Physics

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

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

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

Initially speed is zero, then increases & after some time it becomes constant.

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

$\text{?}\mathrm{}\stackrel{?}{\mathrm{E}}\cdot \mathrm{d}\stackrel{?}{\mathrm{s}}=0\Rightarrow {?}_{\text{net}}={?}_{\text{in}}-{?}_{\text{out}}=0$

$\Rightarrow {?}_{\text{in}}={?}_{\text{out}}$

$\stackrel{?}{\tau }=\stackrel{?}{\mathrm{P}}*\stackrel{?}{\mathrm{E}}$

$\left|\stackrel{?}{\tau }\right|=\mathrm{P}\mathrm{E}\mathrm{s}\mathrm{i}\mathrm{n}?\theta \mathrm{}\Rightarrow 4=\mathrm{q}*2\mathrm{a}*\mathrm{E}\mathrm{s}\mathrm{i}\mathrm{n}?{30}^{?}$

$\mathrm{q}=\frac{4}{\left(2*{10}^{-2}\right)*2*{10}^5*\left(\frac{1}{2}\right)}$

$=2*{10}^{-3}\mathrm{C}$

$=2\mathrm{m}\mathrm{C}$

(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

$\mathrm{T}\mathrm{i}=-{50}^{?}\mathrm{C}$

$\begin{array}{rr}& =223\mathrm{K}\\ & {\mathrm{V}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\propto \sqrt[]{\mathrm{T}}\end{array}$

As ${\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}$  increased by 3 times

So ${\left({\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\right)}_{\mathrm{f}}=4{\left({\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\right)}_{\text{initial}}$

${\mathrm{T}}_{\mathrm{f}}=16{\mathrm{T}}_{\mathrm{i}}$

$=16*223$

$=3568\mathrm{K}$ ${\mathrm{T}}_{\mathrm{f}}=(3568-273{)}^{?}\mathrm{C}$

$={3295}^{?}\mathrm{C}$

15. (c) Fundamental harmonic frequency open pipe

$=\frac{\mathrm{V}}{2\mathrm{L}}={\mathrm{v}}_1$

(say)

Fundamental harmonic frequency of closed pipe $=\frac{\mathrm{V}}{4\mathrm{L}}$

$={\mathrm{v}}_2$ (say)

$\Rightarrow \frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\frac{\frac{\mathrm{V}}{2\mathrm{L}}}{\frac{\mathrm{V}}{4\mathrm{L}}}\mathrm{}2:1$

$\eta =1-\frac{T_{\text{sin}k}}{T_{\text{source}}}=0.5$

$\begin{array}{rr}& \frac{{\mathrm{T}}_{\text{sin}}}{{\mathrm{T}}_{\text{source}}}=0.5\\ & \Rightarrow {\mathrm{T}}_{\text{sin}}=\frac{1}{2}*(273+327)\\ & =\frac{1}{2}*600\\ & =300\mathrm{K}\\ & ={27}^{?}\mathrm{C}\end{array}$