Physics

f = W = 0.5 * 10 = 5 N

N = F

For block not to slide,

  $f\le \mu N$            

$\begin{array}{l}\Rightarrow 5\le 0.2F\\ \Rightarrow F\ge 25N\end{array}$

Ratio of masses on two pistons of the hydraulic lift equals to that of their cross- section area.

$\frac{A_1}{A_2}=\frac{100}{m}$            

Now,   $\frac{4^2{A}_1}{A_2/4^2}=\frac{M}{m}\Rightarrow M=256\frac{A_1}{A_2}.m=25600kg.$  

Percentage Modulation =   $\frac{A_m}{A_c}*100=\frac{20}{80}*100=25\mathrm{\%}$

Potential difference across 2k $\Omega$  is 5V, thus current through it,

$i=\frac{5}{2*10^3}=25*10^{-4}A.$           

Amplitude is proportional to the slit – width, thus,

$\frac{A_1}{A_2}=3$

$\frac{l_{max}}{l_{min}}=\frac{{\left(A_1+A_2\right)}^2}{{\left(A_1-A_2\right)}^2}=\frac{{\left(\frac{A_1}{A_2}+1\right)}^2}{{\left(\frac{A_1}{A_2}-1\right)}^2}=\frac{{\left(3+1\right)}^2}{{\left(3-1\right)}^2}=4.$

If   $\mu$ is Poisson's ratio,

Y = 3K (1 - 2  $\mu$ )      ……… (1)

and Y = 2  $n\left(1+\mu \right)$ ……… (2)

With the help of equations (1) and (2), we can write

$\frac{3}{Y}=\frac{1}{\eta }+\frac{1}{3k}\Rightarrow K=\frac{\eta Y}{9\eta -3Y}$            

$\overrightarrow{F}=10\widehat{i}+5\widehat{j}$

$\overrightarrow{a}=\frac{10}{m}\widehat{i}+\frac{5}{m}\widehat{j}$

$=\frac{10}{0.1kg}\widehat{i}+\frac{5}{0.1}\widehat{j}$

$\overrightarrow{a}=100\widehat{i}+50\widehat{j}$

$a_x=100,a_y=50$

$S_x=ut+\frac{1}{2}a{t}^2$

$=0*2+\frac{1}{2}*100*2*2$

Sx = 200 m

$\frac{a}{b}=\frac{200}{100}=2$

There are three main processes Isothermal, adiabatic and cyclic process. In isothermal, the system is thermally conductive and the temperature remains constant. In adiabatic process, the system is thermally isolated and there is no change in heat temperature. The system returns to its initial stage in the cyclic process.

$W_{AB}=nRTln\frac{2V_1}{V_1}=nRTln2.$

$W_{BC}=P_2\left(V_1-2V_1\right)=-P_2{V}_1=-\frac{1}{2}nRT$                         

[At B, 2P₂ V₁ = nRT]

$W_{CA}=0$   [CA is isochoric process].

$W_{ABCA}=W_{AB}+W_{BC}+W_{CA}=nRT\left(\mathcal{l}n\left(2\right)-\frac{1}{2}\right)$