How to solve???

How to solve??? 21. A man weighs 60 kg at the earth's surface. The radius of the earth is 6400 what height above the earth's surface, his weight becomes 30 kg? 2900 km (A) 100 km (C) 2624 km (B) 1500 km 22. The weight of a man in a lift moving upwards with an 'a' is 600 N _

Dear student
The variation of the weight of the body with the distance from centre of the earth is given as

\frac{W^{'}}{W} = \frac{R^{2}}{(R + h)^{2}} W^{'} = w e i g h t o f t h e b o d y a t h e i g h t h f o r m c e n t r e o f t h e e a r t h W w e i g t o f t h e b o d y o n e a r t h H e r e \frac{W^{'}}{W} = \frac{30}{60} = \frac{1}{2} \frac{1}{2} = \frac{R^{2}}{(R + h)^{2}} \frac{1}{2} = \frac{(6400)^{2}}{(6400 + h)^{2}} \frac{1}{\sqrt{2}} = \frac{6400}{6400 + h} 6400 + h = 6400 \sqrt{2} h = 6400 \sqrt{2} - 6400 \approx 2624 k m h = 2624 k m R e g a r d s