Water is falling down at the rate of 1.8 Kg per minute from a vertical tube. The radius at the bottom of the tube is 0.004 m and pressure around it is 0.76m of Hg. The diameter of the tube at a height of 0.5 m from the bottom is 0.005 m. find the pressure at that point.

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0.724 m of Hg

0.824 m of Hg

0.924 m of Hg

0.624 m of Hg

answer is A.

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Detailed Solution

p₂ = 0.76 m of Hg = 1.013 x 10⁵ N/m²
$large frac mt=Avrho=pi r^2rho v$
$large therefore v_1=frac {(1.8/60)}{(3.14)left [ frac 52times10^{-3} right ]^2times 10^3}=1.5ms^{-1}$
$large A_1v_1=A_2v_2Rightarrow v_2=left ( frac {r_1}{r_2} right )^2 v_1$
$large p_1+frac 12rho v_1^2+rho gh=p_2+frac 12rho v_2^2$

$large p_1=p_2+frac 12rho(v_2^2-v_1^2)-rho gh$
$large =p_2+frac 12rho v_1^2left [ left (frac {r_1}{r_2} right )^2 -1right ]-rho gh$
$large =[1.013times 10^{5}]+frac 12times 10^3times 2.25left [ left (frac {0.005/2}{0.004} right )^2 -1right ]-10^3times 9.8times 0.5=0.72m;of; Hg$