A) \[6.95\text{ }\mu \text{ }sec\]
B) \[\,0.95\text{ }\mu \text{ }sec\]
C) \[1.95\mu \text{ }sec\,\]
D) \[2.15\text{ }\mu \text{ }sec\]
Correct Answer: A
Solution :
[a] Let t = time for the potential of metal sphere to rise by one volt. Now \[\beta -\] particles emitted in this time \[=(6.25\times {{10}^{11}})\times t\] Number of \[\beta \]-particles escaped in this time \[=(80/100)\times (6.25\times {{10}^{10}})t=5\times {{10}^{10}}t\] \[\therefore \] Charge acquired by the sphere in t sec. \[Q=(5\times {{10}^{10}}t)\times (1.6\times {{10}^{-19}})=8\times {{10}^{-19}}t\] ?.(i) (\[\because \] emission of \[\beta \] -particle lends to a charge eon metal sphere) The capacitance C of a metal sphere is given by \[C=4\pi {{\varepsilon }_{0}}\times r\] \[=\left( \frac{1}{9\times {{10}^{9}}} \right)\times \left( \frac{{{10}^{-3}}}{2} \right)=\frac{{{10}^{-12}}}{18}\] farad ?.(ii) we know that \[Q=C\times V\] {Here \[V=1\] volt} \[\therefore \,\,\,\,\,\,\,(8\times {{10}^{-9}})t=\left( \frac{{{10}^{-12}}}{18} \right)\times 1\] Solving it for t, we get \[t=6.95\text{ }\mu sec.\]
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