Physics

$\tau =\eta \frac{v}{h}$  

              Given

              $\tau =10^{-3}N/m^2$  

              (shear stress)

              h =?

              v = 36 km/hr = 10 m/sec

              $10^{-3}=\frac{10^{-2}*10}{h}$

h = 100 m

$\gamma =3\alpha$

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

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

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

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

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

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

by conservation of mechanical energy

              K.E_i + P.E_i = K.E_f + P.E_f

              $\frac{1}{2}*0.5v^2+0=\frac{1}{2}*0.5{\left(\frac{v}{2}\right)}^2+\frac{1}{2}k{x}^2$  

              $\Rightarrow \frac{1}{2}*0.5\left[v^2-\frac{v^2}{4}\right]=\frac{1}{2}k{x}^2$  

              $\Rightarrow \frac{1.5}{4}12*12=\frac{k*9}{100}$  

              $\frac{1.5*36}{9}*100=k$  

              k = 600 N/m¹

A : Series limit of Lyman series

B : Third line of Balmer series

C : Second line of Paschen series

When connected in series, equivalent capacitance,

$C_1=\frac{C*C}{C+C}=\frac{C}{2}$          

When connected in parallel, equivalent capacitance

C₂ = C + C = 2C

$\frac{C_1}{C_2}=\frac{C/2}{2C}=\frac{1}{4}$