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

Q.

A non conducting uniform thin shell of mass m and radius R has uniform charge density +σ  on one half and  σ on another half as shown in figure. It is placed on a rough non conducting horizontal plane. At t = 0 a uniform electric field E=E0j^    N/C is switched on and the solid sphere starts rolling without slipping . The speed of center of the sphere when it rotates through an angle of 530 is nπR3σ0E0m. Find n.
 

Question Image

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 1.92.

(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

As the work done by friction is zero,

Question Image

ΔK=ΔU12mV2[1+23] =PE(cos(π2+53)cosπ/2) =PE×45=σ0×2πR2×R×E0×45 v=48πR3σ0E025m

Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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 non conducting uniform thin shell of mass m and radius R has uniform charge density +σ  on one half and  −σ on another half as shown in figure. It is placed on a rough non conducting horizontal plane. At t = 0 a uniform electric field E→=E0j^    N/C is switched on and the solid sphere starts rolling without slipping . The speed of center of the sphere when it rotates through an angle of 530 is nπR3σ0E0m. Find n.