Numerical

What is the decrease in weight of a body of mass 600kg when it is taken in a mine of depth 5000m?

[ Radius of earth = 6400km, g = 9.8 m/s^{2} ]

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#### Solution

Given that m = 600 kg, d = 5000 m,

R = 6400 km = 6.4 × 10^{6}m

Weight of the body on the surface of the Earth = 600 × 9.8 = 5880 N

At depth d, gravitation acceleration is

`g_d=g[1-d/R]`

`thereforeg_d=g[1-5/6400]=9.8 x 0.999`

`therefore g_d=9.7902`m/s^{2}

Weight on surface = mg

^{ } =600 x 9.8

∴Weight on surface= 5880N

weight of the body at depth=mg_{d}

=600 x 9.7902

=5874N

∴Decrease in weight = mg -mg_{d}

_{ }=5880 N - 5874 N

∴Decrease in weight = 6N

Concept: Acceleration Due to Gravity and Its Variation with Altitude and Depth

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