NCERT Solutions For Class 10 Math Chapter – 13 Exercise – 13.2
Q1. 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 n.
Solution:-
Volume of hemisphere = 2/3∏r3
= 2/3∏(1)3
= 2/3∏ cm3
Volume of cone = 1/3∏r2h
= 1/3∏r2 x 1
= 1/3∏ cm3
Total volume of the solid = Volume of the hemisphere + Volume of the cone
= 2/3∏ cm3 + 1/3∏ cm3 = ∏ cm3
Q2. 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 12 cm If each cone has a height of 2 cm , find the volume of air contained n the model that Rachel made.
Solution:-
Volume of one cone = 1/3 ∏r2h
= 1/3 ∏ (9 x 2/4)
= 3/2 ∏ cm3
Volume of both cones = 2 x 3/2∏ cm3
= 3∏ cm3
Volume of the cylindrical portion = ∏r2h
= ∏(3/2)2 x 8 cm3 = ∏ x 9 x 8/4 cm3 = 18 ∏ cm3
Volume of air contained in the model = Total volume of the solid
= 3∏ cm3 + 18∏ cm3
= 21`x 22/7 cm3
= 66 cm3
Q3. A gulab jamun , contain sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found n 45 gulab jamuns , each shaped like a cylinder with tow hemispherical ends with length 5 cm and diameter 2.8 cm
Solution:-
Volume of one piece of gulab jamun = Volume of the cylindrical portion + Volume of the two hemispherical ends 128 radius of each hemispherical portion = 2.8/2 = 1.4 cm
Volume of hemispherical ends = 2/3 ∏r3
= 2/3 ∏r3
= 2/3 x 22/7 x (1.4)3
= 2/3 x 22/7 x 1.4 x 1.4 x 1.4 cm3
= 2 x 22 x 2 x 14 x 14 / 3 x 10 x 10 x 10 cm3
= 5.74 cm3
Volume of both hemispherical ends = 2 x 5.74 cm3 = 11.48 cm3
Height of the cylindrical portion = (Total height) – (radius of both hemispherical ends)
= 5 cm – 2(1.4) cm
= 5 – 2.8
= 2.2 cm
Volume of the cylindrical portion of gulab jamun = ∏r2h
= 22/7 x (1.4)2 x 2.2 cm3
= 22 x 1.4 x 1.4 x 2.2/7
= 13.55 cm3
Total volume of ne gulab jamun = Volume of the two hemispherical ends + Volume of the cylindrical portion
= 11.48 cm3 + 13.55 cm3
= 25.03 cm3
Volume of sugar syrup = 30% of volume of gulab jamun
= 30/100 x 25.03 cm3 = 7.50 cm3
Volume of sugar syrup is 45 gulab jamuns
= 45(Volume of sugar syrup in one gulab jamun)
= 45 x 7.50
= 337.5 cm3
Q4. A pen stand made of wood is in the shape of a cuboid with four conical depressions o hold pens. The dimensions of the cuboidal are 15 cm by 10 cm by 3.5 cm The radius of each of the depression is 0.5 cm and the depth is 1.4 cm Find the volume of wood in the entire stand
Solution:-
Volume of the cuboidal box = l x b x h
= 15 x 10 x 3.5 cm3
= 525 cm3
Remaining of volume box = Volume of cuboidal box – Volume of four conical depression
= 525 – 1.464
= 523.5 cm3
Q5. 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 cone = 1/3∏r2h
= 1/3∏(5)28 cm3
= 200/3∏ cm3
Volume of one spherical lead shot = 4/3∏r3
= 4/3∏(0.5)3
= 4 x 0.125/3∏
= 0.5/3 ∏ cm3
Volume of n spherical shots =1/4 volume of conical vessel
N(0.5/3∏) = 1/4 (200/3∏)
N(0.5)=50
N = 50 x 10/5
N = 100
Q6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 2 cm , which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of pole, given that 1 cm 3 of iron has approximately 8 g mass.
Solution:-
Volume of 1st cylinder = ∏r2h
= ∏(12)2(220)
= 144 x 220∏
= 99475.2 cm3
Volume of 2nd cylinder = ∏r2h
∏(8)2 (60) cm3
= 64 x 60∏ cm3
= 64 x 60 x 3.14
= 12057.6 cm3
Total volume of solid = Volume of 1st cylinder + Volume of 2nd cylinder
= 99475.2 cm3 + 12057.6 cm3
= 11532.8 cm3
Mass of 11532.8 cm3 of iron = 11532.8 x 8g
= 892.262 kg
Q7. 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 it height is 180 cm.
Solution:-
Volume of hemisphere = 2/3∏(60)3
= 2/3∏(60)3
Volume of conical portion = 1/3∏r2h
= 1/3∏(60)2 x 120
Total volume of hemispherical and conical solid = Volume of hemisphere + Volume of conical portion
= 2/3∏(60)3 cm3 + 1/3∏(60)2 120 cm3
= ∏(60)2 x 80 cm3
Volume of cylindrical solid = ∏r2h
= ∏(60)2 x 180 cm3
Volume of water left in cylinder = Volume of cylinder – (Volume of hemisphere + Volume of cone)
= [∏(60)2 x 180 – ∏(60)2 x 80] cm3
= 1.130 m3
Q8. 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 it volume to be 345 cm3 Check whether she is correct, taking the above as the inside measurements
Solution:-
Volume of cylinder = ∏r2h
= 3.14 x (1)2 x 8 cm3
Volume of spherical part = 4/3 ∏r3
= 4/3 x 3.14 x 85 x 85 x 85/20 x 20 x 20
= 321.39 cm3
Total volume of the glass vessel = Volume of the cylindrical part + Volume of the spherical part
= 25.12 cm3 + 321.39 cm3
= 346.51 cm3