JEE MAIN - Physics (2022 - 26th June Morning Shift - No. 7)
Time period of a simple pendulum in a stationary lift is 'T'. If the lift accelerates with $${g \over 6}$$ vertically upwards then the time period will be :
(Where g = acceleration due to gravity)
$$\sqrt {{6 \over 5}} T$$
$$\sqrt {{5 \over 6}} T$$
$$\sqrt {{6 \over 7}} T$$
$$\sqrt {{7 \over 6}} T$$
توضیح
$T^{\prime}=2 \pi \sqrt{\frac{I}{g_{\text {eff }}}}$
$T^{\prime}=2 \pi \sqrt{\frac{I}{g+\frac{g}{6}}}=2 \pi \sqrt{\frac{6 l}{7 g}}$
$\Rightarrow T^{\prime}=\sqrt{\frac{6}{7}} T$
$T^{\prime}=2 \pi \sqrt{\frac{I}{g+\frac{g}{6}}}=2 \pi \sqrt{\frac{6 l}{7 g}}$
$\Rightarrow T^{\prime}=\sqrt{\frac{6}{7}} T$