Question

We would like to increase the length of a 15 cm long copper rod of cross-section 4 mm² by 1 mm. The energy absorbed by the rod if it is heated is E₁. The energy absorbed by the rod if it is stretched slowly is E₂. The ratio of E₁ and E₂ is $\alpha$. The value of $\frac{\alpha}{100}$ is. [Various parameters of Copper are : Density = 9 × 10³ kg/m³, Thermal coefficient of linear expansion = 16 × 10⁻⁶ K⁻¹, Young's modulus = 135 × 10⁹ Pa, specific heat = 400 J/kg-K] :-

Answer 1

5

Answer 2

Solution:

1. Heating:
   The temperature rise needed for a linear expansion $\Delta L$ is given by

   $$\Delta T = \frac{\Delta L}{L \alpha_{\text{th}}}.$$

   The mass of the rod is

   $$m = \rho \, A \, L.$$

   Thus, the energy absorbed by heating is

   $$E_1 = m c \Delta T = \rho A L c \left(\frac{\Delta L}{L \alpha_{\text{th}}}\right) = \frac{\rho A c \,\Delta L}{\alpha_{\text{th}}}.$$

2. Stretching:
   The elastic energy stored when a rod is stretched by $\Delta L$ is

   $$E_2 = \frac{1}{2}Y A \frac{\Delta L^2}{L}.$$

3. Ratio:
   The ratio is

   $$\alpha =\frac{E_1}{E_2} =\frac{\frac{\rho A c \,\Delta L}{\alpha_{\text{th}}}}{\frac{1}{2}Y A\frac{\Delta L^2}{L}} = \frac{2\rho c L}{Y \alpha_{\text{th}}\, \Delta L}.$$

4. Substitute given values:

   $$\rho = 9\times10^3\,\text{kg/m}^3,\quad c = 400\,\text{J/kg-K},\quad L=15\,\text{cm}=0.15\,\text{m},$$ $$\Delta L = 1\,\text{mm}=0.001\,\text{m},\quad \alpha_{\text{th}} = 16\times10^{-6}\,\text{K}^{-1},\quad Y=135\times10^9\,\text{Pa}.$$

   Now,

   $$\alpha = \frac{2\times 9\times10^3 \times 400 \times 0.15}{135\times10^9 \times 16\times10^{-6} \times 0.001}.$$

   Calculating numerator:

   $$2\times9\times10^3 = 18\times10^{3},$$ $$18\times10^3 \times 400 = 7.2\times10^6,$$ $$7.2\times10^6 \times 0.15 = 1.08\times10^6.$$

   Calculating denominator:

   $$135\times10^9 \times 16\times10^{-6} = 2160\times10^3 = 2.16\times10^6,$$ $$2.16\times10^6 \times 0.001 = 2160.$$

   Thus,

   $$\alpha = \frac{1.08\times10^6}{2160} = 500.$$

   The required value is

   $$\frac{\alpha}{100}=\frac{500}{100}=5.$$

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Core Explanation:

• For heating, energy E₁ = (ρAc ΔL) / (αₜₕ).

• For stretching, energy E₂ = (1/2)YA(ΔL²/L).

• Ratio α = (2ρcL)/(YαₜₕΔL).

• Substitute values to get α = 500; hence, (α/100) = 5.