BSc Nursing GNM 2019 Solved Paper

Time period of a thin bar magnet of vibration magnetometer is given by
T=2π√‌

I

MH

where, I is moment of inertia of a bar magnet of mass mb, length l about an axis passing through its centre and perpendicular to its length,C I=‌

mbl2

12

M is magnetic moment of bar magnet, i.e. M=m×l, where m is magnetic strength of bar magnet and H is the uniform magnetic field in which the magnet is oscillating.
Case I When magnet is cut parallel to its length I′‌‌=‌

mb

2

⋅‌

l2

12

=‌

mbl2

2×12

=‌

I

2

‌ and ‌M′‌‌=‌

m

2

×l=‌

M

2

∴ Time period of each part, when bar magnet is cut parallel to its length is
T′=2π√‌

I′

M′H

=2π√‌

I∕2

(M∕2)H

⇒T′=2π√‌

I

MH

=T
Case II When magnet is cut perpendicular to its length. I′′=‌

mb

2

⋅‌

(l∕2)2

12

= and M′′=m×‌

l

2

=‌

M

2

∴ Time period of each part, when bar magnet is cut perpendicular to its length is
T′′‌‌=2π√‌

I′′

M′′H

=2π√‌

I∕8

(M∕2)H

T′′‌‌=2π√‌

I

4MH

=‌

1

2

×2π√‌

I

MH

⇒T′′‌‌=‌

T

2