An inclined plane of length `5.60` m making an angle of `45^(@)` with the horizontal is placed in a uniform electric field `E=100 Vm^(-1)`. A particle of mass `1 kg` and charge `10^(-2)C` is allowed to slide down from rest position from maximum height of slope. If the coefficient of friction is 0.1, then the time taken by the particle to reach the bottom is

Options

A. 1 s
B. 1.41 s
C. 2 s
D. None of these

Answer

B

Explanation

`mg sin 45^(@) - f - q E sin 45^(@)` = ma
Since f = `mu N`
mg `sin 45^(@) - mu N - q E sin 45^(@) =` ma ….. (i)
and `N = mg cos 45^(@) + q E cos 45^(@)` ….. (ii)
By eq. (i) and (ii)
`ma = mg sin 45^(@) - mu mg cos 45^(@) - mu g E cos 45^(@) - q E sin 45^(@)`
ma = mg `(1)/(sqrt2) - mu mg xx (1)/(sqrt2) - q E mu xx (1)/(sqrt2) - q E xx (1)/(sqrt2)`
`implies a = (9.8)/(sqrt2) - (0.98)/(sqrt2) - (0.1)/(sqrt2) - (1)/(sqrt2)`
`a = (9.8 - 0.98)/(sqrt2) - ((1 + 0.1)/(sqrt2))`
`= 6.24 - 0.78 = 5.46 m//s^(2)`

Now , `S = ut + (1)/(2) at^(2) implies 5.6 = (1)/(2) xx 5.46 xx t^(2)`
`implies t^(2) ~~ 2 implies t = sqrt2 = 1.41` sec