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A thin of length `L` is bent to form a semicircle. The mass of rod is M. What will be the gravitational potential at the centre of the circle ?A. `-(GM)/(L)`B. `-(GM)/(2piL)`C. `-(piGM)/(2L)`D. `-(piGM)/(L)` |
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Answer» Correct Answer - D Since, length if rod is equal to the circumference of semicircle `piR=LrArr R=(L)/(pi)` Therefore, the gravitational at the centre of circle will be `V=-(GM)/(R)=-(piGM)/(L)`. |
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