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Question

A sphere and a cube of same material and same volume are heated up to same temperature and allowed to cool in the same surroundings. The ratio of the amounts of radiations emitted in equal time1 intervals will be

A

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B

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C

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D

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Solution

The correct option is C Q = σ At(T4−T40) If T, T0, σ and t are same for both bodies then QsphereQcube = AsphereAcube = 4πr26a2 But according to problem, volume of sphere = volume of cube ⇒ 43πr3 = a3 ⇒ a = (43π)13 r Substituting the value of a in Eq. 9(i), we get QsphereQcube = 4πr26a2 = 4πr26((43π)13r)2 = 4πr26(43π)13r2 = (π6)13 : 1

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