Class 11th

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

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

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

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

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

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

By property

If first and Last term of Andhra Pradesh and HP are equal (of nn  terms) then $a_r{H}_S=$  first *  x last term

If $r+s=n+1$ r+s=n+1

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,

Single digit factor of $12\to \mathrm{1,2},\mathrm{3,4},6$

  $(x,y,z)$  where , $xyz=12$

$(\mathrm{6,2},1)(\mathrm{4,3},1)(\mathrm{3,2},2)$

$\begin{array}{ccc}\downarrow & \downarrow & \downarrow \\ \text{6Ways}& \text{6 Ways}& \text{3Ways}\end{array}$

 Total 15 Ways

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

$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]$

Equation of normal   $y=mx-2m-m^3$ must pass through centre $\left(\mathrm{0,3}\right)$

$\begin{array}{rr}& m^3+2m+3=0\\ & m^2(m+1)-m(m+1)+3\cdot 1(m+1)=0\\ & (m+1)\left(m^2-m+3\right)=0\\ & m=-1\end{array}$

  $\therefore \mathrm{}$ point of contact of normal at parabola is $\left({\mathrm{a}\mathrm{m}}^2,-2\mathrm{a}\mathrm{m}\right)=\left(\mathrm{1,2}\right)$

So distance between parabola and circle is $=\sqrt[]{2}-1$

Kindly consider the following figure

If $\mu$  is Poisson's ratio,

Y = 3K (1 - 2 $\mu$ )      ……… (1)

and Y = 2 $n\left(1+\mu \right)$  ……… (2)

With the help of equations (1) and (2), we can write

  $\frac{3}{Y}=\frac{1}{\eta }+\frac{1}{3k}\Rightarrow K=\frac{\eta Y}{9\eta -3Y}$