Physics

Energies of given Radiation can have
The following relation
Eγ-Rays > EX-Rays > Emicrowave > EAM Radiowaves
∴ λγ-Rays < X-Rays < microwave < AM Radiowaves
According to tres.
(a) Microwave → 10? ³ m
(b) Gamma Rays → 10? ¹? m (ii)
(c) AM Radio wve → 100 m (i)
(d) X-Rays → 10? ¹? m

M = µ? NiA
Here
µ? = Relative permeability
N = Number of turns
i = Current
A = Area of cross section
M = µ? NiA = µ? nliA
M = µ? niV = 1000 (1000) (0.5) (10? ³) = 500 = 5 * 10²Am²

In adiabatic process
PV? = constant
P (m/ρ)? = constant
As mass is constant
P ∝ ρ?
P_f/P_i = (ρ_f/ρ_i)? = (32)? /? = 2? = 128

Now, using junction analysis
We can say, q? + q? + q? = 0
2 (x - 6) + 4 (x - 6) + 5 (x) = 0
x = 36/11, q? = 36 (5)/11 = 180/11
q? = 16.36µC

At T°C L = L? + L?
At T +? T Leq = L'? + L'?
where L'? = L? (1 + α? T)
L'? = L? (1 + α? T)
Leq = (L? + L? ) (1 + αavg? T)

⇒ (L? + L? ) (1 + αavg? T) = L? + L? + L? α? T + L? α? T
⇒ (L? + L? )αavg = L? α? + L? α?
⇒ αavg = (L? α? + L? α? )/ (L? + L? )

Initially S? L = 2m

S? L = √2² + (3/2)²
S? L = 5/2 = 2.5 m
? x = S? L - S? L = 0.5 m
So since λ = 1 m. ∴? x = λ/2
So white listener moves away from S? Then? x (= S? L − S? L) increases and hence, at? x = λ first maxima will appear.? x = λ = S? L − S? L.
1 = d - 2 ⇒ d = 3 m.

Initially S? L = 2m
S? L = √2² + (3/2)²
S? L = 5/2 = 2.5 m
? x = S? L - S? L = 0.5 m
So since λ = 1 m. ∴? x = λ/2
So white listener moves away from S? Then? x (= S? L − S? L) increases and hence, at? x = λ first maxima will appear.? x = λ = S? L − S? L.
1 = d - 2 ⇒ d = 3 m.

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

${\mathrm{R}}_1=\propto$

${\mathrm{R}}_2=-30\mathrm{c}\mathrm{m}$

$\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{\mathrm{\infty }}-\frac{1}{-30}\right)$

$\frac{1}{f}=\frac{0.5}{30}f=60\mathrm{c}\mathrm{m}$

$\mathrm{}V=\sqrt[]{\frac{B}{?}}$

$\frac{V_{\text{pipe}}}{V_{\text{air}}}=\frac{\sqrt[]{\frac{B}{2?}}}{\sqrt[]{\frac{B}{?}}}=\frac{1}{\sqrt[]{2}}$

${\mathrm{V}}_{\text{pipe}}=\frac{{\mathrm{V}}_{\text{air}}}{\sqrt[]{2}}$

${\mathrm{f}}_{\mathrm{n}}=\frac{(\mathrm{n}+1)V_{\text{pipe}}}{2\mathcal{l}}\mathrm{};\mathrm{}{\mathrm{f}}_1-{\mathrm{f}}_0=\frac{V_{\text{pipe}}}{2\mathcal{l}}=\frac{300}{2\sqrt[]{2}}$

$=105.75\mathrm{H}\mathrm{z}$  (If $\sqrt[]{2}=1.41$ $\text{exception 25:}$ )

$=106.05\mathrm{H}\mathrm{z}$  (If $\sqrt[]{2}=1.414$ $\text{exception 25:}$ )