The time period of a simple pendulum on the surface of the moon compared to its time period on Earth is:
A. The same
B. Decreased
C. Increased
D. Zero
Answer: Option C
Solution (By JKSSB Mock Tests)
The time period of a pendulum is T = 2 * pi * sqrt(L/g). The acceleration due to gravity on the moon (g_moon) is roughly 1/6th of that on Earth (g_earth). Since g decreases, and T is strictly inversely proportional to the square root of g, the time period T must increase. The pendulum will swing much slower.
Explanation:
For spherical mirrors, R = 2f. So R = 2 × 15 = 30 cm. Sign convention: for concave mirror, f and R are negative by Cartesian convention, but magnitude 30 cm. So 30 cm is correct.
Explanation:
Doppler effect: observed frequency higher when approaching, lower when receding. Applications: radar speed guns, astronomical redshift/blueshift, echocardiography. Echo is reflection. Resonance is vibration matching. Diffraction is bending.
Explanation:
Heat travels through spoon from hot end to cold end by conduction (molecular vibration and free electrons). Handle gets hot. Metal good conductor.
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