JEE MAIN - Physics (2025 - 28th January Morning Shift - No. 21)
Two iron solid discs of negligible thickness have radii $R_1$ and $R_2$ and moment of intertia $I_1$ and $I_2$, respectively. For $R_2=2 R_1$, the ratio of $I_1$ and $I_2$ would be $1 / x$, where $\mathrm{x}=$ _______ .
Answer
16
Explanation
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Let surface mass density = $\sigma$
So, $${M_1} = \sigma \times \pi R_1^2$$
$${M_2} = \sigma \times \pi R_2^2 = \sigma \pi {(2{R_1})^2}$$ (As $${R_2} = 2{R_1}$$
$$ = 4\sigma \pi R_1^2$$
$$ \Rightarrow {M_2} = 4{M_1}$$
$${{{I_1}} \over {{I_2}}} = {{{{{M_1}R_1^2} \over 2}} \over {{{{M_2}R_2^2} \over 2}}} = \left( {{{{M_1}} \over {{M_2}}}} \right){\left( {{{{R_1}} \over {{R_2}}}} \right)^2}$$
$$ \Rightarrow {{{I_1}} \over {{I_2}}} = \left( {{{{M_1}} \over {4{M_1}}}} \right){\left( {{{{R_1}} \over {2{R_1}}}} \right)^2} = {1 \over 4} \times {1 \over 4} = {1 \over {16}} = {1 \over x}$$
Hence, $$x = 16$$
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