Class 11th

Electronic configuration of Fe is [Ar] 4s²3d⁶ and in +3 oxidation state it has [Ar] 4s⁰3d⁵ configuration.

$h=\frac{cos\theta +3}{2}$

=> $k=\frac{sin\theta +2}{2}$

=> $cos\theta =2h-3\&sin\theta =2k-2$

${\left(h-3/2\right)}^2+{\left(k-1\right)}^2=\frac{1}{4}$

$\therefore$ circle of radius r = $\frac{1}{2}$

If the block does not slide,

mg sin    $\theta \le \mu mgcos\theta$

  $\Rightarrow tan\theta \le \mu \Rightarrow \frac{dy}{dx}\le \mu \Rightarrow \frac{x}{2}\le 0.5\Rightarrow x\le 1m.$          

Thus, $y\le \frac{1^2}{4}=0.25m=25cm.$

Using conversation of momentum in direction perpendicular to the original direction of motion,

mv₁ sin 30° = mv₂ sin 30°

f = W = 0.5 * 10 = 5 N

N = F

For block not to slide,

  $f\le \mu N$          

$\begin{array}{l}\Rightarrow 5\le 0.2F\\ \Rightarrow F\ge 25N\end{array}$

Given   ${\displaystyle \sum _{i=1}^{18}\left(x_i-\alpha \right)=36i.e}{\displaystyle \sum _{i=1}^{18}{x}_i-18\alpha =36..........\left(i\right)}$

&       ${\displaystyle \sum _{i=1}^{18}{\left(x_i-\beta \right)}^2=90i.e{\displaystyle \sum _{i=1}^{18}{x_i}^2}}-2\beta {\displaystyle \sum _{}^{}{x}_i+18{\beta }^2=90.............\left(ii\right)}$         

(i) & (ii)  ${\displaystyle \sum _{i=1}^{18}{x_i}^2}=90-18\beta +36\beta \left(\alpha +2\right).............\left(iii\right)$

Now variance ${\sigma }^2=\frac{\sum {x_i}^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2$ = 1 given

=> (α - β) (α - β + 4) = 0

Since   $\alpha \ne \beta so\left|\alpha -\beta \right|=4$

Ratio of masses on two pistons of the hydraulic lift equals to that of their cross- section area.

$\frac{A_1}{A_2}=\frac{100}{m}$            

Now,     $\frac{4^2{A}_1}{A_2/4^2}=\frac{M}{m}?M=256\frac{A_1}{A_2}.m=25600kg.$

Given $P_n={\alpha }^n+{\beta }^n,P_{n-1}=11\&P_{n+1}=29$

    $P_n={\alpha }^{n-2}.{\alpha }^2+{\beta }^{n-2}.{\beta }^2.........\left(i\right)$                          

Now quadratic equation having roots α & β will be x² – (α + β)x + αβ = 0

i.e.         x² – x – 1 = 0,     put x = α and put x = β

So          α² = α + 1           & β² = β + 1

(i)     $P_n={\alpha }^{n-2}\left(\alpha +1\right)+{\beta }^{n-2}\left(\beta +1\right)$

$P_n=P_{n-1}+P_{n-2}$          

= > ${P_n}^2=234$  

Given sequence is -16, 8, -4, 2, .

are in G.P. with first term a = -16 & common ratio r =  $\frac{-1}{2}$

Now $t_p=a{r}^{P-1}=-16{\left(-\frac{1}{2}\right)}^{p-1}\&t_q=a{r}^{q-1}=-16{\left(-\frac{1}{2}\right)}^{q-1}$

So A.M. = -8   $\left[{\left(-\frac{1}{2}\right)}^{p-1}+{\left(\frac{-1}{2}\right)}^{q-1}\right]=\alpha \left(let\right)$

Given equation is 4x² – 9x + 5 = 0 gives $x=1,\frac{5}{4}$

From roots we get possible value of b = 1 so

$16{\left(-\frac{1}{2}\right)}^{\frac{P+q-2}{2}}=1OR{\left(-\frac{1}{2}\right)}^{\frac{p+q-2}{2}}=\frac{1}{16}-{\left(-\frac{1}{2}\right)}^4$

$\Rightarrow \frac{p+q-2}{2}=4\Rightarrow p+q=10$

W = 4.5 g

M.W = 90

V = 250 ml

Using M = $\frac{W}{M.W*V}*1000$  

$M=\frac{4.5}{90*250}*1000=0.2$  

M = 2 * 10^-1 M

So, x = 2