InterviewSolution
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InterviewSolution
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In the problem discussed in the text, find the values of `E` and `V` at `p` dua to the remaining mass. |
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Answer» Correct Answer - A::B::C::D Total mass is `M` of volume `(4)/(3) pi R^(3)`. Therefore, mass of cavity of volume `(4)/(3) pi (R/(4))^(3)` will be `(M)/(64)`. Further `C_(1)P = 2 R +( R)/(4) = (9R)/(4)` Using Eq. (i), `E_(R) = (GM)/((3R)^(2)) (-hati) - (G((M)/(64)))/((9R//4)^(2) )(-hati)` `:. E_(R) = (35)/(324) (GM)/(R^(2)) (-hat i)` `:.` Field strength is `(35 GM)/(324 R^(2))` towards `C`. USing Eq. (ii), we have `V_(R) = - (GM)/(3R) - [(-G((M)/(64)))/((9R//4))]` `= - (47)/(144) (GM)/(R)` |
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