JEE Main Physics Class 11 System of Particles and Rotational Motion Part 2 Questions

Given, the length of cylinder,
L = 80 cm = 0.8 m
The radius of cylinder, r = 20 cm = 0.2 m
The moment of inertia about CD,ICD=2.7‌kg−m2
Now, the moment of inertia of cylinder about its axis AB is
IAB=

1

2

Mr2
By parallel axis theorem, the moment of inertia of cylinder about CD will be calculated as
ICD=IAB+M(

L

2

)2
⇒2.7=

Mr2

2

+ML24
⇒2.7=M[

(0.2)2

2

+(0.8)24]
⇒M=15‌kg
Density of cylinder is given as
ρ=

M

V

=

M

πr2L

=

15

π(0.2)2(0.8)

=149.2kgm−3
=1.49×102‌kg‌m−3
Thus, the density of cylinder is 1.49×102kgm−3.