System Of Particles And Rotational Motion Ques 13

13. A wheel of radius $1 m$ rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is

[2002]

(a) $\pi$

(b) $2 \pi$

(c) $\sqrt{2} \pi$

(d) $\sqrt{\pi^{2}+4}$

Show Answer

Answer:

Correct Answer: 13.(d)

Solution:

  1. (d)

Linear distance moved by wheel in half revolution $=\pi r$. Point $P_1$ after half revolution reaches at $P_2$ vertically $2 m$ above the ground.

$\therefore$ Displacement $P_1 P_2$

$=\sqrt{\pi^{2} r^{2}+2^{2}}=\sqrt{\pi^{2}+4} \quad[\because r=1 m]$

Table of Contents
icon  BETA

© 2014-2026 Copyright SATHEE

Privacy Policy

Powered by Prutor@IITK

sathee@iitk.ac.in
Media coverage of SATHEE
goolge

Play Store

goolge

App Store

The Ministry of Education has launched the SATHEE initiative in association with IIT Kanpur to provide free guidance for competitive exams. SATHEE offers a range of resources, including reference video lectures, mock tests, and other resources to support your preparation. Please note that participation in the SATHEE program does not guarantee clearing any exam or admission to any institute.
sathee Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on

Please select your preferred language
कृपया अपनी पसंदीदा भाषा चुनें