Class 11th

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

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 $=\frac{1}{2}{\mathrm{k}\mathrm{x}}^2$

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

${\mathrm{h}}_{\text{max}}=\frac{{\mathrm{u}}^2{\mathrm{s}\mathrm{i}\mathrm{n}}^2??}{2\mathrm{g}}=\frac{280*280}{2*9.8}*\frac{1}{4}$

$=1000\mathrm{m}$

Average speed $=\left(\frac{4{\mathrm{v}}^2}{3\mathrm{v}}\right)$ $\left(\frac{{}^{}}{}\right)$

$=\frac{4\mathrm{v}}{3}$

(d) Initial velocity  $=-v\hat{j}$ $\stackrel{}{}$

Final velocity $=v\hat{i}$ $\stackrel{}{}$

Change in velocity  $=v\hat{i}-(-v\hat{j})$

$\stackrel{}{}\stackrel{}{}$$\stackrel{}{}\stackrel{}{}$ $=v(\hat{i}+\hat{j})$ Momentum gain is along $\stackrel{\mathrm{ˆ}}{\mathrm{i}}+\stackrel{\mathrm{ˆ}}{\mathrm{j}}$ $\stackrel{}{}\stackrel{}{}$

$$ $\Rightarrow$  Force experienced is along $\hat{i}+\hat{j}$ $\stackrel{}{}\stackrel{}{}$

$$  $\Rightarrow$ Force experienced is in North-East direction.

$m=\rho \pi r^{2}1$

$\begin{array}{rr}& \rho =\frac{\mathrm{m}}{\pi {\mathrm{r}}^{2}1}\\ & \frac{\mathrm{\Delta }\rho }{\rho }=\frac{\mathrm{\Delta }\mathrm{m}}{\mathrm{m}}+\frac{2\mathrm{\Delta }\mathrm{r}}{\mathrm{r}}+\frac{\mathrm{\Delta }\mathrm{l}}{1}\\ & \frac{\mathrm{\Delta }\rho }{\rho }*100=\frac{0.002}{0.4}*100+\frac{2*0.001}{0.3}*100+\frac{0.02}{5}*100\\ & =\frac{0.2}{0.4}+\frac{0.2}{0.3}+\frac{2}{5}\\ & =0.5+0.67+0.4\\ & =1.57\\ & =1.6\mathrm{\%}\end{array}$

As the factors controlling temperature and voltage supply are beyond prediction and control so the error occurred due to unpredictable fluctuations of temperature and voltage would be random errors.