Question

A uniform metal rod is used as a bas pendulum. If the room temperature rises by \(10^{\circ} \mathrm{C}\) and the efficient of line as expansion of the metal of the rod is, \(2 \times 10^{-6} 0_{\mathrm{c}}^{-1}\) what will have percentage increase in the period of the pendulum? (A) \(-2 \times 10^{-3}\) (B) \(1 \times 10^{-3}\) (C) \(-1 \times 10^{-3}\) (D) \(2 \times 10^{-3}\)

Short answer

The short answer is: (B) \(1 \times 10^{-3}\%\).

Step-by-step solution

Recall the formula for the period of a simple pendulum

The formula for the period of a simple pendulum is given by: \[T = 2\pi\sqrt{\frac{l}{g}}\] where \(T\) is the period, \(l\) is the length of the pendulum, and \(g\) is the acceleration due to gravity.

Calculate the length of the pendulum after the temperature increase

The increase in length due to temperature change can be calculated using the formula: \[\Delta l = l_0 \alpha \Delta T\] where \(\Delta l\) is the change in length, \(l_0\) is the initial length of the rod, \(\alpha\) is the coefficient of linear expansion, and \(\Delta T\) is the temperature change. In this case, \(\alpha = 2 \times 10^{-6} 0_{\mathrm{C}}^{-1}\) and \(\Delta T = 10^{\circ} \mathrm{C}\). So, we can calculate the change in length as follows: \[\Delta l = l_0( 2 \times 10^{-6})(10) = 2 \times 10^{-5}l_0\] The new length of the pendulum after the temperature increase is: \[l' = l_0 + \Delta l = l_0(1 + 2 \times 10^{-5})\]

Calculate the new period of the pendulum

Now, we substitute the new length value, \(l'\), into the original period formula to get the new period: \[T' = 2\pi\sqrt{\frac{l'}{g}} = 2\pi\sqrt{\frac{l_0(1 + 2 \times 10^{-5})}{g}}\]

Calculate the percentage increase in the period of the pendulum

The percentage increase in the period of the pendulum is given by: \[\frac{T' - T}{T} \times 100\%\] Let's find the ratio \(\frac{T'}{T}\) : \[ \frac{T'}{T} = \frac{2\pi\sqrt{\frac{l_0(1 + 2 \times 10^{-5})}{g}}}{2\pi\sqrt{\frac{l_0}{g}}} = \sqrt{1 + 2 \times 10^{-5}} \] Now, calculate the percentage increase: \[\frac{T'-T}{T} \times 100\% = (\sqrt{1 + 2 \times 10^{-5}} - 1) \times 100\% \approx 1 \times 10^{-3}\%\]

Compare the result with the given options

From the calculation, the percentage increase in the period of the pendulum is \(1 \times 10^{-3}\%\). Comparing this result with the given options, we find that the correct answer is: (B) \(1 \times 10^{-3}\%\)