![JEE Main 2014 Physics, Mechanical Properties of Fluids — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Mechanical_Properties_of_Fluids/648b5a64417cc3fb48d65e90/question_1__q_648b5a64417cc3fb48d65e90__cdn-question-pool.getmarks.app__ca7dd333-725b-4d4a-9c45-f75dbb44784b-image__61bd9f2c2a_final_ppt_sync.png) In the diagram shown, the difference in the two tubes of the manometer is $5\mathrm{cm}$, the cross-section of the tube at $A$ and $B$ is $6{\mathrm{mm}}^{2}$ and $10{\mathrm{mm}}^{2}$ respectively. The rate at which water flows through the tube is $(g=10m{s}^{-2})$

$6{v}_{A}=10{v}_{B}......(1)$ $\frac{1}{2}\rho {v}_{A}^{2}-\rho gh=\frac{1}{2}\rho {v}_{B}^{2}$ .........(2) $\Rightarrow {v}_{A}=\frac{1}{0.8}m{s}^{-1}$ $\frac{dV}{dt}=A{v}_{A}$ $=6\times {10}^{-2}{cm}^{2}\times \frac{100}{0.8}\mathrm{cm}{s}^{-1}=7.5\mathrm{cc}{s}^{-1}$ OR By the concept of the venturi meter tube rate of flow$=Q={A}_{1}{A}_{2}\sqrt{\frac{2g({h}_{1}-{h}_{2})}{({A}_{2}^{2}-{A}_{1}^{2})}}$ $=6\times 10\times {10}^{-4}\sqrt{\frac{2\times 100\times 5}{(-36+100)\times {10}^{-2}}}$ $=7.5\mathrm{cc}{s}^{-1}$

JEE Main 2014 Physics, Mechanical… - JEE Main 2014 Physics PYQ

JEE Main 2014 — Physics Mechanics

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JEE Main 2014 Physics, Mechanical Properties of Fluids — question figure

In the diagram shown, the difference in the two tubes of the manometer is $5\mathrm{cm}$ , the cross-section of the tube at $A$ and $B$ is $6{\mathrm{mm}}^{2}$ and $10{\mathrm{mm}}^{2}$ respectively. The rate at which water flows through the tube is $(g=10m{s}^{-2})$

Official previous-year question

Held on 19 Apr 2014 · Verified 6 Jul 2026.

Options

A
$7.5\mathrm{cc}{s}^{-1}$
B
$12.5\mathrm{cc}{s}^{-1}$
C
$8.0\mathrm{cc}{s}^{-1}$
D
$10.0\mathrm{cc}{s}^{-1}$

Did you get this right?

Solution

$6{v}_{A}=10{v}_{B}......(1)$

$\frac{1}{2}\rho {v}_{A}^{2}-\rho gh=\frac{1}{2}\rho {v}_{B}^{2}$ .........(2)

$\Rightarrow {v}_{A}=\frac{1}{0.8}m{s}^{-1}$

$\frac{dV}{dt}=A{v}_{A}$

$=6\times {10}^{-2}{cm}^{2}\times \frac{100}{0.8}\mathrm{cm}{s}^{-1}=7.5\mathrm{cc}{s}^{-1}$

By the concept of the venturi meter tube

rate of flow $=Q={A}_{1}{A}_{2}\sqrt{\frac{2g({h}_{1}-{h}_{2})}{({A}_{2}^{2}-{A}_{1}^{2})}}$

$=6\times 10\times {10}^{-4}\sqrt{\frac{2\times 100\times 5}{(-36+100)\times {10}^{-2}}}$

$=7.5\mathrm{cc}{s}^{-1}$

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