Question

The surface of water in a water tank of cross-section area \(750\) cm\(^2\) on the top of a house is \(h\) m above the tap level. The speed of water coming out through the tap of cross-section area \(500\) mm\(^2\) is \(30\) cm/s. At that instant, \(dh/dt\) is \(y\times 10^{-3}\) m/s. The value of \(y\) will be

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(5\)

Hint:

Use the equation of continuity: \(A_1\,dh/dt = A_2v\).

Answer 1

The Correct Option is A

Step 1: Plan:
Work in SI units from the start.

Step 2: Steps:
$A_1 = 750\times10^{-4} = 7.5\times10^{-2}$ m$^2$. $A_2 = 500\times10^{-6} = 5\times10^{-4}$ m$^2$. $v = 0.3$ m/s.
$\frac{dh}{dt} = \frac{5\times10^{-4}\times0.3}{7.5\times10^{-2}} = \frac{1.5\times10^{-4}}{7.5\times10^{-2}} = 2\times10^{-3}$ m/s. So $y = 2$.

Final Answer:
The value of $y$ is $2$, option (A). \[ \boxed{2} \]