Physics

In a number without decimal, trailing zeros are insignificant. If a number has a decimal, then preceding zeros are insignificant.

Hence, in 40.40 all zeros are significant

$kx=mg-F_{up}$

$kx=mg-\frac{md}{D}g$

$x=\frac{mg}{k}\left(1-\frac{d}{D}\right)$

$i_b=\frac{5-0.7}{8.6}=0.5mA$

$\Rightarrow I_c=\beta I_b=100*0.5 mA$

By using 

$V_{CE}=V_{CC}-I_C{R}_L=18-50*10^{-3}*100=13 V$

The charge on hole is positive.

$\stackrel{?}{F}=e{E}_0\stackrel{?}{j}$

${\stackrel{?}{a}}_y=\frac{e{E}_0}{m}\stackrel{?}{j}$

$=v_0\stackrel{?}{i}+\left(\frac{e{E}_{0}t}{m}\right)\stackrel{?}{j}$

$\left|\stackrel{?}{v}\right|=\sqrt[]{v_0^2+{\left(\frac{e{E}_{0}t}{m}\right)}^2}=v_0\sqrt[]{1+{\left(\frac{e{E}_{0}t}{m}\right)}^2}$

$\lambda =\frac{{\lambda }_0}{\sqrt[]{1+{\left(\frac{e{E}_{0}t}{m}\right)}^2}}$

Two successive β decays increase the charge no. by 2.

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

$C_2=1\mu F$ , 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 $=\frac{n^{2}R}{5}$  

Given   $F\propto \frac{1}{r^n}$

$m{\omega }^{2}R=\frac{K}{R^n}$

$?\frac{m.4{\pi }^{2}R}{T^2}=\frac{K}{R^n}$

$?T^2\propto R^{n+1}$

$T\propto R^{\frac{n+1}{2}}$