Class 12th

Higher the E.N. difference between hydrogen and other atom then higher be the strength of intermolecular H-bond

Here, order of difference in E.N is

O - H > N – H > C - H

Hence, correct order of H bond strength is,

CH₄ < HCN < NH₃

Leaching involves the given reaction,

$4Au\left(s\right)+8C{N}^{-}\left(aq\right)+2H_{2}O\left(aq\right)+O_2\left(g\right)\to 4{\left[Au{\left(CN\right)}_2\right]}^{-}\left(aq\right)+4O{H}^{-}\left(aq\right)$              

Here, O₂ is required for formation of Au (l) cyanide complex but no complex in absence of O₂.

$2{\left[Au{\left(CN\right)}_2\right]}^{-}\left(aq\right)+Zn\left(s\right)\to {\left[Zn{\left(CN\right)}_4\right]}^{2-}\left(aq\right)+2Au\left(s\right)$                

In above displacement reaction, Zn is oxidized.

For high wavelength, size of antenna should be high.

For Transistor

to act as a switch → Saturation & cut – off state

to act as an amplifier → Active Region

$E_2>E_1$

For statement 1

              hf = E₁  - E₂

Since, E₂ > E₁

So, it should be

hf = E₂ – E₁

Therefore statement 1 is wrong

For statement 2

For the jumping of electron from higher energy orbit

(E₂) to lower energy orbit (E₁)

hf = E₂ – E₁

  $f=\frac{E_2-E_1}{h}$              

Statement (2) is correct

  ${\mu }_2>{\mu }_1$

$\frac{{\mu }_2}{{\mu }_1}>1$               

${\mu }_2=\frac{c}{v_2},{\mu }_1=\frac{c}{v_1}$               

$\frac{{\mu }_2}{{\mu }_1}=\frac{v_1}{v_2}>1\Rightarrow v_1>v_2$               

Frequency remains constant while refraction since energy is constant

$\upsilon =\frac{v}{\lambda }$               

$\lambda \alpha v$               

${\lambda }_1>{\lambda }_2$               

$\lambda \to$ decreases

wavelength and speed decreases but frequency remains constant

$R_1=R$  

$R_2=-R$       

$P=\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$               

$P=\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R}-\left(\frac{-1}{R}\right)\right)$

$P=\frac{\left(\mu -1\right)2}{R}$              

L₂

R₁ = R

$R_2=\infty$               

$R_1=\infty$               

R₂ = ­­-R

Power of  $L_1=\frac{1}{f\text{'}\text{'}}=\left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Power of $L_1=\frac{\left(\mu -1\right)*2}{R}=P$  

Statement 1 is true

Speed of EM ware in vaccum is given by,

$C=\frac{1}{\sqrt[]{{\mu }_0{\in }_0}}$                

Speed of EM wave in a material medium

$v=\frac{1}{\sqrt[]{u_r{\mu }_0{\in }_r{\in }_0}}$                

So, statement (2) is false.

Sol. $m\left(L\right)=m_1{S}_1\left(\mathrm{\Delta }T\right)$

$\Rightarrow \mathrm{m}\left(3.4*{10}^5\right)=\left(200\right)\left(4200\right)\left(25\right)$

$\Rightarrow \mathrm{m}=61.7$