Question

A mass of 5 kg is weighed in a balance at top of a tower 20 m high the mass is then suspended from the pan of the balance by a fine wire 20m long and is rewighed find the change in weight assume that the radous of earth is 6400km

Answer 1

change in weight= 3.0625 × 10^-4 kg-wt

value of g at height h from the surface of the earth is given by, g = go/(1 + h/R)²

here, h = 20m and R = 6400 km

it is clear that, h << R so, g = g0(1 - 2h/R)

⇒g = g0 - 2hg0/R

⇒2hg0/R = g0 - g

⇒change in g = 2hg0/R

= 2 × 20 × 9.8/(6400 × 10³)

= 6.125 × 10^-5 m/s²

so the change in weight , ∆W = mass × change in g

= 5kg × 6.125 × 10^-5 m/s²

= 30.625 × 10^-5 kg - wt

= 3.0625 × 10^-4 kg - wt

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Answer 2

$\huge\bold\purple{Answer:-}$

value of g at height h from the surface of the earth is given by, g = go/(1 + h/R)²

here, h = 20m and R = 6400 km

it is clear that, h << R so, g = g0(1 - 2h/R)

⇒g = g0 - 2hg0/R

⇒2hg0/R = g0 - g

⇒change in g = 2hg0/R

= 2 × 20 × 9.8/(6400 × 10³)

= 6.125 × 10^-5 m/s²

so the change in weight , ∆W = mass × change in g

= 5kg × 6.125 × 10^-5 m/s²

= 30.625 × 10^-5 kg - wt

= 3.0625 × 10^-4 kg - wt