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SITEEE is short for Symbiosis Institute of Technology Engineering Entrance Examination.

Once the ICAR AIEEA Answer key is out, candidates who appeared for the exam get the opportunity for two to three days to raise objections against the provisional answer key.

They need to submit an objection fee along with the valid proof to challenge the answer key. If the objection is valid, there will be changes made accordingly. Also, candidates can calculate their expected ICAR AIEEA score before the result is declared. 

  ${\displaystyle \int \frac{\left(cosx-sinx\right)dx}{\sqrt[]{8-sin2x}}=asi{n}^{-1}\left(\frac{sinx+cosx}{b}\right)+c}$

${\displaystyle \int \frac{\left(cosx-sinx\right)dx}{\sqrt[]{9-{\left(sinx+cosx\right)}^2}}=}$           

Put (sin x + cos x) = t Þ (cos x – sin x) dx = dt

$={\displaystyle \int \frac{dt}{\sqrt[]{3^2-t^2}}=si{n}^{-1}\frac{t}{3}+c=si{n}^{-1}\left(\frac{sinx+cosx}{3}\right)+c}$        

= a = 1, b = 3

$\therefore \left(a,b\right)\equiv \left(1,3\right)$        

Temperature coefficient is 2 that means rate of reaction doubles at every 10°C rise in temperature.
Thus, k? °C = 2? = 64 k? °C

Candidates should know that they will get a total of 120 questions when they appear for the ICAR AIEEA Exam, and they will get a total of 2 hours to complete their exam. The mode of exam is online, a computer-based test, and the language of the question paper is english. 

Number of mol of A reacted = 1×10? = 10?
Number of molecule of A = 10? × 6 ×10²³
Quantum efficiency = (6 ×10¹? ) / (1.2 ×10¹? ) = 0.50

Candidates can find the solved question papers of JEE Advanced exam on the websites of various coaching centres. Also shiksha.com have comoile the solved past 10 year JEE Advanced question papers here: Last 10 Year JEE Advanced Question Papers with Solutions: Download Here.

The SITEEE 2026 exam will be conducted in online mode at designated test centres. The allotted centre details will be available on the admit card.

To decompose 1 mole of H? O, we need two moles of electrons i.e. 2 faraday
To decompose 4 moles of H? O, we need 8 faraday.
Now, Q = I × t? t = (8 × 96500) / 4 = 1.93×10? sec

The previous year JEE Advanced question papers can be downloaded from the official website - jeeadv.ac.in. The official website has the previous year IIT JEE question papers from 2007 to 2025.

According to question,

  $\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{4}$         

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}........(i)$           

Equation of required line is $\frac{x}{a}+\frac{y}{b}=1$  

Obviously B (2, 2) satisfying condition (i)

ΔG° = – 2.303RT log Kp = – 2.303× 2 × 300log10? ³
= – 2.303× 2 × 300 × (– 3) = 4145.4cal mol? ¹

4x – 3y + k₂ = 0

$2r=\frac{40}{5}=8\Rightarrow r=4$

${\left(x-1\right)}^2+{\left(y+2\right)}^2=16$

P? = 1atm
P? = 1atm × 40/100 = 0.40atm
V? = 100 cm³
V? =?
At constant temperature, P? V? = P? V?
So V? = (P? V? ) / P? = (1atm × 100 cm³) / 0.40 atm = 250cm³
Hence, the volume of bulb B = (250 –100) cm³ = 150 cm³

As 500g of tooth-paste has = 0.2g of F?
So, 10 g of toothpaste has = (0.2g × 10? g) / 500g F = 400g of F
Hence, concentration in ppm is 400.

  $\Delta -\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 2\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 0\hfill & \hfill -8\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|,\left[R_2-R_1-2R_3\right]$

= -8 (-3 + k)

For inconsistent $\Delta =0\Rightarrow k=3$  

${\Delta }_x=\left|\begin{array}{ccc}\hfill 10\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 6\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 5m\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=32-40m\ne 0\Rightarrow m\ne \frac{4}{5}$

As 3f-orbital does not exist as value of l cannot be equal to n, hence no electron can be accommodated.