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 $l=\frac{120-60}{4000}=0.015A$

Thus l₂ = l - L_L

= 0.015 – 0.006

= 0.009 = 9mA

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Energy released = Binding energy of product – Binding energy of reacts :

$\Delta E=7.6\times 4-\left(1.1\times 4\right)$

= 26 MeV

For first line of Lyman

$\frac{1}{\lambda }=R\left(1-\frac{1}{4}\right)=R\frac{3}{4}$

$\lambda =\frac{4}{3R}$

3rd line (Paschen)

$\frac{1}{{\lambda }_3}=R\left(\frac{1}{3^2}-\frac{1}{6^2}\right)=\frac{R}{9}\times \frac{3}{4}$

2nd line (B almer)

${\frac{1}{\lambda }}_2=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{R}{4}\times \frac{3}{4}$

$a\left(\frac{4}{3R}\right)=\frac{20}{3R}\Rightarrow a=5$

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For point B, $\frac{1}{u}=-0.10c{m}^{-1},\frac{1}{v}=0$

Thus u = 10 cm, v = $\infty$

i.e., f = 10 cm

$\frac{1}{10}=\left(1.5-1\right)\left(\frac{2}{R}\right)\Rightarrow \frac{1}{R}=\frac{1}{10}$

R = 10 cm

At very high frequencies

$X_C=\frac{1}{\omega C}\to 0$

$X_L=\omega L\to \infty$

Thus equivalent circuit

$Z=1+2+2=5\Omega$

$l=\frac{220}{5}=44A$

1.5 × sin 60 = R × sin 90

$\Rightarrow 1.5\times \frac{\sqrt[]{3}}{2}=n=\frac{\sqrt[]{x}}{4}$

x = 27

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 $k=\sqrt[]{\frac{l}{m}}=\sqrt[]{\frac{m{\mathcal{l}}^2/12}{m}}=\frac{\mathcal{l}}{2\sqrt[]{3}}$

$k=\frac{10\sqrt[]{3}}{2\sqrt[]{3}}=5$

Initially,  $\frac{P}{Q}=\frac{40}{60}=\frac{2}{3}$ - (1)

Finally,  $\frac{P+x}{Q}=\frac{80}{20}=\frac{4}{1}$

(2) (1)

$\frac{P+x}{P}=\frac{4\times 3}{2}=6$

$\Rightarrow 1+\frac{x}{P}=6$

$\frac{x}{P}=5$

x = 5 P = 5 x 4 = 20 $\Omega$

Component of $\overrightarrow{A}$ along $\overrightarrow{B}$ = $\frac{\overrightarrow{A}.\overrightarrow{B}}{\left|\overrightarrow{B}\right|}$

= 2

Capacitance of each capacitor

$C_1=\frac{A3{\epsilon }_0}{\frac{1}{2}}=6A{\epsilon }_0$

$C_2=A4{\epsilon }_0=4A{\epsilon }_0$

Equivalent capacitance

$\Delta v_2=\frac{240A{\epsilon }_0}{4A{\epsilon }_0}=60v$

$v_{foil}=60v$

Breading stress = $\frac{Weight}{Area}$

$\Rightarrow \frac{10}{2.5\times 10.4}=\frac{25}{A}$

$\Rightarrow A=6.25\times 10^{-4}{m}^2$

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