NEET · Physics · STD 11 - 4.1 newtons laws of motion
A block of mass \(m\) is placed on a smooth inclined wedge \(ABC\) of inclination \(\theta\) as shown in the figure. The wedge is given an acceleration \(a\) towards the right. The relation between \(a\) and \(\theta\) for the block to remain stationary on the wedge is

- A \(a = \frac{g}{{cosec\theta }}\)
- B \(a = \frac{g}{{sin\theta }}\)
- C \(a=g tan\)\(\;\theta \)
- D \(a=g cos \) \(\theta \)
Answer & Solution
Correct Answer
(C) \(a=g tan\)\(\;\theta \)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,In\,non - inertial\,frame,\\
N\sin \theta = ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)\\
N\cos \theta = mg\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)\\
From\,\left( i \right)\,and\,\left( {ii} \right),\\
\tan \theta = \frac{a}{g} \Rightarrow a = g\tan \theta
\end{array}\)

\,\,\,\,\,\,\,\,\,\,In\,non - inertial\,frame,\\
N\sin \theta = ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)\\
N\cos \theta = mg\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)\\
From\,\left( i \right)\,and\,\left( {ii} \right),\\
\tan \theta = \frac{a}{g} \Rightarrow a = g\tan \theta
\end{array}\)

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