JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A block of mass \(M\) placed inside a box descends vertically with acceleration \('a'\). The block exerts a force equal to one-fourth of its weight on the floor of the box. The value of \('a'\) will be .............
- A \(\frac{g}{4}\)
- B \(\frac{g}{2}\)
- C \(\frac{3 g }{4}\)
- D \(g\)
Answer & Solution
Correct Answer
(C) \(\frac{3 g }{4}\)
Step-by-step Solution
Detailed explanation
\(m g-N=m a\) \(a=g-\frac{g}{4}\) \(a=\frac{3 g}{4}\)
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