Question

A large tank is filled with water (density = $10^{3}kg/m^3$ ). A small hole is made at a depth $10m$ below water surface. The range of water issuing out of the hole is $R$ on ground. What extra pressure must be applied on the water surface so that the range becomes $2R$
(take 1 atm $=10^{5}Pa$ and $g=10m/s^2$ )

• A. 9 atm
• B. 4 atm
• C. 5 atm
• D. 3 atm

Answer 1

Answer: (D)

Range $=V\sqrt{\frac{2h}{g}}$
To double the range we have to double the velocity of efflux From Bernoulli's theorem
$P_1+\frac{1}{2}\rho V_1^2+\rho gH$
$=P_2+\frac{1}{2}\rho V_2^2+\rho g(H-h)$
$P_1-P_2=\frac{1}{2}\rho \left(V_2^2-V_1^2\right)-\rho gh$
$($ As $V_2>>V_1)\&V_2=2V$
$=\frac{1}{2}10^3\left((2\sqrt{2g.10}{)}^2\right)-10^3\times 10\times 10$
$=2\times 2\times g\times 10\times 10^3-10\times 10\times 10^3$
$=3\times 10^{5}pa$
$=3$ atm $\left(1atm=10^{5}Pa\right)$