A simple pendulum is suspended in a lift which is going up with a… - Vidyadip

MCQ

A simple pendulum is suspended in a lift which is going up with an acceleration $5\ m/s^2$. An electric field of magnitude $5 \ N/C$ and directed vertically upward is also present in the lift. The charge of the bob is $1\ mC$ and mass is $1\ mg$. Taking $g = \pi^2$ and length of the simple pendulum $1\ m$, the time period of the simple pendulum is ......$s$

A
$1$
✓
$2 $
C
$0.5$
D
None of these

✓

Answer

Correct option: B.

$2 $

b
$\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}_{\mathrm{eff}}}}$

$\mathrm{g}_{\mathrm{eff}}=\mathrm{g}-\frac{\mathrm{q} \in}{\mathrm{M}}+5$

$=15-\frac{1 \times 5 \times 10^{-6}}{1 \times 10^{-6}}$

$\mathrm{g}_{\mathrm{eff}}=10=\pi^{2}$

$\mathrm{T}=2 \mathrm{\,sec}$

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