JEE Physics 2023 Question with Solution

Given: Cross-sectional area of tank surface is $A_1 = 750 \, \text{cm}^2$, cross-sectional area of tap is $A_2 = 500 \, \text{mm}^2$, and speed of efflux is $V_2 = 30 \, \text{cm/s}$.

Find: The value of $x$ in $\frac{dh}{dt} = x \times 10^{-3} \, \text{m/s}$.

Using the principle of continuity:

$$A_1 V_1 = A_2 V_2$$

where $V_1$ is the speed of the top surface of water.

Convert units and substitute:

$$A_1 = 750 \times 10^{-4} \, \text{m}^2,$$ $$A_2 = 500 \times 10^{-6} \, \text{m}^2,$$ $$V_2 = 0.3 \, \text{m/s}$$

Therefore,

$$750 \times 10^{-4} \, V_1 = 500 \times 10^{-6} \times 0.3$$ $$V_1 = \frac{500 \times 0.3 \times 10^{-6}}{750 \times 10^{-4}} = 2 \times 10^{-3} \, \text{m/s}$$

The height of water is decreasing, so

$$\frac{dh}{dt} = -V_1 = -2 \times 10^{-3} \, \text{m/s}$$

Comparing with $\frac{dh}{dt} = x \times 10^{-3} \, \text{m/s}$, we get $x = -2$. The solution concludes the asked value is taken in magnitude, so $|x| = 2$.

Therefore, the value of $x$ is $2$.

Given: The water level falls as water leaves the tap.

Find: Whether the sign of $\frac{dh}{dt}$ affects the numerical answer.

From continuity, the top surface moves downward with speed

$$V_1 = 2 \times 10^{-3} \, \text{m/s}$$

Since $h$ decreases with time,

$$\frac{dh}{dt} < 0$$

Hence,

$$\frac{dh}{dt} = -2 \times 10^{-3} \, \text{m/s}$$

So mathematically, $x = -2$ if sign is retained. However, the provided solution and answer key take the required value as the magnitude.

Therefore, the accepted answer is $2$.