**Assignments**
**NCERT Solutions**
Given below are the

**Class 10 Maths** Worksheet for Surface Area and Volume

a) Concepts questions

b) Calculation problems

c) Multiple choice questions

d) Long answer questions

e) Fill in the blank's

**Question 1)**A bucket is in the form of a frustum of a cone ad holds 28.490 liters of water. The radii of the top and bottom are 28cm and 21cm respectively. Find the height of the bucket.

a) 20 cm

b) 15 cm

c) 10 cm

d) None of the above

Solution
capacity of the frustum shaped bucket=28.490 l=28490 cm^{3}

r_{1}= radius of the top of the frustum= 28cm

r_{2}= radius of the bottom of the frustum= 21cm

we know that volume of a frustum= $\frac {\pi h(r_1^2 + r_2^2+ r_1r_2)}{3}$

$28490=\frac {22\times h \times (282+212+28 \times 21)}{7 \times 3} $

h=15 cm

Hence Answer (b)

**Question 2)** Lead spheres of diameter 6cm are dropped into a cylindrical beaker containing some water and are fully submerged. If the diameter of the beaker is 18cm and water rises by 40cm. find the no. of lead sphere dropped in the water.

a) 90

b) 150

c) 100

d) 80

Solution
Here Total Volume of Sphere = Cylinderical volume of water rised
Radius of cylinder(R)= 9 cm

Height of cylinder(H) = 40 cm

Radius of Sphere(r) = 3 cm

So,Number of Lead sphere(n) = Volume of Cylinder/Volume of sphere

=$\frac{\pi r^2h}{\frac{4}{3} \pi r^3}$

Substituting the values

n=90

hence the Answer (a)

**Question 3)** A petrol tank is a cylinder of base diameter 21cm and length 18cm fitted with conical ends each of axis length 9cm. Determine the capacity of the tank

a) 8316cm

^{3}
b) 9456 cm

^{3}
c) 10160 cm

^{3}
d) None of the above

Solution
Volume of the cylindrical portion of the tank= $\pi r^2h$

=22/7 * (10.5)^{2} * 18

= 6237cm^{3}

Volume of 2 conical ends

=$2(\frac {1}{3} \pi r^2h) =2/3 * 22/7 * (21/2)^{2} * 9

=2079cm^{3}

Therefore, capacity of the tank= 6237 + 2079= 8316cm^{3}

Hence the Answer (a)

**Question 4)** A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1: 2: 3.

Solution
The volume of cone is given by formula

V =$ \frac {1}{3}\pi R^{2}h$

Volume of the hemisphere =$ \frac {2}{3}\pi R^{3}$

Volume of the cylinder =$ \pi R^{2}h$

Now Height(H)= Radius (R) as they have height

Volume of Cone=$ \frac {1}{3}\pi R^{3}$

Volume of the hemisphere =$ \frac {2}{3}\pi R^{3}$

Volume of the cylinder =$ \pi R^{3}$

Ratio will be 1:2:3

Hence the Answer (c)

**Question 5)** The radii of the basses of two right circular solid cones of same height are r

_{1} and r

_{2 }respectively. The cones are melted and recast into a solid sphere of radius R. Show that the height of each cone is given by

$h=\frac {4R^3}{r_1^{2} + r_2^{2}}$

Solution
Volume of first cone = $ \frac {1}{3}\pi r_1^{2}h$

Volume of second cone = $ \frac {1}{3}\pi r_2^{2}h$

According to the question, the 2 cones are melted and recast to form a sphere of radius R.

Therefore, volume of first cone + volume of second cone = volume of sphere.

$ \frac {1}{3}\pi r_1^{2}h + \frac {1}{3}\pi r_2^{2}h= \frac {4}{3}\pi R^{3}$

$(r_1^{2} + r_2^{2})h=4R^{3}$

$h=\frac {4R^3}{r_1^{2} + r_2^{2}}$

**Question 6)** Prove that the surface area of a sphere is equal to the curved surface area of the circumscribed cylinder.

Solution
Let the radius of the sphere = r

Surface area of the sphere = $4\pi r^2$

Radius of the cylinder circumscribed over the sphere = r

Height of the cylinder circumscribed over the sphere (h) = 2r

Curved surface area of the cylinder = $2\pi rh$

= $2\pi r(2r)

= $4\pi r^2$

Therefore, the curved surface area of the circumscribed cylinder over a sphere is equal to the surface area of the sphere.

**Question 7)**A bucket of height 8cm and made up of copper sheet is in the form of frustum of a right circular cone with radii of its lower and upper ends as 3cm and 9m respectively.

Calculate –

- the height of cone of which the bucket is a part.
- the volume of water which can be filled in the bucket.
- the area of copper shut required to make the bucket.

Solution
Let h be the height, l the slant height and r_{1} and r_{2} the radii of the circular bases of a frustum of a cone.

We have, h = 8 cm, r_{1}= 9 cm and r_{2} = 3 cm

(i) Let H be the height of the cone of which the bucket is part. Then

$H=\frac {hr_1}{r_1 - r_2}$

Substituting the given values

H=12cm

(ii) Volume of the water which can be filled in the bucket

= Volume of the frustum

=$\frac {\pi}{3}(r_1^2+r_2^2+r_1r_2)h$

=312cm^{3}

(iii) Area of the copper sheet required to make the bucket, where l is the slant height of the frustum is given by

=$\pi (r_1+r_2)l+\pi r_2^2

Where $l=\sqrt {(r_1-r_2)^2 + h^2} = 10cm$

So Area of copper sheet

=$129\pi$ cm^{2}

Hence the Answers

(i) 12cm

(ii) 312 cm^{3}

(iii) $129\pi$ cm^{2}

**Question 8)** Match the column

**Question 9)** 500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0.04m

^{3}?

Solution
The volume of one person = 0.04 m^{3}

Now The amount of water displaced is equal to the volume of the person.

500 people take dip at the same time, then water displaced = 500 * 0.04 = 20 m^{3}

Let h be the height of water raised,then Volume of the water rises will be

=Base X Height = $80 \times 50 h$ = 4000h m^{2}

Now Volume of Water rises = Volume of water displaced by 500 people

4000h=20

h=1/200 m =0.005 m= 5 millimeters

**Question 10)** The slant height of the frustum of a cone is 4cm, and the perimeter of its circular bases is 18cm and 6cm respectively. Find the curved surface area of the frustum ?

Solution
Given
$2\pi r_1 = 18$

$\pi r_1 = 9 $

$2\pi r_2 = 6$

$\pi r_2 = 3$

l = 4 cm

curved surface area = $(\pi r_1 + \pi r_2 )l$

=(9+3)4=48 cm^{2}

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