A pendulum clock is lifted to a height where the gravitational acceleration has a certain value g. Another pendulum clock of same length but of double the mass of the bob is lifted to another height where the gravitational acceleration is g/2. The time period of the second pendulum would be :
(in terms of period T of the first pendulum)
$sqrt{2}$ T
$rac{1}{sqrt{2}}$ T
$2sqrt{2}$ T
T
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC NDA-2 – 2019
For the first pendulum: Length $L_1=L$, gravity $g_1=g$. Time period $T_1 = 2\pi \sqrt{\frac{L}{g}} = T$.
For the second pendulum: Length $L_2=L$ (stated as same length), mass $M_2=2M_1$ (mass does not affect T), gravity $g_2=g/2$. Time period $T_2 = 2\pi \sqrt{\frac{L_2}{g_2}} = 2\pi \sqrt{\frac{L}{g/2}} = 2\pi \sqrt{\frac{2L}{g}}$.
We can rewrite $T_2$ in terms of $T_1$: $T_2 = \sqrt{2} \times (2\pi \sqrt{\frac{L}{g}}) = \sqrt{2} T_1 = \sqrt{2} T$.