Physics Ncert Solutions Class 12th

$C_{eq}=C_1+C_2+C_3+C_4$

= 1 + 2 + 4 + 3

$=10\mu F$

Q = C_eq v

= (10 * 20) $\mu C$

Q = 200 $\mu C$

Path $bc$  is an isochoric process.

$\therefore$ Work done by gas along path $bc$ is zero.

According to given truth table, output is independent on value of $A$

$\therefore$ Output $Y=\overline{B}$

For 90 mA in zener diode, current in R should be greater than 90 mA.

$\frac{45}{R}=90*10^{-3}\Rightarrow R=\frac{45}{90*10^{-3}}=500\Omega$

from newtan's law of cooling :

$\frac{dT}{dt}=-k\left(T-T_0\right)$

Using average value :

$\frac{75-65}{5}=-k\left[\frac{\left(75+65\right)2}{2}-25\right]$

$\frac{\left(65-T\right)}{5}=-k\left[\frac{\left(65+T\right)}{2}-25\right]$

$\Rightarrow \frac{10}{65-T}=\frac{45}{\frac{65+T}{2}-25}$

$\Rightarrow T=57°C$

d = 150 km

$d=\sqrt[]{2hR}$

$h=\frac{d^2}{2R}=\frac{{\left(150*10^3\right)}^2}{2*6.5*10^6}=1731m.$

Population covered

$\pi r^2*\sigma =3.14*150^2*2000=1413*10^5$

$E=\frac{hc}{\lambda }=\frac{1242eV-nm}{621nm}=2eV$

minimum 2eV is required to emit.

Using conservation of momentum

=> 180 V₁ = 4V₂

=> V₂ = 45 V₁

  $Q=K{E}_{180}+K{E}_{\alpha }$              

$Q=\frac{1}{2}180V_1^2+\frac{1}{2}4V_2^2$        

= $\frac{1}{2}*180*{\left(\frac{V_2}{45}\right)}^2+\frac{1}{2}4V_2^2$

$K{E}_{\alpha }=\frac{5.5MeV}{\left(\frac{46}{45}\right)}=5.38MeV$

$P_z=V_z{l}_z\Rightarrow 0.5=8l_z\Rightarrow l_z=l_P=\frac{1}{16}Amp$

When zener is connected across a potential divider arranged with maximum potential drop across zener diode, then

$$ $V_P=V-V_z=20-8=12$  volt Potential difference across protective resistance RP

$R_P=\frac{V_P}{I_P}=\frac{12}{\frac{1}{6}}=192Amp.$

$R=\frac{\Delta V}{\Delta l}=\frac{0.75-0.70}{\left(5-3\right)*10^{-3}}=\frac{0.05*1000}{2}=25\Omega$