A block of mass \(m\) is placed on a smooth wedge of inclination \(\theta\). The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block ((\(g\)) is acceleration due to gravity) will be

According to diagram,
There is no relative motion between and block and wedge,
\(ma \cos⁡\theta=mg \sin⁡\theta\)
Therefore,
\(a=g \tan⁡\theta\)
Now, force exerted by wedge on the block is,
\(N=ma \cos⁡\theta+mg \sin⁡\theta\)
Put the value of a in above equation,
\(N=mg \cos⁡\theta+m(g \tan⁡\theta)\sin⁡\theta\)
\(N=mg \cos⁡\theta+m(g \dfrac{\sin⁡\theta}{\cos⁡\theta}) \sin⁡\theta\)
On solving above equation, we get
\(N=mg\dfrac{\cos^2⁡\theta+\sin^2⁡\theta}{\cos⁡\theta}\)
Hence,
\(N=\dfrac{mg}{\cos\theta}\)⁡
\(\therefore \cos ^2\theta + \sin ^2\theta =1\)

Final answer: (d)