Physics

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

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

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)

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

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

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

Surface energy of bubble $=2*$ charge in surface area $*$  surface tension

$\begin{array}{rr}& =8\pi {\mathrm{R}}^2*\mathrm{T}\\ & =8*3.142*4*{10}^{-4}*3*{10}^{-2}\mathrm{J}\\ & =30.1*{10}^{-5}\mathrm{J}\end{array}$

Stress $=\frac{\mathrm{F}}{\mathrm{A}}=\frac{\mathrm{T}}{\mathrm{A}}=\frac{\mathrm{W}}{\mathrm{A}}$

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

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

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

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

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

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

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

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

Radius of gyration of a solid surface,

${\mathrm{K}}_{\mathrm{S}}=\sqrt[]{\frac{2}{5}\mathrm{R}}$

Radius of gyration of a hollow surface,

${\mathrm{K}}_{\mathrm{H}}=\sqrt[]{\frac{2}{3}}\mathrm{R}$

$\Rightarrow \frac{{\mathrm{K}}_{\mathrm{S}}}{{\mathrm{K}}_{\mathrm{H}}}=\sqrt[]{\frac{3}{5}}=\frac{\sqrt[]{3}}{\sqrt[]{5}}$

The angular acceleration direction is given along angular velocity or opposite to angular velocity depending upon whether angular velocity magnitude is increasing or decreasing and this direction remains along the axis of circular motion.

Potential energy stored in the spring $=\frac{1}{2}{\mathrm{k}\mathrm{x}}^2$

$\begin{array}{rr}& \text{Now}\frac{1}{2}\mathrm{k}(2{)}^2=\mathrm{U}\\ & \&\frac{1}{2}\mathrm{k}(8{)}^2=U^{\mathrm{\text{'}}}\mathrm{}\text{(say)}\\ & \Rightarrow U^{\mathrm{\text{'}}}=\frac{64}{4}U=16U\end{array}$