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The total fees for the MGU Kerala BCom is cumulative of multiple components. These are tuition fee, security deposit, and more. Once the final selection is done, the candidates are required to pay the admission fee amount. The total tuition fees for the BCom course range from INR 2,000 to INR 1 lakh. This fee information is as per the official website/ sanctioning body. It is still subject to changes and hence, is indicative. 

This is a short answer type question as classified in NCERT Exemplar

When the switch is thrown from the off position (open circuit) to the on position (closed circuit), then neither B, nor A nor the angle between B and A change. Thus, no change in magnetic flux linked with coil occur, hence no electromotive force is produced and consequently no current will flow in the circuit.

This is a long answer type question as classified in NCERT Exemplar

International applicants must have an idea of expenses of living at a university when studying overseas. BPP University provides an estimated cost of attendance data that might help students to plan their finances accordingly. An approximate expense table is given below to applicants below:

This is a long answer type question as classified in NCERT Exemplar

$?=B_z\left(\pi l^2\right)=B_0(1+\lambda z)\left(\pi l^2\right)$

According to faraday’s law e= - $\frac{d\varnothing }{dt}$ and also ohm’s law V=IR

So by equating these two

We get B_o( $\pi l^2$ ) $\lambda \frac{dz}{dt}=IR$

I= $\frac{\pi l^2{B}_o\lambda v}{R}$

And energy lost/second = I²R= $\frac{{\left(\pi l^2\lambda \right)}^2{B}_o^2{v}^2}{R}$

As change in PE=mg $\frac{dz}{dt}=mgv$ but if we compare both energy then

Mgv= $\frac{{\left(\pi l^2\lambda \right)}^2{B}_o^2{v}^2}{R}$  or v=

This is a long answer type question as classified in NCERT Exemplar

The conductor of length d moves with speed v, perpendicular to magnetic field B as shown in figure. This produces motional emf across two ends of rod, which is given by= vBd.

Since, switch S is closed at time t= 0. current start growing in inductor by the potential

Difference due to motional emf.

Applying Kirchhoff’s voltage rule, we have

-L $\frac{dI}{dt}$ +vBd=IR

On solving differential equation we get I= $\frac{vBd}{R}$ + Ae^-Rt/2

At t=0 I=0

A= -vBd/R

I= $\frac{vBd}{R}$ (1-e^-Rt/L)

This is a long answer type question as classified in NCERT Exemplar

The conductor of length d moves with speed v, perpendicular to magnetic field B as shown in figure. This produces motional emf across two ends of rod, which is given by= vB d.

Since, switch S is closed at time t= 0. capacitor is charged by this potential difference. Let

Q ( t) is charge on the capacitor and current flows from A to B.

Now, the induced current I= $\frac{vBd}{R}$ - $\frac{Q}{RC}$

On rearranging  = $\frac{Q}{RC}$ + dQ/dt = $\frac{vBd}{R}$

On solving differential equation we get I= $\frac{vBd}{R}$ e^-t/RC

This is a long answer type question as classified in NCERT Exemplar

The component of magnetic field perpendicular the plane = Bcosθ

But when there is a motional emf then it is e = Blv=vBcosθd

And current Is I= $\frac{\mathrm{v}\mathrm{B}\mathrm{c}\mathrm{o}\mathrm{s}\theta d}{R}$ but when discuss about force F= llB

And component of force which only act is fcosθ

So force is IBd(cosθ) = $\frac{\mathrm{v}\mathrm{B}\mathrm{c}\mathrm{o}\mathrm{s}\theta d}{R}$  where v=dx/dt

Also component of weight is mg cosθ

So by applying newton second law of motion

m $\frac{d^{2}x}{d{t}^2}=mgsin\theta -\frac{Bcos\theta d}{R}\left(\frac{dx}{dt}\right)\times \left(Bdcos\theta \right)$

$\frac{dv}{dt}=gsin\theta -\frac{B^2{d}^2}{mR}{\left(cos\theta \right)}^{2}v$

$\frac{dv}{dt}+\frac{B^2{d}^2}{mR}{\left(cos\theta \right)}^{2}v=gsin\theta$

On solving this equation we get

V= $\frac{gsin\theta }{\frac{B^2{d}^2}{mR}{\left(cos\theta \right)}^2}+Aexp(-\frac{B^2{d}^2}{mR}{\left(cos\theta \right)}^{2}t)$

This is a long answer type question as classified in NCERT Exemplar

applying ohm’s law

I=E/R=d $\varnothing /Rdt$

I= dQ/dt or dQ/dt= d $\varnothing /Rdt$

Q (t₁)-Q (t₂)= $\frac{1}{R}$ [ $\varnothing$ t₁- $\varnothing$ t₂]

$\varnothing$ t₁= L_{1 $\frac{{\mu }_0}{2\pi }{\int }_x^{L_2+x}\frac{dx}{x}I$} t₁

= $\frac{{\mu }_0{L}_1}{2\pi }$ I1in [ $\frac{L_2+x}{x}$ ]

The TSTET exam is conducted in two shifts on multiple days. As per the TS TET exam schedule, the exam was held from June 18 to 30, 2025. There are two shifts on each exam day. The TSTET exam timing for shift 1 is 9.00 AM to 11.30 AM and shift 2 is 2.00 PM to 4.30 PM. Have a look at the TS TET shift timings below:

This is a long answer type question as classified in NCERT Exemplar

B= $\frac{{?}_{0}I}{2R}$ (ouside the paper)

Flux= $\frac{{?}_{0}I}{2R}l{?}_{x_0}^x\frac{dr}{r}$ = $\frac{{?}_{0}I}{2R}$ in $\frac{x}{x_0}$

Induced emf= e/R=I

Induced current = $\frac{1}{R}\frac{d?}{dt}$ = $\frac{e}{R}$ = $\frac{{?}_{0}I?}{2?R}$ in $\frac{x}{x_0}$

Yes, canddiaets can join NIELIT Shillong directly as the admission process is merit-based. Candidates must visit the official website of the institute and fill out the form to participate in the merit based admisison process. 
Candidates are shortlisted based on their academic record in Class 10 or Class 12. Candidates are required to pay the course fee if they get the final admission offer.