Class 12th

$C_1=1\mu F{V}_1=100V?U_1=\frac{1}{2}*1*{10}^{-6}*{\left(100\right)}^2=5*{10}^{-3}J$  

$C_2=1\mu F$ , common potential $V=\frac{C_1{V}_1}{C_1{C}_2}=\frac{100}{2}=50V$

Final electrostatic energy stored in both the capacitors.

$=\frac{1}{2}\left(C_1+C_2\right)V^2=\frac{1}{2}*2*{\left(50\right)}^2*{10}^{-6}$

$=2.5*{10}^{-3}J$

$Energylost=2.5*{10}^{-3}$

$\%lossofenergy=\frac{2.5*{10}^{-3}}{5*{10}^{-3}}*100$

$=50\%$

When wire is stretched uniformly

$R\propto \mathcal{R}$

$\frac{R}{R^{\text{'}}}={\left(\frac{l}{nl}\right)}^2=\frac{1}{n^2}$  

$R^{\text{'}}=n^{2}R$

When this elongated wire is cut into 5 parts, resistance of each part  

Effective resistance between A and C $=\frac{n^{2}R}{5}$  

For normal emergence

$r=A$

From Snell's Law

$\frac{\sin ?i}{\sin ?r}=\frac{\mu }{1}$

$?\sin ?i=\sqrt[]{2}\sin ?{30}^0$

$?i={45}^0$

Fraction of molecule having energy $E_a$ or greater than $E_a=e^{\frac{-E_a}{RT}}$  

In case of gas in gas, true solution will be always formed.

$C{a}_3{\left(P{O}_4\right)}_2\to 3C{a}^{2+}+2P{O}_4^{3-}$

It depends on the requirement. But in most cases, AC generator will be preferred over a DC one. DC generator can be useful in cases where you just require a continuous supply of current without any disturbance. But if you want a good quality current to be transmitted over longer distances for significant period of time, go for an AC generator. AC generators are much more reliable and also have longer life durations.

Magnify power of telescope for far point is given as

$m=\frac{{\theta }_i}{{\theta }_0}=\frac{f_0}{f_e}$

$?{\theta }_i={\theta }_0\left(\frac{f_0}{f_e}\right){=1}^0\left(\frac{100}{5}\right)={20}^0$

For thin prism, deviation  $=\left(\mu -1\right)A$

Kindly consider the following figure