Class 11th

  $x^2+y^2-2x+2fy+1=0centre=\left(1,-f\right)$

diameter 2px – y = 1 ………. (i)

2x + py = 4p    ……… (ii)

$x=\frac{5P}{2P^2+2}$          

f = 0

$\left[forP=\frac{1}{2}\right]$                

$\frac{5P}{2P^2+2}=1$            

f = 3 [for P = 2]

substitute (2, 3)

$3=m\pm \sqrt[]{m^2-3}$      

$\therefore m=2$

a₁ = b₁ = 1

$a_n=a_{n-1}+2\left(forn\ge 2\right)b_n=a_n=b_{n-1}$          

Similarly for others

${\displaystyle {\sum }_{n=1}^{11}{a}_n{b}_n={\displaystyle {\sum }_{n=1}^{15}\left(2n-1\right)n^2={\displaystyle {\sum }_{n=1}^{15}2n^3-{\displaystyle {\sum }_{n=1}^{15}{n}^2}}}}$      

$=2{\left[\frac{15*16}{2}\right]}^2-\left[\frac{15*16*31}{6}\right]=27560$

  $?$  Roots of 2ax²  - 8 ax + 1 = 0 are

$\frac{1}{p}and\frac{1}{r}$ and roots of 6bx² + 12bx + 1 = 0 are

  $\frac{1}{q}and\frac{1}{8}$  

Let  $\frac{1}{p},\frac{1}{q},\frac{1}{r},\frac{1}{8}$

as    $\alpha -3\beta ,\alpha -\beta ,\alpha +\beta ,\alpha +3\beta$

$So,\frac{1}{a}-\frac{1}{b}=38$

Yes, it's an important concept to tackle JEE Main questions on rotational mechanics, rigid bodies, and collisions. Learning about the centre of mass simplifies complex motions of objects by treating the entire mass into a single point.  In exams such as JEE, questions on centre of mass are also interrelated with other advanced concepts in physics. So, a thorough conceptual understanding of it is essential.   

Arranging letter in alphabetical order A   D   I   K   M   N   N for finding rank of MANKIND making arrangements of dictionary we get

A …….->

$\frac{6!}{2!}=360$        

D ………->360

l ………. ->360

K ………->360

MAD …….->

$\frac{4!}{2!}=12$        

$\therefore$ Rank of MANKIND = 1440 + 36 + 12 + 2 + 2 = 1492.

Degree of unsaturation = 4 + 1 -   $\left(\frac{5-1}{2}\right)=3$

sp³ hybridized only one carbon

Organic compound -> NH₃

$2N{H}_3+H_{2}S{O}_4\to {\left(N{H}_4\right)}_{2}S{O}_4$  

No. of millimoles of NH₃ = 2 * no. of milimoles of H₂SO₄

= 2 * 2 * 2.5

= 10 milimoles

= No. of milimoles of N

Mass of nitrogen =  $\frac{10}{100}*14=0.14gm$

% of Nitrogen in compound =  $\frac{0.14}{0.25}*100$

= 56%

Pressure of gas = pressure of moist gas – Vapour pressure of water

= 4 atm – 0.4 atm

= 3.6 atm

On doubling the volume, pressure of gas is halved :

Total pressure =  $\left(\frac{3.6}{2}+0.4\right)=2.2atm$

3C (Graphite) + 4H₂ (g) -> C₃H₈ (g)

$\Delta H_f\left(C_3{H}_8\right)=\left[3*\Delta H_{comb}\left(C\right)\right]+\left[4*\Delta H_{comb}\left(H_2\right)\right]-\left[\Delta H_{comb}\left(C_3{H}_8\right)\right]$        

= -10.3.7 kJ/mole

In N₂ & $O_2^{2-}\to$ no. of unpaired electron = 0

In other species ® no. of unpaired electron $\ne$  0