**Important questions for Class 10 Maths Chapter 13 Surface Areas and Volumes** are provided at BYJU’S, which are given as per the new CBSE guidelines for 2019-2020. Students who are preparing for the board exams of CBSE-2020 exams can practice these questions of Surface Areas and Volumes of Class 10 to score full marks for the questions from this chapter.

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## Important Questions & Answers For Class 10 Maths Chapter 13 Surface Areas and Volumes

**Q.1: Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere (see Fig 13.6). The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. (Take π = 22/7)**

**Solution:**

TSA of the toy = CSA of hemisphere + CSA of cone

Curved surface area of the hemisphere = 1/ 2 (4πr^{2}) = 2π r^{2} = 2(22/7)× (3.5/2) × (3.5/2) cm^{2}

Height of the cone = Height of the top – Radius of the hemispherical part

= (5 – 3.5/2) cm = 3.25 cm

Sant height of the cone (l) \(=\sqrt{r^2+h^2}=\sqrt{(\frac{3.5}{2})^2+(3.25)^2}=3.7\ cm\) (approx.)

Therefore, CSA of cone = πrl = (22/7) × (3.5/2) × 3.7 cm^{2}

Hence, the surface area of the top = [2(22/7)× (3.5/2) × (3.5/2) + (22/7) × (3.5/2) × 3.7] cm^{2}

= (22/7) × (3.5/2) (3.5+3.7) cm^{2}

= (11/2) × (3.5 + 3.7) cm^{2}

= 39.6 cm^{2} (approx.)

**Q. 2: Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end as shown in the figure. The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath. (Take π = 22/7)**

**Solution: **

Let h be the height of the cylinder and r be the common radius of the cylinder and hemisphere. Then, the total surface area of the bird-bath = CSA of cylinder + CSA of the hemisphere

= 2πrh + 2πr^{2}

= 2π r(h + r)

= 2 (22/7) × 30 × (145 + 30) cm^{2}

= 33000 cm^{2} = 3.3 m^{2}

**Q. 3:** **2 cubes each of volume 64 cm ^{3} are joined end to end. Find the surface area of the resulting cuboid.**

**Solution: **

The diagram will obtain as below:

Given,

The Volume (V) of each cube is = 64 cm3

This implies that a^{3} = 64 cm^{3}

∴ The side of the cube i.e. a = 4 cm

Also, the breadth and length of the resulting cuboid will be 4 cm each while its height will be 8 cm.

So, the surface area of the cuboid (TSA) = 2(lb + bh + lh)

Now, by putting the values we get,

= 2(8×4 + 4×4 + 4×8) cm^{2}

= (2 × 80) cm^{2}

Hence, TSA of the cuboid = 160 cm^{2}

**Q. 4: A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m ^{2}. (Note that the base of the tent will not be covered with canvas.)**

**Solution:**

It is known that a tent is a combination of a cone and a cylinder as shown below.

From the question, we know that

The diameter = D = 4 m

l = 2.8 m (slant height)

The radius of the cylinder is equal to the radius of the cylinder

So, r = 4/2 = 2 m

Also, we know the height of the cylinder (h) is 2.1 m

So, the required surface area of the given tent = surface area of the cone (the top) + surface area of the cylinder(the base)

= πrl + 2πrh

= πr (l+2h)

Now, substituting the values and solving it we get the value as 44 m^{2}

∴ The cost of the canvas at the rate of Rs. 500 per m^{2} for the tent will be

= Surface area × cost/ m^{2}

= 44 × 500

So, Rs. 22000 will be the total cost of the canvas.

**Q. 5:** **A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference between the volumes of the cylinder and the toy. (Take π = 3.14)**

**Solution:**

Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere as shown in the above figure.

The radius BO of the hemisphere (as well as of the cone) =( ½) × 4 cm = 2 cm.

So, volume of the toy = (⅔) πr^{3} + (⅓) πr^{2}h

= (⅔) × 3.14 × 2^{3} + (⅓)× 3.14 × 2^{2} × 2

= 25.12 cm^{3}

Now, let the right circular cylinder EFGH circumscribe the given solid.

The radius of the base of the right circular cylinder = HP = BO = 2 cm, and its height is

EH = AO + OP = (2 + 2) cm = 4 cm

So, the volume required = volume of the right circular cylinder – volume of the toy

= (3.14 × 2^{2} × 4 – 25.12) cm^{3}

= 25.12 cm^{3}

Hence, the required difference of the two volumes = 25.12 cm^{3}

**Q. 6:** **A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm ^{3}. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.**

**Solution: **

Given,

For the cylinder part, Height (h) = 8 cm and Radius (R) = (2/2) cm = 1 cm

For the spherical part, Radius (r) = (8.5/2) = 4.25 cm

Now, volume of this vessel = Volume of cylinder + Volume of sphere

= π × (1)^{2} × 8 + (4/3) π (1)^{3}

= 346.51 cm^{3}

**Q. 7:** **A cone of height 24 cm and radius of base 6 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.**

**Solution:**

The volume of cone = (⅓) × π × 6 × 6 × 24 cm^{3}

If r is the radius of the sphere, then its volume is (4/3) πr^{3}.

Since the volume of clay in the form of the cone and the sphere remains the same, we

Have (4/3) πr^{3} = (⅓) × π × 6 × 6 × 24 cm^{3}

r^{3 } = 3 × 3 × 24 = 3^{3 }× 2^{3}

So, r = 3 × 2 = 6

Therefore, the radius of the sphere is 6 cm.

**Q. 8:** **Selvi’s house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (an underground tank) which is in the shape of a cuboid. The sump has dimensions 1.57 m × 1.44 m × 95cm. The overhead tank has its radius 60 cm and height 95 cm. Find the height of the water left in the sump after the overhead tank has been completely filled with water from the sump which had been full. Compare the capacity of the tank with that of the sump. (Use π = 3.14)**

**Solution:**

The volume of water in the overhead tank equals the volume of the water removed from the sump.

Now, the volume of water in the overhead tank (cylinder) = πr 2h

= 3.14 × 0.6 × 0.6 × 0.95 m^{3}

The volume of water in the sump when full = l × b × h = 1.57 × 1.44 × 0.95 m^{3}

The volume of water left in the sump after filling the tank

= [(1.57 × 1.44 × 0.95) – (3.14 × 0.6 × 0.6 × 0.95)] m^{3} = (1.57 × 0.6 × 0.6 × 0.95 × 2) m^{3}

Height of the water left in the sump = (volume of water left in the sump)/ (l × b)

= (1.57× 0.6× 0.6× 0.95 ×2)/ (1.57 ×1.44)

= 0.475 m

= 47.5 cm

Capacity of tank / Capacity of sump = (3.14 × 0.6 × 0.6 × 0.95)/ (1.57 × 1.44 × 0.95)

=1/ 2

Therefore, the capacity of the tank is half the capacity of the sump.

**Q. 9: Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.**

**Solution: **

For Sphere 1:

Radius (r_{1}) = 6 cm

∴ Volume (V_{1}) = (4/3) × π× r_{1}^{3}

For Sphere 2:

Radius (r_{2}) = 8 cm

∴ Volume (V_{2}) = (4/3) × π × r_{2}^{3}

For Sphere 3:

Radius (r_{3}) = 10 cm

∴ Volume (V_{3}) = (4/3) × π× r_{3}^{3}

Also, let the radius of the resulting sphere be “r”

Now, Volume of resulting sphere = V_{1} + V_{2} + V_{3}

(4/3)× π× r^{3} = (4/3)× π× r_{1}^{3} + (4/3)× π× r_{2}^{3} + (4/3) × π× r_{3}^{3}

r^{3} = 6^{3} + 8^{3} + 10^{3}

r^{3} = 1728

r = 12 cm

**Q. 10: An open metal bucket is in the shape of a frustum of a cone, mounted on a ****hollow cylindrical base made of the same metallic sheet as shown in the figure. The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket ****is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used ****to make the bucket, where we do not take into account the handle of the bucket. Also, find ****the volume of water the bucket can hold. (Take π = 22/7/)**

**Solution: **

The total height of the bucket = 40 cm, which includes the height of the base.

So, the height of the frustum of the cone = (40 – 6) cm = 34 cm.

Therefore, the slant height of the frustum, \(l=\sqrt{h^2+(r_1-r_2)^2}\)

where r_{1} = 22.5 cm, r_{2} = 12.5 cm and h = 34 cm.

=35.44 cm

The area of metallic sheet used = curved surface area of frustum of cone + area of circular base + curved surface area of cylinder

= [π × 35.44 (22.5 + 12.5) + π × (12.5)^{2} + 2π × 12.5 × 6] cm^{2}

= 22/7 (1240.4 + 156.25 + 150) cm^{2}

= 4860.9 cm^{2}

Now, the volume of water that the bucket can hold (also, known as the capacity

of the bucket)

= (πh)/3 × (r12 + r22 + r1r2)

=(22/7) × (34/3) × [(22.5)2 + (12.5)2 + (22.5)(12.5)]

=(22/7) × (34/3) × 943.75

= 33615.48 cm^{3}

= 33.62 litres (approx.)

**Q. 11: A cylindrical pencil sharpened at one edge is the combination of**

**(A) a cone and a cylinder**

**(B) frustum of a cone and a cylinder**

**(C) a hemisphere and a cylinder**

**(D) two cylinders.**

**Solution: **

(A) a cone and a cylinder

The Nib of a sharpened pencil = conical shape

The rest of the part of a sharpened pencil = cylindrical

Therefore, a pencil is a combination of cylinder and a cone.

**Q. 12: A canal is 300 cm wide and 120 cm deep. The water in the canal is flowing at a speed of 20 km/h. How much area will it irrigate in 20 minutes if 8 cm of standing water is desired?**

**Solution: **

The volume of water flows in the canal in one hour = width of the canal × depth

of the canal × speed of the canal water = 3 × 1.2 × 20 × 1000 m^{3}

= 72000m^{3}

.

In 20 minutes the volume of water = (72000 × 20)/ 60 = 24000 m^{3}

Area irrigated in 20 minutes, if 8 cm, i.e., 0.08 m standing water is required

=24000/0.08

= 300000 m^{2}

= 30 hectares

### Practice Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes

- How many shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9cm × 11cm × 12cm?
- Two solid cones A and B are placed in a cylindrical tube as shown in the figure. The ratio of their capacities is 2:1. Find the heights and capacities of cones. Also, find the volume of the remaining portion of the cylinder.

- Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm.
- How many cubic centimetres of iron is required to construct an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm provided the thickness of the iron is 1.5 cm. If one cubic cm of iron weighs 7.5 g, find the weight of the box.
- A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.
- A solid right circular cone of height 120 cm and radius 60 cm is placed in a right circular cylinder full of water of height 180 cm such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is equal to the radius of the cone.
- A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cubical depression to hold the pens and pins, respectively. The dimension of the cuboid is 10 cm, 5 cm and 4 cm. The radius of each of the conical depressions is 0.5 cm and the depth is 2.1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand.
- From a solid cube of side 7 cm, a conical cavity of height 7 cm and radius 3 cm is hollowed out. Find the volume of the remaining solid.

Very useful and helpful for AISSE 2020 preparation.