Physics

In isothermal process, temperature is constant.

In isochoric process, volume is constant.

In adiabatic process, there is no exchange of heat.

In isobaric process, pressure is constant

For part AM, slope of v – t graph is constant but negative. For part MB, slope of v – t graph is constant but positive.

$\gamma =3\alpha$  

$\Delta V=\gamma V\Delta T=3\alpha .a^3.\Delta T$            

If charge (-Q) at origin is replaced by (+Q), then electric field at the centre of the cube is zero. Thus, electric field at the centre of the cube is as if only (-2Q) charge is present at the origin.

$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\frac{\left(-2Q\right)}{{\left(\frac{\sqrt[]{3}}{2}a\right)}^2}.\frac{1}{\sqrt[]{3}}\left(\widehat{x}+\widehat{y}+\widehat{z}\right)=\frac{|-2Q}{3\sqrt[]{3}\pi {\epsilon }_0{a}^2}\left(\widehat{x}+\widehat{y}+\widehat{z}\right)$

  $F_{1}cos45°+F_{2}cos45°+F_3=\frac{m{v}^2}{r}$

$\Rightarrow \frac{G{m}^2}{{\left(\sqrt[]{2}r\right)}^2}.\frac{1}{\sqrt[]{2}}+\frac{G{m}^2}{{\left(\sqrt[]{2}r\right)}^2}.\frac{1}{\sqrt[]{2}}+\frac{G{m}^2}{{\left(2r\right)}^2}=\frac{m{v}^2}{r}$            

  $\Rightarrow v=\sqrt[]{\frac{Gm}{4r}\left(2\sqrt[]{2}+1\right)}$          

Putting m = 1 kg and r = 1 m,

$v=\frac{1}{2}\sqrt[]{G\left(1+2\sqrt[]{2}\right)}$        

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'

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

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

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

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

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

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

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

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

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

A : Series limit of Lyman series

B : Third line of Balmer series

C : Second line of Paschen series