A) \[\frac{q}{\sqrt{2\pi {{\varepsilon }_{0}}mR}}\]
B) \[\frac{q}{\sqrt{4\pi {{\varepsilon }_{0}}mR}}\]
C) \[\frac{3q}{\sqrt{2\pi {{\varepsilon }_{0}}mR}}\]
D) \[\frac{\sqrt{3}q}{\sqrt{4\pi {{\varepsilon }_{0}}mR}}\]
Correct Answer: D
Solution :
For the bullet to penetrate through the sphere, it must be able to reach the centre of sphere. The potential at the centre of sphere is: \[\therefore \,\,P(r\ge 4)\,{{=}^{5}}{{C}_{4}}{{\left( \frac{1}{4} \right)}^{4}}.\left( \frac{3}{4} \right){{+}^{5}}{{C}_{5}}{{\left( \frac{1}{4} \right)}^{5}}\] For the minimum speed u of the bullet, its speed at the centre is zero. Applying cons. of mechanical energy, have \[=5.\frac{1}{256}.\frac{3}{4}+\frac{1}{45}=\frac{16}{4\times 256}\,=\frac{1}{64}\]
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