries: These industries are engaged in producing goods through the processing of raw materials and thus creating from utilities. These industries may be further divided into four categories on the basis of the method of Operation for production.
• Analytical Industry: Which analyses and separates different elements from the same materials, as in the case of an oil refinery.
• Synthetically Industry: Which combines various ingredients into new products, as in the case of cement.
• Processing Industry: Which involves successive stages for manufacturing finished products, as in the case of sugar and paper.
• Assembling Industry: Which assembles different component parts to make a new product, as in the case of television, car, computer, etc.
• Construction Industries: These industries are involved in the construction of buildings, dams, bridges, roads as well as tunnels and canals.

Question 3.
Write a note on Tertiary industries.
Answer:
The industries which produce utility services and sell them at the profit. These industries help trade, commerce, and industry. The auxiliaries to trade like banking, insurance, warehouse, advertisement, etc are included in this. These industries are further classified as under:

• Personalized service: The individuals and private institutions selling their services to others is called personalized services. E.g Plumber, servant maid, etc.
• Public service: The government provides services to the people without profit motive through Government hospitals, schools, police, government offices, etc.
• Distributive service: Transportation, sales, warehousing, logistics, salesmanship, etc. come under this type of service.
• Financial service: Banking, factoring, accounting, and insurance, etc. are grouped under this type of service.
• Quaternary service: Professional or specialized skills and high technology are used to provide this type of service. E.g. Software development, Auditing, Research, and Development, etc.
• Quinary service: Selective individual experts create new ideas, implement new technologies and implement new policies. These decisions influenced the growth and development of national and international institutions.