A U-tube having a horizontal arm of length 20 | vedclass

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(D) The total length of the water column is $l_w = \frac{V}{A} = \frac{60 \ cm^3}{1 \ cm^2} = 60 \ cm$.When the liquid of density $\rho_l = 4 \ g/cc$

Vedclass · Accepted Answer

$35$

Vedclass · Answer

(D) The total length of the water column is $l_w = \frac{V}{A} = \frac{60 \ cm^3}{1 \ cm^2} = 60 \ cm$.When the liquid of density $\rho_l = 4 \ g/cc$ is poured,it pushes the water out of the horizontal arm.Let the height of the liquid column be $h$. The pressure at the interface of the liquid and water in the $U$-tube must be equal at the same horizontal level.The pressure exerted by the liquid column of height $h$ must balance the pressure exerted by the remaining water column of height $H = l_w - l_{horizontal} = 60 \ cm - 20 \ cm = 40 \ cm$.Using the hydrostatic pressure balance equation: $\rho_l \cdot g \cdot h = \rho_w \cdot g \cdot H$.Given $\rho_l = 4 \ g/cc$,$\rho_w = 1 \ g/cc$,and $H = 40 \ cm$:$4 \cdot h = 1 \cdot 40 \Rightarrow h = 10 \ cm$.The total length of the liquid column in the tube is $l_l = h + l_{horizontal} = 10 \ cm + 20 \ cm = 35 \ cm$.The volume of the liquid required is $V_L = l_l \cdot A = 35 \ cm \cdot 1 \ cm^2 = 35 \ cc$.

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