Question

**** 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:​

Answer 1

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