weight of a body decreases by 1.5%, when it is raised to a height h above the surface of earth. when the same body is taken to same depth h in a mine, its weeight will show

a 0.75% increase

b 0.75% decrease

c 3.0% decrease

d 1.5% decrease

the variation of g with height is given as

g' = g(1 - 2h/R)

so in terms of weight of an object (multiplying both sides by mass 'm')

W' = W(1 - 2h/R)

now as weight reduced by .15%

the new weight will be given a

W' = 0.985W

so,

0.985W = W(1 - 2h/R)

or

0.0985 = (1 - 2h/R)

or

2h/R = 1 - 0.985

or

h = (0.015xR) / 2

thus, the height will be

h = (0.015 x 6.4 x 10^{6}) / 2

thus,

h = 48000 m

now,

the variation of g wit depth will be

g' = g(1- d/R)

here d = h = 48000m

or

W' = W(1 - d/R)

or

W' = W[1 - (48000 / 6.4x10^{6})]

or

W' = 0.9925W

thus, the weight will be reduced by

[(W - W') / W] X 100 = 0.0075 x 100

thus,

**decrease in weight = 0.75 %**

so, the correct answer is **option (b)**.

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