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

|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

Block is not slipping, so

f = mrw² = 2 × 1 × 16 = 32

$S=\frac{F}{L}$  

[S] = [MT^–2]

$F=nA\frac{dv}{dx}$  

$\eta =\frac{ML{T}^{-2}\cdot T}{L^2}=M{L}^{-1}{T}^{-1}$

$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$

L º [ML²T^–1]

$KE=\frac{1}{2}l{\omega }^2$  

KE º ML²T⁻²

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

$\frac{GM}{r}=6.4*10^7$                       . (i)

$\frac{GM}{r^2}=6.4$                   . (ii)

$r=\frac{6.4*10^7}{6.4}$  

= 10⁷ m

= 10,000 km

R + h = 10,000

h = 10,000 – 6400 = 3600 km

$\overrightarrow{E}\times \overrightarrow{B}$ 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

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

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

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

$V_{rms}=\sqrt[]{\frac{3RT}{M}}$  

$\frac{T}{2}=\frac{320}{32}$  

T = 20 K

T = –253°C

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$\frac{V_I}{V_0}=\frac{N_I}{N_0}\Rightarrow \frac{220}{V_0}=\frac{100}{10}$  

V₀ = 22 V

$I_0=\frac{22}{5}=4.4A$  

Young's modulus is property of material of wire and it is independent of geometrical factors.

F = iL B sinq

= $\sqrt[]{2}$  × 1 × 3 × 10^–5 sin90

= 3 $\sqrt[]{2}$  × 10^–5  N

Kindly go through the solution

$\lambda =\frac{12400}{3}=4133.3$  Å

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