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All monosaccharides which differ in configuration at C? and C? gives the same osazone. Since, glucose and fructose differ from each other only in configuration at C? and C? therefore, they give the same osazone. All other options given in the questions do not satisfy this condition and hence, do not from the same osazone.

No. Candidates can only apply for the JEE Advanced exam online on the official website - jeeadv.ac.in. There is no offline for available to apply for the exam. All steps of the application process is online.

Chemistry is all about the scientific knowledge of matter at a molecular level. It makes sense of their reactions for the creation of new products, while Chemical Engineering focuses on applying that very knowledge for the design and optimization of industrial chemical processes and products like chemicals, drugs, and foods. In simple words, Chemistry deals with the Science behind the products, while Chemical Engineering focuses on how to produce those products on a large scale.

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Chemical Engineering Career

The number of students placed over the past three years during JNU UG placements is presented below:

The second ionization energy of K is maximum, because the second electron is removed from fully filled 3p? subshell. The second ionization energies of group 2 elements decrease down the group.
Hence, second I.E. of Ca > second I.E. of Ba.

Candidates can check the JEE Advanced eligibility on the official website - jeeadv.ac.in and also on the official JEE Advanced information brochure released every year.

$\frac{dp}{dt}=0.5p-450andP\left(0\right)=850$           

$\Rightarrow \frac{dp}{P-900}=0.5dt$         

${\displaystyle \underset{850}{\overset{0}{\int }}\frac{dp}{P-900}={\displaystyle \underset{0}{\overset{T}{\int }}0.5dt}}$           

$\Rightarrow ln\left(P-900\right){|}_{805}^0=0.5T$

$\frac{T}{2}=ln\left|\frac{900}{50}\right|=ln18$           

T = 2 ln 18

Nitrogen, sulphur and halogens are tested in an organic compound by lassaigne's test.
The organic compound is fused with sodium metal as to convert these elements into ionisable inorganic substances.
Na + C + N → NaCN
2Na + S → Na? S
2Na + X? → 2NaX
The cyanide, sulphide or halide ions can be confirmed in aqueous solution by usual test.

The median salary trend witnessed over the past three years during JNU placements is presented below:

The age limit to apply for JEE Advanced 2026 is:

• Candidate should be born on or after October 1, 2001.
• Age relaxation of five years is given to SC, ST and PwD candidates.

International candidates coming from different countries must demonstrate their English language proficiency by submitting the scores of IELTS, PTE, and TOEFL. Students who are aiming for Conestoga must submit a minimum PTE score of 58 or above. Students can check the Conestoga College English language requirements below:

The process of moving sodium and potassium ions across the cell membrane is an active transport process involving the hydrolysis of ATP to provide the necessary energy.

${\Delta }_t=\frac{hc}{\lambda }=\frac{6.63\times 10^{-34}\times 3.08\times 10^8}{600\times 10^{-9}}$

$=\frac{6.63\times 3.08\times 10^{-17}}{600}$ J

$=765.765\times 10^{-21}J$

$?766\times 10^{-21}$ J

Conestoga College give an offer letter to Indian students within 6 weeks after they submit their applications. If students are taking admission in peak intakes like January, May, and September, they may receive the offer letter a bit late. But students who apply early and submit all the required documents will receive their offer letter early. 

The official website for JEE Advanced remains the same every year. The official JEE Advanced website is  - jeeadv.ac.in.

  $f\left(x\right)=\frac{4x^3-3x^2}{6}-2sinx+\left(2x-1\right)cosx$

$f\text{'}\left(x\right)=2x^2-x-2cosx+2cosx-\left(2x-1\right)sinx$

= $\left(2x-1\right)\left(x-sinx\right)\ge 0forx\ge 0$  

$f\text{'}\left(x\right)\ge 0\forall x\ge 1/2$           

$\therefore f\left(x\right)$ is increasing in $\left[\frac{1}{2},\infty \right)$  

The given value of µ (spin only)
2.84 = √n (n + 2) BM, So, n = 2
Among the given configurations, d? system in strong field ligand will have 2-unpaired e? in t? g set of orbitals as shown below.

$\frac{d\left[C\right]}{dt}=\left(\frac{20-10}{10}\right)=1mmold{m}^{-3}$

$+\frac{d\left[D\right]}{dt}=\left\{-\frac{d\left(B\right)}{dt}\right\}\times 1.5=\frac{3}{2}\left\{-\frac{d\left[B\right]}{dt}\right\}$

$\left\{-\frac{d\left[B\right]}{dt}\right\}=2\times \left\{\frac{-d\left[A\right]}{dt}\right\}$

$\therefore \frac{1}{6}\left\{\frac{-d\left[B\right]}{dt}\right\}=\frac{1}{3}\left\{\frac{-d\left[A\right]}{dt}\right\}=\frac{1}{9}\left\{\frac{+d\left[D\right]}{dt}\right\}=\frac{d\left[C\right]}{dt}$

$\therefore$ rate of reaction = $\frac{+d\left[C\right]}{dt}$  = 1m.m dm^-3 S^-1

The key highlights of JNU School of Engineering placements 2023 are presented below: