A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other…

Q: A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is 2(5/4) sec. One of the magnets is removed and if the other magnet oscillates in the same field, then the time period in seconds is:

(a) 2(1/4)

(b) 2(1/2)

(c) 2

(d) 2(3/4)

Ans: (c)

Sol: When two identical bar magnets are held perpendicular to each other.

\displaystyle M_1 = \sqrt{M^2 + M^2}

\displaystyle M_1 = \sqrt{2} M ; I1 = I

T1 = 2(5/4) sec ; T2 = ?

M2 = M (as one magnet is removed)

I2 = I/2

\displaystyle \frac{T_2}{T_1} = \sqrt{\frac{I_2}{I_1}\times \frac{M_1}{M_2}}

\displaystyle \frac{T_2}{T_1} = \sqrt{\frac{1}{2}\times \frac{\sqrt{2}M}{M}}

\displaystyle \frac{T_2}{T_1} =  \frac{1}{2^{1/4}}

\displaystyle T_2 = \frac{T_1}{2^{1/4}}

\displaystyle T_2 = \frac{2^{5/4}}{2^{1/4}}

T2 = 2 sec

A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be 2 second…

Q: A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be 2 second. The same needle is then allowed to vibrate in the horizontal plane, and the time period is again found to be 2 seconds. Then the angle of dip is

(a) 0°

(b) 30°

(c) 45°

(d) 90°

Ans: (c)

Sol: \displaystyle t_1 = 2\pi \sqrt{\frac{I}{M B_V}}

\displaystyle 2 = 2\pi \sqrt{\frac{I}{M B_V}}

\displaystyle t_2 = 2\pi \sqrt{\frac{I}{M B_H}}

\displaystyle 2 = 2\pi \sqrt{\frac{I}{M B_H}}

\displaystyle B_V = B_H

\displaystyle tan\delta = \frac{B_V}{B_H} = 1

δ = 45°

The length, breadth and mass of two bar magnets are same but their magnetic moments are 3 M and 2 M respectively…

Q: The length, breadth and mass of two bar magnets are same but their magnetic moments are 3 M and 2 M respectively. These are joined pole to pole and are suspended by a string. When oscillated in a magnetic field of strength B, the time period obtained is 5 s. If the poles of either of the magnets are reversed, then the time period of the combination in the same magnetic field will be

(a) 2√2 s

(b) 5√5 s

(c) 1 s

(d) 3√3 s

Ans: (b)

Time Period \displaystyle T = 2 \pi \sqrt{\frac{1}{M B}}

\displaystyle T = \propto \sqrt{\frac{1}{M }}

Joining pole to pole means sum position, for which T1 = 5 s.

If T2 is time period for difference position, then

\displaystyle \frac{T_2}{T_1} = \sqrt{\frac{M_1 + M_2}{M_1-M_2}}

\displaystyle \frac{T_2}{T_1} = \sqrt{\frac{3 M + 2 M}{3 M - 2 M}} = \sqrt{5}

\displaystyle T_2 = \sqrt{5} T_1 = 5 \sqrt{5} s

In Young’s double slit experiment, using monochromatic light, fringe pattern shifts by a certain distance on the screen when a mica sheet…

Q: In Young’s double slit experiment, using monochromatic light, fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 μm is introduced in the path of one of the two waves.Now mica sheet is removed and distance between slit and screen is doubled, distance between successive maxima or minima remains unchanged. The wavelength of the monochromatic light used in the experiment is

(a) 4000 Å

(b) 5500 Å

(c) 5892 Å

(d) 6071 Å

Ans: (c)

Sol: Here, µ = 1.6 , t =1.964 µm = 1.964 × 10-6 m

Additional path diff. introduced on account of mica sheet

=(µ-1)t = x d/D

When distance between the slit and screen is doubled, then fringe width

β =(2λ D)/d

As β = x

\displaystyle \frac{2 \lambda D}{d} = \frac{(\mu -1)t D}{d}

\displaystyle \lambda = \frac{(\mu - 1) t}{2}

\displaystyle \lambda = \frac{(1.6 - 1) 1.964 \times 10^{-6}}{2}

λ = 5892 × 10-10 m

λ = 5892 A °

In Young’s double slit experiment, the two slits act as coherent sources of equal amplitude a and of wavelength λ…

Q: In Young’s double slit experiment, the two slits act as coherent sources of equal amplitude a and of wavelength λ . In other experiment with the same set up, the two slits are sources of equal amplitude a and wavelength λ , but are incoherent. The ratio of intensity of light at the mid point of the screen in the first case to that in the second case is

(a) 2 : 1

(b) 1 : 2

(c) 2 : 4

(d) 4 : 3

Ans: (a)

Sol : When sources are coherent, intensity at mid point

Imax = (a+a)2 = 4 a2

When source are incoherent, no interference occurs. Intensity at mid point,

I= I1 + I2

I = a2 + a2 = 2 a2

Imax/I = (4a2 )/(2a2 ) = 2∶1