Question

A large tank is filled with water $(\text{density}={10}^3{\text{kg/m}}^3).$ A small hole is made at a depth of $10\text{m}$ below water surface. The range of water coming out of the hole is $R$ on the ground. What extra pressure must be applied on the water surface so that the range become $2R$ ?
$(p_0=1\text{atm}={10}^5\text{Pa}$ and $g=10{\text{m/s}}^2)$

• A. $1\text{atm}$
• B. $2\text{atm}$
• C. $4\text{atm}$
• D. $3\text{atm}$

Answer 1

The correct option is D $3\text{atm}$

Let us suppose extra pressure applied is $p$.
According to question -


For case 1 :
Velocity of efflux,
$v_1=\sqrt{2gh}=\sqrt{2\times 10\times 10}$
$=10\sqrt{2}\text{m/s}$ ........(1)
Range of efflux,
$R=v_{1}t=10\sqrt{2}t$ .....(2) $[$ from (1) $]$
For case 2 :
Range of efflux,
$R_2=2R=20\sqrt{2}t$ $[$ from (2) $]$
$\Rightarrow v_{2}t=20\sqrt{2}t$
$[$ time $t$ will be same as height of the hole is fixed $]$
$\Rightarrow v_2=20\sqrt{2}\text{m/s}$ .......(3)
On applying Bernoulli's Equation for second case,
$p_0+p+\rho gh=p_0+\frac{1}{2}\rho v_2^2$
$\Rightarrow p=\frac{1}{2}\rho v_2^2-\rho gh$
$\Rightarrow p=\frac{1}{2}\times 1000\times (20\sqrt{2}{)}^2-1000\times 10\times 10$
$[$ from (3) $]$
$\Rightarrow p=3\times {10}^5{\text{N/m}}^2=3\text{atm}$
$[1\text{atm}={10}^5{\text{N/m}}^2]$
Extra pressure of $3\text{atm}$ has to be applied to make the range of efflux double.