Two cylindrical vessels of equal crosssectional area 16 cm 2 contain water upto heights 100 cm and 150 cm respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process, is [Take density of water =103 kg / m 3 and .g =10 ms -2]

Question

Two cylindrical vessels of equal crosssectional area 16 cm 2 contain water upto heights 100 cm and 150 cm respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process, is [Take density of water =103 kg / m 3 and .g =10 ms -2] (A) 0.25 J (B) 1 J (C) 8 J (D) 12 J

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/dc898914644affe6c3b3c375338788e6-.png" /> h =( h 1+ h 2/2) Now, W = U i - U f W =(ρ Ah 1) g ( h 1/2)+(ρ Ah 2) g ( h 2/2)-ρ A ( h 1+ h 2) g (( h 1+ h 2/4)) W =(ρ Ag /2)[ h 12+ h 22-(( h 1+ h 2)2/2)] W =1 J

0.25 J

1 J

8 J

12 J

h=2h1​+h2​​ Now, W=Ui​−Uf​ W=(ρAh1​)g2h1​​+(ρAh2​)g2h2​​−ρ A(h1​+h2​)g(4h1​+h2​​) W=2ρAg​[h12​+h22​−2(h1​+h2​)2​] W=1J

$h = \frac{h _{1} + h _{2}}{2}$
Now,
$W = U_{i} - U_{f}$
$W = (ρ A h_{1}) g \frac{h _{1}}{2} + (ρ A h_{2}) g \frac{h _{2}}{2} - ρ$
$A (h_{1} + h_{2}) g (\frac{h _{1} + h _{2}}{4})$
$W = \frac{ρ A g}{2} [h_{1}^{2} + h_{2}^{2} - \frac{( h _{1} + h _{2} ) ^{2}}{2}]$
$W = 1 J$