Question

A rod is found to be 0.05 cm longer at 40°C than it is at 10°C. The length of the rod at 0°C is (coefficient of linear expansion of the material of the rod = \( 1.5 \times 10^{-5} \, \text{°C}^{-1} \))

• A. 101.1 cm
• B. 120.2 cm
• C. 105.1 cm
• D. 111.1 cm

Hint:

To find the original length of an object from its temperature change, use the formula \( \Delta L = L_0 \alpha \Delta T \) and solve for \( L_0 \).

Answer 1

The Correct Option is D

The change in length of the rod due to temperature change is given by the formula: \[ \Delta L = L_0 \alpha \Delta T \] Where: - \( \Delta L = 0.05 \, \text{cm} \) (the increase in length), - \( \alpha = 1.5 \times 10^{-5} \, \text{°C}^{-1} \) (the coefficient of linear expansion), - \( \Delta T = 40 - 10 = 30 \, \text{°C} \) (the temperature change). Substitute the known values: \[ 0.05 = L_0 \times 1.5 \times 10^{-5} \times 30 \] Solving for \( L_0 \): \[ L_0 = \frac{0.05}{1.5 \times 10^{-5} \times 30} = \frac{0.05}{0.00045} \approx 111.1 \, \text{cm} \] Thus, the length of the rod at 0°C is \( \boxed{111.1 \, \text{cm}} \).