Class 11th

U = 3PV + 4

^{$\Rightarrow n{C}_{v}T=3PV+4$}

$\Rightarrow n*\frac{fRT}{2}=3PV+4$

$\Rightarrow f=6+\frac{8}{PV}$

$\Rightarrow f>6\Rightarrow polyatomic$         

$x={\displaystyle \sum _{n=0}^{\infty }co{s}^{2n}\theta =co{s}^0\theta +co{s}^2\theta +co{s}^4\theta +.....=1+co{s}^2\theta +co{s}^4\theta +.....}$

a = 1, r = cos2

$x=S_{\infty }=\frac{a}{1-r}=\frac{1}{1-co{s}^2\theta }=\frac{1}{si{n}^2\theta }$

Similarly, y = $\frac{1}{co{s}^2\theta }\Rightarrow \frac{1}{y}=co{s}^2\theta$ $\frac{}{{}^{}}\frac{}{}{}^{}$

$z=\frac{xy}{xy-1}\Rightarrow xyz-z=xy...........\left(i\right)$

Also, $\frac{1}{x}+\frac{1}{y}=1\Rightarrow x+y=xy..........\left(ii\right)$ $\frac{}{}\frac{}{}$

(i) & (ii) xyz = xy + z (x + y) z = xy + z

  $tan\alpha =\frac{v_{max}}{t_1}=a_1$

$tan\beta =\frac{v_{max}}{t_2}=a_2$

$\Rightarrow \frac{a_1}{a_2}=\frac{t_2}{t_1}\Rightarrow \frac{t_1}{t_2}=\frac{a_2}{a_1}$

Kindly go through the solution

-> $v=\omega \sqrt[]{A^2-x^2}\Rightarrow \frac{v^2}{A^2{\omega }^2}+\frac{x^2}{A^2}=1$ Path is ellipse.

Let the spring is in extension state and   $a_B>a_A,{\overrightarrow{a}}_{AB}=-a_A\widehat{i}-\left(-a_B\widehat{i}\right)=\left(-a_A+a_B\right)\widehat{i}$

Hence we can say that block moves away from block B in the frame of B

F – kx = Ma_B           . (1)

kx = Ma                  . (2)

$\Rightarrow a_B=\frac{F}{M}-a$

According to question,

$\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{4}$

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}........(i)$

Equation of required line is $\frac{x}{a}+\frac{y}{b}=1$ $\frac{}{}\frac{}{}$

Obviously B (2, 2) satisfying condition (i)

Frequency increases on filing. So, initial frequency of A is 335 Hz.

f = 340 – 5 = 335 Hz

Stress = Y * strain

-> $\frac{T_1}{A}=Y*-\frac{\left({\mathcal{l}}_1-\mathcal{l}\right)}{\mathcal{l}}\left(i\right)$  

  $\frac{T_2}{A}=Y*\frac{\left({\mathcal{l}}_2-\mathcal{l}\right)}{\mathcal{l}}\left(ii\right)$                          

$\Rightarrow$ $\frac{T_1}{T_2}=\frac{{\mathcal{l}}_1-\mathcal{l}}{{\mathcal{l}}_2-\mathcal{l}}$                      

$\Rightarrow \mathcal{l}=\frac{T_1{\mathcal{l}}_2-T_2{\mathcal{l}}_2}{T_1-T_2}=\frac{T_2{\mathcal{l}}_1-T_1{\mathcal{l}}_2}{T_2-T_1}$            

$y=\alpha x-\beta x^2$

$\Rightarrow \frac{dy}{dx}=\alpha -2\beta x=0$  

$\Rightarrow x=\frac{\alpha }{2\beta }$           

$y_{max}=\alpha *\frac{\alpha }{2\beta }-\beta *{\left(\frac{\alpha }{2\beta }\right)}^2=\frac{{\alpha }^2}{4\beta }$           

Range = $2x=\frac{\alpha }{\beta }=\frac{2u^{2}sin\theta .cos\theta }{g}$  

On comparing with

y = x tan θ - $\frac{g{x}^2}{2u^{2}co{s}^2\theta }$  

mg – T = ma       . (1)

T * R = l a               . (2)

a = a R                    . (3)

With the help of equations (1), (2) and (3), we get

-> $a=\frac{mg}{m+\frac{l}{R^2}}$  

$v=\sqrt[]{2ah}=\omega R$  

$\Rightarrow {\omega }^2=\frac{2mgh}{l+m{R}^2}$