What is the decrease in weight of a body of mass 600 kg when it is tak

1.	What is the decrease in weight of a body of mass 600 kg when it is taken in a mine of depth 5000 m? (Radius of earth = 6400 km, g = 9.8 m/s2)
Answer» Given: m = 600kg, g = m/s² d = 5000 m, R = 6400 km = 6.4 x 10⁶ m To find: Decrease in weight of body (dW) Formula: g_d = g [1- \(\frac{d}{R}\)] Calculation: Weight of body on the earth surface, W = mg = 600 x 9.8 = 5880 N Since, W_d = mg_d From formula, W_d = mg [1- \(\frac{d}{R}\)] =5880 [1- \(\frac{5\times10^3}{6.4\times10^6}\)] =5880 [\(\frac{6.4-0.005}{6.4}\)] = {antilog{log(5880) + [log(6.395) - log(6.4)]}} = {antilog[3.7694 + (0.8058 – 0.8062)]} = {antilog(3.7694 + 1.9996)} = {antilog(3.7690)} ∴ W_d = 5875 N ∴ Decrease in weight = W - W_d = 5880 – 5875 dW = 5 N Decrease in weight is 5 N.

What is the decrease in weight of a body of mass 600 kg when it is taken in a mine of depth 5000 m? (Radius of earth = 6400 km, g = 9.8 m/s2)

Answer»

Given: m = 600kg, g = m/s² d = 5000 m, R = 6400 km = 6.4 x 10⁶ m

To find: Decrease in weight of body (dW)

Formula: g_d = g [1- \(\frac{d}{R}\)]

Calculation: Weight of body on the earth surface,

W = mg = 600 x 9.8 = 5880 N

Since, W_d = mg_d

From formula,

W_d = mg [1- \(\frac{d}{R}\)]

=5880 [1- \(\frac{5\times10^3}{6.4\times10^6}\)]

=5880 [\(\frac{6.4-0.005}{6.4}\)]

= {antilog{log(5880) + [log(6.395) - log(6.4)]}}

= {antilog[3.7694 + (0.8058 – 0.8062)]}

= {antilog(3.7694 + 1.9996)}

= {antilog(3.7690)}

∴ W_d = 5875 N

∴ Decrease in weight = W - W_d = 5880 – 5875

dW = 5 N

Decrease in weight is 5 N.

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