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 \(g =10 m / s ^{2}\) )

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

(C) \(2\)

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 \(B \rightarrow C\)

\(s =u t _{2}+\frac{1}{2} at _{2}^{2}\)

\(\frac{10}{\sin 30^{\circ}}=\frac{1}{2}\left(10 \sin 30^{\circ}\right) t _{2}^{2}\)

\(t _{2}=2 \sqrt{2}\)

So total time

\(T=t_{1}+t_{2}\)

\(=2 \sqrt{2}+2\)

\(=2(\sqrt{2}+1)\sec\)

Same subject

List \(I\)	List \(II\)
\((A)\) Young's Modulus \((Y)\)	\((I)\) \(\left[ M L ^{-1} T ^{-1}\right]\)
\((B)\) Co-efficient of Viscosity \((\eta)\)	\((II)\) \(\left[ M L ^2 T ^{-1}\right]\)
\((C)\) Planck's Constant \((h)\)	\((III)\) \(\left[ M L ^{-1} T ^{-2}\right]\)
\((D)\) Work Function \((\phi)\)	\((IV)\) \(\left[ M L ^2 T ^{-2}\right]\)

From JEE Mains