Glycerine of density 1.25 times 10^3 , kg , m | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given: Density $ho = 1.25 	imes 10^3 \, kg \, m^{-3}$, $A_1 = 10 \, cm^2 = 10 	imes 10^{-4} \, m^2$, $A_2 = 5 \, cm^2 = 5 	imes 10^{-4} \, m^

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Given: Density $ho = 1.25 	imes 10^3 \, kg \, m^{-3}$, $A_1 = 10 \, cm^2 = 10 	imes 10^{-4} \, m^2$, $A_2 = 5 \, cm^2 = 5 	imes 10^{-4} \, m^2$, $\Delta P = P_1 - P_2 = 3 \, N \, m^{-2}$.By the equation of continuity, $A_1 v_1 = A_2 v_2$.Therefore, $v_1 = \frac{A_2}{A_1} v_2 = \frac{5}{10} v_2 = 0.5 v_2$.Using Bernoulli's equation for horizontal flow: $P_1 + \frac{1}{2} ho v_1^2 = P_2 + \frac{1}{2} ho v_2^2$.$\Delta P = P_1 - P_2 = \frac{1}{2} ho (v_2^2 - v_1^2)$.Substituting $v_1 = 0.5 v_2$:$3 = \frac{1}{2} 	imes (1.25 	imes 10^3) 	imes (v_2^2 - (0.5 v_2)^2)$.$3 = 0.625 	imes 10^3 	imes (v_2^2 - 0.25 v_2^2) = 0.625 	imes 10^3 	imes 0.75 v_2^2$.$3 = 468.75 v_2^2$.$v_2^2 = \frac{3}{468.75} = 0.0064$.$v_2 = \sqrt{0.0064} = 0.08 \, m \, s^{-1}$.The rate of flow (discharge) $Q = A_2 v_2 = (5 	imes 10^{-4} \, m^2) 	imes (0.08 \, m \, s^{-1}) = 40 	imes 10^{-6} \, m^3 \, s^{-1} = 4 	imes 10^{-5} \, m^3 \, s^{-1}$.Comparing with $x 	imes 10^{-5} \, m^3 \, s^{-1}$, we get $x = 4$.

Similar Questions

In this figure,an ideal liquid flows through a tube of uniform cross-section. The liquid has velocities $v_A$ and $v_B$,and pressures $P_A$ and $P_B$ at points $A$ and $B$ respectively.

$A$ wind with speed $40 \ m/s$ blows parallel to the roof of a house. The area of the roof is $250 \ m^2$. Assuming that the pressure inside the house is atmospheric pressure,the force exerted by the wind on the roof and the direction of the force will be $(\rho_{air} = 1.2 \ kg/m^3)$.

Water is pumped through the hose shown below, from a lower level to an upper level. Compared to the water at point $1$, the water at point $2$:

The Bernoulli's equation is given by $p + \frac{1}{2} \rho v^{2} + h \rho g = k$,where $p =$ pressure,$\rho =$ density,$v =$ speed,$h =$ height of the liquid column,$g =$ acceleration due to gravity,and $k$ is a constant. The dimensional formula for $k$ is the same as that for:

The velocity of kerosene in a horizontal pipe is $5 \ m/s$. What is the velocity head in $m$? (Take $g = 10 \ m/s^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module