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The first year tuition fee of an MBBS course at Ural State Medical University is around INR 3.15 L, while the annual fee for the following five years is around INR 3 L.

This takes the overall course fee for a duration of six years to INR 18.2 L.

Students must also account for the annual costs of living in the region, which varies depending on the kind of accommodation, a student pursues.

Prestige Institute of Management & Research, Bhopal offers eminent placements. The institute has a placement record, with over 100+ organisations visiting the campus every year. The top recruiters are Aditya Birla Group, Fullerton India, HDFC Bank, Parle, Reliance, Red FM 93.5, Kotak Mahindra, Dainik Bhaskar. Sundaram Mutual, Mahindra, Welspun, and others. 

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

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

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The Ural State Medical University ranks among the top 100 colleges & universities in Russia per rankings rolled out by a number of prominent university ranking publications.

The institute ranks #79 per the National University Rankings given by a prominent Russian ranking publication, while also securing a global rank #1001 per Times Higher Education (THE) Rankings.

Here are some important documents you must carry during BA in Mass Communication application process:

• Class 10 and 12 marksheets
• Passport size photographs (at least two)
• Valid Photo ID (Passport, Adhaar, PAN Card, etc.)
• Migration Certificate

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

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

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

$3R=a$

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°.

After pursuing a Seoul National University program, graduates secure employment at the university campus and in the following companies:

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 $Q^{-2}$ ] $\times \left[Q^2\right]\left[M^a\right]\times {\left[L{T}^{-1}\right]}^b{\left[L\right]}^c$

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

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

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

Explanation- the magnetic field induction at a pojnt z distance from dipole moment is

B= $\frac{{\mu }_o}{4\pi }\frac{2M}{z^3}$

Along z axis from p to q

${\int }_p^{Q}B.dl={\int }_P^{Q}Bdlcos0$ = ${\int }_P^{Q}Bdz$

= ${\int }_a^R\frac{{\mu }_o}{2\pi }\frac{M}{z^3}dz=\frac{{\mu }_{o}M}{2\pi }\left(-\frac{1}{2}\right)\left(\frac{1}{R^2}-\frac{1}{a^2}\right)$

= $\frac{{\mu }_{o}M}{4\pi }\left(\frac{1}{a^2}-\frac{1}{R^2}\right)$

(b) the point A lies on the equatorial line of the magnetic dipole of moment Msin $\theta$

B= $\frac{{\mu }_0}{4\pi }\frac{Msin\theta }{R^3}$ dl= Rd $\theta$

$\int B.dl=\int Bdlcos\theta ={\int }_0^{\frac{\pi }{2}}\frac{{\mu }_0}{4\pi }\frac{Msin\theta }{R^3}Rd\theta$

Circular arc = $\frac{{\mu }_o}{4\pi }\frac{M}{R}(-cos\theta )$ = $\frac{{\mu }_o}{4\pi }\frac{M}{R^2}$

(c) along x axis over the path ST

From figure every point lies on the equatorial line of magnetic dipole so

B= $\frac{{\mu }_o}{4\pi }\frac{M}{x^3}$   the value of B is zero as the angle between M and dl is 90 

(d)along the quarter circle TP of radius a

B along circular arc $\int B.dl={\int }_{\pi /2}^0\frac{{\mu }_0}{4\pi }\frac{Msin\theta }{a^3}$ d $\theta$

B = $-\frac{{\mu }_o}{4\pi }\frac{M}{a^2}$

Net magnetic field through PQST  is zero

Yes, Seoul National University has a Career Services department that focuses on supporting students throughout their careers by providing advice and opportunities to them. Some of the services offered at the are as follows:

Career Survey & Assessment

You can take the PTE test for as many times as you want with only 1 condition, i.e. you can take the test only 12 times within a period of 12 months. This simply means that if you already took the PTE test 12 times in 7 months of time then, you will have to wait 5 months to take the test again. 

If you need to take the test for more thn 12 times in a 12 month window then, you may request for one as an exception. For that purpose, you will have to connect with the Pearson Customer Support team. 

4 days before of exam