Question

A stone is rotated steadily in a horizontal circle with a time period $T$ by means of a string of length $l$ . If the tension in the string is kept constant and length $l$ increase by $1\% $ , then percentage change in time period $T$ is
A. $1\% $
B. $0.5\% $
C. $2\% $
D. $0.25\% $

Answer 1

Hint For motion of stone in a horizontal circle the required tension is equal to centripetal force. We use this equality to find the expression for time period and then differentiate it to calculate the percentage change in time period.

Complete Step by step solution
For motion in a horizontal circle:
 Tension=centripetal force
i.e. $F = ml{\omega ^2}......(1)$
where, $F$ is tension
$m$ is mass of stone
$l$ is length of string
And, $\omega $is angular velocity
Now we know that,
$\omega = \dfrac{{2\pi }}{T}......(2)$
From equation (1) and (2) we get,
$F = ml \times {\left( {\dfrac{{2\pi }}{T}} \right)^2}$
On solving we get,
$T = 2\pi \sqrt {\dfrac{{ml}}{F}} ......(3)$
Now differentiating equation (3) we get percentage change in time period is,
$\dfrac{{\Delta T}}{T} \times 100 = \dfrac{1}{2} \times \dfrac{{\Delta l}}{l} \times 100$
Given, percentage increase in length is $1\% $
Therefore,
$
  \dfrac{{\Delta T}}{T} \times 100 = \dfrac{1}{2} \times 1\% \\
  \dfrac{{\Delta T}}{T} \times 100 = 0.5\% \\
 $
Hence percentage change in time period is $0.5\% $

Option (B) is correct.

Note Alternate method,
We know the time period is given by, $T = 2\pi \sqrt {\dfrac{l}{g}} $
Hence differentiating above result we get,
$\dfrac{{\Delta T}}{T} = \dfrac{1}{2} \times \dfrac{{\Delta l}}{l}$
Now percentage change is,
$\dfrac{{\Delta T}}{T} \times 100 = \dfrac{1}{2} \times \dfrac{{\Delta l}}{l} \times 100$
Hence,
 $
\dfrac{{\Delta T}}{T} \times 100 = \dfrac{1}{2} \times 1\% \\
\dfrac{{\Delta T}}{T} \times 100 = 0.5\% \\
$
Hence percentage change in time period is $0.5\% $
We get the same result.