Class 11th

W = 4.75 g

$n=\frac{4.75}{26}$ T = 323 K, R = 0.0826

  $P=\frac{740}{760}atm$          

 V = $\frac{nRT}{P}=\frac{4.75*0.0826*323}{26\left(\frac{740}{760}\right)}=5litre$

$L_1=\frac{x-\lambda }{1}=\frac{y-\frac{1}{2}}{\frac{1}{2}}=\frac{z}{-\frac{1}{2}}$           

$\therefore SD=\frac{\left|-2\lambda +3\left(2\lambda +\frac{1}{2}\right)+\lambda \right|}{\sqrt[]{14}}=\frac{\left|5\lambda +\frac{3}{2}\right|}{\sqrt[]{14}}$           

$5\lambda +\frac{3}{2}=-\frac{7}{2}\Rightarrow 5\lambda =-5\Rightarrow \lambda =-1$           

$\therefore \left|\lambda \right|=1$

$C_x{H}_y\left(g\right)+\left(x+\frac{y}{4}\right)O_2\left(g\right)\to xC{O}_2\left(g\right)+\frac{y}{2}{H}_{2}O\left(\mathcal{l}\right)$

V 6V - -

- - 4V

Volume of CO2 = 4 * $V_{C_x{H}_y}$ ${}_{{}_{}{}_{}}$

Vx = 4V

x = 4

Volume of O2 = 6 * $V_{C_x{H}_y}$ ${}_{{}_{}{}_{}}$

$V\left(x+\frac{y}{4}\right)=6V$

$4+\frac{y}{4}=6$

y = 8

S = 432 km/hr

tan 60° =   $\frac{h}{y}$

$h=\sqrt[]{3}$ y

$tan30°=\frac{h}{x+y}$           

$?x=432*\frac{1000}{3600}*20$           

x = 2400 m

y = 1200 m

$\therefore h=1200\sqrt[]{3}m$          

y = ax² + bx + c

a + b + c = 2

$\therefore \frac{dy}{dx}=2ax+b\Rightarrow {\frac{dy}{dx}|}_{\left(0,0\right)}=b=1$    

It passes through (0, 0)

0 = 0 + 0 + c -> c = 0

a = 1

$\therefore a=1,b=1,c=0$           

$tan\left(\frac{1}{4}si{n}^{-1}\frac{\sqrt[]{63}}{8}\right)$

$Put\frac{1}{4}si{n}^{-1}\frac{\sqrt[]{63}}{8}=\theta$

$sin4\theta =\frac{\sqrt[]{63}}{8}\therefore cos4\theta =\frac{1}{8}$

$co{s}^2\theta =\frac{7}{8}\Rightarrow se{c}^2\theta =\frac{8}{7}$

$ta{n}^2\theta =\frac{8}{7}-1=\frac{1}{7}$

$tan\theta =\frac{1}{\sqrt[]{7}}$

y = x² + 4

x² = y – 4

y = 4x – 1

$PQ=\frac{\left|4*\frac{1}{2}t-\frac{1}{4}{t}^2\right|-4-1}{\sqrt[]{17}}$           

$p=\frac{\left|t^2-8t-2\right|}{4\sqrt[]{17}}$          

$\frac{dp}{dt}=\frac{1}{4\sqrt[]{17}}$           (2t – 8) for maximum / minimum, $\frac{dp}{dt}=0\Rightarrow t=4$  

$also\frac{d^{2}p}{d{t}^2}>0att=4$           

Hence the closest point becomes at t = 4 is (2, 8)

Kindly consider the following figure

Linear (180°)

Li -> Crimson Red

Na -> Yellow

Rb -> Red violet

Cs -> Blue

VIBGYOR

Wavelength

Red violet > Crimson Red >   Yellow   >   Blue

      $\downarrow$                           $\downarrow$             $\downarrow$             $\downarrow$           

780 nm             670.8 nm     589.2 nm    455.5 nm