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Yes, JIIT, Online admits students based on Class 12 scores. The institute admits students based on merit. Students can check the table below to know the detailed course eligibility:

To secure admission at JIIT, Online, candidates must complete graduation as the basic eligibility criteria. Aspirants can check the points mentioned below to know more:

1. Aspirants can visit the official website
2. Apply online for admission
3. Students are selected based on merit

IMT CDL provides PG and certificate programmes to students. At the postgraduate level, IMT CDL offers a two-year PGDM, a 15-month Executive PGDM, and a 13-month PGCM course to students. Some of the certificate programmes available at the IMT CDL include Proficiency Certificate Programme in Financial Management, Proficiency Certificate Programme in Human Resource Management, and others. Below mentioned are some of the IMT CDL courses and their respective eligibility and IMT CDL fee:

Each year, Duke receives over 50,000 applications, of which the university admits a mere 2,800-3,000 students. The UG admission trends at Duke over the years can be seen in the table given below:

Who is good? Students are misbehaved by professors, staff, principals.

${\lambda }_e=\frac{h}{mv}=\frac{h}{p_e}$    -(1)

${\lambda }_{Ph}=\frac{h}{P_{Ph}}$              -(2)

A/C to question

${\lambda }_e={\lambda }_{Ph}$  

$P_e=P_{Ph}\Rightarrow \frac{P_P}{P_{Ph}}=1$              -(3)

$k.E_e=E_e=\frac{1}{2}m{v}^2$  

$=\frac{1}{2}Pev$              -(4)

$k.E_{Ph}=E_{Ph}=m{c}^2$  

   = mc c

q = P_Ph c               -(5)

$\left(4\right)÷\left(5\right)$  

$\frac{E_e}{E_{Ph}}=\frac{v}{2c}$  

If size of object is very small as compare to wave length of EM wave in free space then, scattering will happen.

Since

$\overrightarrow{E}.\overrightarrow{B}=0\left(\overrightarrow{E}\perp \overrightarrow{B}\right)$  

$\&\frac{E_0}{B_0}=C=3\times 10^{8}m/sec$  

for option D

$\frac{E_0}{B_0}=\frac{\sqrt[]{301.6^2+452.4^2}}{10^{-6}\sqrt[]{1.5^2+1^2}}\begin{array}{cc}\hfill =301.6\times 10^6\hfill & \hfill =3.01\times 10^{8}m/sec\hfill \end{array}$  

No, right now admissions are not open in the Siddharth College of Law. But I have an existing opportunity available for you right now only if you are a female student. You can apply for admission at Shanti Devi arya mahila college, Dinanagar, District Gurdaspur. Right now is the examination time but still you can get admission here. Consult me for more details.

In an oxidation-reduction reaction or redox reaction, the process of oxidation (addition of electrons) and reduction (removal of electrons) happens at the same time.

$r=\frac{mv}{qB}$

$r_{\alpha }=\frac{m_{\alpha }v}{q_{\alpha }B}$

$r_P=\frac{m_{P}v}{q_{P}B}$

$\frac{r_{\alpha }}{r_P}=\frac{m_{\alpha }{q}_P}{m_P{q}_{\alpha }}=\frac{m_{\alpha }}{m_P}\left(\frac{q_P}{q_{\alpha }}\right)$

$=\frac{4m}{m}\left(\frac{q}{2q}\right)=2:1$

Redox reaction does not have a fixed formula but is an equation that forms when oxidation and reduction occur simultaneously. For example, 2Fe + 3Cl? 2FeCl? (Fe loses electrons and Cl gains).

Redox reactions are of four kinds: Decomposition, Combination, Displacement, and disproportionation reactions.

Volume = constant

$A\mathcal{l}=C$  

$Ad\mathcal{l}+\mathcal{l}dA=0$  

$\frac{d\mathcal{l}}{\mathcal{l}}+\frac{dA}{A}=0$  

$\frac{dR}{R}\times 100=\left(\frac{d\mathcal{l}}{\mathcal{l}}-\frac{dA}{A}\right)\times 100$  

= 0.4 - (-0.4)

= 0.8 %

$V=\frac{-d?}{dt}=-\left(15t^2+8t+2\right)$

at t = 2 sec, V = - (15 × 4 + 8 × 2 + 2)

 = 78 V

$I=\frac{V}{R}=\frac{78}{5}=15.6A$

Due to H- bond in water, it has high melting point and melting point of other hydrides of the group are depending upon the molecular weight.

Due to high crystallity Be has the highest M.P.

Be = 1560 K

Mg = 925 K

Ca = 1120 K

Sr = 1062 K

During the electrolysis of dilute H₂SO₄

$2H_{2}S{O}_4\stackrel{\left(1\right)electrolysis}{\to }2HS{O}_4^{-}+2H^{+}$  

$H_2{S}_2{O}_8+2H_{2}O\stackrel{hydrolysis}{\to }2H_{2}S{O}_4+\underset{\left(A\right)}{H_2{O}_2}$  

$2H^{+}+2e\to H_2\uparrow$  

In the solid form of  $H_2{O}_2$ dihedral angle is equal to 90.2°.