Class 11th

Using Ideal gas Equation :

PV = nRT

$\Rightarrow n=\frac{PV}{RT}=\frac{400*10^3*500*10^{-6}}{8.3*100}=0.008$    

=> $m_1=0.12,m_2=0.64$

=> $\frac{m_2}{m_1}=\frac{16}{3}$

For circular motion F_{net   $=\frac{M{V}^2}{R}$}

_{$\sqrt[]{2F}+F\text{'}=\frac{M{V}^2}{R}$}

$\sqrt[]{2}\frac{GMM}{{\left(\sqrt[]{2}R\right)}^2}+\frac{GMM}{{\left(2R\right)}^2}=\frac{M{V}^2}{R}$                

$\frac{GM}{R}\left[\frac{1}{\sqrt[]{2}}+\frac{1}{4}\right]=V^2$              

$V=\frac{1}{2}\sqrt[]{\frac{GM}{R}\left(2\sqrt[]{2}+1\right)}$

Using F = MA = $m\frac{V}{T}$  

=> m = FTV^-1

Load of mass will be equally distributed among the four colours so force on each columns will be 125 * 10³ N.

Cross section area of the column = $\pi \left[{\left(1\right)}^2-{\left(0.5\right)}^2\right]=2.355m^2$

Using young's modulus :  $\epsilon =\frac{\sigma }{Y}=\frac{F}{AY}=\frac{125*10^3}{2.355*2*10^{11}}=2.65*10^{-7}$

In SHM sum of kinetic and potential energy will be constant and average kinetic energy & average potential energy in one time will be remains same.

Length of perpendicular from origin to xcosec a - y sec a = k cot2 a and x sin a + ycos a = k sin 2a are

$p=kcot\alpha sin\alpha cos\alpha =\frac{k}{2}.\frac{cos2\alpha }{sin2\alpha }sin2\alpha =\frac{k}{2}cos2\alpha \Rightarrow \frac{2p}{k}=cos2\alpha$

and $q=ksin2\alpha \Rightarrow \frac{q}{k}=sin2\alpha$

on solving these two we get 4p² + q² = k²

Heisenberg uncertainty principle :

$\begin{array}{l}\Rightarrow \Delta x\Delta p\ge ?\\ \Rightarrow \Delta x\left(m\Delta v\right)\ge ?\end{array}$

$\Rightarrow \Delta x\left(m\sqrt[]{\frac{3RT}{m}}\right)\ge ?$

$\Rightarrow \Delta x\propto \frac{1}{\sqrt[]{m}}$

$e^{4x}+2e^{3x}-e^x-6=0$

$e^x=t\in \left(0,\infty \right)$

$t^4+2t^3-t-6=0$

Let f (t) = t⁴ + 2t³ – t – 6

f' (t) = 4t³ + 6t² – 1f

Þf' (0) = -1, f' (+ $\infty )$ = + $\infty$

For t > 0

Þ f' (t) = 0 has only one root.

One solution of f (t) = 0 is possible

Acceleration due to gravity at r distance above the surface = $\frac{GM}{{\left(R+r\right)}^2}$  

Acceleration due to gravity at r distance below the surface = $\frac{GM}{R^3}\left(R-r\right)$

So, ratio = $\frac{\left(R-r\right){\left(R+r\right)}^2}{R^3}=\frac{\left(R-r\right)\left(R^2+r^2+2Rr\right)}{R^3}=1+\frac{r}{R}-\frac{r^2}{R^2}-\frac{r^3}{R^3}$

Equation of plane is 3x – 2y + 4z – 7 + $\lambda$ (x + 5y – 2z + 9) = 0. (i)

It passes through (1, 4, -3) and we get $\lambda =\frac{2}{3}$

$\therefore$ from (i) we get 11x + 4y + 8z – 3 = 0 Þ -11x – 4y – 8z + 3 = 0

$\therefore \alpha +\beta +\lambda =-11-4-8=-23$