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Let t = time for the potential of metal sphere to rise by 2 volt.
Now beta particles escaped in this time = (5x10^10)xt
No. of beta particles escaped in this time = 40% x (5x10^10)x t
= 40/100 x 5 x 10^10 x t
=2 x 10^10 t
So Charge acquired by the sphere in t sec Q = (2 x 10^10 t) x (1.6x10^-19) since we know Q = ne
Q= 3.2 x 10^-9 t coulomb
Radius r = 10^-2 m
The Capacitance C of a metal sphere is given by C = (4 *pie *epsilon )*r
C = [1/(9x10^9)*]*10^-2 = 10^-11/9 farad
We know that Q = C*V (where V = 2 volt)
3.2 x 10^-9 t = 10^-11/9 * 2
t = 700 x 100^-6 sec
Answer (ii) is correct
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