QJEEM23Phc0aa: JEE Main 2023 Physics - Class 11 JEEM23Phc0aa Solution: Glycerin of density is flowing through the conical section of…

We can use the Bernoulli equation and continuity equation to solve this problem. The Bernoulli equation is given by:

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

The continuity equation is given by:

$A_1 v_1 = A_2 v_2$

From the given data, we have:

$P_1 - P_2 = 3 \mathrm{~Nm}^{-2}$ $A_1 = 10 \mathrm{~cm}^2 = 10 \times 10^{-4} \mathrm{~m}^2$ $A_2 = 5 \mathrm{~cm}^2 = 5 \times 10^{-4} \mathrm{~m}^2$ $\rho = 1.25 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$

Rearrange the continuity equation to solve for $v_2$ :

$v_2 = \frac{A_1}{A_2} v_1 = 2v_1$

Substitute $v_2$ and rearrange the Bernoulli equation:

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

$3 = \frac{1}{2} \times 1.25 \times 10^3 (4v_1^2 - v_1^2)$

Now, solve for $v_1$ :

$3 = \frac{1}{2} \times 1.25 \times 10^3 \times 3v_1^2$ $v_1^2 = \frac{3}{1.875 \times 10^3}$

$v_1 = \sqrt{\frac{3}{1.875 \times 10^3}}$

$v_1 \approx 0.0400 \mathrm{~m} \mathrm{~s}^{-1}$

Now, calculate the rate of flow of glycerin through the pipe (volume flow rate) using $v_1$ and $A_1$ :

$Q = A_1 v_1$

$Q = 10 \times 10^{-4} \times 0.0400$

$Q = 4 \times 10^{-5} \mathrm{~m}^{3} \mathrm{~s}^{-1}$

So, the rate of flow of glycerin through the pipe is $4 \times 10^{-5} \mathrm{~m}^{3} \mathrm{~s}^{-1}$ , and the value of $x$ is 4.