The area of cross-section of a large tank is $0.5 m^{2}$ , it has a narrow opening near the bottom having area of cross-section $1 {cm}^{2}$ . A load of $25 kg$ is applied on the water at the top in the tank. Neglecting the speed of water in the tank, the velocity of water, coming out of the opening at the time when water level in the tank is $40 cm$ above the bottom, will be ${cms}^{- 1} [Take g = 10 {ms}^{- 2}]$

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 300.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Using Bernoulli's principle :

$P_{A} + \frac{1}{2} {ρV}_{A}^{2} = P_{B} + \frac{1}{2} ρ {V_{B}}^{2}$

$P_{0} + \frac{250}{0.5} + (0.4 \times 1000 \times 10) + \frac{1}{2} {ρ (0)}^{2} = P_{0} + \frac{1}{2} ρ {V_{B}}^{2}$

$4500 = \frac{1}{2} (1000) {V_{B}}^{2}$

$V_{B} = 3 m / s = 300 cm / s$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE