Verified Solution • 2025–26 Session moderate

QJEEM19Ph8c13: JEE Main 2019 Physics - Class 11 JEEM19Ph8c13 Solution: A simple pendulum oscillating in air has period T. The bob of…

A simple pendulum oscillating in air has period T. The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is 1/16 th of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is :

Multiple Choice Options

  • A 2T1102T\sqrt {{1 \over {10}}}
  • B 4T1144T\sqrt {{1 \over {14}}}
  • C 4T1154T\sqrt {{1 \over {15}}}
  • D 2T1142T\sqrt {{1 \over {14}}}

Step-by-Step Verified Solution

For a simple pendulum T = 2πLgerr2\pi \sqrt {{L \over {{g_{err}}}}}

Situation 1: when pendulum is in air →\to geff = g
Situation 2:when pendulum is in liquid
→\to geff = (1−ρliquidρbody)=g(1−116)=15g16\left( {1 - {{{\rho _{liquid}}} \over {{\rho _{body}}}}} \right) = g\left( {1 - {1 \over {16}}} \right) = {{15g} \over {16}}

So, T′T=2πL15g/162πLg{{{T^{'}}} \over T} = {{2\pi \sqrt {{L \over {15g/16}}} } \over {2\pi \sqrt {{L \over g}} }}

⇒T′=4T15\Rightarrow {T^{'}} = {{4T} \over {\sqrt {15} }}

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