JEE Physics 2024 Question with Solution

Given: Initial pressure reading is $P_1 = 4.5 \times 10^4 \, \text{N/m}^2$, final pressure reading is $P_2 = 2.0 \times 10^4 \, \text{N/m}^2$, and for water $\rho = 1000 \, \text{kg/m}^3$.

Find: The value of $V$ if the speed is $\sqrt{V} \, \text{m/s}$.

Using Bernoulli's equation for steady flow,

$$P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2$$

When the valve is closed, the water is at rest, so $v_1 = 0$. Also, taking the two points at the same height, the height terms cancel. Hence,

$$P_1 = P_2 + \frac{1}{2}\rho v_2^2$$

Substituting the given values,

$$4.5 \times 10^4 = 2.0 \times 10^4 + \frac{1}{2} \times 1000 \times v_2^2$$

So,

$$\frac{1}{2} \rho v_2^2 = 4.5 \times 10^4 - 2.0 \times 10^4 = 2.5 \times 10^4$$

Therefore,

$$500v_2^2 = 2.5 \times 10^4$$ $$v_2^2 = \frac{2.5 \times 10^4}{500} = 50$$

Thus,

$$v_2 = \sqrt{50} \, \text{m/s}$$

So $V = 50$.

Therefore, the required value is $50$.

Given: The pressure drops from $4.5 \times 10^4 \, \text{N/m}^2$ to $2.0 \times 10^4 \, \text{N/m}^2$ when water starts flowing.

Find: The value of $V$ in $v = \sqrt{V} \, \text{m/s}$.

The pressure difference is used to provide kinetic energy per unit volume of the flowing water:

$$P_1 - P_2 = \frac{1}{2}\rho v^2$$

Now,

$$P_1 - P_2 = 4.5 \times 10^4 - 2.0 \times 10^4 = 2.5 \times 10^4 \, \text{N/m}^2$$

Using $\rho = 1000 \, \text{kg/m}^3$,

$$2.5 \times 10^4 = \frac{1}{2} \times 1000 \times v^2$$ $$2.5 \times 10^4 = 500v^2$$ $$v^2 = 50$$

Hence the speed is $\sqrt{50} \, \text{m/s}$, so the required value of $V$ is $50$.

The final answer is $50$.