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JEE Mains · Physics · STD 11 - 5. work,energy,power and collision

Two inclined planes are placed as shown in figure. A block is projected from the Point \(A\) of inclined plane \(A B\) along its surface with a velocity just sufficient to carry it to the top Point \(B\) at a height \(10 m\). After reaching the Point \(B\) the block slides down on inclined plane \(BC\). Time it takes to reach to the point \(C\) from point \(A\) is \(t (\sqrt{2}+1) s\). The value of \(t\) is........(use \(g =10 m / s ^{2}\) )

  1. A \(8\)
  2. B \(4\)
  3. C \(2\)
  4. D \(6\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2\)

Step-by-step Solution

Detailed explanation

From E.C. \(=\frac{1}{2} mv _{0}^{2}= mgh\) \(v _{0}=10 \sqrt{2}\) For \(A \rightarrow B\) at \(B , v =0\) \(a =- g \sin 45^{\circ}=\frac{-10}{\sqrt{2}}\) \(v = u + at _{1} \Rightarrow 0=10 \sqrt{2}-\frac{10}{\sqrt{2}} t _{1} \Rightarrow t _{1}=2\,sec\) For \(B \rightarrow C\)…
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