Class 10 - Mathematics
Surface Areas and Volumes - Exercise 13.2
Top Block 1
Exercise 13.2
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.
Answer :
Given that,
Radius(r) of hemispherical part = Radius of conical part (r) = 1 cm
Volume of solid = Volume of conical part + Volume of hemispherical part
= πr2 h/3 + 2πr3 /3
= (π(1)2 * 1)/3 +( 2π * 13 )/3
= π/3 + 2 π/3
= 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.)
Answer :
From the figure, it can be observed that
Height (h1) of each conical part = 2 cm
Height (h2) of cylindrical part = 12 − 2 * Height of conical part
= 12 − 2 * 2
= 8 cm
Mddle block 1
Volume of air present in the model = Volume of cylinder + 2 * Volume of cones
= πr2 h2 + 2 * (πr2 h1)/3
= π(3/2)2 * 8 + {2π * (3/2)2 * 2}/3
= π * 9/4 * 8 + 2π/3 * 9/4 * 2
= 18π + 3 π
= 21π
= 21 * 22/7
= 3 * 22
= 66 cm2
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 hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig. 13.15).
Answer :
Radius (r) of cylindrical part = Radius (r) of hemispherical part = 2.8/2 = 1.4 cm
Length of each hemispherical part = Radius of hemispherical part = 1.4 cm
Length (h) of cylindrical part = 5 − 2 * Length of hemispherical part
= 5 − 2 * 1.4
= 5 – 2.8 = 2.2 cm
Volume of one gulab jamun = Volume of cylindrical part + 2 * Volume of hemispherical part
= πr2 h + 2 * (2πr3)/3
= πr2 h + 4πr3/3
= π(1.4)2 * 2.2 + {4π * (1.4)3}/3
= 22/7 * 1.4 * 1.4 * 2.2 + (4 * 22/7 * 1.4 * 1.4 * 1.4)/3
= 13.552 + 11.498
= 25.05 cm3
Volume of 45 gulab jamuns = 45 * 25.05 = 1127.25 cm3
Volume of sugar syrup = 30% of volume
= 30/100 * 1127.25
= 338.17
= 338 cm3
Question : 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. 13.16).
Answer :
Radius (r) of each conical depression = 0.5 cm
Volume of wood = Volume of cuboid − 4 * Volume of cones
= lbh – 4 * πr2 h/3
= 15 * 10 * 3.5 – 4 * 1/3 * 22/7 * (1/2)2 * 1.4
= 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.
Answer :
Height (h) of conical vessel = 8 cm
Radius (r1) of conical vessel = 5 cm
Radius (r2) of lead shots = 0.5 cm
Volume of water spilled = Volume of dropped lead shots
⇒ 1/4 * Volume of cone = n * 4πr23/3
⇒ 1/4 * πr12 h/3 = n * 4πr23/3
⇒ r12 h = n * 16r23
⇒ 52 * 8 = n * 16(0.5)3
⇒ 25 * 8 = n * 16(1/2)3
⇒ 25 * 8 = n * 16 * 1/8
⇒ 25 * 8 = n * 2
⇒ 2n = 200
⇒ n = 200/2 = 100
Hence, the number of lead shots dropped in the vessel is 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 π = 3.14)
Answer :
From the figure, it can be observed that
Radius (r1) of larger cylinder = = 12 cm
Height (h2) of smaller cylinder = 60 cm
Radius (r2) of smaller cylinder = 8 cm
Mass of 1 iron = 8 g
Total volume of pole = Volume of larger cylinder + volume of smaller cylinder
= πr12 h1 + πr22 h2
= π(12)2 * 220 + π(8)2 * 60
= π[144 * 220 + 64 * 60]
= π[31680 + 3840]
= 3.14 * 35520
= 111532.8 cm3
Mass of 1 cm3 iron = 8 g
Mass of 111532.8 cm3 iron = 111532.8 * 8 = 892262.4 g = 892262.4/1000 kg
= 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 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.
Answer :
Height (h2) of conical part of solid = 120 cm
Height (h1) of cylinder = 180 cm
Radius (r) of cylinder = 60 cm
Volume of water left = Volume of cylinder − Volume of solid
= Volume of cylinder – (Volume of cone + Volume of Hemisphere)
= πr2 h1 – (πr2 h2/3 +2 πr3/3)
= π(60)2 * 180 – {π(60)2 * 120/3 +2 π(60)3/3}
= π(60)2 [180 – (40 + 40)]
= π(60)2 [180 – 80]
= 3.14 * 3600 * 100
= 1131428.57 cm3
= 1131428.57/1000000 m3
= 1.131 m3
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 π = 3.14.
Answer :
Radius (r2) of cylindrical part = 2/2 = 1 cm
Radius (r1) spherical part = 8.5/2 = 4.25 cm
Volume of vessel = Volume of sphere + Volume of cylinder
= 4πr13/3 + πr22 h
= {4π(4.25)3}/3 + π(1)2 * 8
= {4 * 3.14 * 76.765625}/3 + 3.14 * 8
= 321.392 + 25.12
= 346.512 cm3
Hence, she is wrong.
