Physics

 $E=\frac{\lambda }{4\pi {\epsilon }_{0}a}\left[sin\alpha +sin\alpha \right]=\frac{\lambda sin\alpha }{2\pi {\epsilon }_{0}a}$

$\Rightarrow E=\frac{Q}{L2\pi {\epsilon }_0*\left(\frac{\sqrt[]{3}L}{2}\right)}*\frac{1}{2}=\frac{Q}{2\sqrt[]{3}\pi {\epsilon }_0{L}^2}$

$tan\alpha =\frac{\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\sqrt[]{3}L/2}\Rightarrow \alpha =\frac{\pi }{6}\Rightarrow sin\alpha =\frac{1}{2}$

Method – I:

$F=M_{sphere}{E}_{ring}=M*\frac{Gmx}{{\left(R^2+x^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=M*\frac{Gmx\sqrt[]{8}R}{{\left(R^2+8R^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{\sqrt[]{8}GMm}{27R^2}$

Method – II:

$F={\displaystyle \int dFcos\theta =cos\theta {\displaystyle \int \left(dm\right)\frac{GM}{\left(R^2+8R^2\right)}=\frac{\sqrt[]{8}GMm}{27R^2}}}$

As height of image is less than height of image and has same orientation as that of object, so mirror must be convex.

The best conductor of electricity is Silver (Ag) at room temperature. It is not used in normal wiring even though being the best conductor because of its high cost.

• Electrical Conductivity ($\sigma$): $6.3 \times 10^ {7} \ \text {S/m}$

• Electrical Resistivity ($\rho$): $1.59 \times 10^ {-8} \ \Omega \cdot \text {m}$ (the lowest among all metals)

A -> m_A = 1 kg, P_A = P

B ->  m_B = 2 kg,   P_B = P

$?kE=\frac{P^2}{2m}$

$\therefore \frac{k{E}_A}{k{E}_B}=\frac{\left(\frac{P^2}{2mA}\right)}{\left(\frac{P^2}{2m_B}\right)}=\frac{m_B}{m_A}=\frac{2}{1}$

$7.2*\frac{5}{18}=2m/s$ [V_A = V_B = 2m/s]

$f{\text{'}}_B=f_A\left(\frac{v+v_B}{v-v_A}\right)$

$\downarrow$     

App freq heard by car B

$f{\text{'}}_B=676\left[\frac{340+2}{340-2}\right]=676*\frac{342}{338}=684Hz$  

Similarly, $f{\text{'}}_A=f_B\left(\frac{v+v_B}{v-v_B}\right)=684Hz$

$\therefore$ Beat frequency heard by both = 684 – 676 = 8Hz

radius r = 0.5 mm

$\sigma =5*10^{7}s/m$

$\overrightarrow{E}=10*10^{-3}V/m=10^{-2}V/m$

As we know

$J=\sigma E$

$J=\frac{i}{A}=\sigma E$

$i=\sigma EA=5*10^7*10^{-2}*\pi *{\left(0.5*10^{-3}\right)}^2=1.25\pi *10^{-1}A=125\pi mA=x^3\pi mA$

$$ $\therefore$  x = 5

f = 3GHz = 3 * 10⁹ Hz

${\lambda }_{vacuum}=\frac{v}{f}=\frac{3*10^{8}m/s}{3*10^9/s}=10^{-1}m=0.1m$

${\lambda }_{vacuum}=\frac{{\lambda }_{vacuum}}{{\mu }_{medium}}=\frac{0.1}{{\mu }_{medium}}$

${\mu }_{medium}=\sqrt[]{{\mu }_r{\in }_r}=\sqrt[]{1*2.25}=1.5$

$\therefore {\mu }_{medium}=\frac{0.1m}{1.5}=0.0667m=6.67cm=667*10^{-2}cm$

Sound level is given in dB = 10 log $\frac{{}_{}}{{}_{}}\frac{}{{}_{}}$ $\left(\frac{P_0}{P_i}\right)or10log\left(\frac{l}{l_0}\right)$

As sound level decreases 5 dB every km,

So, in 20 km sound level will decrease by 20 * 5 = 100 dB.

$\Delta \beta ={\beta }_2-{\beta }_1=10log\left(\frac{l_2}{l_1}\right)$

$-100=10log\left(\frac{l_2}{l_1}\right)$

$10^{-10}=\frac{l_2}{l_1}\Rightarrow l_2=10^{-10}{l}_1$

$P_2=10^{-10}{P}_1$

$P_2=10^{-10}*\left(0.1*10^3\right)\Rightarrow P_2=10^{-8}W=10^{-x}W$

$$ $\therefore$  x = 8

${\omega }_0=10^{5}rad/sec$

P = 16w, 120v at resonance

$P=\frac{v^2}{R}\Rightarrow 16=\frac{{\left(120\right)}^2}{R}\Rightarrow R=\frac{14400}{16}=900\Omega$