Question

The time period of a freely suspended magnet is a 4 seconds. If it is broken in length into two equal parts and one part is suspended in the same way, then its time period will be 

(a) 4 sec 

(b) 2 sec 

(c) 0.5 sec 

(d) 0.25 sec

Answer 1

The correct option  (b) 2 sec

Explanation:

T = 2π√(I/M_B)

For a rod I = (Mℓ²/I²)

For half length

I' = [{(m/2)(ℓ/2)²}/12]

= (mℓ²/12) × (1/8) = (I/8)

and magnetic moment = M' = (M/2)

Hence new time period  = T' = 2π√(I'/M_B)

T' = 2π√[(I/8)/{(M/2)B}]

= 2π√(I/MB) ∙ √(2/8)

= (2π/2) √(I/M_B)

= (1/2)[2π√(I/M_B)]

T' = (1/2)T

T' = (T/2)

Given T = 4 sec

T' = (4/2) = 2sec