Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

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

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

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

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

[At B, 2P₂ V₁ = nRT]

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

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

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

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

$m{\omega }^2.\frac{2d}{3}=\frac{G.m.2m}{d^2}\Rightarrow \omega =\sqrt[]{\frac{3Gm}{d^3}}$

$T=\frac{2\pi }{\omega }=2\pi \sqrt[]{\frac{d^3}{3Gm}}$

$\alpha =\frac{\Delta l_C}{\Delta l_E}=\frac{3.5}{4}=\frac{7}{8}$

$\beta =\frac{\alpha }{1-\alpha }=\frac{7/8}{1-7/8}=7$

Focus of a spherical convex mirror is in the same side of centre of curvature. Thus, f = + $\frac{1}{2}r.$

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.

$\gamma =3\alpha$

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