Two rigid walls at a distance of 1m act as heat reservoirs at $100 º C$ and $0 º C$ . A well lagged uniform rod having length of 1m at $0 º C$ is held between both the walls. It is found that when the system reaches steady state, the supporting force can be removed and limiting friction acts on the rod at both the ends. Here $α$ for rod $= 10^{- 5} / ° C$ , $Y = 2 \times 10^{11} N / m^{2}$ , and $μ_{walls} = 0.3$ what is the mass of the rod in Kg? Cross sectional area A = 4mm²

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Detailed Solution

In steady state,

$d ℓ = d x α Δ T$
T = ax + b
x = 0, T = 0
x = 1m, T = 100°C
T = 100x
$\int d ℓ = \int 100 x \times d x \times 10^{- 5} Δ ℓ = 10^{- 3} \times \frac{1}{2} F = \frac{Y A}{ℓ} Δ ℓ = \frac{2 \times 10^{11} \times 4 \times 10^{- 6} \times 10^{- 3}}{1} \times \frac{1}{2} = 400 N 2 μ N = m g m = \frac{2 \times 0.3 \times 400}{10} = 24 k g$