A stone is rotated steadily in a horizontal circle with a period T by a string of length l. If the tension in the string is kept constant and l increases by 1%, what is the percentage change in T

The correct answer is 0.5%.

Step-by-Step Explanation:

Let a stone of mass ^m be rotated in a horizontal circle of radius equal to the length of the string ^l with a constant tension ^F. The tension in the string provides the necessary centripetal force for this circular motion.

The formula for the centripetal force (tension ^F) is given by:
$F=m{\omega }^{2}l$
where $\omega$ is the angular velocity of the stone.

The relationship between angular velocity $\omega$ and the time period ^T is:
$\omega =\frac{2\pi }{T}$

Substituting this value of $\omega$ into the tension equation:
$F=m{\left(\frac{2\pi }{T}\right)}^{2}l$
$F=\frac{4{\pi }^{2}ml}{T^2}$

Rearranging the equation to solve for ^T:
$T^2=\frac{4{\pi }^{2}ml}{F}$
Taking the square root on both sides:
$T=2\pi \sqrt{\frac{ml}{F}}$

Since the mass of the stone ^m and the tension ^F are kept constant, the time period ^T is directly proportional to the square root of the length of the string ^l:
$T\propto l^{1/2}$

For a small percentage change (typically less than 10%), we can use the binomial approximation or error formula to relate the fractional change in ^T to the fractional change in ^l:
$\frac{\Delta T}{T}=\frac{1}{2}\left(\frac{\Delta l}{l}\right)$

Expressing this relation in percentage form:
Percentage change in ^T = $\frac{1}{2}$ × Percentage change in ^l

Given that the length of the string ^l increases by 1%:
Percentage change in ^T = $\frac{1}{2}$ × 1% = 0.5%

Therefore, the percentage change in the period of rotation ^T is 0.5%.