Glycerin of density $1.25 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ is flowing through the conical section of pipe The area of cross-section of the pipe at its ends are $10 \mathrm{~cm}^{2}$ and $5 \mathrm{~cm}^{2}$ and pressure drop across its length is $3 ~\mathrm{Nm}^{-2}$. The rate of flow of glycerin through the pipe is $x \times 10^{-5} \mathrm{~m}^{3} \mathrm{~s}^{-1}$. The value of $x$ is ___________.

<p>We can use the Bernoulli equation and continuity equation to solve this problem. The Bernoulli equation is given by:</p> <p>$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$</p> <p>The continuity equation is given by:</p> <p>$A_1 v_1 = A_2 v_2$</p> <p>From the given data, we have:</p> <p>$P_1 - P_2 = 3 \mathrm{~Nm}^{-2}$ $A_1 = 10 \mathrm{~cm}^2 = 10 \times 10^{-4} \mathrm{~m}^2$ $A_2 = 5 \mathrm{~cm}^2 = 5 \times 10^{-4} \mathrm{~m}^2$ $\rho = 1.25 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$</p> <p>Rearrange the continuity equation to solve for $v_2$:</p> <p>$v_2 = \frac{A_1}{A_2} v_1 = 2v_1$</p> <p>Substitute $v_2$ and rearrange the Bernoulli equation:</p> <p>$P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2)$<br/><br/> $3 = \frac{1}{2} \times 1.25 \times 10^3 (4v_1^2 - v_1^2)$</p> <p>Now, solve for $v_1$:</p> <p>$3 = \frac{1}{2} \times 1.25 \times 10^3 \times 3v_1^2$ $v_1^2 = \frac{3}{1.875 \times 10^3}$<br/><br/> $v_1 = \sqrt{\frac{3}{1.875 \times 10^3}}$<br/><br/> $v_1 \approx 0.0400 \mathrm{~m} \mathrm{~s}^{-1}$</p> <p>Now, calculate the rate of flow of glycerin through the pipe (volume flow rate) using $v_1$ and $A_1$:</p> <p>$Q = A_1 v_1$<br/><br/> $Q = 10 \times 10^{-4} \times 0.0400$<br/><br/> $Q = 4 \times 10^{-5} \mathrm{~m}^{3} \mathrm{~s}^{-1}$</p> <p>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.</p>