Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
11.9k views
in Physics by (51.1k points)

The U-tube shown has a uniform cross section. A liquid is filled in the two arms up to heights h1 and h2, and then the liquid is allowed to move. Neglect viscosity and surface tension. When the levels equalize in the two arms, the liquid will

(a) be at rest

(b) be moving with an acceleration of g (h1-h2 /h1+h2+h)

(c) be moving with a velocity of (h1-h2(g/2(h1+h2+h))

(d) exert a net force to the right on the tube

1 Answer

+1 vote
by (50.3k points)
selected by
 
Best answer

Correct Answer is: (c) be moving with a velocity of (h1-h2(g/2(h1+h2+h))

When the levels equalize,

the height of the liquid in each arm = h1+h2/2

We may then visualize that a length h1 - h1+ h2 / 2 = h- h2 / 2 of the liquid has been transferred from the left arm to the right arm. Then,

mass of this liquid = (h1 - h2 / 2) aρ,

where, A = area of tube, ρ = density of the liquid.

Distance through which it moves down = h1 - h2 /2.

∴ loss in gravitational potential energy = (h1 - h2 / 2) Aρ.

The mass of the entire liquid = (h1 + h2 + h) Aρ. 

If this moves with a velocity v,

its kinetic energy = 1/2 (h1 + h2 + h) Aρv2

Equating energies, we get v. 

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...