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The NEET registrations for the year 2025 have been closed. The NEET registration for the year 2025 were scheduled between 7th February till 7th March. The NEET exam was conducted during 4th May this year. The counseling rounds have been postponed with the new dates yet to be revealed. The timeline for the NEET exam 2026 will be revealed at a later date.

$R_1={?}_1\frac{\mathcal{l}}{A}{R}_2={?}_2\frac{\mathcal{l}}{A}$

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{A}{\mathcal{l}}\left[\frac{1}{{?}_1}+\frac{1}{{?}_2}\right]$

$=\frac{3*10^{?6}}{\frac{1}{4}}\left[\frac{1}{1.7*10^{?8}}+\frac{1}{2.6*10^{?8}}\right]$

$R=0.858*10^{?3}?$

Yes, UGI Greater Noida registeration form is currently available on the official website. Aspirants of the BBA programme who meet the course-specific eligibility criteria can visit the official website to apply. The applications must be submitted timely to get admission and for application acceptance. 

$\overrightarrow{r}.\left(\widehat{i}-\widehat{j}+2\widehat{k}\right)=2$

&  $\overrightarrow{r}.\left(2\widehat{i}+\widehat{j}-\widehat{k}\right)=2$ are two planes the direction ratio of the line of intersection of then is collinear to   

$\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill -1\hfill \end{array}\right|=-\widehat{i}+5\widehat{j}+3\widehat{k}$

Any point on the line in given by x – y = 2

& 2x + y = 2

$\Rightarrow x=\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.,y=\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.,z=0$

$\therefore equationoflineL:\frac{x-\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{-1}=\frac{y+\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{5}=\frac{z}{3}=r$

$\therefore pointP\left(\frac{33}{35},\frac{45}{35},\frac{41}{35}\right)\equiv \left(\alpha ,\beta ,\gamma \right)$

$\therefore 35\left(\alpha +\beta +\gamma \right)=119$                                           

A + B + C = 18500…. (1)

A = B + 25% of   $B=\frac{5B}{4}$ $\frac{}{}$ …. (2)

B = C + 20% of $C=\frac{6C}{5}$ $\frac{}{}$ …. (3)

Equation (1) gives,

$\frac{5B}{4}+\frac{6C}{5}+C=18500$

$\Rightarrow \frac{3C}{2}+\frac{6C}{5}+C=18500$

37C = 185000 C = 5000

B = 6000, A = 7500.

A + B = 13500

The Bihar BEd CET (Combined Entrance Test) is a 120-mark exam, with each question carrying one mark. There is no negative marking.

Using conservation of energy:

$mgh=\frac{1}{2}{l}_{cm}{\omega }^2+\frac{1}{2}m{v}^2=\frac{1}{2}{l}_{cm}\frac{V^2}{r^2}+\frac{1}{2}m{v}^2$

$v^2=\frac{mgh}{\frac{1}{2}\frac{l_{cm}}{r^2}+m{v}^2}$

So, sphere has the greatest velocity of COM and ring has least.

The NEET entrance exam has been set for May 3, 2026, for admission to UG-level medical courses. Students need to submit an online application form to enter the admission process.

Total energy of monoatomic gas at equilibrium

$=\frac{3}{2}{K}_{B}T$

Yes, a three-year BBA course is offered at the Greater Noida campus of the United Group of Institutions. It is offered through the United College of Education. Candidates seeking admission in the UG course must meet the eligibility and selection criteria. For selection, the applicants performance in the IPU CET examination is evaluated. Students who meet the eligibility requirements can visit the official website to register and apply.

Let the total work by LCM of 20, 24 and 30 i.e. 120 units

Work done per day by A and B = a + b = $\frac{120}{20}=6$ $\frac{}{}$

Work done per day by B and C = b + c = $\frac{120}{24}=5$ $\frac{}{}$

Work done per day by A and C = a + c = $\frac{120}{30}=4$ $\frac{}{}$

2 (a + b + c) = 15

So, a + b + c = 7.5

Required number of days = $\frac{120}{7.5}=16\text{?}days$

$\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}-\left(i\right)$              

$\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}-\left(ii\right)\left|\overrightarrow{a}\right|=2$              

Taking dot product with  $\overrightarrow{c}\&\overrightarrow{a}$ respectively in (i) & (ii) we have 

  $\left[\overrightarrow{c}\overrightarrow{a}\overrightarrow{b}\right]={\left|\overrightarrow{c}\right|}^2$            

Again (i) & (ii) $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0$ Þ    

Opt 1.   Projection of $\overrightarrow{a}on\overrightarrow{b}\times \overrightarrow{c}=\frac{\overrightarrow{a}.\left(\overrightarrow{b}\times \overrightarrow{c}\right)}{\left|\overrightarrow{b}\times \overrightarrow{c}\right|}=\frac{4}{2}=2$  

 Opt 2.   $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]+\left[\overrightarrow{c}\overrightarrow{a}\overrightarrow{b}\right]=2\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=8$ (2)

(1) $\overrightarrow{b}\times \left(\overrightarrow{a}\times \overrightarrow{b}\right)=\overrightarrow{b}\times \overrightarrow{c}$ Þ

Opt 3.   ${\left|3\overrightarrow{a}+\overrightarrow{b}-2\overrightarrow{c}\right|}^2=9{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+4{\left|\overrightarrow{c}\right|}^2+2\left(3\overrightarrow{a}.\overrightarrow{b}-6\overrightarrow{a}.\overrightarrow{c}-2\overrightarrow{b}.\overrightarrow{c}\right)$  

Opt 4.  $\overrightarrow{a}\times \left(\overrightarrow{c}\times \overrightarrow{b}-\overrightarrow{b}\times \overrightarrow{c}\right)$  

x (t) = A sin $\omega$ $$ t + B cos $\omega t$ $$

$\Rightarrow x\left(t\right)=\sqrt[]{A^2+B^2}sin\left(\omega t+\delta \right)=\sqrt[]{A^2+B^2}cos\left(\omega t-\left(\frac{\pi }{2}-\delta \right)\right)$

At t = 0:

$x\left(0\right)=\sqrt[]{A^2+B^2}sin\left(\delta \right)$

$v\left(0\right)=\omega \sqrt[]{A^2+B^2}cos\left(\delta \right)$

$\Rightarrow {\left(x\left(0\right)\right)}^2+{\left(\frac{v\left(0\right)}{\omega }\right)}^2=A^2+B^2$

$tan?=tan\left(\frac{\pi }{2}-\delta \right)=cot\delta =\frac{v\left(0\right)}{x\left(0\right)\omega }$

$\Rightarrow ?=ta{n}^{-1}\left(\frac{v\left(0\right)}{x\left(0\right)\omega }\right)$

Let the numbers be x & y

(mean proportion)2 = xy.

So, xy = 784

x = $\frac{784}{y}$ $\frac{}{}$ .… (1)

x, y and 224 are in continued proportion

y2 = 224 x .… (2)

y2 = 224 * $\frac{784}{y}$ $\frac{}{}$

y3 = 14 * 16 * 28 * 28

y = 56 and x = 14

Alternate Method

Use options and check.

Yes, a candidate can apply for admission in the M.Tech course at Techno Main Salt Lake if graduation result is pending. The institute may offer them provisional admission in this case. However, the candidate must submit the marksheet and certificate for their graduation once the results are declared. After completing this step, the institute will confirm their admission in the M.Tech course.