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Surface Areas and Volumes Exercise - 12.2

Hi Friends and Champs!

 In this chapter, we will learn to find the resulting volumes by joining or scooping of different shape bodies in the same way as in previous exercise. Let's start. 
Exercise 12.2
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:
radius of both cone and hemisphere r = 1 cm
height of the cone h = r = 1 cm
Volume "v" of the solid = volume of the hemisphere + volume of the cone
v = 2/3 πr³ + 1/3 πr²h
=2/3 π1³ +  1/3 π1²×1
=2/3 π + 1/3 π
=π cm³ Ans.
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:
Radius of the model r = 3/2 cm
Length of the model L = 12 cm
height of each cone h = 2 cm
height of the cylinder H= L - 2h = 12 -4 = 8 cm
Total volume of the model
=volume of both cone + volume of the cylinder
=2 × 1/3 πr²h + πr²H
= 2/3 π(3/2)²×2 + π(3/2)²×8
=3π + 18π
=21π= 21 × 22/7 =66 cm³ Ans.
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 hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig. 12.15). 
Solution:
One gulab jamun radius r= 2.8/2 =1.4 cm
One gulab jamun cylinder height h
= total length of gulab jamun - 2 hemisphere radius 
= 5 - 2 × 1.4 
=5 - 2.8
=2.2 cm
One gulab jamun volume
=volume of gulab jamun cylinder + 2 hemisphere volume 
= πr²h + 2 × 2/3 πr³
= πr² ( h + 4/3 r )
= 22/7 × (1.4)² × [2.2 + 4/3 ×1.4 ] 
= 22/7 × 14/10×14/10 × [22/10 + 4/3 ×14/10 ]
=154/25 × 122/30
=9394/375
volume of 45 gulab jamun = 45 × 9394/375
volume sugar syrup = 30/100 × 45 × 9394/375=338.184 ≈ 338 cm³ Ans.

4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions 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 (see Fig. 12.16). 
Solution:
Volume of the cuboid = 15 × 10 × 3.5 = 525 cm³
volume of one depression = 1/3 πr²h
=1/3 × 22/7 × (0.5)² × 1.4
=1/3 × 22/7 × 0.25 × 1.4
=11/30 cm³
volume of 4 conical depressions = 4 × 11/30 =22/15 cm³
Volume of wood in the entire stand 
= volume of the cuboid - volume of 4 conical depressions 
= 525 - 22/15
=523.53 cm³ Ans.
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:
Volume of the vessel = 1/3πr²h
= 1/3 π×5²×8
=200π/3
volume of one lead shot = 4/3 πr³
=4/3 π × (0.5)³
=π/6 
Let number of lead shots dropped = x
since 1/4 th water is flown out from the vessel , hence
x × volume of one lead shot = 1/4 th volume of vessel
⇒ x × π/6 = 1/4 × 200π/3 
⇒ x = (200π/12) ÷ π/6 
⇒ x = 100 Ans.
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 = π = 3.14) 
Solution:
Volume of first cylinder = πr²h
=3.14 × 12 × 12 × 220
= 99475.2 cm³
Volume of second cylinder = πr²h
=3.14 × 8 × 8 × 60
=12057.6 cm³
Total volume of the pole = volume of first cylinder + volume of second cylinder
=99475.2 + 12057.6
=111532.8 cm³
Hence mass of the pole = 8 × 111532.8 = 892262.4 g or 892.2624 Kg Ans
7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 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:
Volume of water left in the cylinder V=
Volume of the cylinder - volume of the hemisphere - volume of the right circular cone
=π- 2/3 πr³ - 1/3 πr²h
where H = 180 cm height of the cylinder
and h = 120 cm height of the right circular cone
Hence, volume of water left in the cylinder
V = πr²(H - 2/3 r - 1/3 h)
=22/7 × 60 × 60 ×(180 - 2/3 × 60 - 1/3 × 120 )
=1131428.571 cm³
=1.131429 m³ Ans.
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 π = 3.14.
Solution:
volume of cylindrical neck = πr²h
= 3.14 × 1² × 8
=25.12 cm³
volume of spherical part = 4/3 πr³
=4/3 × 3.14 × (8.5/2)³
=321.3921
total volume of the vessel = volume of neck + volume of spherical part
=321.3921 + 25.12
=346.5121 cm³ Ans.
hence the child is not correct by some error.


I hope all the questions are properly readable and understandable, in case of some confusion, kindly let me know in comment section 🙏🙏🙏

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