A bakelite beaker has volume capacity of 500 cc at 30ᵒC. When it is partially filled with Vm volume (at 30ᵒC) of mercury, it is found that the unfilled volume of the beaker remains constant as temperature is varied. If γ (beaker) = 6 × 10 –6 ᵒC –1 and γ (mercury) = 1.5 × 10 –4 ᵒC –1 , where γ is the coefficient of volume expansion, then Vₘ (in cc) is close to ____.

Solution

Physics sir · 3 steps

Empty space stays fixed only if the beaker and the mercury grow by exactly the same volume.

1. Set the condition. The unfilled volume is V₀ - Vₘ. For it to stay constant, the increase in the beaker's capacity must equal the increase in the mercury's volume for every temperature change Δ T.

   V₀ γ_b Δ T = Vₘ γₘ Δ T ⇒ Vₘ = V₀ γ_b/γₘ

2. Put in numbers. γ_b/γₘ = (6 × 10⁻⁶)/(1.5 × 10⁻⁴) = 0.04, so Vₘ = 500 × 0.04 cc.
3. So. Vₘ = 20 cc, so the answer is 20.

The common slip: 480. Reporting the unfilled volume 500 - 20 gives 480, not 20. The question asks for the mercury volume Vₘ.

The idea to keep. If a gap must stay constant as temperature changes, set the two expansions equal: V₁γ₁ = V₂γ₂. The material that expands more must be present in a smaller amount.

Answer from the NTA final answer key. Solution by PrepWiser.