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Question: The volume of a block of metal changes by \(0.12\% \) when heated through \({20^ \circ }C\). Then \(...

The volume of a block of metal changes by 0.12%0.12\% when heated through 20∘C{20^ \circ }C. Then α\alpha is
A. 2.0×10−5/∘C2.0 \times {10^{ - 5}}{/^\circ }C
B. 4.0×10−5/∘C4.0 \times {10^{ - 5}}{/^\circ }C
C. 6.0×10−5/∘C6.0 \times {10^{ - 5}}{/^\circ }C
D. 8.0×10−5/∘C8.0 \times {10^{ - 5}}{/^\circ }C

Explanation

Solution

To solve this question, we have to use the formula for the volumetric expansion in terms of the change in temperature. By putting the values given in the question into that formula, we will get the value of the coefficient of the volumetric expansion of the block. Finally, using the relation between the linear coefficient and the volumetric coefficient of expansion, we will get the final answer.

Complete step-by-step solution:
We know that the change in volume of a material is proportional to the original volume of the material, that is,
ΔV∝V\Delta V \propto V
Also, we know that the change in volume of a material is proportional to the change in temperature of the material, that is,
ΔV∝ΔT\Delta V \propto \Delta T
On combining the above two proportionalities, we can write
ΔV∝VΔT\Delta V \propto V\Delta T
Replacing the proportionality sign with the equality sign, we get
ΔV=γVΔT\Delta V = \gamma V\Delta T
Here γ\gamma is the coefficient of volumetric expansion.
Dividing by VV on both the sides, we get
ΔVV=γΔT\dfrac{{\Delta V}}{V} = \gamma \Delta T
Multiplying both the sides by 100100 we get
ΔVV×100=100γΔT\dfrac{{\Delta V}}{V} \times 100 = 100\gamma \Delta T_______(1)
According to the question, the percentage change in the volume of the metal block is equal to 0.12%0.12\% .
This means that
ΔVV×100=0.12\dfrac{{\Delta V}}{V} \times 100 = 0.12(2)
Also, the change in temperature of the block is equal to 20∘C{20^ \circ }C, that is,
ΔT=20∘C\Delta T = {20^ \circ }C(3)
Substituting (2) and (3) in (1) we get
0.12=2000γ0.12 = 2000\gamma
⇒γ=0.122000\Rightarrow \gamma = \dfrac{{0.12}}{{2000}}__(4)
Now, we know that the coefficient of volumetric expansion is related with the coefficient of linear expansion by the relation
γ=3α\gamma = 3\alpha
⇒α=γ3\Rightarrow \alpha = \dfrac{\gamma }{3}
Substituting (4) in the above relation, we finally get
α=0.126000\alpha = \dfrac{{0.12}}{{6000}}
On solving we get,
α=2.0×10−5/∘C\alpha = 2.0 \times {10^{ - 5}}{/^ \circ }C
Hence, the correct answer is option A.

Note: Carefully note that we are given the percentage change in the volume of the metal block and not the fractional change. Therefore, we had to multiply with a hundred on both sides of the relation of the volumetric expansion.