Magnetism> a thin rectangular magnet suspended freel...

a thin rectangular magnet suspended freely has a period of oscillations equal to T. now it is broken into two halves eachhaving half of the original length and one piece is made to oscillate freely in the same field. if its period of oscillation is T' . than the ratio of T'/T is ----a magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60 degrees. the torque needed to maintain the needle in its position will be-----an ammeter reads upto 1 ampere. its internal resistance is upto 0.81 ohm. to increase the range to 10 ampere.the value of the required shunt-----

Aditi Chauhan , 11 Years ago

Grade 10

1 Answers

ROSHAN MUJEEB

Last Activity: 4 Years ago

Time period for oscillating magnet is
T=2πMBI.........(1)
where I is the moment of inertia,

M is the magnetic moment
M=x×l
x=polestrength
When the magnet is broken in two equal parts, the new magnetic dipole moment is
M′=2x×l=2M
I′=2ML12
=122M×(2L)2
=12×8ML2
=8I
New time period is
T′=2πM′BI′
TT′=M′I′
Putting all the values we get
TT′=41=21