An inclined plane of length 5.60 m making an angle of \(45^{\circ}\) with the horizontal is placed in an 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 slops. It the coefficient of friction is 0.1 the time taken by the particle to reach the bottom is _________. (Charge on of particle = \(10^{-2} C\))

(B)
\(\begin{aligned} \theta & =45^{\circ} & & d=5.60 m \\ E & =100 \frac{V}{ m } & & m =1 kg \\ \mu & =0.1 & & t=? \\ q & =10^{-2} C & & \end{aligned}\)

from figure,
\(N=m g \cos 45^{\circ}+q E \sin 45^{\circ}\)
\(\begin{aligned} \therefore N=1 \times 10 \times \frac{1}{\sqrt{2}} & +10^{-2} \\ & \times 100 \times \frac{1}{\sqrt{2}}\end{aligned}\)
\(\begin{aligned} \therefore N & =\frac{10}{\sqrt{2}}+\frac{1}{\sqrt{2}} \\ N & =7.77 N\end{aligned}\)
\(\begin{array}{r} F =m g \sin 45^{\circ}-q E \cos 45^{\circ}-\mu N \\ F =1 \times 10 \times \frac{1}{\sqrt{2}}-\frac{10^{-2} \times 100}{\sqrt{2}} \\ -0.1 \times 7.77\end{array}\)
\(\begin{aligned} F & =\frac{10}{\sqrt{2}}-\frac{1}{\sqrt{2}}-0.777 \\ F & =5.58 N \\ & \approx 5.6 N\end{aligned}\)
\(\begin{array}{l} F =m a \\ a=\frac{ F }{m}=\frac{5.6}{1} \\ a=5.6 ms^{-2}\end{array}\)
Now, \(d=v_0 t+\frac{1}{2} a t^2\)
\(\begin{aligned} \therefore & & d & =\frac{1}{2} a t^2 \quad\left(\because v_0=0\right) \\ \therefore & & t^2 & =\frac{2 d}{a} \\ \therefore & & t & =\sqrt{\frac{2 \times 5.6}{5.6}} \\ & & t & =\sqrt{2}=1.41 s\end{aligned}\)