JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 24)
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 :
$$2T\sqrt {{1 \over {10}}} $$
$$4T\sqrt {{1 \over {14}}} $$
$$4T\sqrt {{1 \over {15}}} $$
$$2T\sqrt {{1 \over {14}}} $$
Explanation
For a simple pendulum T = $$2\pi \sqrt {{L \over {{g_{err}}}}} $$
Situation 1: when pendulum is in air $$ \to $$ geff = g
Situation 2:when pendulum is in liquid
$$ \to $$ geff = $$\left( {1 - {{{\rho _{liquid}}} \over {{\rho _{body}}}}} \right) = g\left( {1 - {1 \over {16}}} \right) = {{15g} \over {16}}$$
So, $${{{T^{'}}} \over T} = {{2\pi \sqrt {{L \over {15g/16}}} } \over {2\pi \sqrt {{L \over g}} }}$$
$$ \Rightarrow {T^{'}} = {{4T} \over {\sqrt {15} }}$$
Situation 1: when pendulum is in air $$ \to $$ geff = g
Situation 2:when pendulum is in liquid
$$ \to $$ geff = $$\left( {1 - {{{\rho _{liquid}}} \over {{\rho _{body}}}}} \right) = g\left( {1 - {1 \over {16}}} \right) = {{15g} \over {16}}$$
So, $${{{T^{'}}} \over T} = {{2\pi \sqrt {{L \over {15g/16}}} } \over {2\pi \sqrt {{L \over g}} }}$$
$$ \Rightarrow {T^{'}} = {{4T} \over {\sqrt {15} }}$$
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