$A$ fluid is flowing through a horizontal pipe of varying cross-section,with speed $v \ m/s$ at a point where the pressure is $P \ Pa$. At another point where pressure is $\frac{P}{2} \ Pa$,its speed is $V \ m/s$. If the density of the fluid is $\rho \ kg/m^3$ and the flow is streamline,then $V$ is equal to

  • A
    $\sqrt{\frac{P}{2\rho} + v^2}$
  • B
    $\sqrt{\frac{P}{\rho} + v^2}$
  • C
    $\sqrt{\frac{2P}{\rho} + v^2}$
  • D
    $\sqrt{\frac{P}{\rho} + v}$

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The platelets are drifting with the blood flowing in a streamline flow through a horizontal artery as shown below. The artery is contracted in region $II$. Choose the correct statement.

The figure shows a liquid of given density flowing steadily in a horizontal tube of varying cross-section. The cross-sectional areas at $A$ is $1.5 \, cm^2$,and at $B$ is $25 \, mm^2$. If the speed of the liquid at $B$ is $60 \, cm/s$,then find $(P_A - P_B)$ in $Pa$. (Given: $P_A$ and $P_B$ are liquid pressures at points $A$ and $B$ respectively. Density $\rho = 1000 \, kg/m^3$. $A$ and $B$ are on the axis of the tube.)

$A$ horizontal pipe carries water in a streamlined flow. At a point along the pipe,where the cross-sectional area is $10 \text{ cm}^2$,the velocity of water is $1 \text{ ms}^{-1}$ and the pressure is $2000 \text{ Pa}$. What is the pressure of water at another point where the cross-sectional area is $5 \text{ cm}^2$ (in $Pa$)? [Density of water $= 1000 \text{ kgm}^{-3}$]

$Assertion :$ The pressure of water increases when it flows from a narrow pipe to a wider pipe.
$Reason :$ According to the equation of continuity,$A_1v_1 = A_2v_2$,and by Bernoulli's principle,where velocity is high,pressure is low.

The area of cross-section of a large tank is $0.5 \; m^{2}$. It has a narrow opening near the bottom having an area of cross-section $1 \; cm^{2}$. $A$ load of $25 \; kg$ is applied on the water at the top in the tank. Neglecting the speed of water in the tank,the velocity of the water coming out of the opening at the time when the height of water level in the tank is $40 \; cm$ above the bottom will be $\dots \; cm \; s^{-1}$. [Take $g = 10 \; m \; s^{-2}$]

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