JEE Class main Answered
Please Solve the following questions.
Asked by Mrinal | 07 Aug, 2019, 14:48: PM
Expert Answer
Electrostatic force, which makes the ring to rotate, is in the direction of electric field and is parallel to ground.
We do not have any force that has component normal to ground,
hence rotating ring will not experience any normal reaction force and firction force.
Weight of the ring, gravitational force due to mass of ring acts normally.
But in this case it can be assumed mass of ring is negligible.
Hence electrostatic force is much greater than gravitational force.
Hence it can be concluded, friction force is zero
Answered by Thiyagarajan K | 08 Aug, 2019, 10:31: AM
Application Videos
Concept Videos
JEE main - Physics
Asked by sumalathamadarapu9 | 23 Oct, 2024, 22:06: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by py309649 | 13 Oct, 2024, 13:39: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by coolskrish | 13 Oct, 2024, 12:50: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by midnightmoon3355 | 09 Oct, 2024, 09:09: AM
ANSWERED BY EXPERT
JEE main - Physics
Asked by rambabunaidu4455 | 03 Oct, 2024, 16:03: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by ratchanavalli07 | 17 Sep, 2024, 07:46: AM
ANSWERED BY EXPERT
JEE main - Physics
Asked by yayashvadutta45 | 15 Sep, 2024, 19:47: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by adithireddy999 | 03 Sep, 2024, 09:35: AM
ANSWERED BY EXPERT
JEE main - Physics
Asked by vaishalinirmal739 | 29 Aug, 2024, 18:07: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by vradhysyam | 26 Aug, 2024, 17:17: PM
ANSWERED BY EXPERT