JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A block '\(A\)' takes \(2\,s\) to slide down a frictionless incline of \(30^{\circ}\) and length ' \(l\) ', kept inside a lift going up with uniform velocity ' \(v\) '. If the incline is changed to \(45^{\circ}\), the time taken by the block, to slide down the incline, will be approximately \(........\,s\)
- A \(2.66\)
- B \(0.83\)
- C \(1.68\)
- D \(0.70\)
Answer & Solution
Correct Answer
(C) \(1.68\)
Step-by-step Solution
Detailed explanation
\(a = g \sin \theta\) \(\ell=\frac{1}{2} g \sin 30^{\circ}(2)^{2}\) \(\ell=\frac{1}{2} g \sin 45^{\circ} t ^{2}\) \(\left(\frac{1}{2}\right)(4)=\frac{1}{\sqrt{2}} t ^{2} \Rightarrow t =\sqrt{2 \sqrt{2}}=1.68\)
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