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# Escape velocity at earth's surface is 11.2 km/s. Escape velocity at the surface of a planet having mass 100 times and radius 4 times that of earth will be56 km/s112 km/s305 km/s11.2 km/s

A
112 km/s
B
305 km/s
C
56 km/s
D
11.2 km/s
Solution
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#### Ve=√2gRVe=√2GMR[∴g=GMR2]Now, 11.1Km/hr=√2GMR V1e=√2G(100M)4R 11.1V1e=15 V1e=55.5 Km/hr

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