Question

Length of second pendulum is decreased by $1$ then the gain or loss in time per day will be nearly
A.$0.44\;{\rm{s}}$
B. $4.4\;{\rm{s}}$
C. $44\;{\rm{s}}$
D. $440\;{\rm{s}}$

Answer 1

In this question, we need to calculate the gain or loss in time per day. If the length of the second pendulum is decreased by some percentage, then there will be a loss of time per day. But if the length of the second pendulum is increased, then in that case there will be gain in the time per day. In this question, there will be a loss in time per day. To find the value of this loss in time per day, we will use the relation for the time period of oscillation. Then we will write another expression of the changed length. On dividing both the expressions, we can find the change in time per day.

Complete step by step answer:
Given:
The length of the second pendulum is decreased by $1$.
Let us assume the length of the first pendulum as $l$.
Then the length of second pendulum can be expressed as
${l'} = l - \left( {\dfrac{1}{{100}} \times l} \right)\\\ \Rightarrow{l'} = \dfrac{{99}}{{100}}l\\\ \Rightarrow{l'} = 0.99l$…….(i)
We will write the expression for the time period of oscillation of pendulum as:
$T = 2\pi \sqrt {\dfrac{l}{g}}$
Where $T$ is the time period of oscillation, $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.
We will rewrite the above expression as:
$l = \dfrac{{{T^2}g}}{{4{\pi ^2}}}$……(ii)
Change in length can be expressed as:
$\Delta l = l - {l'}$
We will substitute the value of $l'$ from the equation (i).
$\Delta l = l - 0.99l\\\ \Rightarrow\Delta l = 0.01l$
We will differentiate equation (ii) as expressed:
$\Delta l = \dfrac{{2T\Delta Tg}}{{4{\pi ^2}}}$……(iii)
where $\Delta T$ is the change in time period.
Now, we will divide equation (iii) by equation (ii).
$\dfrac{{\Delta l}}{l} = \dfrac{{2T\Delta T}}{{{T^2}}}\\\ \Rightarrow\dfrac{{\Delta l}}{l} = \dfrac{{2\Delta T}}{T}$
We will substitute $0.01l$ for $\Delta l$ and $86400\;{\rm{s}}$ for $T$ in the above expression.
$\dfrac{{0.01l}}{l} = \dfrac{{2\Delta T}}{{86400\;{\rm{s}}}}\\\ \therefore\Delta T = 432\;{\rm{s}}$

This is approximately equal to $440\;{\rm{s}}$. Hence, option D is the correct answer.

Note: We know that the length of the pendulum and the time per day are directly proportional. Hence, in our question, when the length is decreased, it will lead to a decrease in the time per day. We are using the expression of the time period of oscillation to express the relationship between time period and length.