A large tank is filled with water (density `= 10^(3) kg//m^(3)`). A small hole is made at a depth of 10 m below water surface. The range of water issuing out of the hole is R on ground. What extra pressure (in atm) 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)`)

Options

A. 1 atm
B. 2 atm
C. 4 atm
D. 3 atm

Answer

D

Explanation

`Deltap=(1)/(2)rho v^(2) " or" v=sqrt((2(Deltap))/(rho))`

R=vt or `R prop sqrt(Deltap)`
To make R two times `Deltap`should be 4times. Initially
`Deltap=rho_(w)gh`
`=10^(3)xx10xx10`
`10^(5) Nm^(-2)=1 "atm"`
In second case, `rho_(w)gh+` extra pressure `=4(Deltap)=4` atm
or 1 atm+extra pressure =4atm
`therefore` Extra pressure =3 atm.