Physics NCERT Exemplar Solutions Class 11th Chapter Five

$Bcos\delta =B_H$

B cos 30° = 0.5

$B=\frac{0.*2}{\sqrt[]{3}}=\frac{1}{\sqrt[]{3}}$

$M_A={\rho }_A*\frac{4}{3}\pi R_A^3$ ${\rho }_B=4{\rho }_A$

$M_B={\rho }_B*\frac{4}{3}\pi R_B^3$ $R_B=\frac{R_A}{2}$

$\frac{M_A}{M_B}=2,\frac{R_A}{R_B}=2$ $\frac{V_{EA}}{V_{EB}}=\sqrt[]{\frac{2G_1{M}_A}{R_A}*\frac{R_B}{2G_1{M}_B}}$

$v_{EA}=v_{EB}=12kmse{c}^{-1}$

$\left|\overrightarrow{A}+\overrightarrow{B}\right|=2\left|\overrightarrow{A}-\overrightarrow{B}\right|$

Given A = B

Squaring equation (1) both side.

${\left|\overrightarrow{A}+\overrightarrow{B}\right|}^2={\left(2\left|\overrightarrow{A}-\overrightarrow{B}\right|\right)}^2$

$A^2+B^2+2\overrightarrow{A}.\overrightarrow{B}=4\left(A^2+B^2-2\overrightarrow{A}.\overrightarrow{B}\right)$

2A² + 2A² cosq = 4 (2A² – 2A² cosq)

   2A² + 2A² cosq = 8A² – 8A² cosq

 10A² cosq = 6A²

cosq =   $\frac{3}{5}$

 q = cos^-1 (3/5)

dimension of a Pv²

        $\frac{a}{b}=pv$      dimension of b = v

$a=\frac{F}{m}=\frac{mv}{m}=\frac{1*10}{2}=5m/se{c}^2$

At w_r impedance of the circuit is equal to the resistance of the circuit.

Left of w_r, circuit is mainly capacitive

 Right of w_r, the circuit is mainly Inductive.

$\sqrt[]{k\cdot E}$

$\frac{r_f}{r_0}=\frac{\sqrt[]{k.E_f}}{k.E_i}$

$\frac{r_f}{r_0}=\frac{\sqrt[]{4k.E_i}}{\sqrt[]{k.E_i}}$

$\frac{r_f}{r_0}=\frac{2}{1}=2:1$

$\omega =\Delta k.E=\frac{1}{2}m\left[v_t^2-v_i^2\right]$

$=\frac{1}{2}m{\left[b*4^{5/2}\right]}^2-\frac{1}{2}m{\left[0\right]}^2$

$=\frac{1}{2}*0.5\left[0.25^2*4^5\right]=2^4=16J$

Spring constant of wire (k) =   $\frac{AE}{L}$

Case 1

      $k\left[\Delta \right]=1g$          

                      $k\left[L_1-L\right]=1g$  - (1)

              Case 2         $k\left[L_2-L\right]=2g$             - (2)

          $\left(1\right)÷\left(2\right)$      

$\frac{L_1-L}{L_2-L}=\frac{1}{2}$

L = 2L₁ – L₂