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For admission to the BSc in Hospitality and Hotel Administration at IHM Gwalior, candidates must submit documents such as the Class 12 marksheet, Class 10 certificate, valid NCHMCT JEE scorecard, category certificate (if applicable), passport-sized photographs, and a government-issued photo ID. Originals and photocopies are required for verification during counseling.

The key highlights recorded during I.K. Gujral Punjab Technical University M.Tech Placements are given in the data tabulated below:

During the NSB Academy placements 2025, the highest international and domestic packages offered to the PGDM programme were INR 43.5 LPA and INR 21 LPA, respectively. Check out the table for more clarity:

The final step of BSc admission at IHM Gwalior involves paying the tuition fees, which range between INR 2.9 Lakh, and submitting original documents for verification at the campus. Once completed, the admission is finalised, allowing students to start their academic journey.

The KCET exam for Adichunchanagiri University BSc admission 2025 is typically conducted in April, and results are announced in May. Following the results, the counselling and seat allotment process is carried out. Only students who fulfill the eligibility criteria will be considered in final merit list.

A (g) -> B (g)        K_P = 100

$\Delta G$ at 300 K and 1 atm

Using

$\Delta G=-RTlnK{}_P$           

$\Delta G=-R\times 300ln100$            

=-R × 300 × 2 × 2.3

$\Delta G=-1380R$            

So; x = 1380.

BSc selection is based on the candidate’s entrance exam score in NCHMCT JEE. After the IHM application submission, candidates participate in counseling and seat allotment rounds conducted by the authorities. Document verification and seat confirmation are also part of the selection process.

I.K. Gujral Punjab Technical University average package for MBA has not been disclosed yet. However, students can refer to the median package given in the table below:

$\Delta T_f=K_f\times m\times i$

0.93 = 1.86 × 1 × i

i = $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ $i=1+\left(\frac{1}{n}-1\right)$ $\raisebox{1ex}{$$}\!\left/ \!\raisebox{-1ex}{$$}\right.$

$\therefore n=2$

A fluctuating trend can be observed in the median package recorded at IKGPTU. The median package for MBA stood at INR 3.50 LPA in 2025. Check out the  data tabulated below to know more:

Atma Ram Sanatan Dharma College provides the flagship programme of Bachelor of Science (BSc), which is a popular choice for students in Science. ARSD College BSc admissions are held on the basis of entrance test, and the accepted exam is CUET. Interested candidates can visit the official website and begin the application process for ARSD College BSc.

$Mn{O}_4^{-}+S_2{O}_3^{2-}\stackrel{O{H}^{-}}{\to }Mn{O}_2+S{O}_4^{2-}$

O.S of S in $S{O}_4^{2-}=+6$

The key highlights recorded during I.K. Gujral Punjab Technical University M.Tech Placements are:

The NSB Academy provides amazing placements and packages to the PGDM students. Check out the table for the NSB Academy placements recorded during season 2025:

BSc students can apply online by visiting the official IHM Gwalior website. They must register by providing personal details, verify their email, complete the application form, upload necessary documents, pay the application fee, and submit the form to complete the process for BSc admission.

Effective no. of octahedral voids in a lattice = n

Effective no. of lattice point in a lattice = n

$$ $\therefore$  Ratio = n/n = 1

Stability constant are :

K₁ = 10⁴

K₂ = 1.58 × 10³

K₃ = 5 × 10²

K₄ = 10²

Overall stability constant K will be

K = K₁ × K₂ × K₃ × K₄

 = 7.9 × 10¹¹

Now, overall equilibrium constant for dissociation of [Cu (NH₃)₄]²⁺ is

$K\text{'}=\frac{1}{K}=\frac{1}{7.9\times 10^{11}}=0.126\times 10^{-11}$  

= 1.26 × 10^-12

So; x = 1 (Rounded off to the nearest integer)

$S\left(g\right)+6F\left(g\right)\to S{F}_6\left(g\right)$

$\Delta {H^0}_R=\Delta {H^0}_f\left(S{F}_6\right)-\Delta {H^0}_f\left(S\right)-6\times \Delta {H^0}_f\left(F\right)$

= (-1100) (275) 6 (80) = 1855

$\Delta {H^0}_{S-F}=\frac{1855}{6}=309.16KJ/mol\cong 309$

Hi, the table below shows the specialisations offered by top Engineering colleges in Karnataka accepting KCET score along with their number of colleges: