Verified Solution • 2025–26 Session moderate

QJEEM21Ph685a: JEE Main 2021 Physics - Class 11 JEEM21Ph685a Solution: A bob of mass 'm' suspended by a thread of length l undergoes…

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 14{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 :-

Multiple Choice Options

  • A T
  • B 32{3 \over 2} T
  • C 34{3 \over 4} T
  • D 43{4 \over 3} T

Step-by-Step Verified Solution

T=2πl/gT = 2\pi \sqrt {l/g}

When bob is immersed in liquid

mgeff = mg −- Buoyant force

mgeff = mg −- v σ\sigma g ( σ\sigma = density of liquid)

=mg−vρ4g= mg - v{\rho \over 4}g

=mg−mg4=3mg4= mg - {{mg} \over 4} = {{3mg} \over 4}

∴\therefore geff=3g4{g_{eff}} = {{3g} \over 4}

T1=2πl1geff{T_1} = 2\pi \sqrt {{{{l_1}} \over {{g_{eff}}}}}

l1=l+l3=4l3, leff=3g4{l_1} = l + {l \over 3} = {{4l} \over 3},\,{l_{eff}} = {{3g} \over 4}

By solving

T1=432πl/g{T_1} = {4 \over 3}2\pi \sqrt {l/g}

T1=4T3{T_1} = {{4T} \over 3}

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