wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A container filled with liquid up to height h is place on a smooth horizontal surface. The container is having a small hole at the bottom. As the liquid comes out from the hole, the container moves in a backward direction with acceleration a and finally, when all the liquid is drained out, it acquires a velocity v. Neglect mass of the container. In this case
119627.png

A
both a and v depend on h
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
only a depends on h
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
only v depends on h
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
neither a nor v depends on h
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C only v depends on h
The velocity with which the liquid comes out is v0=2gh.
Let cross-sectional area of the hole be A. Then the force exerted by the ejecting fluid due to change in momentum on the container is F=ρAv2
F=ρAv2=ma, where m is the mass of the liquid inside the container.
ρAv2=ρAh×a
a=2g
Differentiating the above equation, we can have v=2gt where t is the time in which the liquid comes out completely which is dependent on h.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Bernoulli's Principle
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon