The loss in weight of a solid immersed in a liquid at 9 ∘ C is W 0 , and at t ∘ C is W . If the cubical expansion coefficients of solid and liquid are γ s and γ l , respectively, then:

At 9 ∘ C the buoyant force is W 0 Δ T = t − 9 W = W 0 1 + γ l ( t − 9 ) 1 + γ s ( t − 9 ) To first order, W ≈ W 0 [ 1 + ( γ s − γ l ) ( t − 9 )]

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The loss in weight of a solid immersed in a liquid at $9^{\circ} C$ is $W_{0}$ , and at $t^{\circ} C$ is $W$ . If the cubical expansion coefficients of solid and liquid are $γ_{s}$ and $γ_{l}$ , respectively, then:

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Answer:

Solution:

At $9^{\circ} C$ the buoyant force is $W_{0}$
$Δ T = t - 9$
$W = W_{0} \frac{1 + γ _{s} ( t - 9 )}{1 + γ _{l} ( t - 9 )}$
To first order,
$W \approx W_{0} [1 + (γ_{s} - γ_{l}) (t - 9)]$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Thermal Expansion

⏱ 2mℹ️ Source: QB

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