Students can download Maths Chapter 7 Mensuration Ex 7.5 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 7 Mensuration Ex 7.5

Multiple Choice Questions

Question 1.

The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is

(1) 60π cm^{2}

(2) 68π cm^{2}

(3) 120π cm^{2}

(4) 136π cm^{2}

Solution:

(4) 13671 cm^{2}

Hint:

Here, h = 15 cm, r = 8 cm

C.S.A of a cone = πrl sq. units. = π × 8 × 17 = 136π cm

^{3}

Question 2.

If two solid hemispheres of same base radius r units are joined together along with their bases, then the curved surface area of this new solid is

(1) 4πr^{2} sq. units

(2) 6πr^{2} sq. units

(3) 3πr^{2} sq. units

(4) 8πr^{2} sq. units

Answer:

(1) 4πr^{2} sq. units

Hint:

When you joined two hemispheres together, the solid sphere is formed

C.S.A of the new solid = C.S.A of a sphere = 4πr^{2} sq. units.

Question 3.

The height of a right circular cone whose radius is 5 cm and slant height is 13 cm will be

(1) 12 cm

(2) 10 cm

(3) 13 cm

(4) 5 cm

Solution:

(1) 12 cm

Hint:

Here r = 5 cm and l = 13 cm

Question 4.

If the radius of the base of a right circular cylinder is halved keeping the same height, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is _________

(1) 1 : 2

(2) 1 : 4

(3) 1 : 6

(4) 1 : 8

Answer:

(2) 1 : 4

Hint:

Let the radius of the cylinder be “r” and the height be “h”

Radius of the new cylinder = \(\frac{r}{2}\) (Height will be same)

Volume of the new cylinder : Volume of the original cylinder

= \(\pi r_{1}^{2} h: \pi r_{2}^{2} h\) (πh is same)

= \(r_{1}^{2}: r_{2}^{2}\)

= \(\left(\frac{r}{2}\right)^{2}: r^{2}\)

= \(\frac{r^{2}}{4}: r^{2}=\frac{1}{4}: 1\)

= 1 : 4

Question 5.

The total surface area of a cylinder whose radius is \(\frac{1}{3}\) of its height is _______

(1) \(\frac{9 \pi h^{2}}{8}\) sq. units

(2) 24πh^{2} sq.units

(3) \(\frac{8 \pi h^{2}}{8}\) sq.units

(4) \(\frac{56 \pi h^{2}}{8}\) sq.units

Answer:

(3) \(\frac{8 \pi h^{2}}{8}\) sq.units

Hint:

Let the height of the cylinder be “h”

Radius of the cylinder = \(\frac{1}{3}\) h

T.S.A of the cylinder = 2πr(h + r)

Question 6.

In a hollow cylinder, the sum of the external and internal radii is 14 cm and the width is 4 cm. If its height is 20 cm, the volume of the material in it is

(1) 5600π cm^{3}

(2) 11200π cm^{3}

(3) 56π cm^{3}

(4) 3600π cm^{3}

Solution:

(2) 112007π cm^{3}

Hint:

Here, let the external radius be “R” and the internal radius be “r”

R + r = 14 ……(1)

Width (R – r) = 4 ……(2)

Height of the hollow cylinder = 20 cm

Volume of the hollow cylinder = πh × (R^{2} – r^{2})

= πh(R + r) (R – r)

= π × 20 (14) × 4

= π × 1120

= 1120π cm^{3}

Question 7.

If the radius of the base of a cone is tripled and the height is doubled then the volume is ______

(1) made 6 times

(2) made 18 times

(3) made 12 times

(4) unchanged

Answer:

(2) made 18 times

Hint:

Radius of a cone = r

Height of a cone = h

Volume of the cone = \(\frac{1}{3}\) πr^{2}h cu. units

When the radius is increased three-time (tripled) and the height is doubled

Radius is 3r and the height is 2h

Volume of the new cone

Volume is increased 18 times.

Question 8.

The total surface area of a hemisphere is how many times the square of its radius.

(1) π

(2) 4π

(3) 3π

(4) 2π

Solution:

(3) 3π

Hint:

T.S.A of the hemisphere = 3πr^{2}

The square of the radius is 3π times.

Question 9.

A solid sphere of radius x cm is melted and cast into a shape of a solid cone of the same radius. The height of the cone is _______

(1) 3x cm

(2) x cm

(3) 4x cm

(4) 2x cm

Answer:

(3) 4x cm

Hint:

Radius of a sphere = Radius of a cone = x cm

Volume of a cone = Volume of a sphere

Question 10.

A frustum of a right circular cone is of height 16cm with radii of its ends as 8cm and 20cm. Then, the volume of the frustum is

(1) 3328π cm^{3}

(2) 3228π cm^{3}

(3) 3240πcm^{3}

(4) 3340π cm^{3}

Solution:

(1) 3328π cm^{3}

Hint:

Here, h = 16 cm, r = 8 cm, R = 20 cm

Volume of the frustum

Question 11.

A shuttlecock used for playing badminton has the shape of the combination of ______

(1) a cylinder and a sphere

(2) a hemisphere and a cone

(3) a sphere and a cone

(4) frustum of a cone and a hemisphere

Answer:

(4) frustum of a cone and a hemisphere

Hint:

Question 12.

A spherical ball of radius r_{1} units is melted to make 8 new identical balls each of radius r_{2} units. Then r_{1} : r_{2 }is _______

(1) 2 : 1

(2) 1 : 2

(3) 4 : 1

(4) 1 : 4

Answer:

(1) 2 : 1

Hint:

Volume of the first sphere : Volume of second sphere = 8 : 1

Question 13.

The volume (in cm^{3}) of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 1 cm and height 5 cm is ________

(1) \(\frac{4}{3} \pi\)

(2) \(\frac{10}{3} \pi\)

(3) 5π

(4) \(\frac{20}{3} \pi\)

Answer:

(1) \(\frac{4}{3} \pi\)

Hint:

Radius of the sphere = 1 cm

Volume of the Sphere = \(\frac{4}{3}\) πr^{3} cu. units

= \(\frac{4}{3}\) × π × 1 × 1 × 1 cm^{3}

= \(\frac{4}{3}\) π cm^{3}

Question 14.

The height and radius of the cone of which the frustum is a part are h_{1} units and r_{1} units respectively. Height of the frustum is h_{2} units and the radius of the smaller base is r_{2} units. If h_{2} : h_{1} = 1 : 2 then r_{2} : r_{1} is ______

(1) 1 : 3

(2) 1 : 2

(3) 2 : 1

(4) 3 : 1

Answer:

(2) 1 : 2

Hint:

h_{2} : h_{1} = 1 : 2

h_{1} : h_{2} = 2 : 1

Ratio of their volumes

Volume is 2 : 1 the ratio of their radius also 2 : 1

r_{1} : r_{2} = 2 : 1 But r_{2} : r_{1} = 1 : 2

Question 15.

The ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is

(1) 1 : 2 : 3

(2) 2 : 1 : 3

(3) 1 : 3 : 2

(4) 3 : 1 : 2

Solution:

(4) 3 : 1 : 2

Hint:

Volume of (cylinder : cone : sphere)