JEE MAIN - Physics (2021 - 31st August Evening Shift - No. 9)
A bob of mass 'm' suspended by a thread of length l undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density $${1 \over 4}$$ times that of the bob and the length of the thread is increased by 1/3rd of the original length, then the time period of the simple harmonic oscillations will be :-
T
$${3 \over 2}$$T
$${3 \over 4}$$T
$${4 \over 3}$$T
Explanation
$$T = 2\pi \sqrt {l/g} $$
When bob is immersed in liquid
mgeff = mg $$-$$ Buoyant force
mgeff = mg $$-$$ v$$\sigma$$g ($$\sigma$$ = density of liquid)
$$ = mg - v{\rho \over 4}g$$
$$ = mg - {{mg} \over 4} = {{3mg} \over 4}$$
$$\therefore$$ $${g_{eff}} = {{3g} \over 4}$$
$${T_1} = 2\pi \sqrt {{{{l_1}} \over {{g_{eff}}}}} $$
$${l_1} = l + {l \over 3} = {{4l} \over 3},\,{l_{eff}} = {{3g} \over 4}$$
By solving
$${T_1} = {4 \over 3}2\pi \sqrt {l/g} $$
$${T_1} = {{4T} \over 3}$$
When bob is immersed in liquid
mgeff = mg $$-$$ Buoyant force
mgeff = mg $$-$$ v$$\sigma$$g ($$\sigma$$ = density of liquid)
$$ = mg - v{\rho \over 4}g$$
$$ = mg - {{mg} \over 4} = {{3mg} \over 4}$$
$$\therefore$$ $${g_{eff}} = {{3g} \over 4}$$
$${T_1} = 2\pi \sqrt {{{{l_1}} \over {{g_{eff}}}}} $$
$${l_1} = l + {l \over 3} = {{4l} \over 3},\,{l_{eff}} = {{3g} \over 4}$$
By solving
$${T_1} = {4 \over 3}2\pi \sqrt {l/g} $$
$${T_1} = {{4T} \over 3}$$
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