Exam

Admission to MBA at Gateway Institute of Engineering and Technology, Gateway Education is granted on the basis of the interview taken. It also becomes crucial for students to fulfill the set eligibility criteria in order to get admission.

Candidates get admitted to GITAM School of Pharmacy, Visakhapatnam  programs based on their GITAM GAT results. To take part in the BPharm and MPharm admissions procedure, candidates might take the entrance exam. 
Those whose entrane scores put them on the short list can pay the course fee t

There are multiple factors that help universities abroad make decisions about admissions for international students using SAT. Some of the key factors include:

• SAT score of the applicant versus the applicant pool score.
• Syncing SAT score with previous adacemic background.
• Based on SAT score assess appl

There are 63 countries which accept SAT scores for admission and use them to assess applicants. These countries are as follows:

Shanti Business School accepts candidates who have scored 50% at graduation level (45% in case of specified categories) and have a valid score of CAT/ MAT/ XAT/ ATMA/ CMAT that is not earlier than 12 months from the date of applying. The final selection of the candidate is based on 35% Entrance Exam

Explanation- velocity of a freely falling body is v= $\sqrt[]{2gh}$

And $?=\frac{h}{mv}=\frac{h}{m\sqrt[]{2gh}}$

$?=h$ ^-1

Explanation-number of photon emitted per second n= $\frac{p}{\frac{hc}{?}}=\frac{p?}{hc}=\frac{20\times 5000\times {10}^{-10}}{6.62\times {10}^{-34}\times 3\times {10}^8}$

$=5\times {10}^{19}{s}^{-1}$

(ii) E=hc $/?$ = $\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{5000\times {10}^{-10}\times 1.6\times {10}^{-19}}=2.48eV$  this enegy is greater than 2 so emission is possible

(iii) work function $?$ = $\frac{p}{4?d^2}\times ?r^2?t$ = ${?}_o$

$?t$ = $\frac{4?d^2}{p{r}^2}$ = $\frac{4\times 2\times 16\times 1.6\times {10}^{-19}\times 2^2}{20\times {(1.5\times {10}^{-10})}^{-2}}=28.4s$

(iv) N= $\frac{n?r^2}{4?d^2}\times ?t$

 = $\frac{5\times {10}^{19}\times {(1.5\times {10}^{-10})}^2\times 28.4}{4\times ({2)}^2}$ =2

(v) as the time of emission is 11.04s so

Explanation- according to law of conservation of momentum

S0 m_Av+m_b0=m_Av₁+m_Bv₂

So m_A(v-v₁)= m_Bv₂

according to law of conservation of kinetic energy

1/2m_Av²=1/2m_Av₁²+1/2m_Bv₂²

So m_A(v²-v₁²)= m_Bv₂²

From above eqn we can say that v+v₁=v₂ or v=v₂-v₁

So v₁= $\frac{m_A{-m}_B}{m_A{+m}_B}$ v  and v₂= $\frac{2m_A}{m_A{+m}_B}$ v

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