Question

If a man weighs 90 N on the surface of the earth, then the height above the surface of the earth of radius R, where the weight is 30 N is (A. 0.73R , B. R/√3 , C. 3R , D. √3R)

Answer 1

answer : option (1) 0.73R

weight of man at the surface of earth, $W_0$ = 90N

weight of man at h height above the surface of earth, W = 30N

using formula,

W = $\frac{W_0}{\left(1+\frac{h}{R}\right)^2}$

⇒ 30 = 90/(1 + h/R)²

⇒30/90 = 1/(1 + h/R)²

⇒3 = (1 + h/R)²

⇒√3 = 1 + h/R

⇒(√3 - 1) = h/R

⇒h = (√3 - 1)R = (1.73 - 1)R = 0.73R

hence, option (1) is correct choice.

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Answer 2

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

weight of man at the surface of earth, W_0 = 90N

weight of man at h height above the surface of earth, W = 30N

using formula,

W = \frac{W_0}{\left(1+\frac{h}{R}\right)^2}

⇒ 30 = 90/(1 + h/R)²

⇒30/90 = 1/(1 + h/R)²

⇒3 = (1 + h/R)²

⇒√3 = 1 + h/R

⇒(√3 - 1) = h/R

⇒h = (√3 - 1)R = (1.73 - 1)R = 0.73R

hence, option (1) is correct choice.