Answer:
(a) P.E. at \[1\,km\,=mgh={{10}^{-3}}\times
10\times {{10}^{3}}=10\,J\]
P.E
at ground = 0
\ Loss in P.E. = 10 J ? 0 = 10 J
(b)
K.E. of the drop at 1 km = 0
K.E.
of the drop when it just hit the ground
\[=\,\frac{1}{2}\,m{{\upsilon
}^{2}}=\frac{1}{2}\,\times {{10}^{-3}}\,=\,2500\]
\[=1.25\,J\]
Gain
in K.E. \[=12.5\,J\]
(c)
Gain in K.E. is greater than loss is P.E. This is because, a part of P.E. is
used to overcome the force of friction of air.
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