Class 11th

m₁ = 5kg                                                                      

m₂ = 3kg

As m₁ is in rest -

T = m_gg = 30

N = m₁ g cosq

T =    $m_{1}gsin\theta \Rightarrow 50sin\theta =30$

->sinθ  $=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\Rightarrow \theta =37°$  

$\therefore N=50cos\theta =50cos37°=50*\frac{4}{5}=40N$

Xenon has 8 valence electrons. In ${\mathrm{X}\mathrm{e}\mathrm{F}}_2$ , it forms two bonds with fluorine, leaving three lone pairs. The steric number =2 bonds +3 lone pairs $=5$ .
As per the VSEPR Theory  the electron? pair geometry due to SN = 5 will be trigonal bipyramidal. The hybridization type will be ${\mathrm{s}\mathrm{p}}^3\mathrm{d}$ hybridisation. However, the three lone pairs occupy equatorial positions, resulting in a linear molecular geometry.

Time to reach at max height t = $\frac{u}{g}no.$  of balls thrown in 1 sec = n.            

So, time taken by each ball to reach max^m height, = $\frac{1}{n}sec$  

i.e.   $\frac{u}{g}=\frac{1}{n}\Rightarrow u=\frac{g}{n}$                                       So, h_max = $\frac{u^2}{2g}$  

=  $\frac{g^2}{2g{n}^2}$  

  $=\frac{g}{2n^2}$  

$y=0.5sin\frac{2\pi }{\lambda }\left(400t-x\right)m$

$y=y_{0}sin\left(\underset{\omega }{\underset{?}{\frac{2\pi }{\lambda }*400}}\underset{k}{\underset{?}{t-\frac{2\pi }{\lambda }}}x\right)m$

Wave speed, $V=\frac{\omega }{k}=400m/s$

$V=\sqrt[]{\frac{3RT}{M}}$ $V\text{'}=\sqrt[]{\frac{3R2T}{\left(\raisebox{1ex}{$M$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}}=2\frac{\sqrt[]{3RT}}{M}=2v$

At steady state :-

$\frac{F}{A/2}=Y\frac{\Delta \mathcal{l}}{\mathcal{l}}\Rightarrow F=\frac{Y\Delta \mathcal{l}A}{2\mathcal{l}}$

$=\frac{YA}{2}=\frac{2*10^{11}*10^{-4}}{2}=10^{7}N$

A. Torque   $\underset{}{\overset{}{\to }}$   (iii) Nm

B. Stress  $\underset{}{\overset{}{\to }}$ (iv) Nm^-2

C. Latent Heat  $\underset{}{\overset{}{\to }}$ (ii) J kg^-1

D. Power   $\underset{}{\overset{}{\to }}$ (i) Nm S^-1

$g=\frac{GM}{r^2}$

$dg=-2\frac{GM}{r^3}dr$

$\Rightarrow dg*100=-2\left(\frac{GM}{r^2}\right)\frac{dr}{r}*100$

$\Rightarrow \frac{dg}{g}*100=\underset{}{\underset{?}{-2*2}}=-4\mathrm{\%}$

$Ft=m\Delta v$

$?F?=\frac{m\Delta v}{t}=\frac{10*10^{-3}*4.5}{100*5}$ 9 * 10^-4 N.

= 9 * 10^-5 N

= 9 dyne.