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

Option: 1
2s

Option: 2
-

Option: 3
--

Option: 4
-

Answers (1)

best_answer

For journey from A to B,

$\mathrm{V_B^2=V_A^2+2 a S_1 }$

$\mathrm{0=V_A^2+2\left(-g \sin 45^{\circ}\right) \times \frac{10}{ \sin 45^{\circ}} }$

$\mathrm{V_A=10 \sqrt{2} \mathrm{~m} / \mathrm{s} }$

$\mathrm{V_B=V_A+a t }$

$\mathrm{0=10 \sqrt{2}+\left(-g \sin 45^{\circ}\right) t_1 }$

$\mathrm{t_1=2s}$

For journey from B to C

$\mathrm{S =u t_2+\frac{1}{2} a t_2^2 }$

$\mathrm{\frac{10}{\sin 30^{\circ}} =\frac{1}{2} \times g \sin 30^{\circ} t_2^2 }$

$\mathrm{t^2 _{2}=8 }$

$\mathrm{t_2 =2 \sqrt{2} s }$

$\mathrm{\text { Total time } =t_1+t_2 }$

$\mathrm{=2+2 \sqrt{2} }$

$\mathrm{t=2 s}$

Posted by

Gautam harsolia

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