JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 10)
A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of radio m/M is close to :
0.77
0.57
0.37
0.17
Explanation
Initially :
_9th_January_Evening_Slot_en_10_1.png)
After putting 2 masses of each 'm' at a distance $${L \over 2}$$ from center :
_9th_January_Evening_Slot_en_10_2.png)
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
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
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