Class 12th

We can define enzymes as biocatalysts. It is important in biological processes like digestion, metabolism, and the synthesis of cellular components. If enzymes stop their functions, then the body can not generate enough proteins, which is important for survival. The examples of enzymes are lactase and amylase.

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.

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

Optical path $=\mu x$ =μx

Path-difference $=\mu x$ =μx

Phase-difference $=\frac{2\pi }{\lambda }*$ =2πλ*  path-difference $=\frac{2\pi }{\lambda }\mu x$ =2πλμx

Area of ellipse =  $\pi ab$

Magnetic dipole moment =    $\frac{\pi abi}{2}$

Torque    $\overrightarrow{\tau }=\overrightarrow{M}*\overrightarrow{B}$

  $\tau =MBsin\theta$          

$?\overrightarrow{M}\left|\right|\overrightarrow{B}$ $\boxed{\tau =0}$               

${\left(n_i\right)}^2=n_e*n_h$ where $n_e=\frac{5*{10}^{28}}{{10}^6}$

$\begin{array}{rr}\therefore \mathrm{}{n}_h& =\frac{{\left(n_i\right)}^2}{n_e}=\frac{\left(1.5*{10}^{16}\right)\left(1.5*{10}^{16}\right)*{10}^6}{5*{10}^{28}}\\ n_h& =4.5*{10}^9/{\mathrm{m}}^3\end{array}$