JEE MAIN - Physics (2022 - 25th June Evening Shift - No. 22)
Moment of Inertia (M.I.) of four bodies having same mass 'M' and radius '2R' are as follows:
I1 = M.I. of solid sphere about its diameter
I2 = M.I. of solid cylinder about its axis
I3 = M.I. of solid circular disc about its diameter
I4 = M.I. of thin circular ring about its diameter
If 2(I2 + I3) + I4 = x . I1, then the value of x will be __________.
Answer
5
Explanation
$$2\left( {{1 \over 2} + {1 \over 4}} \right) \times M{(2R)^2} + {1 \over 2}M{(2R)^2} = x{2 \over 5}M{(2R)^2}$$
$$ \Rightarrow 1 + {1 \over 2} + {1 \over 2} = x \times {2 \over 5}$$
$$ \Rightarrow x = 5$$
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