A)
B)
C)
D)
Correct Answer: B
Solution :
Potential is at any point inside the sphere is constant and is equal to that on the surface \[{{n}_{oct}}=\frac{35}{114}\] At outside point \[{{\chi }_{oct}}=\frac{\frac{25}{100}}{\frac{25}{100}+\frac{35}{114}}=0.45\]i.e.,\[{{\chi }_{hep}}=1-0.45=0.55\] Hence, (b) is correct.
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