Question

A copper rod of \(88 cm \) and an aluminium rod of unknown length have their increase in length independent of increase in temperature. The length of aluminium rod is: (\(\alpha_{cu}-1.7\times 10^{-5}K^{-1} \)and \(\alpha_{Al}=2.2×10^{-5}K^{-1}\)) 

• A. 6.8 cm
• B. 113.9 cm
• C. 88 cm
• D. 68 cm

Answer 1

The Correct Option is D

To solve this problem, we need to determine the length of the aluminum rod such that its increase in length with temperature change is independent, or the same as that of the copper rod. This involves using the concept of linear expansion.

The formula for linear expansion is given by:

\(\Delta L = L \cdot \alpha \cdot \Delta T\)

where:

• \(\Delta L\) is the change in length,
• \(L\) is the original length of the material,
• \(\alpha\) is the coefficient of linear expansion,
• \(\Delta T\) is the change in temperature.

For the lengths to be independent of the change in temperature, the increase in length \(\Delta L\) due to temperature should be equal for both rods. Hence, we set the equations of change in length for copper and aluminum rods equal:

\(\Delta L_{Cu} = \Delta L_{Al}\)

Since the temperature change \(\Delta T\) will cancel out, we have:

\(L_{Cu} \cdot \alpha_{Cu} = L_{Al} \cdot \alpha_{Al}\)

We know:

• \(L_{Cu} = 88 \, cm\)
• \(\alpha_{Cu} = 1.7 \times 10^{-5} \, K^{-1}\)
• \(\alpha_{Al} = 2.2 \times 10^{-5} \, K^{-1}\)

Substitute these values into the equation:

\(88 \cdot 1.7 \times 10^{-5} = L_{Al} \cdot 2.2 \times 10^{-5}\)

Solve for \(L_{Al}\):

\(L_{Al} = \frac{88 \cdot 1.7 \times 10^{-5}}{2.2 \times 10^{-5}}\)
\(L_{Al} = \frac{88 \cdot 1.7}{2.2}\)
\(L_{Al} = \frac{149.6}{2.2}\)
\(L_{Al} \approx 68 \, cm\)

Therefore, the length of the aluminum rod is approximately 68 cm.