JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 16)
A car is moving with a constant speed of 20 m/s in a circular horizontal track of radius 40 m. A bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be : (Take g = 10 m/s$$^2$$)
$$\frac{\pi}{2}$$
$$\frac{\pi}{6}$$
$$\frac{\pi}{4}$$
$$\frac{\pi}{3}$$
Explanation
_25th_January_Morning_Shift_en_16_1.png)
In car’s frame, FBD of bob
_25th_January_Morning_Shift_en_16_2.png)
where $a_{P}=$ Pseudoforce or centrifugal force
$\theta=\tan ^{-1}\left(\frac{a_{P}}{g}\right)=\tan ^{-1}\left(\frac{v^{2}}{R g}\right)=\tan ^{-1}\left(\frac{400}{40 \times 10}\right)$ $=45^{\circ}$
Comments (0)
