Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.2 Pdf

Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.2 Pdf

TN Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.2: Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.2 – Mensuration on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 7 Exercise 7.2 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 7.2

TN Samacheer Kalvi 10th Maths Solutions Chapter 7 – Mensuration

Board TNSCERT Class 10th Maths
Class 10
Subject Maths
Chapter 7 (Exercise 7.2)
Chapter Name Mensuration

TNSCERT Class 10th Maths Pdf | all Exercise Solution

 

Exercise – 7.2

 

(1) A 14 m deep well with inner diameter 10 m is dug and the earth taken out is evenly spread all around the well to form an embankment of width 5 m. Find the height of the embankment.

Solution:

Given that, diameter of the well = 10m

∴radius of the well = 10/2 = 5cm

∴height (h1) of the well = 14m

Width of the embankment = 5 m

∴internal radius (r) = 5cm

External radius (R) = (5+5) m

= 10m

Let the height of the embankment = h m.

Now, vol of the dug = vol of the embankment

πr2 h1 = πh (R2 – r2)

r2 – h1 = (R2 – r2) h

5×5×14 = {(10)2 – (5)2} h

h = 5×5×14/100-25

h = 5×5×14/75

h = 14/3

h = 4.66m

Therefore, the height of the embankment is 4.66m

 

(2) A cylindrical glass with diameter 20 cm has water to a height of 9 cm. A small cylindrical metal of radius 5 cm and height 4 cm is immersed it completely. Calculate the raise of the water in the glass?

Solution:

Given that, cylinder glass diameter is = 20cm

Cylinder glass radius (r1) = 10cm

∴ Height of the water (h) = 9cm

 

Cylindrical metal radius (r2) = 5cm

Cylindrical metal height (h1) = 4cm

Let, the height of the raise of the water in the glass is h2 cm.

Now, volume of the cylindrical metal

= πr22 h1

= π × (5)2 × 4 cm3

= 5 × 5 × 4 × π cm3

= 100 cm3

Volume of the water raises in the cylindrical glass = πr12 h2

= π × (10)2 × h2 cm3

= 100π h2 cm3

Now volume of the cylindrical metal and volume of the water noised in the cylindrical glass are equal.

∴ 100π = 100πh2

h2 = 1 cm

Therefore, the height of the noise of the water in the glass is 1 cm.

 

(3) If the circumference of a conical wooden piece is 484 cm then find its volume when its height is 105 cm.

Solution:

Given conical wooden height (h) = 105 cm.

Let the conical wooden piece radius be r cm.

Now, Circumference of the conical wooden = 484 cm

2πr = 484

r = 484×7/2×22

∴ r = 77cm

Now, volume of the conical wooden piece is = 1/3 πr2 h

= 1/3 × 22/7 × 77 × 77 × 105 cm3

= 110 × 77 × 77 cm3

= 652190 cm3

Thus, the volume of the conical wooden piece is 652190 cm3

 

(4) A conical container is fully filled with petrol. The radius is 10m and the height is 15 m. If the container can release the petrol through its bottom at the rate of 25 cu. meter per minute, in how many minutes the container will be emptied. Round off your answer to the nearest minute.

Solution:

Given the conical container radius (r) = 10cm

The conical container height (h) = 15cm

Now, volume of the conical container is = 1/3 πr2h m3

= 1/3 × 22/7 × 10 × 10 × 15 m3

= 22×100×5/7 cm3

Now, given that the container can release the petrol rate is 25 m3/minute.

Now, taken time = (22×100×5/7)/25 minute

= 22×100×5/25×7 minute

= 440/7 minute

= 62.86 minute

= 63 minute

Therefore, the container fully filled time taken 63 minute.

 

(5) A right angled triangle whose sides are 6 cm, 8 cm and 10 cm is revolved about the sides containing the right angle in two ways. Find the difference in volumes of the two solids so formed.

Solution:

Now, if the triangle is revolved about side 6cm. Then, the formed is cone.

Now, cone radius (r1) = 6cm

Cone height (h1) = 8cm

∴ Volume of the cone = 1/3 πr12 h1

= 1/3 × 6 × 6 × 8 cm3

= 2112/7 cm3

If the triangle is revolved about side 8cm, then formed is cone.

Cone radius (r2) = 8cm

height (h2) = 6cm

Volume of the cone = 1/3 π r22 h2

= 1/3 × 22/7 × 8×8×6 cm3

= 2816/7 cm3

Difference in volume of the two solids

= (2816/7 – 2112/7) cm3

= 2816-2112/7 cm3

= 704/7 cm3

= 100.57 cm3

Therefore, the difference in volume of the two solids is 100.57 cm3.

 

(6) The volumes of two cones of same base radius are 3600 cm3 and 5040 cm3. Find the ratio of heights.

Solution:

Let two cones radius = r cm

One cone height = h1

and other cone height = h2

Now, (1/3 πr2 h1)/(1/3 π r2h2) = 3600/5040

h1/h2 = 5/7

∴h1 : h2 = 5:7

Therefore the required ratio is 5:7.

 

(7) If the ratio of radii of two spheres is 4:7, find the ratio of their volumes.

Solution:

Let, one spheres radius be 4x and other spheres radius 7x.

Now, v1/v2 = (v/3 π r13)/(4/3 π r23) = (r1)3/(r2)3

v1/v2 = (4x)3/(7x)3 = 64/343

v1/v2 = 64/343

∴v1 : v2 = 64:343

Therefore the volume ratio is 64:343

 

(8) A solid sphere and a solid hemisphere have equal total surface area. Prove that the ratio of their volume is 3√3: 4.

Solution:- 

Let the sphere radius = r1 units

Let the hemisphere radius = r2 units

Now, sphere total surface are and hemisphere total surface area are equal.

4πr12 = 3π r22

r12/r22 = 3π/4π

r1/r2 = r3/2

∴r1 : r2 = r3 : 2

Now, volume of sphere (v1) = 4/3 πr13

Volume of hemisphere (v2) = 2/3 πr23

∴v1 : v2 = 4/3 πr13: 2/3 πr23

v1 : v2 =  (r1)3 : (r2)3

= 2 × (√3)3 : (2)3

= 6√3 : 8

= 3√3 : 4

∴The ratio of their volume is 3√3:4 [Proved]

 

(9) The outer and the inner surface areas of a spherical copper shell are 576π cm2 and 324π cm2 respectively. Find the volume of the material required to make the shell.

Solution:

Let the outer radius = R cm

and the inner radius = r cm

Now, outer surface area = 576π cm2

4π R2 = 576π

R2 = 144

R = 12 cm

Now, inner surface area = 324π cm2

4π r2 = 324π

r2 = 81

r = 9

Volume of the material required to make the shell

= 4/3 π (R3 – r3) cm3

= 4/3 × 22/7 × {(12)3 – (9)3} cm3

= 4/3 × 22/7 × (1728 – 729) cm3

= 4/3 × 22/7 × 999

= 4×22×333/7 cm3

= 4186.29 cm3

Therefore the volume of the material is 4186.29 cm3.

 

(10) A container open at the top is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends are 8 cm and 20 cm respectively. Find the cost of milk which can completely fill a container at the rate of₹40 per litre.

Solution:

Given that, the height of the frustum (h) = 16 cm

The upper radius (R) = 20cm

The lower radius (r) = 8cm

Volume of the frustum

= πh/3 (R2 + Rr + r2) cm3

= 22/7 × 1/3 × 16 × {(20)2 + 20 × 8 + (8)2} cm3

= 22×16/7×3 (400 + 160 + 64) cm3

= 22×16×624/7×3 cm3

= 352×208/7 cm3

= 73216/7 cm3

= 10459.43 cm3

= 10.45943 litre

∴ Cost of milk in the container

₹ = 10.45943×40

₹ 418.36

Therefore the cost of the milk is ₹418.36

 

Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 7 Mensuration Exercise 7.2

Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.

Updated: March 3, 2022 — 2:11 pm

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