Consider a completely full cylindrical water tank of height 1.6 m and cross-sectional area 0.5 ~m^2 . It has a small hole in its side at a height 90 cm from the bottom. Assume, the cross-sectional area of the hole to be

Correct answer: D. 4 ~m / s Apply Bernouli equation between points 1 \& 2 aligned & P_1+(1)/(2) v_1^2+ g h=P_2+(1)/(2) v_2^2+0 & P_0+ mg ~A + g (70)/(100)=P_0+(1)/(2) v_2^2 & (5000)/(0.5)+10^3 × 10 (70)/(100)=(1)/(2) × 10^3 v_2^2 aligned aligned & 10^3+10^3 × 7=(10^3)/(2) v_2^2 & v_2^2=16 & v_2=4 ~m / s aligned As the tank area is large v_1 is negligible compared to v_2

JEE Main 20253 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Morning Shift
Subject: Physics
Chapter: Mechanical Properties of Fluids

JEE Main 2025 · 3 April · Morning Shift · Physics · Mechanical Properties of Fluids

Consider a completely full cylindrical water tank of height 1.6 m...

Consider a completely full cylindrical water tank of height 1.6 m and cross-sectional area

0.5 \mathrm{~m}^2

. It has a small hole in its side at a height 90 cm from the bottom. Assume, the cross-sectional area of the hole to be negligibly small as compared to that of the water tank. If a load 50 kg is applied at the top surface of the water in the tank then the velocity of the water coming out at the instant when the hole is opened is :

\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)

Consider a completely full cylindrical water tank of height 1.6 m and cross-sectional area 0.5 ~m^2 . It has a small

Options

A $3 \mathrm{~m} / \mathrm{s}$
B $5 \mathrm{~m} / \mathrm{s}$
C $2 \mathrm{~m} / \mathrm{s}$
D $4 \mathrm{~m} / \mathrm{s}$

Correct answer

D. $4 \mathrm{~m} / \mathrm{s}$

Step-by-step solution

Apply Bernouli equation between points

1 \& 2

\begin{aligned}<br/>& P_1+\frac{1}{2} \rho v_1^2+\rho g h=P_2+\frac{1}{2} \rho v_2^2+0 \\ & P_0+\frac{\mathrm{mg}}{\mathrm{~A}}+\rho g \frac{70}{100}=P_0+\frac{1}{2} \rho v_2^2 \\ & \frac{5000}{0.5}+10^3 \times 10 \frac{70}{100}=\frac{1}{2} \times 10^3 v_2^2<br/>\end{aligned}

\begin{aligned}<br/>& 10^3+10^3 \times 7=\frac{10^3}{2} \mathrm{v}_2^2 \\ & \mathrm{v}_2^2=16 \\ & \mathrm{v}_2=4 \mathrm{~m} / \mathrm{s}<br/>\end{aligned}

As the tank area is large

\mathrm{v}_1

is negligible compared to

\mathrm{v}_2

Practice Mechanical Properties of Fluids on Quantrex Academy →

More from Mechanical Properties of Fluids

Passage: A container of height 2 m, length 2 m and breadth 1 m is made of insulating… JEE Advanced 2026 A liquid of density 600 kg/m ^3 flowing steadily in a tube of varying cross-section. The… JEE Main 2026 A spherical liquid drop of radius R acquires the terminal velocity v_1 when falls through… JEE Main 2026 A cylindrical vessel of 40 cm radius is completely filled with water and its capacity is… JEE Main 2026 Eight mercury drops, each of radius r , coalesce to form a bigger drop. The surface… JEE Main 2026 A water spray gun is attached to a hose of cross sectional area 30 cm ^2 . The gun… JEE Main 2026 The surface tension of a soap solution is 3.5 × 10^ -2 N/m. The work required to increase… JEE Main 2026 The surface tension of a soap bubble is 0.03 N/m. The work done in increasing the… JEE Main 2026 Full Mechanical Properties of Fluids list All JEE Main PYQs