Class 11th

Let inclination of required line is,

So the coordinates of point B can be assumed as

$\left(4-\frac{\sqrt[]{29}}{3}cos\theta ,3-\frac{\sqrt[]{29}}{3}sin\theta \right)$

which satisfies x – y – 2 = 0

$sin\theta -cos\theta =\frac{3}{\sqrt[]{29}}$

By squaring

$sin2\theta =\frac{20}{29}=\frac{2tan\theta }{1+ta{n}^2\theta }$

Point B : $\left(\frac{10}{3},\frac{4}{3}\right)$ $\frac{}{}\frac{}{}$

which also satisfies x + 2y = 6

Kindly go through the solution

 $a=\frac{v^2}{R}$

$\overrightarrow{a}=-\frac{v^2}{R}cos\theta \widehat{i}-\frac{v^2}{R}sin\theta \widehat{j}$

  $\frac{C_P}{C_v}=y=1+2/f$

Heat Released by block = Heat gain by large Ice block

  $\Rightarrow mc\Delta T=M_{ice}L$              

->5 * 0.39 * 500 = m_ice * 335

  $m_{ice}=\frac{5*0.39*500}{335}$              

= 2.91 kg

Given

6a² = 24

a² = 4

a = 2m

$\Delta v=v\left(\gamma \right)\Delta T$                

$=3\alpha v\Delta T$              

$=3*5*10^{-4}*\left(a^3\right)*10$                

$=3*5*10^{-4}*8*10$              

$\begin{array}{l}=120*10^3\\ =1.2*10^{5}c{m}^3\end{array}$                

F. B. D of Beaker

By NLM2

$\begin{array}{l}f=ma=m{\omega }^{2}R\\ \Rightarrow m{\omega }^{2}R\le \mu N\\ R\le \mu g/{\omega }^2\end{array}$                              

$\underset{n??}{lim}\left(\sqrt[]{n^2?n?1}+n?+?\right)=0$

$=\underset{n??}{lim}\left[\sqrt[]{1?\frac{1}{n}?\frac{1}{n^2}}+?+\frac{?}{n}\right]=0$

$??=?1$

Now, $8\left(?+?\right)=8\left(?\frac{1}{2}\right)=?4$

First angle, θ₁ = θ, $R=\frac{u^{2}sin2\theta }{g}$   

Another angle (θ₂ = 90 - θ) for which range will be

Same as that of θ₁ = θ

$R^1=\frac{u^{2}sin2\left(90-\theta \right)}{g}=\frac{u^2}{g}sin2\theta =R$

at  ${\theta }_1=\theta ,h_1=\frac{u^{2}si{n}^2\theta }{2g}$

$\&{\theta }_2=90-\theta ,h_2=\frac{u^{2}si{n}^2{\theta }_2}{2g}=\frac{u^{2}co{s}^2\theta }{2g}$

$\Rightarrow h_1{h}_2=\frac{1}{4^2}{\left[\frac{u^{2}sin2\theta }{g}\right]}^2$  

$R=4\sqrt[]{h_1{h}_2}$  

So, Both statement is true & Reason is correct explanation for statement 1.

  $C{N}^{-},N{O}^{+}and{O}_2^{2+}$ have bond order = 3