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Question: If the volume of a block of metal changes by \(0.12\%\), when it is heated through \(20{}^\circ C\),...

If the volume of a block of metal changes by 0.12%0.12\%, when it is heated through 20C20{}^\circ C, then the coefficient of linear expansion is then
4×105 per C4\times {{10}^{-5}}\text{ per }{}^\circ C
2×105 per C2\times {{10}^{-5}}\text{ per }{}^\circ C
0.5×105 per C0.5\times {{10}^{-5}}\text{ per }{}^\circ C
4×104 per C4\times {{10}^{-4}}\text{ per }{}^\circ C

Explanation

Solution

To find the coefficient of linear expansion, we first need to find the coefficient of volumetric expansion. The formula to find the volumetric expansion coefficient is similar to the linear coefficient. Using the relation between volumetric and linear coefficient, we will find our answer.
Formula Used:
ΔVV=γΔθ\dfrac{\Delta V}{V}=\gamma \Delta \theta
γ=3α\gamma =3\alpha

Complete step-by-step solution:
We know that the coefficient of linear expansion is nothing but the percentage linear change in the body divided by the temperature change. This can be denoted as:
Δll=αΔθ\dfrac{\Delta l}{l}=\alpha \Delta \theta --------(i)
Where, Δl\Delta l is the change in the length of the body, ll is the original length of the body, α\alpha is the coefficient of linear expansion and Δθ\Delta \theta is the change in temperature of the body.
But here, we are not given with the percentage linear change of the body. Here, percentage volumetric change is given. Thus, we will have to apply the same formula (i), with slight alteration so that instead of percentage linear change, we can accommodate percentage volumetric change.
Here, for volumetric change, formula (i) can be written as:
ΔVV=γΔθ\dfrac{\Delta V}{V}=\gamma \Delta \theta -------(ii)
Where, VV is the original volume of the body and ΔV\Delta V is the change in volume of the body. γ\gamma is the coefficient of volumetric expansion.
In the question, the percentage change in volume is given as 0.120.12%. This can be written as:
ΔVV=0.12100\dfrac{\Delta V}{V}=\dfrac{0.12}{100}
Also, the temperature change given is:
Δθ=20C\Delta \theta =20{}^\circ C
Using equation (i)
0.12100=γ×20\dfrac{0.12}{100}=\gamma \times 20
γ=6×105 per C\Rightarrow \gamma =6\times {{10}^{-5}}\text{ per }{}^\circ \text{C} ----(iii)
We know that the relation between coefficient of linear expansion and the coefficient of volumetric expansion is given by:
γ=3α\gamma =3\alpha
Using equation (iii), we can find α\alpha as:
α=6×1053\alpha =\dfrac{6\times {{10}^{-5}}}{3}
α=2×105 per C\Rightarrow \alpha =2\times {{10}^{-5}}\text{ per }{}^\circ \text{C}
Hence, the correct option is (b).

Note: In some cases, the coefficient of linear expansion may come negative. This does not mean that your answer is wrong. The negative sign, simply, implies that as the temperature increases, the linear dimensions of the body decreases. The unit of these coefficients are per C\text{per }{}^\circ \text{C}.