JEE Physics 2023 Question with Solution

Given: radius of ball $r = 1 \, \text{mm} = 10^{-3} \, \text{m}$, density of ball $\rho_b = 10.5 \times 10^3 \, \text{kg/m}^3$, density of glycerine $\rho_\ell = 1.5 \times 10^3 \, \text{kg/m}^3$, and the ball attains terminal velocity so $a = 0$.

Find: the value of $x$ in the viscous force expression.

At terminal velocity, the net force is zero, so the viscous force balances the effective weight of the ball:

$$F_v = V \rho_b g - V \rho_\ell g$$

Therefore,

$$F_v = Vg(\rho_b - \rho_\ell)$$

The volume of the spherical ball is

$$V = \frac{4}{3}\pi r^3$$

Substituting,

$$F_v = \frac{4}{3}\pi r^3 g(\rho_b - \rho_\ell)$$

Now put the given values:

$$F_v = \frac{4}{3} \times \frac{22}{7} \times (10^{-3})^3 \times 9.8 \times (10.5 \times 10^3 - 1.5 \times 10^3)$$ $$F_v = \frac{4}{3} \times \frac{22}{7} \times 10^{-9} \times 9.8 \times 9 \times 10^3$$ $$F_v = \frac{4}{3} \times \frac{22}{7} \times 10^{-9} \times 88.2 \times 10^3$$ $$F_v = 3696 \times 10^{-7} \, \text{N}$$

Thus, comparing with the given form, the value of $x$ is $7$.

Therefore, the answer is $7$.

Given: the ball moves in glycerine and attains constant velocity, which means terminal velocity.

Find: the unknown exponent-related value $x$.

When the velocity becomes constant, acceleration is zero. Hence viscous force equals weight minus buoyant force:

$$F_v = mg - F_B$$

Using mass and buoyancy in terms of volume,

$$F_v = V\rho_b g - V\rho_\ell g$$ $$F_v = Vg(\rho_b - \rho_\ell)$$

For a sphere,

$$V = \frac{4}{3}\pi r^3$$

So,

$$F_v = \frac{4}{3}\pi (10^{-3})^3 \times 9.8 \times (10.5 - 1.5) \times 10^3$$ $$F_v = 3696 \times 10^{-7} \, \text{N}$$

Hence, $x = 7$.

The final answer is $7$.