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The UPESEAT 2026 application process comprises steps such as registration, form-filling, uploading the documents and paying the registration fee.

Kindly consider the following figure

$9*9*3=81*3=243$

Kindly consider the following figure

$\overrightarrow{L}=\overrightarrow{r}\times m\overrightarrow{v}$

= $\left(3\widehat{i}-\widehat{j}\right)\times \left(3\widehat{j}+\widehat{k}\right)$

$=9\widehat{k}+3\left(-\widehat{j}\right)-\widehat{i}$

$=-\widehat{i}-3\widehat{j}+9\widehat{k}$

$\left|\overrightarrow{L}\right|=\sqrt[]{1+9+81}=\sqrt[]{91}$

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The requirements for the SRMJEEE mock tests are mentioned below:

1. Laptop with functional camera and microphone.
2. Intel i3 (4TH Gen) or above for no lags
3. 4gb RAM or above so that the functioning is smooth
4. High internet Speed
5. Any the latest browser - Mac 10.14 Mojave and above

Clearly all equations can be $\lambda x-2y-6z=0$  form where $\lambda =1,\frac{4}{3},\frac{6}{5}$  

So for $x=0y=-3z$

So $x^2+y^2+z^2=10z^2$  (as  ) $x=0\mathrm{}y=-3z$

ATQ $10\le 10z^2\le 200$

$1\le z^2\le 20\Rightarrow z=\pm 1,\pm 2,\pm 3,\pm 4$

m = 0.3g density,   ${\rho }_{\left(ball\right)}=8g/cc$                   $\therefore V=\frac{m}{\rho }=\frac{0.3}{8}cc$                     

${\rho }_{glycerene}=1.3g/cc=1.3\times 10^{3}kg/m^3$

F_v + F_B = mg.

Þ F_v = mg – F_B = .3 × 10^-3 × 10 – 1.3 ×  $\frac{0.3}{8}\times 10^{-6}\times 10\times 10^3$

  $=3\times 10^{-3}-\frac{.39}{8}\times 10^{-2}=3\times 10^{-3}-.5\times 10^{-3}$

= 2.5 × 10^-3N

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If you can't access the SRMJEEE mock tests, you don't have worry as you can always access to third party mock tests. Although they are paid, but the costing is pretty reasonable. Students can also download the past year question papers and solve them by keeping a timer clock, this way they will be able to track their progress.

By property

If first and Last term of Andhra Pradesh and HP are equal (of nn  terms) then $a_r{H}_S=$  first *  x last term

If $r+s=n+1$ r+s=n+1

we know, = $\frac{\lambda }{d}$ $\frac{}{}$ $\mu =\frac{C}{V}=\frac{C}{\upsilon \lambda }$ $\frac{}{}\frac{}{}$

${}_{}\frac{}{{}_{}{}_{}}$  ${\lambda }_1=\frac{C}{{\mu }_1{\upsilon }_1}$ [with change in medium frequency remain constant]

${\lambda }_2=\frac{C}{{\mu }_2{\upsilon }_2}$

$\frac{{\lambda }_2}{{\lambda }_1}=\frac{{\mu }_1}{{\mu }_2}=\frac{5}{7}$ ${\lambda }_2=\frac{5}{7}{\lambda }_1$

${\theta }_1=\frac{{\lambda }_1}{d_1}$

2= $\frac{{\lambda }_2}{d_2}$

$\frac{{\theta }_2}{{\theta }_1}=\frac{{\lambda }_2}{{\lambda }_1}[?d_1=d_2]$

2 $=\frac{5}{7}\times {\theta }_1$

${\theta }_2=\frac{5}{7}\times 0.35$

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$\frac{B_0^2}{2{\mu }_0}C$

$B_0^2=\frac{I2{\mu }_0}{C}$

$B_0^2=\frac{0.22\times 2\times 4\times 10^{-7}}{9\times 108}$

B₀ = 42.9 × 10^-9

42.9 Ans

Equation of tangent of at $y=\mathrm{l}\mathrm{o}\mathrm{g}?x$  at $\left(e^{\lambda },\lambda \right)$  is $y-\lambda =\frac{1}{e^{\lambda }}\left(x-e^{\lambda }\right)$

It cuts y   axis at $(0,\lambda ,-1)$

Also normal to $y^2=8x$    at $\left(\mathrm{7,4}\right)$    is $y-4=-(x-2)$  

  $y=-x+6\mathrm{}$  Cuts y axis at $\left(\mathrm{0,6}\right)\mathrm{}$  SO ATQ $6=-1+\lambda \Rightarrow \lambda =7$

To access the SRMJEEE mock tests, candidates will have to follow the steps mentioned below - 

1. Visit the official website or just google "SRMJEEE Mock Test official portal"
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4. Take picture of yourself and your ID
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