42. The magnet suspended in uniform magnetic field
is heated so as to reduce its magnetic moment by
19%. By doing this, the time period of the magnet
will
1) increases by 11%
3) increases by 19%
2) decreases by 19%
4) decreases by 4%
a
.
Answers
answer : option (1) increase by 11%
it is given that The magnet suspended in uniform magnetic field is heated so as to reduce its magnetic field by 19%.
we know, time period of the magnet is given by,
T = 2π√{I/MB}
where I is moment of inertia of magnet, M is magnetic moment and B is magnetic field.
if I and B is constant
T ∝ 1/√M
i.e., T₁/T₂ = √{M₂/M₁}
if we assume M₁ = 100 ,
M₂ = 100 - 100% of 19 = 100 - 19 = 81
so, T₁/T₂ = √{81/100} = 9/10
T₂ = (10/9) T₁
so, % increase in Time period = (T₂ - T₁)/T₁ × 100
= (10/9 - 1)/1 × 100
= 1/9 × 100.
= 11.11 %
therefore, the time period of the magnet will be increased by 11%
The time period of the magnet will 1) increases by 11%
Explanation:
The time period of the magnet is given by the formula:
T = 2π √(I/(MB))
Where,
I = Moment of inertia
B = Magnetic field
M = Magnetic moment
⇒ T = 1/√M
The time period before and after the reduction of magnetic moment:
T₁/T₂ = (1/√M₁)/(1/√M₂)
T₁/T₂ = √M₂/√M₁
T₁/T₂ = √(M₂/M₁)
On substituting the values, we get,
T₁/T₂ = √(81/100)
T₁/T₂ = 9/10
T₂ = T₁ × 10/9
∴ T₂ = 1.11 T₁ = 11.% of T₁