Physics

Explanation- When a diamagnetic material is dipped in liquid nitrogen, it again behaves as a diamagnetic material. Thus, superconducting material will again behave as a diamagnetic material. When this diamagnetic material is placed near a bar magnet, it will be feebly magnetised opposite to the direction of magnetising field.

(i) Thus, it will be repelled.

(ii) Also its direction of magnetic moment will be opposite to the direction of magnetic field of magnet.

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Explanation -  $\mathit{\aleph }$ = I/H

Diamagnetism is due to orbital motion of electrons in an atom developing magnetic moments opposite to applied field. Thus, the resultant magnetic moment of the diamagnetic material is zero and hence, the susceptibility χ of diamagnetic material is not much affected by temperature.

Paramagnetism and ferromagnetism is due to alignments of atomic magnetic moments in the direction of the applied field. As temperature is raised, the alignment is disturbed, resulting decrease in susceptibility of both with increase in temperature.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Answer- density of nitrogen = 28g/22.4L= 28g/22400cc

density of copper = 8g/22.4L= 8g/22400cc

on comparing $\frac{{\rho N}_2}{\rho Cu}$ = $\frac{28}{22400}*\frac{1}{8}$  = 1.6 $*$  10^-4

$\frac{{\aleph N}_2}{\aleph Cu}$ = $\frac{5*{10}^{-9}}{{10}^{-5}}$  = 5 $*$  10^-4

As  we know $\aleph \propto$ $\rho$

$\frac{{\rho N}_2}{\rho Cu}=$  = $\frac{{\aleph N}_2}{\aleph Cu}$ 1.6 $*$ 10^-4

This is a Short Answer Type Questions as classified in NCERT Exemplar

Answer- M (intensity of magnetisation) = 10⁶A/m

Length = 10cm = 10 $*$ 10^-2= 0.1m

M= I_m/l

I_m= M $*$ l= 10^{6 $*$} 0.1 = 10⁵A

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Answer – M= $\frac{eh}{4\pi m}$  or M $\propto$ 1/m

$\frac{M_p}{M_e}$ = $\frac{m_e}{m_p}$  = $\frac{M_e}{{1837M}_e}$ <<1

M_p<e

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- as we know n = L/2 $\pi R$

Magnetic moment of circle = m₁= n₁IA₁= L.I $\pi R$ .²/2 $\pi$ R = LIR/2………(1)

Magnetic moment of square = m₂= n₂IA₂= $\frac{L}{4a}$ .I.a²= Lia/4…………….(2)

Moment of inertia of circle = MR²/2…………….(3)

Moment of inertia of square = Ma²/12…………….(4)

Frequency of circle f₁= 2 $\pi \sqrt[]{\frac{I_1}{m_{1}B}}$

Frequency of square f₂= 2 $\pi \sqrt[]{\frac{I_2}{m_{2}B}}$

As f₁=f₂

2 $\pi \sqrt[]{\frac{I_1}{m_{1}B}}$ =2 $\pi \sqrt[]{\frac{I_2}{m_{2}B}}$

So m₂/m₁= I₂/I₁

From eqn 1,2,3 and 4

$\frac{LIa.2}{4*LIR}$ = $\frac{M{a}^{2}2}{12M{R}^2}$

$\frac{a}{2R}=\frac{a^2}{6R^2}$

$3R=a$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- P is also on the magnetic equator, so the angle of dip= 0, because the value of angle of Dip at equator is zero. Q is also on the magnetic equator, thus the angle of dip is zero.

As earth tilted on its axis by 11.3°, thus the declination at Q is11.3°.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- B_v= $\frac{{\mu }_o}{4\pi }\frac{2mcos\theta }{r^3}$

B_H=  $\frac{{\mu }_o}{4\pi }\frac{msin\theta }{r^3}$ adding and squaring both equations we get

B_v²+B_H²= $\frac{{\mu }_o}{4\pi }\frac{m^2}{r^6}[4{cos}^2\theta +{sin}^2\theta ]$

B= $\sqrt[]{B_v^2+B_v^2}=$ = $\frac{{\mu }_o}{4\pi }\frac{m}{r^3}[3{cos}^2\theta +1]$ ^1/2

The value of B is minimum if cos $\theta =\frac{\pi }{2}$

(b) tan $\delta =\frac{B_V}{B_H}=\frac{\mathrm{}\frac{{\mu }_o}{4\pi }\frac{2mcos\theta }{r^3}}{\frac{{\mu }_o}{4\pi }\frac{msin\theta }{r^3}\mathrm{ }}$ = 2cot $\theta$

For dip angle is zero

cot $\theta$  =0 and $\theta$ = $\frac{\pi }{2}$

(c) tan $\delta =\frac{B_V}{B_H}$ if angle of dip is 45

tan  $45=\frac{B_V}{B_H}$ so B_v=B_H

2cot $\theta$ =1

Cot $\theta$ =1/2 and tan $\theta$  = 2

 So $\theta =$ tan^-12

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- as  is dimensionless quantity , it should have no involvement of charge Qin its dimensional formula

Let $\chi ={\mu }_0{e}^2{m}^a{v}^b{R}^c$

[ $M^0{L}^0{T}^0{Q}^0$ ]= [ML $Q^{-2}$ ] $*\left[Q^2\right]\left[M^a\right]*{\left[L{T}^{-1}\right]}^b{\left[L\right]}^c$

   = [ $M^{1+a}+L^{1+b+c}{T}^{-b}{Q}^0$ ]

After solving we get a=-1 , b= 0, c =-1

Putting these values in equations

$\chi ={\mu }_o{e}^2{m}^{-1}{v}^2{R}^{-1}=\frac{{\mu }_0{e}^2}{mR}$

By using their standard values we get $\chi$ =10 which proves it is dimensionless quantity