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This is a multiple choice answer as classified in NCERT Exemplar

(c) E=hv

V= $\frac{11\times 1.6\times {10}^{-19}}{6.62\times {10}^{-34}}$ = 2.65 $\times$  10¹⁵hz this belongs to uv region

At the time of registration for NEET MDS counselling, a candidate needs to pay two kinds of fee, refundable and non refundable.  Non-refundable: A candidate who wishes to participate in the counselling of NEET MDS has to pay a non-refundable registration fee for the submission of choices and Refundable: During the registration process a candidate is also required to pay a part of the tuition fee. This will be refunded to the candidate if he/ she is not allocated a seat in the counselling process. Candidates do have to pay both fees during registration for NEET MDS counselling.

Although it is looking tough to get admission but you can go for CSAB round, there are possibilities that you will get your seat here. 

To apply for BE admission at Vidyavardhini's College of Engineering and Technology, candidates must register through the official portal of Maharashtra CET Cell (DTE). The MHT CET cell conducts the Centralized Admission Process (CAP) to allocate BE seats. The CAP is conducted in various rounds, usually around July/ August. The registration process starts after the MHT CET results are released, which is usually in June. Candidates must have eligible JEE (Main)/ MHT CET scores to be eligible to participate in the counselling process. Hence, they must also stay updated on the application dates of the mentioned exams.

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

Since, electric field of an EM wave is an oscillating field and so is the electric force caused by it on a charged particle. This electric force averaged over an integral number of cycles is zero, since its direction changes every half cycle.

Hence, electric field is not responsible for radiation pressure.

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

As the distance is doubled, the area of spherical region ( 4πr² )

will become four times, so the intensity becomes one fourth the initial value but in case of laser it does not spread, so its intensity remain same.

Geometrical characteristic of LASER beam which is responsible for the constant intensity are as following

(i) Unidirectional

(ii) Monochromatic

(iii) Coherent light

(iv) Highly collimated

These characteristic are missing in the case of light from the bulb.

The entrance exam accepted by NIFT Kangra is NIFTEE. Candidates are required to follow the entrance-based admission process to stand a chance at recieving the final admission offer. Candidates who do get the final admission offer must pay the course fee and secure their seat in the programme. 
Candidates are shortlisted based on their entrance performance and the interview round performance. 

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

Pressure = force/area

Force =dp/dt

U=p.C or p= U/c

Pressure = $\frac{1}{A}.\frac{U}{C.dt}$

Pressure = I/C

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

q=CV

And we know q=It

Then I_ddt= cdV

I_d= Cdv/dt

1 $\times {10}^{-3}$ = 2 $\times$ 10^{-6 $\times$} dV/dt

DV/dt= 500V/s, this will produce the desired result.

Recently, UCMS has released the round-wise seat allotment result for the admission to DM in Endocrinology for the candidates belonging to General All India quota. Considering the UCMS NEET SS cutoff 2024, the required rank to get into the institute stood at 51, which was slightly than that of the 2023. The NEET SS 2023 cutoff for the same course stood at 54 for the General AI quota candidates.

Yes. NEET PG 2025 application form was reopened on June 13, 2025. The application form window was opened to choose exam city as NBE revised list of cities following exam postponement. Candidates had to choose exam city from June 13 to June 17, 2025. The correction window was opened from June 20 to June 22, 2025. NEET PG 2025 admit cards will be uploaded on July 31, 2025.

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

As we know radiant flux density S=  $\frac{1}{{\mu }_0}$ (E $\times B$ )= c^{2 $?$} ₀(E $\times B$ )

Suppose electromagnetic waves be propagating along x-axis. The electric field vector of

Electromagnetic wave be along y-axis and magnetic field vector be along z-axis. Therefore,

E=E₀cos (kx-wt)

B=B₀cos (kx-wt)

E $\times$ B= (E₀B₀)cos²(kx-wt)

S= c^{2 ${\epsilon }_0$} (E $\times B$ )

Radiant flux density over a complete cycle is

S_av= c^{2 $?$} ₀|E_{0 $\times$} B₀| $\frac{1}{T}{\int }_0^T{cos}^2(kx-wt)dt$

     = c^{2 $?$} ₀|E₀B₀| $\frac{1}{T}\times \frac{T}{2}$

     = $\frac{C^2}{2}\epsilon$ ₀E₀( $\frac{E_0}{c}$ )

     = $\frac{c}{2}\epsilon$ ₀E₀²= $\frac{c}{2}\times$ $\frac{1}{c^2{\mu }_0}$  E₀²

     = $\frac{E_0^2}{2{\mu }_{0}c}$