Q50. Time period of oscillation of a bar magnet in a uniform magnetic field is T₀. The magnet is cut

Q: What is the correct answer for CUET UG Physics Q50?

The correct answer is c. The time period of oscillation of a bar magnet in a uniform magnetic field (B) is given by T = 2π√(I/(MB)), where I is the moment of inertia and M is the magnetic dipole moment.Initially, T₀ = 2π√(I/(MB)).When the bar magnet is cut into 3 equal parts transverse to its length (i.e., perpendicular to its length), each part has:1. Length L' = L/3.2. Mass m' = m/3.3. Assuming the pole strength 'p'

Q: Q50. Time period of oscillation of a bar magnet in a uniform magnetic field is T₀. The magnet is cut into 3 equal parts transverse to its length. The new time period of oscillation of one part will…

The correct answer is c. The time period of oscillation of a bar magnet in a uniform magnetic field (B) is given by T = 2π√(I/(MB)), where I is the moment of inertia and M is the magnetic dipole moment.Initially, T₀ = 2π√(I/(MB)).When the bar magnet is cut into 3 equal parts transverse to its length (i.e., perpendicular to its length), each part has:1. Length L' = L/3.2. Mass m' = m/3.3. Assuming the pole strength 'p'

Q: What are the options for this CUET UG Physics question?

A. T₀ · B. T₀/√3 · C. T₀/3 · D. T₀/4

Question

Q50. Time period of oscillation of a bar magnet in a uniform magnetic field is T₀. The magnet is cut into 3 equal parts transverse to its length. The new time period of oscillation of one part will be:
(a) T₀
(b) T₀/√3
(c) T₀/3
(d) T₀/4

FirstInTest · Accepted Answer

The time period of oscillation of a bar magnet in a uniform magnetic field (B) is given by T = 2π√(I/(MB)), where I is the moment of inertia and M is the magnetic dipole moment.Initially, T₀ = 2π√(I/(MB)).When the bar magnet is cut into 3 equal parts transverse to its length (i.e., perpendicular to its length), each part has:1.  Length L' = L/3.2.  Mass m' = m/3.3.  Assuming the pole strength 'p' remains unchanged for each part, the new magnetic dipole moment M' = p * L' = p * (L/3) = (pL)/3 = M/3.4.  The moment of inertia of a uniform bar magnet of mass 'm' and length 'L' about its center (perpendicular to its length) is I = mL²/12. For one part, the mass is m' = m/3 and length L' = L/3. So, the new moment of inertia I' = m'(L')²/12 = (m/3) * (L/3)² / 12 = (m/3) * (L²/9) / 12 = mL²/(12 * 27) = I/27.Now, calculate the new time period T':T' = 2π√(I'/(M'B)) = 2π√((I/27) / ((M/3)B)) = 2π√(I/(MB) * (3/27)) = 2π√(I/(MB) * (1/9)).T' = (1/√9) * 2π√(I/(MB)) = (1/3) * T₀.So, the new time period of oscillation for one part is T₀/3.

Answer

T₀

Answer

T₀/√3

Answer

T₀/3

Answer

T₀/4

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