Physics

Vertical component of velocity just after collision $=\frac{u}{2\sqrt[]{2}}$

$k_i=\frac{m{u}^2}{2}{k}_f=\frac{1}{2}m\frac{u^2}{2}+\frac{1}{2}m\frac{u^2}{8}=\frac{5m{u}^2}{16}$ $\frac{}{\text{exception 25:}}$

Fraction  $=\frac{k_i-k_f}{k_i}=1-\frac{\frac{5m{u}^2}{16}}{\frac{m{u}^2}{2}}=1-\frac{5}{8}=\frac{3}{8}$

Heat lost $=$ =  Heat gained

$\Rightarrow \mathrm{}m*540+m*s_{\omega }*(100-5)=10*80+(10+74+10)*s_{\omega }*5$

$\Rightarrow \mathrm{}m(540+95)=800+94*5$

$\Rightarrow \mathrm{}m635=1270$

$\therefore \mathrm{}m=2gm$

If the multimeter shows zero or low resistance reading for forward bias and does not change even on reversing the connection then the diode is defective. It is short.

If the diode shows a high resistance under both forward and reverse biased conditions, it is defective. It is open.

If the multimeter shows zero or low resistance reading for forward bias and shows high resistance reading on reversing the connection then the diode is not defective.

For gas $\mathrm{A}:P_A($ ${}_{}$ initial $)={\left(\frac{v_f}{v_i}\right)}^{\gamma }P=(2{)}^{3/2}P$ ${\left(\frac{{}_{}}{{}_{}}\right)}^{}{}^{}$

$P_A=2\sqrt[]{2}P$

For gas B: $P_B($ ${}_{}$ initial $)=P$ $$

For gas $\mathrm{C}:P_C($  initial $)=\frac{2V_0}{V_0}P=2P\mathrm{}\therefore$  Ratio $=2\sqrt[]{2}:1:2$

  $\eta =\frac{W.D.bygas}{Heatabsorb}$

$\eta =\frac{Heat\left(Net\right)}{Heat\left(Absorb\right)}$

$\eta =\frac{Areaunderthecurve}{HeatAbsorb}$

$\eta =\frac{\frac{1}{2}*S_0{T}_0}{\frac{1}{2}\left(3T_0\right)S_0}=\frac{1}{3}$

Voltmeter readings

  $\begin{array}{ll}\Rightarrow & 20=I_{1}R\\ \Rightarrow & 20=\left(\frac{R_v}{R+R_v}\right)4R\\ \Rightarrow & R_{v}R=5R+5R_v\Rightarrow \mathrm{}R=\frac{5R_v}{R_v-5}\\ \therefore & R=\frac{5}{1-\frac{5}{R_v}}\end{array}$

For $5<{\mathrm{R}}_v<\mathrm{\infty },\mathrm{R}>5\mathrm{\Omega }$

Stopping potential increases and number of photons decreases.

$I=\frac{hc{n}_p}{\lambda t}$

$\frac{1}{0.9}=+\frac{1}{\left(10?2t\right)}+\frac{1}{2t}$  

t = 0.5 sec and t = 4.5 sec.

Here the width of principal maxima is 2.5 mm

Then its half width is =  $\frac{\beta }{2}$  

$=\frac{2.5}{2}=1.25*10^{-3}m$             

Diffraction angle q = $\frac{\left(\frac{\beta }{2}\right)}{D}=\frac{1.25*10^{-3}}{2}$       …(i)

  $\therefore a\theta =\lambda \therefore \theta =\frac{\lambda }{a}$   …(ii)

From (i) and (ii)

  $\lambda =\theta a$          

$\frac{1.25*10^{-3}}{2}*10^{-3}$            

$\therefore \lambda =6250$ = 6.25 * 10^-7 m