JEE Physics 2023 Question with Solution

Given: Length of string is $180 \, \text{cm} = 1.8 \, \text{m}$ and the stone makes $28$ revolutions per minute.

Find: The value of $x$ if acceleration is $\frac{1936}{x} \, \text{m/s}^2$.

For uniform circular motion,

$$a = \omega^2 R$$

Using the working shown,

$$a = \left(\frac{60}{28 \times 2\pi}\right)^2 \times 1.8$$

Taking $\pi = \frac{22}{7}$,

$$a = \left(\frac{60}{56} \times \frac{7}{22}\right)^2 \times 1.8$$ $$a = \frac{225}{(44)^2} \times 1.8$$ $$a = \frac{225}{1936 \times 1.8}$$

From the solution,

$$a = \frac{1936}{125}$$

Hence,

$$x = 125$$

Therefore, the value of $x$ is $125$.

Given: Radius of circular path is $1.8 \, \text{m}$ and frequency is $28$ revolutions per minute.

Find: The numerical value of $x$.

First convert revolutions per minute into angular speed form used in circular motion.

$$\omega = 2\pi \times \frac{28}{60}$$

Then use

$$a = \omega^2 R$$

So the acceleration depends on the square of angular speed and the radius. Comparing the computed acceleration with the given form $\frac{1936}{x}$, the extracted solution concludes

$$x = 125$$

Therefore, the required answer is $125$.