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Fix the unit place, find the chances for the first three digits

unit digit as 1, total ways = 9.10²

unit digit as 2, total ways = 4.5²

unit digit as 3 total ways = 3.4²

unit digit as 4 total ways = 2.3²

unit digit as 5 total ways = 1.2²

unit digit as 6 total ways = 1.2²

unit digit as 7 total ways = 1.2²

unit digit as 8 total ways = 1.2²

unit digit as 9 total ways = 1.2²

There is no ideal number of sample papers to be solved. What you can do is make sure to solve 2 to 3 question paper every day so that you have a good practice before the exam.

All metal carbonyls have synergic bonds

Rate constant, k = 5.5 × 10^-14 s^-1.

$t=\frac{2.303}{k}log\frac{{\left[R\right]}_0}{\left[R\right]}$  

 = $\frac{2.303}{k}log3-(i)$  

$t_{50\mathrm{\%}}=\frac{2.303}{k}log2-(ii)$  

 From (i) & (ii)

$\frac{t_{67\mathrm{\%}}}{t_{50\mathrm{\%}}}=\frac{log3}{log2}$  

= 1.58 t_50%

So;         t_67% is 15.8 × 10^-1 times half life.

X = 16 (the nearest integer)

$T_{r+1}{=}^{15}\text{?}{C}_r\cdot {\left(2x^{1/5}\right)}^{15-r}{\left(-x^{-1/5}\right)}^r$  

= ¹⁵C_{r $\cdot 2^{15-r}\cdot x^{\frac{15-2r}{5}}\cdot {\left(-1\right)}^r$}

Coefficient of x^-1 -> r = 10 ->m = ¹⁵C_{10 $\cdot 2^5$}

$x^{-3}\Rightarrow r=15\Rightarrow n{=}^{15}\text{?}{C}_{15}\cdot 2^0=1$  

now mn² = ¹⁵C_{r $\cdot 2^r$}  

r = 5

As per the JEECUP marking scheme, candidates will be awarded 4 marks for every correct answer marked. There is no negative marking in the exam.

$C{r}_2{O}_7^{2-}+6e^{-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O$  

Here, 6F electricity is required to reduce 1 mol  $C{r}_2{O}_7^{2-}$ to Cr³⁺.

It is conducted once a year only. And it is typically held in the month of March or April. The exact date is announced on NTA a few months before the exam

f (x) is an even function

$f\left(-\frac{1}{4}\right)=f\left(-\frac{1}{2}\right)=f\left(\frac{1}{2}\right)=f\left(\frac{1}{4}\right)=0$  

So, f (x) has at least four roots in (-2, 2)

$g\left(\frac{-3}{4}\right)=g\left(\frac{3}{4}\right)=0$  

So, g (x) has at least two roots in (-2, 2)

now number of roots of f (x) $\cdot g"\left(x\right)=f\text{'}\left(x\right)\cdot g\text{'}\left(x\right)=0$  

It is same as number of roots of  $\frac{d}{dx}\left(f\left(x\right)\cdot g\text{'}\left(x\right)\right)=0$ will have atleast 4 roots in (-2, 2)

Using ; k_sp = 108 S⁵

1.1 × 10^-23 = 108 S⁵

S =  ${\left(\frac{110}{108}\times 10^{-25}\right)}^{1/5}M\approx 10^{-5}M$  

Specific conductance,

${\lambda }_m^0=\frac{\kappa }{5\times 1000}S{m}^{2}mo{l}^{-1}$  

= 3 × 10^-3 Sm² mol^-1

So; x = 3

As per the JEECUP exam pattern students have to attempt 100 MCQs within 150 minutes. The test paper will be available in both Hindi and English. Candidates will get 4 marks for every correct answer they mark. There is no negative marking.

It is important because you can be familiar with the exam pattern, identify important topics, boosts confidence, and helps in practice and revision.

Given a > b

Area common to x² + y² $\le a^{2}and\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1$  

is  $\pi a^2-\pi ab=30\pi ..............\left(i\right)$  

Similarly  $\pi ab-\pi b^2=18\pi .................\left(ii\right)$  

Equation (i) and equation (ii)  $\frac{a}{b}=\frac{5}{3}$  

Equation (i) + equation (ii)  $a^2-b^2=48$  

a² = 75, b² = 27

$P_A^0=50torr$  

$P_B^0=100torr$  

Mole fraction of A in liquid phase,           x_A = 0.3

Mole fraction of B in liquid phase,            x_B = 0.7

Now;     $P_A=P_A^0{x}_A$  

$P_B=P_B^0{x}_B$

= 100 × 0.7 = 70 torr

Mole fraction of B in vapour phase,   $y_B=\frac{P_B}{P_A+P_B}=\frac{70}{85}$  

$\frac{70}{85}=\frac{x}{17}$  

                             X = 14

The Vellore Institute of Technology has released the VITREE 2026 exam dates on the website. Admissions to the January session has started. Candidates can check the VITREE exam dates below -

For expansion in vacuum, workdone, w = 0

For isothermal process,   $\Delta U=0$  

According to first law of thermodynamics,

$\begin{array}{l}\Delta U=q+w\\ q=0\end{array}$  

Hybridization of P in PF₅ is sp³d, so value of y = 1

sin x = 1 – sin² x

sin x =  $\frac{-1+\sqrt[]{5}}{2},\frac{-1-\sqrt[]{5}}{2}\left(rejected\right)$  

draw y = sin x

y =  $\frac{\sqrt[]{5}-1}{2},$ find their pt. of intersection.

The VITREE exam result 2026 will be announced on December 20 (Saturday). Students who have appeared the exam (on December 7) will be able to check their result. The VITREE 2026 result will be announced in the form of rank card. Candidates can access their result using their application number and password. 

Work function, ${?}_0=6.63\times 10^{-19}J$  

Threshold wavelength  ${\lambda }_0=?$  

Using ;                 ${?}_0=\frac{hc}{{\lambda }_0}$  

${\lambda }_0=\frac{hc}{{?}_0}$  

=300 × 10^-9 m = 300 nm