## If θ1 and θ2 be the apparent angles of dip observed in two vertical planes at right angles to each other

Q: If θ1 and θ2 be the apparent angles of dip observed in two vertical planes at right angles to each other , then the true angle of dip θ is given by :

(a) tan2θ = tan2θ1 + tan2θ2

(b) cot2θ = cot2θ1 – cot2θ2

(c) tan2θ = tan2θ1 – tan2θ2

(d) cot2θ = cot2θ1 + cot2θ2

Ans: (d)

Sol: $\large tan\theta_1 = \frac{tan\theta}{cos\alpha}$

$\large cos\alpha = \frac{tan\theta}{tan\theta_1}$ …(i)

$\large tan\theta_2 = \frac{tan\theta}{cos(90-\alpha)}$

$\large tan\theta_2 = \frac{tan\theta}{sin\alpha}$

$\large sin\alpha = \frac{tan\theta}{tan\theta_2}$ …(ii)

$\large sin^2\alpha + cos^2\alpha = 1$

cot2θ = cot2θ1 + cot2θ2

## 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

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

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

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

## The horizontal component of earth’s magnetic field at place is 0.36× 10^-4 weber/m². If the angle of dip at that placed is…..

Q. The horizontal component of earth’s magnetic field at place is 0.36× 10-4 weber/m². If the angle of dip at that placed is 60° then the value of vertical component of earth’s magnetic field will be (in wb/m²)

(a) 6 × 10-5 T

(b) 6√2 × 10-5 T

(c) 3.6√3 × 10-5 T

(d) √2 × 10-5 T

Ans:(c)

Sol: BH = 0.36 × 10-4 weber/m² ; BV = ? ; δ = 60°

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

$\displaystyle B_V = B_H tan\delta$

$\displaystyle B_V = 0.36 \times 10^{-4} tan60$

$\displaystyle B_V = 0.36 \times 10^{-4} \sqrt{3}$

BV = 3.6√3 × 10-5 T