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Velocity of block in equilibrium, in first case,

$v=A\omega =A.\sqrt[]{\frac{k}{M}}$

Velocity of block in equilibrium, is second case,

$v\text{'}=A\text{'}\omega \text{'}=A\text{'}\sqrt[]{\frac{k}{M+m}}$

From conservation of momentum,

Mv = (M + m) v’

$\Rightarrow MA\sqrt[]{\frac{k}{M}}=\left(M+m\right)A\text{'}\sqrt[]{\frac{k}{M+m}}\Rightarrow A\text{'}=A\sqrt[]{\frac{M}{M+m}}$

${\overrightarrow{a}}_1=x\widehat{i}-\widehat{j}+\widehat{k}\&{\overrightarrow{a}}_2=\widehat{i}+y\widehat{j}+z\widehat{k}$           

given  ${\overrightarrow{a}}_1\&{\overrightarrow{a}}_2$ are collinear then ${\overrightarrow{a}}_1=\lambda {\overrightarrow{a}}_2$  

$\Rightarrow \left(x\widehat{i}-\widehat{j}+\widehat{k}\right)=\lambda \left(\widehat{i}+y\widehat{j}+z\widehat{k}\right)$         

Since $\widehat{i},\widehat{j}\&\widehat{k}$ are not collinear so

  $Sox\widehat{i}+y\widehat{j}+z\widehat{k}=\lambda \widehat{i}-\frac{1}{\lambda }\widehat{j}+\frac{1}{\lambda }\widehat{k}$         

Hence possible unit vector parallel to it be  $\frac{1}{\sqrt[]{3}}\left(\widehat{i}-\widehat{j}+\widehat{k}\right)$ for $\lambda$ =

Since, $\frac{x^2}{\alpha kT}$ $\frac{{}^{}}{}$ should be dimensionless.

So, dimension of $\alpha ,\left[\alpha \right]=\frac{L^2}{M{L}^2{T}^{-2}}=M^{-1}{T}^2$ $\frac{{}^{}}{{}^{}{}^{}}{}^{}{}^{}$

Dimension of $\alpha {\beta }^2$ ${}^{}$ should be that of W.

So, $\left[\alpha {\beta }^2\right]=M{L}^2{T}^{-2}$ ${}^{}{}^{}{}^{}$

$\Rightarrow \left[{\beta }^2\right]=\frac{M{L}^2{T}^{-2}}{M^{-1}{T}^2}=M^2{L}^2{T}^{-4}\Rightarrow \left[\beta \right]=ML{T}^{-2}$

$\frac{dq}{dt}=t=20t+8t^2$

$\Rightarrow {\displaystyle \underset{0}{\overset{q}{\int }}dq={\displaystyle \underset{0}{\overset{15}{\int }}\left(20t+8t^2\right)dt}}$

$\Rightarrow q=20\times \frac{15^2}{2}+8.\frac{15^3}{3}=11250C$

Given f(x) = ${\displaystyle {\int }_e^x{e}^{t}f\left(t\right)dt+e^x..........\left(i\right)}$  

using Leibniz rule then

f’(x) = e^xf(x) + e^x

  _{$\Rightarrow \frac{dy}{dx}=e^{x}y+e^{x}wherey=f\left(x\right)then\frac{dy}{dx}=f\text{'}\left(x\right)$}                

P = -e^x, Q = e^x

Solution be y. (I.F.) =   ${\displaystyle \int Q\left(I.F.\right)dx+c}$

I. f. =   $e^{{\displaystyle \int -e^{x}dx}=e^{-e^x}}$

$\Rightarrow y.\left(e^{-e^x}\right)={\displaystyle \int e^x.e^{-e^x}dx+c}$          

  $y.e^{-e^x}=-{\displaystyle \int dt+c=-t+c=-e^{-e^x}+c..........\left(ii\right)}$          

Put x = 0 , in (i) f (0) = 1

$From\left(ii\right),\frac{1}{e}=-\frac{1}{e}+cgivenc=\frac{2}{e}$        

Hence f(x) = 2. $e^{\left(e^x-1\right)}-1$  

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