Class 11th

 $\frac{dU}{dr}=0\Rightarrow \frac{d}{dr}\left[\frac{A}{r^{10}}\right]-\frac{d}{dr}\left[\frac{B}{r^5}\right]$

$\Rightarrow \frac{d}{dr}\left[A{r}^{-10}\right]-\frac{d}{dr}\left[B{r}^{-5}\right]=0$

$\Rightarrow -10A{r}^{-11}+5B{r}^{-6}=0$

$r={\left(\frac{2A}{B}\right)}^{\frac{1}{5}}$

$\underset{\_}{\begin{array}{l}A\left(g\right)?B\left(g\right)+\frac{1}{2}C\left(g\right)\\ t=0amoles00\\ -a\alpha moles+a\alpha moles+\frac{a\alpha }{2}moles\end{array}}$

Eq.   a(1 - α)          aα           (aα/2)

Moles           moles       moles

Total no. of moles at equilibrium

= nA + nB + nC

$=a\left(1-\alpha \right)+a\alpha +\frac{a\alpha }{2}=a\left[1+\frac{\alpha }{2}\right]$

${\left(P_A\right)}_{eq}={\left(x_A\right)}_{eq}*P_{eq}=\frac{a\left(1-\alpha \right)}{\left(1+\alpha /2\right)}P=\frac{1-\alpha }{\left(1+\alpha /2\right)}P$

${\left(P_B\right)}_{eq}={\left(x_B\right)}_{eq}*P_{eq}=\frac{a\alpha }{a\left(1+\alpha /2\right)}P=\frac{\alpha }{\left(1+\alpha /2\right)}P$

${\left(P_C\right)}_{eq}={\left(x_C\right)}_{eq}*P_{eq}=\frac{a\alpha /2}{a\left(1+\alpha /2\right)}P=\frac{\alpha /2}{\left(1+\alpha /2\right)}P$

$K_P=\frac{{\left(P_B\right)}_{eq}*{\left(P_C\right)}_{eq}^{1/2}}{{\left(P_A\right)}_{eq}}=\frac{{\left(\alpha \right)}^{3/2}{P}^{1/2}}{\left(1-\alpha \right){\left(2+\alpha \right)}^{1/2}}$

  a + a₁……a_n, 100               n arithmetic mean 100

a + n = 33 ……. (i)

$\left(\frac{a_1}{a_n}=\frac{1}{7}\right)$

$\frac{\left(a+d\right)}{\left(100-d\right)}=\frac{7}{7}$

7a + 8d = 100…………… (ii)

a + (n + 1)d = 100………………. (iiI)

Solving these equations (i), (ii) & (iii), we get

n = 23 & d =    $\frac{15}{4}$

a = 10

  $\left(1-x^2+3x^3\right){\left(\frac{5}{2}{x}^3-\frac{1}{5x^2}\right)}^{11},x\ne 0$

(r + 1)^th term for expansion of   ${\left(\frac{5}{2}{x}^3-\frac{1}{5x^2}\right)}^{11}$

  n – r = x₃

r = y   $0\le x,y\le 11shouldbenaturalnumber.$

= ¹¹C_y   $*{\left(\frac{5}{2}\right)}^x*{\left(-\frac{1}{5}\right)}^y*x^{\left(3x-2y\right)}$

for term to be independent of x, power of x should be zero.

(3x – 2y) = 0 (when 1 is multiplied).

3x + 3y = 33

y = $\frac{33}{5}\left(Nosolution\right)$  

3x – 2y + 2 = 0 where (-x²) is multiplied).

3x +  3y = 33

y = 7, x = 4

$\begin{array}{cc}\hfill 3x-2y+3=0\hfill & \hfill 3x+3y=33\hfill \end{array}\}$

$\therefore$   coefficient of term independent of x.

$\left(-1\right){*}^{11}\text{?}{C}_7*{\left(\frac{5}{2}\right)}^4*{\left(\frac{-1}{5}\right)}^7$

$=\frac{33}{200}$

${\left[PtC{l}_4\right]}^{2-}\to Pt$ has dsp² hybridization

BrF₅ -> Br has sp³d² hybridization

PCl₅ -> P has sp³d hybridization

[Co (NH₃)₆]³⁺ -> Co has d²sp³ hybridization.

Degenerate orbitals must have same value of energy

Orbitals with same n and   values are degenerate orbitals.

Let the number of chocolates given to C₁, C₂, C₃ & C₄ be a, b, c, d respectively.

Given 4   $\le b\le 7$

$2\le c\le 6$

Now using these the maximum number of chocolates that can be given to C₁ or C₄ is 24 (where b & c are given 2 & 4 chocolates).

$\therefore 0\le a\le 24$

$0\le d\le 24$

& a + b + c + d = 30

So, total possible solution to the above equation.

Coefficient of x³⁰ in.

$\left(x^0+x+x^2+....+x^{24}\right)\left(x^4+x^5+....+x^7\right)\left(x^2+x^3+....+x^6\right)\left(x^0+x+x^2+....+x^{24}\right)$

=  ${\left(1+....+x^{24}\right)}^2\left(x^4\right)\left(1+x+....+x^3\right)*x^2\left(1+x+....+x^4\right)$

$=\left(x^{56}-2x^{31}+x^6\right)*\left(x^9-x^4-x^5+1\right)*{\left(x-1\right)}^{-4}$

x⁵⁶ & x³¹ can never give x³⁰ so we discard them.

$x^6*x^9*{\left(x-1\right)}^{-4}{\to }^{15+4-1}\text{?}{C}_{4-1}{=}^{18}\text{?}{C}_3$

Coefficient x³⁰ ® ¹⁸C₃ – ²³C₃ – ²²C₃ + ²⁷C₃

=   $\frac{18*17*16}{6}-\frac{23*22*21}{6}-\frac{22*21*20}{6}+\frac{27*26*25}{6}$

= 430

Mass of C₁₅H₃₀ = volume * Density

= 1000 $m\mathcal{l}$  * 0.756 gm/ $m\mathcal{l}$  

C₁₅H₃₀ + 22.5 O₂   $\stackrel{}{\to }$ 15CO₂ + 15H₂O

No. of moles of C₁₅H₃₀ = $\frac{756}{210}$ moles

No. of moles of O₂ required =   $\left(22.5*\frac{756}{210}\right)moles$

Mass of O₂ required = 22.5 *   $\frac{756}{210}*32=2592gm$

No. of moles of CO₂ liberated = 15 *   $\left(\frac{756}{210}\right)$ moles

Mass of O₂ liberated =   $15*\frac{756}{210}*44=2376gm$

${\omega }_1=\sqrt[]{\frac{k}{m}}=\sqrt[]{\frac{2k}{0.05}}$

$\begin{array}{l}\Rightarrow A_1{\omega }_1=A_2{\omega }_2\\ \frac{A_1}{A_2}=\frac{{\omega }_2}{{\omega }_1}=\sqrt[]{\frac{9k}{0.1}}*\sqrt[]{\frac{0.05}{2k}}\end{array}$

$\frac{A_1}{A_2}=\frac{3}{2}$

T₁ = 727 + 273 = 1000k

T₂ = 127° + 273 = 400 k

Q₁ = 5 * 10³ k cal

$\Rightarrow \frac{600}{1000}=\frac{w}{5000}$                

w = 12.6 * 10⁶ J