JEE Physics 2026 Question with Solution

Given: $t = A \rho^a r^b \eta^c \sigma^d$ for the time to attain terminal velocity.

Find: The value of $\frac{b+c}{a+d}$.

From the solution, the characteristic time to reach terminal velocity is

$$\tau = \frac{m}{6\pi \eta r}$$

For the ball,

$$m = \frac{4}{3}\pi r^3 \rho$$

Substituting in the expression for time,

$$t \propto \frac{r^3 \rho}{\eta r}$$

Therefore,

$$t \propto \frac{\rho r^2}{\eta}$$

Comparing with $t = A \rho^a r^b \eta^c \sigma^d$,

$$a = 1, \quad b = 2, \quad c = -1, \quad d = 0$$

Now,

$$\frac{b+c}{a+d} = \frac{2 + (-1)}{1 + 0} = 1$$

Therefore, the required numerical value is $1$.

Given: A sphere moves through a viscous liquid and attains terminal velocity after a characteristic time $t$.

Find: The indices in $t = A \rho^a r^b \eta^c \sigma^d$ and then evaluate $\frac{b+c}{a+d}$.

For Stokes flow, the damping coefficient is proportional to

$$6\pi \eta r$$

Hence the time constant is mass divided by damping coefficient:

$$t \propto \frac{m}{6\pi \eta r}$$

The mass of the ball is proportional to volume into density:

$$m = \frac{4}{3}\pi r^3 \rho$$

So,

$$t \propto \frac{\left(\frac{4}{3}\pi r^3 \rho\right)}{6\pi \eta r}$$

Ignoring numerical constants,

$$t \propto \frac{r^3 \rho}{\eta r}$$

Thus,

$$t \propto \rho r^2 \eta^{-1}$$

There is no dependence on liquid density $\sigma$ in the extracted solution, so

$$\sigma^0$$

Therefore,

$$a=1, \; b=2, \; c=-1, \; d=0$$

Now compute

$$\frac{b+c}{a+d} = \frac{2-1}{1} = 1$$

So the answer is $1$. The solution's lists $16$, but the worked solution clearly gives $1$, so the solution has been followed.