Two uniform brass rods A and B of lengths l and 2 l and radii 2 r and r respectively are heated to the same temperature. The ratio of the increase in the volumes of A to that of B is

Step-by-step solution

For brass rod A Volume V 1 = π 2 r 2 × l ...(i) For volume expansion V 1 ′ = V 1 ( 1 + γ ∆ t ) ⇒ V 1 ′ - V 1 ∝ V 1 Or ∆ V 1 ∝ V 1 ...(ii) Similarly, for brass rod B Volume V 2 = π ( r ) 2 × 2 l ...(iii) and ∆ V 2 ∝ V 2 ...(iv) Dividing Eq. (i) by Eq. (ii), we get V 1 V 2 = π 4 r 2 l π r 2 2 l = 2 1 From eqs. (ii) and (iv), ∆ V 1 ∆ V 2 = 2 1