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(C) Let the original water level be at line $D$. When oil is poured into the left arm,the water level in the right arm rises by $65\, mm$ to level $E$

Vedclass · Accepted Answer

$928$

Vedclass · Answer

(C) Let the original water level be at line $D$. When oil is poured into the left arm,the water level in the right arm rises by $65\, mm$ to level $E$. Consequently,the water level in the left arm drops by $65\, mm$ from the original level $D$ to level $B$.The height of the water column above the interface level $BC$ in the right arm is $h_{water} = 65\, mm + 65\, mm = 130\, mm = 0.13\, m$.The oil column height $h_{oil}$ is the distance from the top of the oil $A$ to the interface $B$. Since the top of the oil is $10\, mm$ above the water level $E$,the total height is $h_{oil} = 65\, mm + 65\, mm + 10\, mm = 140\, mm = 0.14\, m$.Equating the pressures at the interface level $BC$:$P_{atm} + ho_{oil} g h_{oil} = P_{atm} + ho_{water} g h_{water}$$ho_{oil} h_{oil} = ho_{water} h_{water}$$ho_{oil} = ho_{water} 	imes \frac{h_{water}}{h_{oil}}$Given $ho_{water} = 1000\, kg/m^3$:$ho_{oil} = 1000 	imes \frac{130}{140} = 1000 	imes \frac{13}{14} \approx 928.57\, kg/m^3$.Rounding to the nearest integer,we get $928\, kg/m^3$.

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