# Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.5 Pdf

## Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.5 Pdf

TN Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.5: Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.5 – Mensuration on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 7 Exercise 7.5 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 7.5

### TN Samacheer Kalvi 10th Maths Solutions Chapter 7 – Mensuration

 Board TNSCERT Class 10th Maths Class 10 Subject Maths Chapter 7 (Exercise 7.5) Chapter Name Mensuration

#### TNSCERT Class 10th Maths Pdf | all Exercise Solution

Exercise – 7.5

(*) Multiple choice questions

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

(A) 60 π cm2

(B) 68 π cm2

(C) 120 π cm2

(D) 136 π cm2

Solution:

Cone diameter (2r) = 16cm

Cone height = 15cm

l = √r2 + h2= √82 + 152 = √64+2225

l = √289 + 7cm

Cone curved surface area = πrl cm2

= π × 80 × 17 cm

= 136π cm2 (D)

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

(A) 4πr2 sq. units

(B) 6πr2 sq. units

(C) 3πr2 sq. units

(D) 8πr2 sq. units

Solution:

We know that, hemisphere curves surface area = 2πr2, where r = radius of hemisphere.

Two hemisphere curved surface area = 2×2πr2 sq. units

= 4πr2 sq. units (A)

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

(A) 12cm

(B) 10cm

(C) 13cm

(D) 5 cm

Solution:

We know that,

h = height

l = slant height = 1.3cm

l = √r2 + h2

h = √l2 – r2

= √(13)2 – (15)2 = √169 – 25

h = √144 = 12cm (A)

(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

(A) 1:2

(B) 1:4

(C) 1:6

(D) 1:8

Solution:

Let, cylinder height = h units

New cylinder radius = r/2 units

Original cylinder volume = πr2h sq. units

New cylinder volume = π × (r/2)2 × h sq. units

= πr2h/h sq. units

New cylinder volume/orginal cylinder = (πr2h/h)/πr2h = 1/4  = 1:4 (B)

(5) The total surface area of a cylinder whose radius is 1/3 of its height is

(A) 9πh2/8 sq. units

(B) 24πh2 sq. units

(C) 8πh2/9 sq. units

(D) 56πh2/9 sq. units

Solution:

Let, cylinder height = h units

Cylinder total surface area = 2πr (h + r) sq. units

= 2π h/3 (h + h/3) sq. units

= 2π h/3 (4h/3)

= 8πh2/9 sq. units (C)

(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

(A) 5600π cm3

(B) 11200π cm3

(C) 56π cm3

(D) 3600π cm3

Solution:

Let external radius = R cm

Let internal radius = r cm

Given R + r = 14 —- (i) R – r = 4 —– (ii)

(i) + (ii), we get

R + r + R – r = 14+4

2R = 18

R = 9 cm

From (i), R = 9 putting, we get

R + r = 14

r = 14 – R

r = 14 – 9

r = 5 cm

∴ Volume of the hollow cylinder

= π (R2 – r2) h cm3

= π (9)2 – (5)2 20 cm3 [h = 20cm]

= π {81 – 25} × 20 cm3

= π × 56 × 20 cm3

= 11200 π cm3 (B)

(7) If the radius of the base of a cone is tripled and the height is doubled then the volume is

(D) Unchanged

Solution:

Let, cone radius = r units

Cone height = h units

Volume of cone = 1/3 πr2h cube. Units

New cone radius = 3r units

New cone height = 2h units.

New cone volume = 1/3 π× (3r)2 × 2h cube unit

= 1/3 π × 3r × 3r × 2h cube unit

= 18 × 1/3 πr2h cube unit

= 18 times original cone volume (B)

(8) The total surface area of a hemi-sphere is how much times the square of its radius.

(A) π

(B) 4π

(C) 3π

(D) 2π

Solution:

Total surface area of hemisphere

= 3πr2 sq. units

= 3π × (radius)2 sq. units

= 3π times the square of radius

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

(A) 3x cm

(B) x cm

(C)4x cm

(D)2x cm

Solution:

Given, sphere radius (r) = x cm.

∴ Volume of sphere = 4/3 πr3 cm3

Cone radius (r) = x cm

Cone height = h cm

Now, volume of cone = volume of sphere

1/3 πr2h = 4/3 πr3

r2h = 4r3

h = 4r

h = 4x cm (C)

(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

(A) 3328π cm3

(B) 3228π cm3

(C) 3240π cm3

(D) 3340π cm3

Solution:

Frustum of a right circular cone height (h) = 16cm

r = 8cm and R = 20cm

Volume of the frustum = 1/3 πh (R2 + Rr + r2) cm3

= 1/3 π × 16 × (400 + 160 + 64) cm3

= 1/3π × 16 × 624 cm3

= 3328 π cm3 (A)

(11) A shuttle cock used for playing badminton has the shape of the combination of

(A) A cylinder and a sphere

(B) A hemisphere and a cone

(C) A sphere and a cone

(D) Frustum of a cone and a hemisphere

Solution:

(d) Frustum of a cone and a hemisphere

(12) A spherical ball of r1 units is melted to make 8 new identical balls each of radius r2 units. Then r1 : r2 is

(A) 2:1

(B) 1:2

(C) 4:1

(D) 1:4

Solution:

Volume of the sphere/Volume of each sphere = 8/1

(4/3 π r13)/(y/3π r23 = 8/1

r13/r23 = 8/1

(r1/r2)3 = (2/1)3

r1:r2 = 2:1 (A)

(13) The volume (in cm3) 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

(A) 4/3 π

(B) 5π

(C) 5π

(D) 20/3π

Solution:

Volume of the sphere = 4/3π r3 cube units

= 4/3 π× 1 × 1 × 1 cm3

= 4/3π cm3 (A)

(14) The height and radius of the cone of which the frustum is a part are h1 units and r1 units respectively. Height of the frustum is h2 units radius of the smaller base is r2 units. If h2:h1 = 1:2 then r2:r1 is

(A) 1:3

(B) 1:2

(C) 2:1

(D) 3:1

Solution:

Given h2:h1 = 1:2

Ratio of their volumes

= 1/3 π h1 (r12 + r22 + r1r2): 1/3 π h2 (r12 + r22 + r1r2)

= 1/3 πh1 : 1/3 π h2

h1 : h2 = 2:1

Then r1 : r2 = 2:1 [∵ volume ratio and radius ratio same]

r2 : r1 = 1:2 (B)

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

(A) 1:2:3

(B) 2:1:3

(C) 1:3:2

(D) 3:1:2

Solution:

Cylinder : cone : sphere = πr2h : 1/3 πr2h : 4/3 πr3

= r2h : 1/3 r2h  : 4/3 r3

= h: h/3 : 4r/3

=h : h/3 : 4/3 × h/2 [r = h/2]

= 3 : 1 : 2 (D)

Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 7 Mensuration Exercise 7.5

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Updated: March 3, 2022 — 2:40 pm