Physics

Kindly consider the following figure

Kindly consider the following figure

$\overrightarrow{L}=\overrightarrow{r}*m\overrightarrow{v}$

= $\left(3\widehat{i}-\widehat{j}\right)*\left(3\widehat{j}+\widehat{k}\right)$

$=9\widehat{k}+3\left(-\widehat{j}\right)-\widehat{i}$

$=-\widehat{i}-3\widehat{j}+9\widehat{k}$

$\left|\overrightarrow{L}\right|=\sqrt[]{1+9+81}=\sqrt[]{91}$

m = 0.3g density,   ${\rho }_{\left(ball\right)}=8g/cc$                   $\therefore V=\frac{m}{\rho }=\frac{0.3}{8}cc$                     

${\rho }_{glycerene}=1.3g/cc=1.3*10^{3}kg/m^3$

F_v + F_B = mg.

Þ F_v = mg – F_B = .3 * 10^-3 * 10 – 1.3 *  $\frac{0.3}{8}*10^{-6}*10*10^3$

  $=3*10^{-3}-\frac{.39}{8}*10^{-2}=3*10^{-3}-.5*10^{-3}$

= 2.5 * 10^-3N

we know, = $\frac{\lambda }{d}$ $\frac{}{}$ $\mu =\frac{C}{V}=\frac{C}{\upsilon \lambda }$ $\frac{}{}\frac{}{}$

${}_{}\frac{}{{}_{}{}_{}}$  ${\lambda }_1=\frac{C}{{\mu }_1{\upsilon }_1}$ [with change in medium frequency remain constant]

${\lambda }_2=\frac{C}{{\mu }_2{\upsilon }_2}$

$\frac{{\lambda }_2}{{\lambda }_1}=\frac{{\mu }_1}{{\mu }_2}=\frac{5}{7}$ ${\lambda }_2=\frac{5}{7}{\lambda }_1$

${\theta }_1=\frac{{\lambda }_1}{d_1}$

2= $\frac{{\lambda }_2}{d_2}$

$\frac{{\theta }_2}{{\theta }_1}=\frac{{\lambda }_2}{{\lambda }_1}[?d_1=d_2]$

2 $=\frac{5}{7}*{\theta }_1$

${\theta }_2=\frac{5}{7}*0.35$

$\frac{B_0^2}{2{\mu }_0}C$

$B_0^2=\frac{I2{\mu }_0}{C}$

$B_0^2=\frac{0.22*2*4*10^{-7}}{9*108}$

B₀ = 42.9 * 10^-9

42.9 Ans

Least  count of Vernier = 0.1mm

$\Rightarrow$ Reading of Vernier Scale = 5 * 0.1 = 0.5mm

The corrected diameter of sphere = Main Scale Reading + Vernier Scale reading + Zero correction = 1.7 + 0.05 + 0.05 = 1.8cm = 180 * 10^-2 cm.

Assume the length of train be I and its acceleration be a.

$v^2=u^2+2al\Rightarrow al=\frac{v^2-u^2}{2}$

Velocity when middle point crosses the post,

$T=2\pi \sqrt[]{\frac{r^3}{G{M}_e}}$

$T_B-T_A=\frac{2\pi }{\sqrt[]{G{M}_e}}\left(r_B^{3/2}-r_A^{3/2}\right)=\frac{2*3.14}{\sqrt[]{6.67*10^{-11}*6*10^{24}}}\left[{\left(8*10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left(7*10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$\lambda =\frac{h}{p}or\lambda =\frac{h}{mv}$

${\lambda }_p={\lambda }_{\alpha }\left[Given\right]$

$\frac{h}{m_p{V}_p}=\frac{h}{m_{\alpha }{v}_{\alpha }}\Rightarrow \frac{v_p}{v_{\alpha }}=\frac{m_{\alpha }}{m_p}$

$\Rightarrow \frac{v_p}{v_{\alpha }}=\frac{4m}{m}=\frac{4}{1}$ or 4 : 1