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 radius (r) = 8cm

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,

r = radius = 5cm

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 radius = r radius

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

radius (r) = h/3 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

(A) Made 6 times

(B) made 18 times

(C) made 12 times

(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:

Sphere radius (r) = 1cm

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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