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The string of a pendulum is horizontal.The mass of bob attached to it is m.Now, the string is released.The tension in the string in the lowest position is
Options
(a) mg
(b) 2 mg
(c) 3 mg
(d) 4 mg
Correct Answer:
3 mg
Explanation:
When string is released, then according to conservation of energy,
(1/2) mv² = mgh
v² = 2gh = 2gR
Now, when it is at the bottom,
T – mg cos 0² = mv²/R
or T – mg = (m/R)(2gR)
T = mg + 2mg = 3mg.
Related Questions: - For a series L-C-R circuit, the rms values of voltage across various components
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Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- For a series L-C-R circuit, the rms values of voltage across various components
- A radioactive element forms its own isotope after 3 consecutive disintegrations.
- The de-Broglie wavelength of neutron in thermal equilibrium at temperature T is
- A spherical drop of capacitance 1μF is broken into eight drops of equal radius.
- The amount of heat energy required to raise the temperature of 1 g of Helium
Topics: Laws of Motion (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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