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The official website to download the CMA Final admit card is the ICMAI student portal. Candidates need to visit the website and navigate to the CMA Students tab. From there, they can access the admit card download link listed on the student dashboard or sidebar. Admit cards for the CMA Final course are released separately for each exam term. ICMAI provides separate links to download admit cards for Foundation, Intermediate, and Final levels.

CMA Final admit card download link is available for all candidates who submit the exam form on the official student website. Candidates can also find the CMA final admit card on the hompage of ICMAI website.

No, the CMA Final admit card is not available offline. The Institute of Cost Accountants of India (ICMAI) releases the admit card only through its official website. Candidates must visit the ICMAI website and log in using their registration number to download the admit card. It is not sent by post or provided at any ICMAI office. Students must download and print the admit card before the exam, as it is mandatory for entry into the exam hall. Regularly checking the official website for updates is highly recommended.

NIIT mainly offers skill-based courses and IT training rather than traditional B.Tech degrees. If you want to do B.Tech in Computer Science, NIIT might not be the right place.

For B.Tech CS, you should look at engineering colleges or universities that offer degree programs. With 68% in 12th, you can apply to many private colleges or universities that have moderate eligibility criteria.

CMA June Final admit card will release a week before the exam date. The Institute of Cost Accountants of India (ICMAI) does not release the schedule for issuance of the CMA Final admit card. The admit card is downloadable from the official exam website a week before the exam date. Since the CMA June  exam is scheduled to commence in the second week of June, candidates can expect the release of the CMA June admit card for Final course subjects in the first week of June.

CMA admit card for all three courses are expected to release in the first week of June. For real-time updates, stay tuned to Shiksha's CMA admit card article.

 To be eligible for the General Nursing and Midwifery (GNM) degree program at Satguru Institute of Nursing Education in Ludhiana, candidates must have completed their 10+2 examination in any stream with a minimum aggregate of 45%.So you have high chances with 93% aggregate.You need not to worry about it.

Hi, you can choose the best hotel management college in West Bengal on the basis of:

Ranking:

• Check if the college is affiliated with:

  • NCHMCT

  • UGC / AICTE approved

Placement:

• • Top Recruiters

  • Average salary packages

  • Alumni placed in top hotel chains (Taj, Oberoi, Marriott, etc.)

Faculty:

• Seasoned faculty

• international exposure or hotel management certifications

Internship:

• Strong internship tie-ups with 5-star hotels or international brands can boost your career

Location:

• • Safe and well-connected

  • Offers opportunities for part-time jobs/internships

Hi, professionals can move into managerial roles with experience and promotions and salaries ranging from INR 8 LPA to INR 20+ LPA.

But as freshers you can earn:

• Government: INR 3.5 to INR 5 LPA (Lakhs Per Annum)

• Private Institutes:  INR 2.5 to INR 4 LPA

5.23 Number of atomic dipoles, n = 2 $\times {10}^{24}$

Dipole moment of each atomic dipole, M = 1.5 $\times {10}^{-23}$ J/T

Magnetic field, $B_1$ = 0.64 T

The sample is cooled to the temperature, $T_1$ = 4.2 K

Total dipole moment of the atomic dipole, $M_{tot}$ = n $\times$ M

= 2 $\times {10}^{24}$ $\times 1.5\mathrm{}\times {10}^{-23}\mathrm{}=\mathrm{}\mathrm{}30\mathrm{}\mathrm{}\mathrm{J}/\mathrm{T}$

Magnetic saturation is achieved at 15 %

Hence, effective dipole moment, = 15 % $\times 30$ = 4.5 J/T

Yes, companies like Taj, Oberoi, Marriott, Hyatt, and Hilton often recruit from reputed institutes like IIIHM and IIHM, etc.

Hi, the top ranked Hotel Management colleges in West Bengal according to Outlook for the year 2024 are given below:

5.20 Number of turns in the circular coil, n = 30

Radius of the circular coil, r = 12 cm = 0.12 m

Current in the coil, I = 0.35 A

Angle of dip, $\beta$ = 45 $°$

The magnetic field due to current I at a distance r is given by

B = $\frac{{\mu }_{0}2\pi nI}{4\pi r}$ , where $B_1$

${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

Then B = $\frac{4\pi \times {10}^{-7}\times 2\pi \times 30\times 0.35}{4\pi \times 0.12}=5.50\times {10}^{-5}$ T

5.18 Current in the wire, I = 2.5 A

Angle of dip at the given location of earth, $\beta$ = 0 $°$

Earth’s magnetic field, H = 0.33 G = 0.33 $\times {10}^{-4}$ T

The horizontal component of earth’s magnetic field is given as:

$H_H$ = H $\cos ?\beta$ = 0.33 $\times {10}^{-4}\cos ?0°$ = 0.33 $\times {10}^{-4}$ T

The magnetic field at the neutral point at a distance R from the cable is given by the relation:

$H_H=\frac{{\mu }_{0}I}{2\pi R}$

${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

R = $\frac{{\mu }_{0}I}{2\pi H_H}$ = $\frac{4\pi \times {10}^{-7}\times 2.5}{2\pi \times 0.33\mathrm{}\times {10}^{-4}}$ = 15.15 $\times {10}^{-3}$ m = 1.52 cm