A current of 5A flows in a resistor of 2 ohms. Calculate the energy dissipated in 300 seconds in the resistor.

Options

A. `15KJ`
B. `15000kJ`
C. `1500J`
D. `150J`

Answer

A

Explanation

Due to rotation about axis OO. a force is acting on the liquid column AB onward, pushing up the liquid column on the right. The centrifugal force acting on the liquid column AB onwards, pushing up the liquid column on the right. The centrifugal force acting on the small element dx of area of cross section dx, distance x from the axis is

`dmomega^(2)x=Adxrhoomega^(2)x`
The total force due to column of length `L(=AB)` is
`=underset(0)overset(L)intArhoomega^(2).xdx`
`=Arhoomega^(2)(L^(2))/(2)`
Pressure at B = atmospheric pressure `+h_(1)rhog+(Arhoomega^(2))/(A).(L^(2))/(2)`
Pressure at B due to liquid column on the right hand side = atmospheric pressure `+h_(2)rhog`
Net pressure due to the left hand side = pressure at B due to liquid column
`therefore` atm pressure `+hrhog+rhoomega^(2)(L^(2))/(2)`
atm pressure `+h_(2)rhog`
`thereforeh_(2)-h_(1)=H_(0)=(omega^(2)L^(2))/(2g)`
According to this formula in our fig.

`H_(1)=(omega^(2)L_(1)^(2))/(2g),H_(2)=(omega^(2)L_(2)^(2))/(2g)`
As `omega` is the same, and `L_(2)gtL_(1),H_(2)gtH_(1)`
But in both A and B surface of water will go up.