A) Zero
B) \[\frac{GM}{8{{R}^{2}}}\]
C) \[\frac{GM}{2{{R}^{2}}}\]
D) \[\frac{GM}{{{R}^{2}}}\]
Correct Answer: C
Solution :
Mass of cavity (i.e., of the portion removed), \[M=\rho \times \frac{4}{3}\pi \frac{{{R}^{3}}}{8}=\frac{M}{8}\] So, \[{{\vec{E}}_{\text{resultant}}}={{\vec{E}}_{M}}-{{\vec{E}}_{M/8}}\] \[=0-\frac{G(M/8)}{{{(R/2)}^{2}}}=\frac{-GM}{2{{R}^{2}}}\] \[-\text{ve}\]sign gives the direction i.e., it is away from the centre of cavity.You need to login to perform this action.
You will be redirected in
3 sec