A) 5 kV
B) 50 kV
C) 5 V
D) 50 V
Correct Answer: B
Solution :
| [b] Key Idea: When bullet of mass m and charge q is accelerated through potential difference of V volt, then it attains a kinetic energy equal to qV. |
| Kinetic energy of bullet \[=qV\] |
| i.e., \[\frac{1}{2}m{{v}^{2}}=qV\] |
| \[\Rightarrow \] \[V=\frac{m{{v}^{2}}}{2q}\] |
| Given, \[m=2g=2\times {{10}^{-3}}kg,\,\,v=10\,m/s,\] |
| \[q=2\mu \,C=2\times {{10}^{-6}}C\] |
| Substituting the values in relation for V, we obtain |
| \[V=\frac{2\times {{10}^{-3}}\times {{(10)}^{2}}}{2\times 2\times {{10}^{-6}}}\] |
| \[=50\times {{10}^{3}}\,V\,=\,50\,kV\] |
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