**** Hestion Paper :*****
Q.(1): From a solid cylinder whose height is
2.4cm and diameter 1.4cm,a conical cavity of
the same height and diameter is hollowed
out. Find the volume of the remaining solid ta
the nearest cm^3.(use te(pi) =22/7)
Ans:
Answers
Answer:
2 cm
Step-by-step explanation:
given
height of the cylinder = 2.4cm
diameter of the cylinder = 1.4 cm
radius of the cylinder = 1.4/2
= 0.7 cm
height and diameter of the cone is equal to the height and diameter of the solid cylinder
for solid cylinder
volume of solid cylinder = πr^2h
= 22/7*(0.7)^2*2.4
= 22/7*0.7*0.7*2.4
= 22*0.1*0.7*2.4
= 22*1/10*7/10*24/10
= 3696/1000
volume of cone = 1/3πr^2h
= 1/3*22/7*(0.7)^2*2.4
= 1/3*22/7*0.7*0.7*2.4
= 22*7/10*1/10*8/10
= 1232/1000
volume of the remaining solid = volume of solid cylinder - volume of cone
=3696/1000-1232/1000
= 2464/1000
= 2.464
therefore nearest cm^2 = 2 cm
hope it will help you