JEE Physics 2023 Question with Solution

Given: Density of glycerine $\rho = 1.25 \times 10^3 \, \text{kg/m}^3$, areas $A_1 = 10 \times 10^{-4} \, \text{m}^2 = 1 \times 10^{-3} \, \text{m}^2$ and $A_2 = 5 \times 10^{-4} \, \text{m}^2$, and pressure drop $\Delta P = P_1 - P_2 = 3 \, \text{N/m}^2$.

Find: The value of $x$ in $Q = x \times 10^{-5} \, \text{m}^3\text{/s}$.

Using continuity equation,

$$A_1 v_1 = A_2 v_2$$

So,

$$v_2 = \frac{A_1}{A_2} v_1 = \frac{10 \times 10^{-4}}{5 \times 10^{-4}} v_1 = 2v_1$$

Now apply Bernoulli's equation,

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

Therefore,

$$P_1 - P_2 = \frac{1}{2}\rho \left(v_2^2 - v_1^2\right)$$

Substituting $\Delta P = 3$ and $v_2 = 2v_1$,

$$3 = \frac{1}{2}\rho \left((2v_1)^2 - v_1^2\right)$$ $$3 = \frac{1}{2}\rho \left(4v_1^2 - v_1^2\right) = \frac{3}{2}\rho v_1^2$$

Putting $\rho = 1.25 \times 10^3$,

$$3 = \frac{3}{2}(1.25 \times 10^3)v_1^2$$ $$v_1^2 = \frac{2}{1.25 \times 10^3} = \frac{1}{625}$$ $$v_1 = \sqrt{\frac{1}{625}} = \frac{1}{25} = 0.04 \, \text{m/s}$$

Now the rate of flow is,

$$Q = A_1 v_1 = (1 \times 10^{-3})(0.04) = 4 \times 10^{-5} \, \text{m}^3\text{/s}$$

Comparing with $Q = x \times 10^{-5} \, \text{m}^3\text{/s}$, we get $x = 4$.

Therefore, the value of $x$ is $4$.

Given: The liquid flows through a conical pipe section with different cross-sectional areas, so the speed changes from one end to the other.

Find: The numerical value of $x$.

Because the pipe narrows from $10 \, \text{cm}^2$ to $5 \, \text{cm}^2$, the fluid speed at the smaller area must increase. From continuity,

$$A_1 v_1 = A_2 v_2$$

Hence,

$$v_2 = 2v_1$$

The pressure drop is used to supply the increase in kinetic energy per unit volume. So Bernoulli's equation gives,

$$\Delta P = \frac{1}{2}\rho\left(v_2^2 - v_1^2\right)$$

Substitute $v_2 = 2v_1$,

$$3 = \frac{1}{2}(1.25 \times 10^3)\left(4v_1^2 - v_1^2\right)$$ $$3 = \frac{3}{2}(1.25 \times 10^3)v_1^2$$ $$v_1^2 = \frac{2}{1250} = \frac{1}{625}$$ $$v_1 = 0.04 \, \text{m/s}$$

Now use discharge formula,

$$Q = A_1 v_1 = (1 \times 10^{-3})(0.04)$$ $$Q = 4 \times 10^{-5} \, \text{m}^3\text{/s}$$

So the required coefficient is $x = 4$.

Therefore, the final answer is $4$.