Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

A circular disc of mass M and radius R is rotating about its axis with angular speed  ω1. If another stationary disc having radius  R2 and same mass M is dropped co-axially on to the rotating disc. Gradually both discs attain constant angular speed  ω2. The energy lost in the process is p% of the initial energy. Value of p is __________ .

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

answer is 0020.00.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Conservation of Angular Momentum, ​MR22ω1=MR22+MR242ω2​⇒ω2=45ω1Given : energy lost=p% of initial energy​⇒Ki−Kf=p100×Ki⇒p=Ki−KfKi×100 = 12MR22ω12−12MR22+MR242ω2212MR22ω12×100⇒p=1−54ω2ω12×100​⇒p=1−45×100=20
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring
A circular disc of mass M and radius R is rotating about its axis with angular speed  ω1. If another stationary disc having radius  R2 and same mass M is dropped co-axially on to the rotating disc. Gradually both discs attain constant angular speed  ω2. The energy lost in the process is p% of the initial energy. Value of p is __________ .