AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A large container (with open top) of negligible mass and uniform cross-sectional area $A$ has a small hole of cross-sectional area $a$ in its side wall near the bottom. The container is keep on a smooth horizontal platform and contains a liquid of density $ρ$ and mass $m$ . If the liquid starts flowing out of the hole at time $t = 0$ , the velocity of the liquid when $75 %$ of the liquid has drained out is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\sqrt{\frac{3 m g}{4 A ρ}}$

$\sqrt{\frac{2 m g}{A ρ}}$

$2 \sqrt{\frac{m g}{A ρ}}$

$\sqrt{\frac{m g}{2 A ρ}}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

When 75% of the liquid has drained out, the height of the liquid in the container will be h' = h/4. For this height,

Velocity of liquid flowing out (v')

$= \sqrt{2 g h'} = \sqrt{2 g \times \frac{h}{4}} = \sqrt{\frac{g h}{2}}$

Now, h = $\frac{m}{A ρ}$ .Hence

$v' = \sqrt{\frac{g m}{2 A ρ}}$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

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

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!