A) 0
B) \[\frac{GM}{8{{R}^{2}}}\]
C) \[\frac{GM}{2{{R}^{2}}}\]
D) \[\frac{GM}{{{R}^{2}}}\]
Correct Answer: C
Solution :
Let, \[\vec{E}=\]filed at the center after hollowing \[{{\vec{E}}_{1}}=\]field at the center before hollowing \[{{\vec{E}}_{2}}=\]field at the center due to the removed portion. By principle of superposition of fields: \[\vec{E}={{\vec{E}}_{1}}-{{\vec{E}}_{2}}\] Let us plant the origin at the center. \[\vec{E}=0-\frac{Gm\hat{i}}{{{(R/2)}^{2}}}=-\frac{4Gm\hat{i}}{{{R}^{2}}}\] Where, m is the mass of the removed portion. \[m=\rho \frac{4}{3}\pi {{\left( \frac{R}{2} \right)}^{3}};\]where \[\rho =\frac{M}{\frac{4\pi {{R}^{3}}}{3}}\] Inserting the value of \[\rho ,\]we get \[m=\frac{M}{8}\] \[\therefore \]\[\vec{E}=\frac{-4G}{{{R}^{2}}}\times \frac{M\hat{i}}{8}=-\frac{GM\hat{i}}{2{{R}^{2}}}\] \[\therefore \]\[\left| {\vec{E}} \right|=\frac{GM}{2{{R}^{2}}}\] Hence, the correction option is [c].
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