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
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
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Use the equation of continuity: \(A_1\,dh/dt = A_2v\).
Updated On: Oct 1, 2026
- \(2\)
- \(3\)
- \(4\)
- \(5\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
The volume leaving through the tap per second equals the volume lost from the tank per second: \(A_1\left|\frac{dh}{dt}\right| = A_2v\).
Step 2: Convert units:
Tank area \(A_1 = 750\) cm\(^2\). Tap area \(A_2 = 500\) mm\(^2 = 5\) cm\(^2\). Speed \(v = 30\) cm/s.
Step 3: Calculate:
\[ \left|\frac{dh}{dt}\right| = \frac{5\times30}{750} = 0.2\ \text{cm/s} = 2\times10^{-3}\ \text{m/s} \]
So \(y = 2\).
Final Answer:
The value of \(y\) is \(2\), option (A). \[ \boxed{2} \]
The volume leaving through the tap per second equals the volume lost from the tank per second: \(A_1\left|\frac{dh}{dt}\right| = A_2v\).
Step 2: Convert units:
Tank area \(A_1 = 750\) cm\(^2\). Tap area \(A_2 = 500\) mm\(^2 = 5\) cm\(^2\). Speed \(v = 30\) cm/s.
Step 3: Calculate:
\[ \left|\frac{dh}{dt}\right| = \frac{5\times30}{750} = 0.2\ \text{cm/s} = 2\times10^{-3}\ \text{m/s} \]
So \(y = 2\).
Final Answer:
The value of \(y\) is \(2\), option (A). \[ \boxed{2} \]
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