Class 12th

The inertness of ? subshell electrons towards bond formation is known as inert pair effect. This effect increases down the group thus, for Sn,  oxidation state is more stable, whereas, for Pb,  oxidation state is more stable, i.e.,  is reducing while  is oxidising.

Kindly go through the solution 

Charge on C₁ = CV

            Charge on C₂ = 4CV

When connected in parallel

$V_C=\frac{5V}{3}$  

$Q{\text{'}}_1=\frac{5}{3}CV$ $Q{\text{'}}_2=\frac{10}{3}CV$

$?$ $E=\frac{1}{2}C{V}^2$

$E\text{'}=\frac{25}{18}C{V}^2+\frac{25}{9}C{V}^2$  

$\frac{25}{6}C{V}^2=\frac{50E}{6}$

x = 50

$-0.85=\frac{-13.6}{n^2}$  

n = 4

Number of transitions = $\frac{4*3}{2}=6$  

Current in circuit (I) = $\frac{8-2}{6}$  = 1 A.

So, V_C – 4 (1) – 2 + 8 – 2 (1) = V_A

V_C – 6 – 2 + 8 = V_A

V_C – V_A = 0 V

COAM gives Iw₀ = 2Iw

$\omega =\frac{{\omega }_0}{2}$

Loss in KE =  $\frac{1}{2}I{\omega }_0^2-\frac{1}{2}\left(2I\right){\left(\frac{{\omega }_0}{2}\right)}^2$

$=\frac{1}{2}I{\omega }_0^2$

$=\frac{1}{4}*5*\frac{2}{2}*100=250J$  

$\left|m\right|=\left|\frac{v}{u}\right|=2$  

|v| = |2u|

n + 2n = 45

n = 15 cm

u = –15

v = 30

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$  

$\frac{1}{30}+\frac{1}{-15}=\frac{1}{f}$  

$\frac{1-2}{30}=\frac{-1}{30}=\frac{1}{f}$  

f = 30 cm

Final resistance $\left(R\right)=\frac{V}{I}=80\Omega$

then, R = R₀ (1 + µDT)

80 = 60 (1 + 2 * 10^–4 DT)

DT = 1666.67

T – 27

T = 1693.66

= 1694°C

$\overrightarrow{E}*\overrightarrow{B}$ is along +z axis

$B_0=\frac{E_0}{C}$  

$B=\frac{E_0}{C}sin\left(\omega t-kz\right)\widehat{i}$

Lω = 1000 L                            Lω = R.

New

R = 1000 L

2000 L

$cos\theta =\frac{R}{Z}$

$=\frac{1000L}{\sqrt[]{{\left(1000L\right)}^2+{\left(2000\text{?}L\right)}^2}}$

$=\frac{1}{\sqrt[]{1+4}}=\frac{1}{\sqrt[]{5}}$