A particle of mass $100g$ is projected at time $t=0$ with a speed $20m{s}^{–1}$ at an angle $45^{\circ}$ to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time $t=2s$ is found to be $\sqrt{K}\mathrm{kg}{m}^{2}{s}^{-1}$. The value of $K$ is ______. (Take $g=10m{s}^{-2}$) ![JEE Main 2023 Physics, Motion In Two Dimensions — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Motion_In_Two_Dimensions/648b5a6f417cc3fb48d68182/question_1__q_648b5a6f417cc3fb48d68182__cdn-question-pool.getmarks.app__9242892d-1de7-42ac-888e-b6d32f29a7bb-9895103__497aa4c715_final_ppt_sync.png)

A particle of mass 100g is… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Mechanics

2023integereasy

A particle of mass $100g$ is projected at time $t=0$ with a speed $20m{s}^{–1}$ at an angle $45^{\circ}$ to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time $t=2s$ is found to be $\sqrt{K}\mathrm{kg}{m}^{2}{s}^{-1}$ . The value of $K$ is ______.

(Take $g=10m{s}^{-2}$ )

JEE Main 2023 Physics, Motion In Two Dimensions — question figure

Official previous-year question

Held on 29 Jan 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

As torque due to gravitational force will be variable, therefore we can use

$\displaystyle \Delta L={\int }_{0}^{t}\tau dt$ .

Now, torque of $mg$ at time $t$ $=({v}_{x}t)mg$ and horizontal velocity of the projectile will remain constant $({v}_{x}=v\mathrm{cos}45^{\circ}=20\times \frac{1}{\sqrt{2}}=10\sqrt{2})$ .

${L}_{0}={\int }_{0}^{2}mg({v}_{x}t)dt$

$=mg{v}_{x}\frac{{t}^{2}}{2} =(0.1)(10)(10\sqrt{2})\frac{{2}^{2}}{2}$

$=20\sqrt{2}$

$=\sqrt{800}\mathrm{kg}{m}^{2}{s}^{-1}$

Therefore, $K=800$ .

Your note

JEE Main Mechanics in other years

Mechanics 2026 Mechanics 2025 Mechanics 2024 Mechanics 2022 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

As torque due to gravitational force will be variable, therefore we can use

$\displaystyle \Delta L={\int }_{0}^{t}\tau dt$ .

Now, torque of $mg$ at time $t$ $=({v}_{x}t)mg$ and horizontal velocity of the projectile will remain constant $({v}_{x}=v\mathrm{cos}45^{\circ}=20\times \frac{1}{\sqrt{2}}=10\sqrt{2})$ .

${L}_{0}={\int }_{0}^{2}mg({v}_{x}t)dt$

$=mg{v}_{x}\frac{{t}^{2}}{2} =(0.1)(10)(10\sqrt{2})\frac{{2}^{2}}{2}$

$=20\sqrt{2}$

$=\sqrt{800}\mathrm{kg}{m}^{2}{s}^{-1}$

Therefore, $K=800$ .

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