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Flux passing through the shaded surface of a sphere when a point charge q is placed at the center is (radius of the sphere is R) |
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Answer» Solution :K.E = K.E of sphere DUE to rolling + K.E due to rotation about 0 `=[(1//2)mv^(2) +1//2 IOMEGA^(2)] + 1/2 I xx v^(2)//R^(2) = (7//10)mv^(2) + 1/2 xx 2/5 mr^(2) xx v^(2)//R^(2) =(7//10) mv^(2) [1+(2R^(2))/(7R^(2))]` |
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