Class 12th

Kindly consider the following Image

$\left. [\mathrm{C}\mathrm{o}{\left({\mathrm{N}\mathrm{H}}_3\right)}_6\right]{\mathrm{C}\mathrm{l}}_3+3{\mathrm{A}\mathrm{g}\mathrm{N}\mathrm{O}}_3?3\mathrm{A}\mathrm{g}\mathrm{C}\mathrm{l}$

$\frac{\text{Mole of}\left. [\mathrm{C}\mathrm{o}{\left({\mathrm{N}\mathrm{H}}_3\right)}_6\right]{\mathrm{C}\mathrm{l}}_3}{1}=\frac{\text{Mole of}{\mathrm{A}\mathrm{g}\mathrm{N}\mathrm{O}}_3}{3}$

$\frac{0.3}{267.46}=\frac{0.125*\mathrm{V}*{10}^{-3}}{3}$

$V=26.92\mathrm{m}\mathrm{L}$

Till input voltage reaches 4 V. No zener is in breakdown region. So V? = V? then now hen V? changes between 4 V to 6 V one zener with 4 V will breakdown are P.D. across this zener will become constant and remaining potential will drop acro resistance in series with 4 V zener.

Now current in circuit increases Abruptly and source must have an internal resistance due to which. Some potential will get drop across the source also so correct graph between V? and t will be
We have to assume some resistance in series with source.

Number of bond between sulphur and oxygen $=8$

Number of bond between sulphur and sulphur $=8$

Before inserting slab
C_i = ε? A/d
After inserting dielectric slab
C_i = ε? lw/d
C_f = C? + C?
C_f = (Kε? A? /d) + (ε? A? /d)
C_f = 2C_i ⇒ (Kε? wx/d) + (ε? w (l-x)/d) = 2ε? lw/d
4x + l - x = 2l
x = l/3

Potential of centre, = V =
Vc = K (Σq)/R
Vc = K (0)/R = 0
Electric field at centre E_B = E_B = ΣE
Let E be electric field produced by each

charge at the centre, then resultant electric field will be
Ec = 0, since equal electric field vectors are acting at equal angle so their resultant is equal to zero.

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

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

Without catalyst: $K=A{e}^{-Ea/RT}$

$\begin{array}{rr}& \text{Presence catalyst:}{10}^6\mathrm{K}=A{\mathrm{e}}^{-\mathrm{E}\mathrm{c}/\mathrm{R}\mathrm{T}}\\ & \mathrm{E}\mathrm{q}\left(\mathrm{B}\right)-\mathrm{E}\mathrm{q}\left(\mathrm{A}\right)\mathrm{}{10}^6={\mathrm{e}}^{-(\mathrm{E}-\mathrm{E}\mathrm{c})/\mathrm{R}\mathrm{T}}\\ & \frac{\mathrm{E}-\mathrm{E}\mathrm{c}}{\mathrm{R}\mathrm{T}}=2.303*6\\ & \mathrm{\Delta }\mathrm{E}=\mathrm{E}\mathrm{c}-\mathrm{E}=(-2.303)*6\mathrm{R}\mathrm{T}\end{array}$