JEE Physics 2023 Question with Solution

Given: mass of the vehicle = $200 \, \text{kg}$, radius of the road = $70 \, \text{m}$, angular velocity = $0.2 \, \text{rad/s}$.

Find: The centripetal force acting on the vehicle.

The centripetal force required to keep an object of mass $m$ moving in a circular path of radius $r$ with angular velocity $\omega$ is given by

$$F = m\omega^2 r$$

Substituting the given values,

$$F = (200)(0.2)^2(70)$$ $$F = 200 \times 0.04 \times 70 = 560 \, \text{N}$$

Therefore, the centripetal force acting on the vehicle is $560 \, \text{N}$. The correct option is B.

Given: $m = 200 \, \text{kg}$, $\omega = 0.2 \, \text{rad/s}$, $r = 70 \, \text{m}$.

Find: $F_c$.

Use the formula

$$F_c = m\omega^2 r$$

Now calculate the square of angular velocity first:

$$(0.2)^2 = 0.04$$

Then,

$$F_c = 200 \times 0.04 \times 70$$ $$F_c = 8 \times 70 = 560 \, \text{N}$$

Thus, the centripetal force is $560 \, \text{N}$.