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Question

The energy of a 100 μF capacitor charged to 6 kV is used to lift a 50 kg mass (by powering a motor by energy of capacitor) without incurring any losses. What would be the greatest vertical height through which mass could be raised?
(Take g=10 m/s2)

A
1.6 m
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B
1.2 mm
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C
3.6 m
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D
0.6 mm
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Solution

The correct option is C 3.6 m
Given:
C=100 μF; m=50 kg; V=6 kV

According to question 100% energy of capacitor has been used to lift the mass.

So, applying energy conservation

Loss of Electrostatic Energy stored in capacitor = Gain in gravitational potential energy of mass

12CV2=mgh

12×(100×106)×(6×103)2=50×10×h

50×106×36×106=500 h

h=36×50500=3.6 m

Hence, option (b) is correct.

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