Correct Bernoulli's equation is (symbols have their usual meaning) :
$P+m g h+\frac{1}{2} m v^2=$ constant
$P+\rho g h+\frac{1}{2} \rho v^2=$ constant
$P+\rho g h+\rho v^2=$ constant
$P+\frac{1}{2} \rho g h+\frac{1}{2} \rho v^2=$ constant
Water is flowing steadily through a horizontal tube of non uniform cross-section. If the pressure of water is $4$ $\times $ $10^4$ $N/m^2$ at a point where cross-section is $0.02$ $m^2$ and velocity of flow is $2$ $m/s$, what is pressure at a point where cross-section reduces to $0.01$ $m^2$.
Different heads are in Column - $\mathrm{I}$ and its formulas are given in Column - $\mathrm{II}$. Match them appropriately.
Column - $\mathrm{I}$ | Column - $\mathrm{II}$ |
$(a)$ Velocity head | $(i)$ $\frac{P}{{\rho g}}$ |
$(b)$ Pressure head | $(ii)$ $h$ |
$(iii)$ $\frac{{{v^2}}}{{2g}}$ |
A wooden block of volume $1000\, cm^3$ is suspended from a spring balance. It weighs $12$ $N$ in air. It is suspended in water such that half of the block is below the surface of water. The reading of the spring balance is ...... $N$
Water flows steadily through a horizontal pipe of variable cross-section. If the pressure of water is $P$ at a point where flow speed is $v$ , the pressure at another point where the flow speed is $2v$ , is (Take density of water as $\rho $ )
Prove Bernoulli’s Principle.