Question

Q. Two cylindrical vessels of equal cross-sectional area A contain water up to heights h1 and h2. The vessels are interconnected so that the levels in them are equal. Calculate the work done by the force of gravity during the process. The density of water is p.
​

Answer 1

Answer:

Question:-

Two cylindrical vessels of equal cross-sectional area A contain water up to heights h1 and h2. The vessels are interconnected so that the levels in them are equal. Calculate the work done by the force of gravity during the process. The density of water is p.

Answer:-

P.E. of liquid in cylinder 1

U1 (m) g h1/2= (p×A×h1) g h1/2 = pAgh21/2

Note:-

[The total mass can be supposed to be concentrated at the center of the filled part which will be at height h1/2]

Similarly P.E. of liquid in cylinder 2 U2 = pAgh22/2

Total P.E. initially U = U1+ U2 (h21 + h22)

After the equalising of levels.

P.E. of liquid in cylinder 1 U1= mg h/2= pAg/2 h2

P.E. of liquid in cylinder 2 U2' = mg h/2 = PAg/2 h2

Total P.E. finally U' = U1' +U2' = PAgh2

The change in P.E.

AU = U-U'= pAg [h21/2 + h22/2- h2]

Total volume remains the same.

Ah1+ Ah2 = Ah + Ah

⇒h = h1+ h2/2

Therefore,

ΔU PAg [h21/2+ h22/2- (h1+ h2/ 2)2]

=PAg/4(h1- h2)2

This change in P.E. is the work done by

gravity.

Explanation:

$Hope \: it \: helps.$

Answer 2

★Given:-

• Two cylindrical vessels are of equal cross-sectional area.
• They contain water up to heights h₁ and h₂.
• The vessels are interconnected so that the levels in them are equal.
• The density of water is р.

★To find:-

• The work done by the force of gravity during the process.

★Solution:-

We know,

✦Mass = density × volume

Also,

✦Volume = area × length

Therefore,

➜m = p.v

➜m = p.A.h____(1)

The total volume of water is constant.

Hence,after interconnection the height of water in each vessel becomes (h₁ + h₂)/2.

Total change in length:-

➜h = h₁-(h₁+h₂)/2

➜h = (2h₁-h₁-h₂)/2

➜h = (h₁ - h₂)/2

In one vessel the water level falls to (h₁ - h₂)/2 and in other vessel the water level rises to (h₁ - h₂)/2.

Substituting the value of h in equation(1),

Mass of water = p.A.(h₁ - h₂)/2  

We know:

✦Work done by gravity = mgh

Where we have,

• m = p.A.(h₁ - h₂)/2  
• g = gravity constant
• h =(h₁ - h₂)/2  

Putting values,

➜W = mgh

➜{p.A.(h₁ - h₂)/2}g{(h₁ - h₂)/2}

➜A.p.g((h₁ - h₂)/2)²

Hence, the work done by the force of gravity during the process is

A.p.g((h₁ - h₂)/2)².

_______________