JEE Physics 2023 Question with Solution

Given: Radius of the spherical body is $r = 5 \, \text{mm}$ and error in radius is $\Delta r = 0.1 \, \text{mm}$.

Find: Whether Assertion $A$ and Reason $R$ are true.

For a sphere falling through a viscous liquid, terminal velocity is proportional to the square of radius:

$$V_t \propto r^2$$

So the fractional error is:

$$\frac{\Delta V_t}{V_t} = 2\frac{\Delta r}{r}$$

Substituting the given values:

$$\frac{\Delta V_t}{V_t} \times 100 = 2 \times \frac{0.1}{5} \times 100 = 4\%$$

Hence, Assertion $A$ is true.

Reason $R$ says terminal velocity is inversely proportional to radius, but from the relation above, $V_t \propto r^2$, not inverse of $r$. So Reason $R$ is false.

Therefore, the correct option is D.

The solution also labels the correct option as C and finally writes $\boxed{(3)}$, but this contradicts the working. The working clearly shows Assertion $A$ is true and Reason $R$ is false, so D is the defensible answer.

Given: $V_t \propto r^2$, $r = 5 \, \text{mm}$, and $\Delta r = 0.1 \, \text{mm}$.

Find: Percentage error in terminal velocity.

If a quantity depends on radius as

$$V_t \propto r^2$$

then percentage error multiplies by the power of $r$. Therefore,

$$\frac{\Delta V_t}{V_t} = 2\frac{\Delta r}{r}$$

Now,

$$\frac{\Delta r}{r} = \frac{0.1}{5} = 0.02$$

So,

$$\frac{\Delta V_t}{V_t} = 2 \times 0.02 = 0.04$$

Thus, percentage error is

$$0.04 \times 100 = 4\%$$

Therefore, the statement about percentage error is correct, but the stated reason is incorrect because terminal velocity varies as $r^2$, not as $\frac{1}{r}$.