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## Frequently Asked Questions on Chapter 13- Surface Areas and Volumes

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass?

p>Given, the height of the big cylinder (H) = 220 cmRadius of the base (R) = 24/12 = 12 cm

So, the volume of the big cylinder = πR2H

= π(12)2 × 220 cm3

= =99565.8 cm3

Now, the height of smaller cylinder (h) = 60 cm

Radius of the base (r) = 8 cm

So, the volume of the smaller cylinder = πr2h

= π(8)2 × 60 cm3

= 12068.5 cm3

∴ Volume of iron = Volume of the big cylinder + Volume of the small cylinder

99565.8 + 12068.5

=111634.5 cm3

We know,

Mass = Density x volume

So, mass of the pole = 8×111634.5

= 893 Kg (approx.)

A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder?

It is given that radius of the sphere (R) = 4.2 cm

Also, Radius of cylinder (r) = 6 cm

Now, let height of cylinder = h

It is given that the sphere is melted into a cylinder.

So, Volume of Sphere = Volume of Cylinder

∴ (4/3)× π×R3 = π× r2 × h.

=> h = 2.74 cm

A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform?

It is given that the shape of the well is in the shape of a cylinder with a diameter of 7 m

Also, Depth (h) = 20 m

Volume of the earth dug out will be equal to the volume of the cylinder

∴ Volume of Cylinder = π × r2 × h

= 22 × 7 × 5 m3

Let the height of the platform = H

Volume of soil from well (cylinder) = Volume of soil used to make such platform

π × r2 × h = Area of platform × Height of the platform

We know that the dimension of the platform is = 22 × 14

So, Area of platform = 22 × 14 m2

∴ π × r2 × h = 22 × 14 × H

Or, H = 2.5 m

How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cn × 10 cm × 3.5 cm?

It is known that the coins are cylindrical in shape.

So, height (h1) of the cylinder = 2 mm = 0.2 cm

Radius (r) of circular end of coins =1.75/2 = 0.875 cm

Now, the number of coins to be melted to form the required cuboids be “n”

So, Volume of n coins = Volume of cuboids

n × π × r2 × h1 = l × b × h

n × π × (0.875)2× 0.2 = 5.5 × 10 × 3.5

Or, n = 400

The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the surface area of the frustum?

Given,

Slant height (l) = 4 cm

Circumference of upper circular end of the frustum = 18 cm

∴ 2πr1 = 18

Or, r1 = 9/π

Similarly, circumference of lower end of the frustum = 6 cm

∴ 2πr2 = 6

Or, r2 = 6/π

Now, CSA of frustum = π (r1 + r2) × l

= π (9/π + 6/π) × 4

= 12 × 4 = 48 cm2

Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Height of cylinder = 12 – 4 = 8 cm

Height of cone = 2 cm

Now, the total volume of the air contained will be = Volume of cylinder + 2 × (Volume of cone)

Total volume = πr^2h + [2 × (⅓ πr^2h )]

= 18 π + 2(1.5 π)

= 66 cm^3.

Hence, the volume of air contained in the model that Rachel made was 66 cm^3.

Find the height of the cylinder, if a metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm.

Radius of the sphere (R) = 4.2 cm

Also, Radius of cylinder (r) = 6 cm

Now, let height of cylinder = h

It is given that the sphere is melted into a cylinder.

So, Volume of Sphere = Volume of Cylinder

(4/3)× π× R^3 = π× r^2 × h.

h = 2.74 cm

Hence, the height of the cylinder is 2.74cm.

Find the height of the platform. When a 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m.

Shape of the well is in the shape of a cylinder with a diameter of 7 m

Also, Depth (h) = 20 m

Volume of the earth dug out will be equal to the volume of the cylinder

Volume of Cylinder = π × r^2 × h

= 22 × 7 × 5 m^3

Let the height of the platform = H

Volume of soil from well (cylinder) = Volume of soil used to make such platform

π × r^2 × h = Area of platform × Height of the platform

We know that the dimension of the platform is = 22 × 14

So, Area of platform = 22 × 14 m^2

π × r^2 × h = 22 × 14 × H

Or, H = 2.5 m

Hence, the height of the platform is 2.5m.