Class 11th

  $L=\underset{x\to \infty }{lim}\frac{{\left(2+\frac{1}{x}\right)}^{40}{\left(4-\frac{1}{x}\right)}^5}{{\left(2+\frac{3}{x}\right)}^{45}}=32$

Any point on x + y = 1 can be taken as (t, 1 – t) The equation of chord with this as mid-point is y (1 – t) -2a (x + t) = (1 – t)² – 4at

It passes through (a, 2a)

So, t² – 2t + 2a² – 2a + 1 = 0

This should have two distinct real roots.

So D > 0

$\begin{array}{l}\Rightarrow a^2-a<0\\ \Rightarrow 0<a<1\end{array}$          

So, length of latus rectum < 4 and 0 < a < 1

$\Rightarrow LR=1,2,3$            

Out of numbers $\frac{1}{2},\frac{2}{4},\frac{3}{6},\frac{1}{3},\frac{2}{6},\frac{2}{3},\frac{4}{6}$ will result only 3 distinct rational numbers.

Total numbers =   

Total number of electron in Ti = 22
Total number of electron in Ti? = 22 – 4 = 18 So EAN value of Ti = 18 + 12 + 4 = 34

Let P is a root   $\left(\in I\right)$

Case-I : p is odd

p¹⁰ + ap⁹ + b   $\ne 0$

Because LHS odd

Case-II : p is even

$p^{10}+a{p}^9+b\ne 0$  

because, LHs is odd

Initial m equivalent of Cu²? = 200 * 0.5 * 2 = 200 m eq
So electricity passed = (0.965*3600)/96500 = 36 * 10? ³ eq. = 36 m eq
m eq CuBr? remaining = 200 – 36 = 164 ∴ N = meq / V (in ml) = 164/200 = 0.82

From ΔT? = K? * m, and ΔT? = 15°C
∴ m = ΔT? / K? = 15 / 1.86 = 8.06
So the amount of propyl alcohol to be added.
= m * molwt = 8.06 * 60 = 483.6 g

Factual on lyophilic colloids
(A) It is easy to prepare
(B) Stable
(C) Revisable
(D) Viscosity is high and surface tension is low for DP. and DM.

Sm ⇒ +2, +3 (4f? , 6 s²); Gd ⇒ +3 (4f?5 d¹6 s²)
Tm ⇒ +2 + 3 (4f¹³,6 s²); Nd ⇒ +2, +3, +4 (4f? , 6 s²)

${\eta }_1=1-\frac{T_2}{T_1},{\eta }_2=1-\frac{T_3}{T_2}$

as ${\eta }_1={\eta }_2\Rightarrow \frac{T_2}{T_1}=\frac{T_3}{T_2}\Rightarrow T_2={\left(T_1{T}_3\right)}^{\frac{1}{2}}=\sqrt[]{400*900}=600\mathrm{K}$