Physics

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].

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

Given, L = 2H

l = 2 sin (t²)

$u={\displaystyle \underset{0}{\overset{2}{\int }}Lidi=\frac{L}{2}{\left[i^2\right]}_0^2=\frac{L}{2}\left[4-0\right]}$

$u=\frac{2}{2}*4=4J$

Power calculations can also differ based on the nature of forces, which are as follows:

• Conservative Forces: Here, work done is path-independent. e.g., gravity, spring force
• Non-Conservative Forces: Work depends on the path. e.g., friction

$\equiv Req=2.5\Omega$

$l=\frac{v}{R_{eq}}=\frac{5}{2.5}=2A$

$l_1=\frac{1}{2}m{r}^2$

$\begin{array}{l}{l}_2=\frac{1}{2}m{r}^2\\ l_3=\frac{1}{2}m{r}^2\\ l_4=\frac{2}{5}m{r}^2\end{array}$