QJEEM25Ph9d61: JEE Main 2025 Physics - Class 11 JEEM25Ph9d61 Solution: Two iron solid discs of negligible thickness have radii and and …

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, $x = 16$