A body falls through a distance 'h' in certain time on the earth. Then if the same body is related on another planet having mass and radius twice as that of the earth, the distace through which it falls in the same time is
(a) h/2
(b) 2h
(c) h
(d) 4h
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Answer:
Explanation:
It is h/2 .
Just calculate g in both the cases by the formula g=GM/R^2
and then substitute it in s=ut + 1/2 at^2 where u is 0
On comparingetting we get h'=h/2
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xoSHOAIBxo:
If the radius of the earth were to be increased by a factor of 3 by what factor would its density have to be changed to keep (g) the same
I think the density will remain same . Let me not sureally
options are
a)3 b)1/3 c)6 d)1/6
I think it is 1/3
how ?
i am sorry i am wasting your time
Nothing as such
Density×volume= mass. So substitute M by d×v and like the previous answer compare
U will get it as 1/3
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