Question 136: Properties of Matter

Question

Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density
d. The area of the base of both vessels is S but the height of liquid in one vessel is x1
 and in the
other, x2
. When both cylinders are connected through a pipe of negligible volume very close to the
bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height.
The change in energy of the system in the process is:

Accepted Answer

$${u_i} = \left[ {dS{x_1}.{{{x_1}} \over 2} + dS{x_2}.{{{x_2}} \over 2}} ight]g\left\{ {dS{x_1} 	o m,\,{{{x_1}} \over 2} 	o h(C.O.M)} ight\}$$
Here total volume remains same.
$$ 	herefore $$ V
i
= V
f
$$ \Rightarrow $$ S(x
1
+ x
2
) = S(h + h)
$$ \Rightarrow $$ h = $${{{x_1} + {x_2}} \over 2}$$
u
f
= (dSh)g$${h \over 2} 	imes 2$$
$$ \Rightarrow $$ $${u_f} = \left[ {dS\left( {{{{x_1} + {x_2}} \over 2}} ight) 	imes \left( {{{{x_1} + {x_2}} \over 4}} ight) 	imes 2} ight]g$$
$$ 	herefore $$ $${u_i} - {u_f} = dsg\left[ {{{x_1^2} \over 2} + {{x_2^2} \over 2} - {{{{\left( {{x_1} + {x_2}} ight)}^2}} \over 4}} ight]$$
$$ = dsg{{{{\left( {{x_1} - {x_2}} ight)}^2}} \over 4}$$

Answer

gdS(x2 + x1)2

Answer

gdS$$\left( {x_2^2 + x_1^2} \right)$$

Answer

$${1 \over 4}gdS{\left( {{x_2} - {x_1}} \right)^2}$$

Answer

$${3 \over 4}gdS{\left( {{x_2} - {x_1}} \right)^2}$$

Properties of Matter

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