Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A second's pendulum is mounted in a rocket. Its period of oscillation decreases when the rocket:

Comes down with uniform acceleration

Moves around the earth in a geostationary orbit

Moves up with a uniform velocity

Moves up with the uniform acceleration

Formula: The time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$ , where $L$ is the length and $g_{eff}$ is the effective acceleration due to gravity .
Analysis: To decrease the period $T$ , the effective gravity $g_{eff}$ must increase (since $T \propto \frac{1}{\sqrt{g_{eff}}}$ ).
Case 1 (Moving up with acceleration $a$ ): When the rocket accelerates upwards, the pseudo-force acts downwards, adding to gravity. Thus, $g_{eff} = g + a$ . Since $g_{eff} > g$ , the period $T$ decreases.
Case 2 (Moving down with acceleration $a$ ): Pseudo-force acts upwards. $g_{eff} = g - a$ . $T$ increases.
Case 3 (Uniform velocity): Acceleration $a = 0$ . $g_{eff} = g$ . Period remains unchanged.
Case 4 (Geostationary orbit): Inside a satellite orbiting the earth, the effective gravity is zero ( $g_{eff} \approx 0$ ). The pendulum would not oscillate ( $T \to \infty$ ).

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