Class 11th

$dm=\left(\frac{m}{L}\right)dx$

$\therefore T=\frac{m{\omega }^2}{2L}\left(L^2-x^2\right)$

$\therefore ?L={\int }_0^L\frac{m{\omega }^2}{2L\pi r^{2}Y}\left(L^2-x^2\right)dx$

$=?L=\frac{m{\omega }^2{L}^2}{3\pi r^{2}Y}$

Number of moles of C = Number of moles of CO₂ = $\frac{330}{44}$ moles

Number of moles of H = 2 * no. of moles of H₂O = $\left(\frac{270}{18}*2\right)$ moles

 Mass of C = $\frac{330}{44}*12$ gm = 90 gm

 Mass of H = $\frac{270}{18}*2*1gm=30gm$

$\begin{array}{l}\mathrm{\%}ofC=\frac{90}{120}*100\mathrm{\%}=75\mathrm{\%}\\ \mathrm{\%}ofH=\frac{30}{120}*100\mathrm{\%}=25\mathrm{\%}\end{array}$  

$y=A\sin ?\frac{2\pi t}{T}$

$t_2-t_1=\frac{T}{2\pi }\left. [{\sin }^{-1}?\frac{x_1}{A}-{\sin }^{-1}?\frac{x_2}{A}\right]$

Deformation energy per unit volume,

$U=\int dU=\int \frac{P^2}{2B}Ady$

$=\int \frac{{\rho }^2{g}^{2}y}{2B}Ady$

$U=\frac{{\rho }^2{g}^2}{2B}\frac{h^3}{3}A$

$=\frac{{\rho }^2{g}^2{h}^2}{6B}\left(volume\right)$

$=1.5\mu J$

$U=a{P}^4$

$\frac{PV}{\gamma -1}=a{P}^4$

$P^{-3}V=a\left(\gamma -1\right)$

$P{V}^{-1/3}=constant$

$ComparingwithP{V}^x=constant$

$x=-\frac{1}{3}$

$C=\frac{R}{\gamma -1}+\frac{R}{1-x}=\frac{R}{\frac{5}{3}-1}+\frac{R}{1+\frac{1}{3}}=\frac{3R}{2}+\frac{3R}{4}$

$C=\frac{9R}{4}$

$b=a\sqrt[]{1-e^2}$

$\frac{dA}{dt}=\frac{L}{2m}$

$\frac{A}{T}=\frac{L}{2m}$

$T=\frac{2mA}{L}=\frac{2m\pi ab}{L}=\frac{2m\pi a^2\sqrt[]{1-e^2}}{L}$

x = t² – t + 1        … (1)

y = t² + t + 1    … (2)

y – x = 2t & x + y = 2 (t² + 1)

__________on eliminating 't' we get

$\Rightarrow \left(x+y-2\right)=2{\left(\frac{y-x}{2}\right)}^2$

${\left(x-y\right)}^2=2\left(x+y-2\right)$

Axis : x – y = 0

Tangent at vertex : x + y – 2 = 0

Vertex : (1, 1) = (x, y)

Angular retardation,

In disproportionation reaction, one element of a compound will simultaneously get reduced and oxidised. In ClO₄ -, oxidation number of Cl is +7 and it can not increase it further. So, ClO₄- will not get oxidised and so will not undergo disporportionation reaction.

Let S be the distance between Two ends a be the constant acceleration

As we know $v^2-u^2=2aS$

or, $aS=\frac{v^2-u^2}{2}$

Let $v_c$  be velocity at mid point.

Therefore, $v_c^2-u^2=2a\frac{S}{2}$

$v_c^2=u^2+aS$

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

$v_c=\sqrt[]{\frac{u^2+v^2}{2}}$