Question

# A uniform metal rod is used as a bar pendulum. If the room temperature rises by $10^\circ \,{\text{C}}$ and the coefficient of linear expansion of the metal rod is $2 \times {10^{ - 6}}$per $^\circ \,{\text{C}}$, the period of the pendulum will have percentage increases of:A. $- 2 \times {10^{ - 3}}$B. $- 1 \times {10^{ - 3}}$C. $2 \times {10^{ - 3}}$D. $1 \times {10^{ - 3}}$

Verified
129k+ views
Hint: Use the formula,
$\dfrac{{\Delta T}}{T}\,\, = \,\,\dfrac{1}{2}\alpha \Delta t$ and find the value in percentage.

Complete step by step solution:
A pendulum is a weight hanging from a pivot so it can swing freely. When a pendulum is displaced sideways from its position of rest, equilibrium, due to gravity it is subjected to a restore force which will accelerate it back to the position of equilibrium.

The formula which expresses the time period of a pendulum at a temperature ${\theta _o}$ is given below:
$T = 2\pi \sqrt {\dfrac{{{l_o}}}{g}}$ …… (1)
When the temperature is ${\theta _o}$, the formula can be re-written a:
$T' = 2\pi \sqrt {\dfrac{l}{g}}$ …… (2)

In this problem, you are asked to find the percentage increase of the period of the pendulum. That means you need to find fractional increment first, then convert it into percentage.

The ratio of final length and initial length of the pendulum is:

$\dfrac{{T'}}{T} = \sqrt {\dfrac{{l'}}{l}}$ …… (3)
Where,
$l'$ indicates final length.
$l$ indicates initial length.

Reproducing the equation (3), we get:

$\dfrac{{T'}}{T} = \sqrt {\dfrac{{l\left( {1 + \alpha \Delta t} \right)}}{l}} \\ = \sqrt {\dfrac{{l + \alpha l\Delta t}}{l}} \\ = \sqrt {\dfrac{l}{l} + \dfrac{{\alpha l\Delta t}}{l}} \\ = \sqrt {1 + \alpha \Delta t} \\$
$\dfrac{{T'}}{T}\,\, = \,\,1 + \dfrac{1}{2}\alpha \Delta t$

So, the change (loss or gain) in time per unit time is:

$\dfrac{{T' - T}}{T}\,\, = \,\,\dfrac{1}{2}\alpha \Delta t \\ \dfrac{{\Delta T}}{T}\,\, = \,\,\dfrac{1}{2}\alpha \Delta t \\$
Where,
$\alpha$ indicates coefficient of linear expansion of the metal rod.
$\Delta t$ indicates the change (rise or fall) in temperature.

Substitute, $\alpha = 2 \times {10^{ - 6}}$ and $\Delta t = \,\,10^\circ \,{\text{C}}$in the above equation:
$\dfrac{{\Delta T}}{T}\,\, = \,\,\dfrac{1}{2}\alpha \Delta t \\ = \dfrac{1}{2} \times 2 \times {10^{ - 6}} \times 10 \\ = 1 \times {10^{ - 5}} \\$
Hence, the percentage increase in the period of the pendulum:

$= \dfrac{{\Delta T}}{T} \times 100\% \\ = 1 \times {10^{ - 5}} \times 100\% \\ = 1 \times {10^{ - 3}}\% \\$

Note: In this problem, you are asked to find the percentage increase in the period of the pendulum. For this, first you need to find the ratio of the final length and the initial length. Final length is calculated by applying the formula which includes coefficient of linear expansion of the metal rod. While finding the fraction of increase in length, use the temperature in centigrade, but not in kelvin. Using kelvin, it may affect the result.