A cylindrical tank has a hole of area 2 `cm^(2)` at its bottom, if water is poured into the tank from a tube above it at the rate of 100 `cm^(3) s^(-1)`, then the maximum height upto
(Acceleration due to gravity, g = 10 `ms^(-2)`)

Options

A. 2.5 cm
B. 1.25 cm
C. 5.5 cm
D. 3.5 cm

Answer

B

Explanation

Given, area of hole in tank, A = `2cm^(-2)`
`rArr " "A = 2 x 10^(-4) m^(2)`,
Volume flow rate = 100 `cm^(2)`/sec
= 100 `xx 10^(-6) m^(3)//"sec" = 10^(-4) m^(2) sec^(-1)`
At maximum height h velocity of water flowing through hoel, v = `sqrt(2gh)`
`therefore` From principle of continuity of flow of liquid, volume flow rate = Av
`10^(-2) = Av = 2xx 10^(-4) sqrt(2gh) ( because v = sqrt(gh))`
`rArr" " 10^(-4) = 2 xx 10^(-4) sqrt(2gh)`
`rArr " " (10^(4))/(2 xx 10^(-4)) = sqrt(2gh)`
`rArr " " (1)/(2) = sqrt(2gh)`
`rArr " " ((1)/(2))^(2) = 2gh`
`rArr " " (1)/(4) xx (1)/(2g) = h`
`rArr " " h = (1)/(80) = 0.125` m
`rArr " " h = 1.25` cm