Question

A large tank is filled with water (density = $10^{3}\, kg / m ^{3}$ ). A small hole is made at a depth $10\, m$ 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 $g =10\, m / s ^{2}$ )

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

Answer 1

Answer: (D)

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