AP 10th Class Maths 12th Chapter Surface Areas and Volumes Exercise 12.2 Solutions

Well-designed AP 10th Class Maths Guide Chapter 12 Surface Areas and Volumes Exercise 12.2 offers step-by-step explanations to help students understand problem-solving strategies.

Surface Areas and Volumes Class 10 Exercise 12.2 Solutions – 10th Class Maths 12.2 Exercise Solutions

Unless stated otherwise take π = \(\frac{22}{7}\)

Question 1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
Solution:
Given radius of hemisphere (or) cone r = 1 cm
height of the cone h = r = 1 cm

Volume of the solid = Volume of hemisphere + Volume of cone
= \(\frac{2}{3}\) πr³ + \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) πr²(2r+h)
= \(\frac{1}{3}\) πr²(2×1 + 1)
= \(\frac{1}{3}\) π(3)
= π cubic cm

Therefore, volume of the solid = π cm³.

Question 2.
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.)
Solution:

Given radius of cylinder (or) cone
= \(\frac{d}{2}\) = \(\frac{3}{2}\) = 1.5 cm
height of cone h₁ = 2 cm
height of model = 12 cm
height of cylinder h₂ = 12 – 2 × 2
= 12 – 4 = 8 cm
Volume of model = Volume of cylinder + 2 x Volume of cone

Volume of the model = 66 cm

Question 3.
A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemi-spherical ends with length 5 cm and diameter 2.8 cm (as shown in figure).

Solution:

Given length of the Gulab Jamun = 5 cm
radius of cylinder (or) hemisphere (r) d 2.8
= \(\frac{d}{2}\) = \(\frac{2.8}{2}\) = 1.4 cm
height of cylinder part = 5 – 2 × r
= 5 – 2 × 1.4 h
= 5 – 2.8 = 2.2 cm

Volume of each gulab jamun = Volume of cylinder + Volume of 2 hemispheres

Volume of each gulab jamun = \(\frac{75.152}{3}\) cm³
Volume of 45 gulab jamun = 45 × \(\frac{75.152}{3}\)
= 1127.28 cm³
Syrup in the container
= 30% of the volume
= 30% of 1127.28 30
= \(\frac{30}{100}\) × 1127.28
Therefore, volume of sugar syrup = 338.184 cm³

Question 4.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimen-sions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (as shown in figure.)

Solution:
Given dimensions of cuboid are
15 × 10 × 3.5 cm

Volume of cuboid = 15 × 10 × 3.5 = 525 cm³.
radius of the cone r = 0.5 cm
height of the cone h = 1.4 cm
Volume of each depression (or) cone
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 0.5 × 0.5 × 1.4
Volume of 4 depressions

∴ Volume of wood in the entire stand = Volume of cuboid – Volume of 4 depressions
= 525 – 1.47 = 523.53 cm³.

Question 5.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:

Given radius of the cone r = 5 cm
height of the cone h = 8 cm
radius of the sphere r = 0.5 cm
Volume of sphere = \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) π(0.5)³ cm³

Volume of cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\)π (5)² × 8
= \(\frac{1}{3}\)π × 25 × 8
= \(\frac{200}{3}\)π cm³
\(\frac{1}{4}\) of water flows out of the cone
So, volume of water flows out of the cone = \(\frac{1}{4}\) × \(\frac{200}{3}\)π = \(\frac{50}{3}\)π cm³
Let number of shots = n
Volume of x shots = n × \(\frac{4}{3}\) π (0.5)³
Volume of x shots = Volume of water flows out

n = 100
Therefore, number of shots dropped in the vessel = 100.

Question 6.
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 8g mass. (Use n = 3.14)
Solution:

Given radius of the large cylinder
R = \(\frac{d}{2}\) = \(\frac{24}{2}\) = 12 cm
height of large cylinder H = 220 cm
radius of small cylinder r = 8 cm
height of small cylinder h = 60 cm
Volume of large cylinder V₁ = πR²H
V₁ = π × 12² × 220

Volume of small cylinder V₂ = πR²H
V₂ = π × 8² × 60

Volume of iron pole = V₁ + V₂ = π × 12² × 220 + π × 8² × 60 = π (144 × 220 + 64 × 60)
= π (31680 + 3840)
= 3.14 × 35520
Volume of iron pole = 111532.8 cm³
Mass of 1 cm³ iron = 8g
Mass of 111532.8 cm³ iron = 111532.8 × 8 = 892262.4 grams
∴ Total mass of iron pole = 892.262 kg.

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of ra-dius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:

Given radius of the cylinder R = 60 cm
Height of cylinder H = 180 cm
Radius of the cone r₁ = 60 cm
Height of the cone h₁ = 120 cm
Radius of hemisphere r₂ = 60 cm
Volume of cylinder V = πR²H
= π ⋅ 60² × 180

Volume of cone V₁ = \(\frac{1}{3}\) πr₁²h₁
= \(\frac{1}{3}\) π × (60)² × 120

Volume of hemisphere V₂ = \(\frac{2}{3}\) πr₂³
= \(\frac{2}{3}\) π × (60)³

Volume of water left out = Volume of cylinder – Volume of cone – Volume of hemisphere

Volume of water left = 1.1314 m

Question 8.
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 cm3. Check whether she is correct, taking the above as the inside measurements, and n = 3.14.
Solution:
Given radius of cylinder r₁ = \(\frac{d}{2}\) = \(\frac{2}{2}\) =1 cm
height of cylinder h₁ = 8 cm
radius of the sphere r₂ = \(\frac{d}{2}\) = \(\frac{8.5}{2}\) cm
Volume of cylindrical part
= πr₁²h₁ = π × 1² × 8 = 8π

Volume of spherical part

Total volume of water in the vessel = Volume of cylinder + Volume of sphere

Total volume of water in the vessel = = 346.5 cm³
But, child found that as 345 cm3 So, child is not correct.
Hence, the volume found by child is not correct.