Class 11th

Given    n = 2^x. 3^y. 5^z . (i)

$y+z=5\&\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$

On solving we get y = 3, z = 2

So, n = 2^x. 3³. 5²

So that no. of odd divisor = (3 + 1) (2 + 1) = 12

Hence no. of divisors including 1 = 12

${\displaystyle \sum _{n=1}^{\infty }\frac{n^2+6n+10}{\left(2n+1\right)!}}$

put 2n + 1 = r, r = 3, 5, 7,.

so n = $\frac{r-1}{2}$

$Now{\displaystyle \sum _{r=\left(3,5,7,.....\right)}^{\infty }\frac{n^2+6n+10}{\left(2n+1\right)!}}={\displaystyle \sum _{r=\left(3,5,7.....\right)}^{\infty }\left(\frac{\frac{{\left(r-1\right)}^2}{4}+3r-3+10}{r!}\right)}={\displaystyle \sum _{r=\left(3,5,7,......\right)}^{\infty }\left(\frac{r^2+10r+29}{4.r!}\right)}$

$=\frac{1}{8}\{41e-\frac{19}{e}-80)=\frac{41}{8}e-\frac{19}{8}{e}^{-1}-10$

N > O > Be > B

Due to half filled configuration N has more I.E than oxygen and due to fully filled configuration Be has more I.E than B.

(a) → (ii)                (b) → (iii)            (c) → (iv)            (d) → (i)

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