A light bulb of power 100 W is placed at the centre of a hollow sphere of radius 10 cm . If the 66 % of the energy is converted into light, then the pressure exerted by the light on the surface of the sphere will be (Assume the surface of sphere to be perfectly absorbing)

Question

A light bulb of power 100 W is placed at the centre of a hollow sphere of radius 10 cm . If the 66 % of the energy is converted into light, then the pressure exerted by the light on the surface of the sphere will be (Assume the surface of sphere to be perfectly absorbing), 1 . 0 × 10 - 5 N m - 2, 1 . 5 × 10 - 7 N m - 2, 1 . 75 × 10 - 6 N m - 2, 7 . 5 × 10 - 5 N m - 2

MARKS App · Accepted Answer

Intensity of bulb at r = 10 cm , we have I = P 4 π r 2 Intensity of bulb, that is converted to light is, I ' = 0 . 66 I ⇒ I ' = 0 . 66 × P 4 π r 2 Radiation pressure in this case is given by, P r = I ' c , as light is absorbed at the surface of sphere. ⇒ P r = 0 . 66 c × P 4 π r 2 ⇒ P r = 0 . 66 3 × 10 8 × 100 4 π × 0 . 10 2 = 1 . 75 × 10 - 6 N m - 2

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