bp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0 auto;max-height:40vh;vertical-align:baseline" alt="JEE Main 2025 (Online) 28th January Morning Shift Physics - Rotational Motion Question 41 English Explanation">

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, $$x = 16$$

19:T4be,

JEE Main 2025 (Online) 28th January Morning Shift Physics - Rotational Motion Question 41 Hindi Explanation

माना पृष्ठ द्रव्यमान घनत्व = $\sigma$

अतः, $${M_1} = \sigma \times \pi R_1^2$$

$${M_2} = \sigma \times \pi R_2^2 = \sigma \pi {(2{R_1})^2}$$ (चूँकि $${R_2} = 2{R_1}$$

$$ = 4\sigma \pi R_1^2$$

$$ \Rightarrow {M_2} = 4{M_1}$$

$${{{I_1}} \over {{I_2}}} = {{{{{M_1}R_1^2} \over 2}} \over {{{{M_2}R_2^2} \over 2}}} = \left( {{{{M_1}} \over {{M_2}}}} \right){\left( {{{{R_1}} \over {{R_2}}}} \right)^2}$$

$$ \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} \time

Rotational Motion

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