Two inclined planes are placed… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Mechanics

2022integerhard

Two inclined planes are placed as shown in figure.

A block is projected from the Point $A$ of inclined plane $AB$ along its surface with a velocity just sufficient to carry it to the top Point $B$ at a height $10m$ . 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 $g=10m{s}^{-2}$ )

JEE Main 2022 Physics, Work Power Energy — question figure

Official previous-year question

Held on 27 Jul 2022 · Verified 6 Jul 2026.

Did you get this right?

Solution

JEE Main 2022 Physics, Work Power Energy — solution figure

From energy conservation at point $A$ and $B$ ,

$\frac{1}{2}m{v}_{0}^{2}=mgh$

$\Rightarrow {v}_{0}=\sqrt{gh}=\sqrt{10\times 10}$

$\Rightarrow {v}_{0}=10\sqrt{2}m{s}^{-1}$

For $A\rightarrow B$

At $B,v=0$

Acceleration of a particle moving on a smooth incline is $g\mathrm{sin}\theta$ . Therefore, along $AB$ , $a=-g\mathrm{sin}45^{\circ}=\frac{-10}{\sqrt{2}}m{s}^{-2}$ .

Using equation of motion,

$v=u+a{t}_{1}$

$0=10\sqrt{2}-\frac{10}{\sqrt{2}}{t}_{1}\Rightarrow {t}_{1}=2s$

For $B\rightarrow C$

Using second equation of motion,

$s=u{t}_{2}+\frac{1}{2}a{t}_{2}^{2}$

$\Rightarrow \frac{10}{\mathrm{sin}30^{\circ}}=\frac{1}{2}(10\mathrm{sin}30^{\circ}){t}_{2}^{2}$

${t}_{2}=2\sqrt{2}s$

So total time $T={t}_{1}+{t}_{2}$ $=2\sqrt{2}+2$

$=2(\sqrt{2}+1)s$

Your note

JEE Main Mechanics in other years

Mechanics 2026 Mechanics 2025 Mechanics 2024 Mechanics 2023 Mechanics 2021 Mechanics 2020 Mechanics 2019 Mechanics 2018 Mechanics 2017 Mechanics 2016 Mechanics 2015 Mechanics 2014

Keep practicing Mechanics

Work through every JEE Main Mechanics PYQ, year by year.

Browse all Physics topics

Solution

JEE Main 2022 Physics, Work Power Energy — solution figure

From energy conservation at point $A$ and $B$ ,

$\frac{1}{2}m{v}_{0}^{2}=mgh$

$\Rightarrow {v}_{0}=\sqrt{gh}=\sqrt{10\times 10}$

$\Rightarrow {v}_{0}=10\sqrt{2}m{s}^{-1}$

For $A\rightarrow B$

At $B,v=0$

Acceleration of a particle moving on a smooth incline is $g\mathrm{sin}\theta$ . Therefore, along $AB$ , $a=-g\mathrm{sin}45^{\circ}=\frac{-10}{\sqrt{2}}m{s}^{-2}$ .

Using equation of motion,

$v=u+a{t}_{1}$

$0=10\sqrt{2}-\frac{10}{\sqrt{2}}{t}_{1}\Rightarrow {t}_{1}=2s$

For $B\rightarrow C$

Using second equation of motion,

$s=u{t}_{2}+\frac{1}{2}a{t}_{2}^{2}$

$\Rightarrow \frac{10}{\mathrm{sin}30^{\circ}}=\frac{1}{2}(10\mathrm{sin}30^{\circ}){t}_{2}^{2}$

${t}_{2}=2\sqrt{2}s$

So total time $T={t}_{1}+{t}_{2}$ $=2\sqrt{2}+2$

$=2(\sqrt{2}+1)s$

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics