Class 12th

The minimum eligibility for BTech admission at NIT Jalandhar is passing Class 12 with atleast 50% aggregate. Hence, with 60% in Class 12, the candidate is eligible for applying for admission in BTech at NIT Jalandhar. The selection criteria for the same is JEE Main and JoSAA counselling. Hence, the candidate must aslo clear JEE Main and Register and attend JoSAA counselling to be allotted a seat at the college for BTech on merit basis.

$\mathrm{R}=22*{10}^3\pm 5\mathrm{\%}$ First band  Red

2nd band $=$  Red

3rd band =Orange (Multiplier)

4th band = Gold (Tolerance)

$\mathrm{d}\stackrel{?}{\mathrm{B}}=\frac{{\mu }_0(\mathrm{I}\stackrel{?}{\mathcal{l}}\stackrel{?}{\mathcal{l}}*\stackrel{?}{\mathrm{r}})}{4\pi {\mathrm{r}}^3}$

The expression for magnetic field depends on current carrying element $\mathrm{I}\mathrm{d}?\stackrel{?}{\mathcal{l}}$ $\stackrel{}{}$ , which is a vector quantity, therefore, statement-I is correct and statement-II is wrong.

No current through '  $G$  '

So potential difference across $\mathrm{R}$ is $2\mathrm{V}$

$\begin{array}{rr}& \mathrm{i}=\frac{8}{400}\\ & \mathrm{R}=\left(\frac{2}{8}*400\right)=100\mathrm{\Omega }\end{array}$

The circuit can be redrawn as an equivalent circuit given below

$\Rightarrow \mathrm{i}=\frac{5}{10}=0.5\mathrm{A}$

  $3\mu \mathrm{F}$ and $3\mu \mathrm{F}$ $$ in parallel

(bonus)

Using the equation

$\mathrm{e}\mathrm{V}=\mathrm{h}v-?$

or ${}_{}$

$\frac{\mathrm{e}\mathrm{V}}{2}=\frac{hv}{2}-h{v}_{\mathrm{T}\mathrm{h}}$

${\mathrm{e}\mathrm{V}}_{\mathrm{s}}=\mathrm{h}\mathrm{u}-{\mathrm{h}\mathrm{u}}_{\mathrm{T}\mathrm{h}}$

Data Incorrect

$\mathrm{n}=\sqrt[]{{?}_{\mathrm{r}}{\mathrm{u}}_{\mathrm{r}}}$

$\mathrm{n}=\frac{\mathrm{c}}{\mathrm{v}}\Rightarrow \mathrm{v}=\frac{\mathrm{c}}{\mathrm{n}}$

$\mathrm{V}=\left(\frac{\mathrm{c}}{\sqrt[]{{?}_{\mathrm{r}}{\mu }_{\mathrm{r}}}}\right)$

$\mathrm{B}=0.5\mathrm{T}$

Angle between $\stackrel{?}{\mathrm{B}}$  & A $\stackrel{?}{\mathrm{A}}$  is zero

$?=$ B.A. $\mathrm{c}\mathrm{o}\mathrm{s}?0$

$=0.5*\left(1\right)*1$

$=0.5\mathrm{W}\mathrm{b}$

(P? ¹AP - I)²
= (P? ¹AP - I) (P? ¹AP - I)
= P? ¹A (PP? ¹)AP - P? ¹AP - P? ¹AP + I
= P? ¹A²P - 2P? ¹AP + I
= P? ¹ (A² - 2A + I)P = P? ¹ (A - I)²P
| (P? ¹AP - I)²| = |P? ¹ (A - I)²P| = |P? ¹| | (A - I)²| |P| = | (A - I)²| = |A - I|²
A - I = [1, 7, w²], [-1, w², 1], [0, -w, -w]
|A - I| = 1 (-w³ + w) - 7 (w) + w² (w) = -w³ + w - 7w + w³ = -6w.
|A - I|² = (-6w)² = 36w².