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 \(\alpha_{Al}=2.2×10^{-5}K^{-1}\))

Question

The temperature of a metallic sphere of radius \(R\) is increased by a small amount \(\Delta T\). If the linear coefficient of thermal expansion of the metal is \(\alpha\), the approximate increase in the volume of the sphere is:

Answer 1

To find the length of the aluminium rod such that its increase in length is independent of temperature changes, given a copper rod and their respective coefficients of linear expansion, we can use the relationship for thermal expansion:

The formula for the change in length due to temperature is:

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

Where:

\(\Delta L\) = Change in length
\(L\) = Original length of the rod
\(\alpha\) = Coefficient of linear expansion
\(\Delta T\) = Change in temperature

For the increase in lengths to be independent of the increase in temperature, the changes in length of both rods must be equal:

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

Here, the temperature changes (\Delta T) are the same for both rods, so they cancel out, leaving us with:

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

Substituting the given values, we have:

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

Solving for \(L_{Al}\) (length of aluminium rod):

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

Calculating:

\[L_{Al} = \frac{88 \times 1.7}{2.2} = \frac{149.6}{2.2} = 68 \, cm\]

Therefore, the length of the aluminium rod that ensures the increase in length is independent of the temperature change is 68 cm.

Answer 2