QJEEM19Phd75a: JEE Main 2019 Physics - Class 11 JEEM19Phd75a Solution: A rod of mass 'M' and length '2L' is suspended at its middle by …

Initially :



After putting 2 masses of each 'm' at a distance ${L \over 2}$ from center :



We know,

Time period (T) = 2 $\pi$ $\sqrt {{{\rm I} \over C}}$

$\therefore$   T $\propto$ $\sqrt {\rm I}$

$\therefore$    Frequency (f) $\propto$ $\sqrt {{1 \over {\rm I}}}$

$\therefore$     ${{{f_1}} \over {{f_2}}}$ = $\sqrt {{{{{\rm I}_2}} \over {{{\rm I}_1}}}}$

Also given that,
After putting two masses 'm' at both end new frequency becomes 80% of initial frequency.

$\therefore$    f₂ = 0.8f₁

$\therefore$     ${{{f_1}} \over {0.8{f_1}}}$ = $\sqrt {{{{{\rm I}_2}} \over {{{\rm I}_1}}}}$

$\therefore$     ${{{{{\rm I}_1}} \over {{{\rm I}_2}}}}$ = 0.64

Initial moment of inertia of the system,

${{{\rm I}_1}}$ = ${{M{{\left( {2L} \right)}^2}} \over {12}}$

Final moment of inertia of the system,

I₂ = ${{M{{\left( {2L} \right)}^2}} \over {12}}$ + 2 $\left( {m{{\left( {{L \over 2}} \right)}^2}} \right)$

$\therefore$     ${{M{{\left( {2L} \right)}^2}} \over {12}}$ = 0.64 $\left[ {{{M{L^2}} \over 3} + {{m{L^2}} \over 2}} \right]$

$\Rightarrow$     ${{M{L^2}} \over {3 \times 0.64}}$ = ${{M{L^2}} \over 3}$ + ${{M{L^2}} \over 2}$

$\Rightarrow$     ${M \over {1.92}} - {M \over 3} = {m \over 2}$

$\Rightarrow$     ${{1.08M} \over {3 \times 1.92}}$ = ${m \over 2}$

$\Rightarrow$     ${m \over M}$ = ${{1.08 \times 2} \over {3 \times 1.92}}$ = 0.37