CLASS-10-VOLUME-AND-SURFACE-AREA-EXERCISE-SOLUTION-NCERT
EXERCISE 12.2 SURFACE AREA AND VOLUMES OF THE SOLIDS
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 pi.
Question2 : . 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.)
Putting the value of the pie, we can get the final answer = 21*22/7 = 66 cubic 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 hemispherical ends, with length 5 cm and a diameter 2.8 cm
Height of the cylinder = 5-1.4-1.4 = 5 - 2.8 = 2.2
Volume of the one jamun = volume of the cylindrical part +2* volume of the hemispherical part
= П r^2h + 2*(2/3) Пr^3
= Пr^2 (h +(4/3)r)
= П*(1.4)*1.4(2.2 +(2/3)*1.2)
= 2* П*1.96(9/3)= П*5.88 cubic cm
Volume of the 45 gulab jamuns = 45* 5.88* cubic cm
=2*264.6 П cubic cm
Volume of the syrup = 30% of the 264.6*2 П cubic cm
= 0.3*(22/7)*264.6*2 cubic cm
= 2*249.48 cubic cm
=498.96
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).
Solution: For the cuboid l = 15 cm, b = 10cm, h = 3.5 cm
Volume of the cuboid = l*b*h = 15*10*3.5= 525 cubic cm
Radius of the depressions = 0.5cm and the height = 1.4 cm
Number of depression in the figure = 4
Volume of the 4 conical depression = 4*(1/3)*π r*r*h = 4*(1/3)*(22/7)*0.5*0.5*1.4 = 4.4 cubic cm
The volume of the remaining solid = 525- 4.4 = 520.6 cubic cm
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: The vessel s in the form of a cone;
radius = 5cm
height = 8cm
Volume of the vessel = (1/3)*π*R*R*h = (1/3)*π*5*5*8 cubic cms
Volume of the Lead = (4/3)π r*r*r cubic cms = (4/3)*π*0.5*0.5*0.5
When the lead is shot, the volume that came out = 1/4 *volume of the container = (1/4)* (1/3)*π*5*5*8 cubic cms
Number of the lead shot = (Volume of the container )*(1/4)) /(Volume of the one lead)
= (1/3)*π*5*5*8*(1/4) / ((4/3)*π*0.5*0.5*0.5 )
= (25*8*2*2*2)/4*4 (0.5 = 1/2)
= 100 leads
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)
Solution: Given that the cylinder's height = 220 cm
radiusof the cylinder = 12 cm
Height of the other cylinder = 60 cm
Radius of the other cylinder = 8cm
Volume of the combined solid = Volume of the first cylinder + volume of the other cylinder
=3.14 * ( 220 *12*12 + 60*8*8) = 3.14(31680 + 3840)
= 111532.8 cubic cm
Given that the mass of the 1cm^3 = 8 gm
T0tal mass = 892262.4 gm = 892.262 Kg ( 1000 gm = 1Kg)
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.
Solution: Height of the right circular cone = 120 cm
radius of the right circular cone = 60cm
radius of the hemisphere = 60cm
Height of the bigger cylinder = 180 cm
and radius = 60 cm
Solution: Volume of the bigger cylinder = π*r*r*H = π*60*60*220 cubic cms
Volume of the smaller cylinder hemisphere = π r*r*h + (2/3)π r*r*r
= πr*r*r (h +(2/3)r)
= πr*60*60 (120 +(2/3)60) cubccms
Remaining volume = Volume of the bigger cylinder - Volume of the (cylinder and hemisphere)
= π*60*60*220 cubic cms _πr*60*60 (120 +(2/3)60) cubic cms
π*60*60(220-160) cubic cms = 3.14*60*60*60
= 678240 cubic cm
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.
Solution: Diameter of the spherical part = 8.5cm
Radius of cylinder = 2/2 = 1cm
Height of the cylindrical neck = 8cm
volume of the container = Volume of the cylinder +volume of the sphere
= π r*r*h + (4/3)π R*R*R
= π(r*r*h +4/3 R)
=3.14*(1*1*8 + (4/3)*(8.5/2)^3 )
= 3.14(8 +2543.2/24)
=3.14*(113.96)
= 357.85 cubic cm
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