Albert Einstein theoretically proposed the existence of gravitational waves in the year 1915.

Proton and neutron are made up of quarks. Later, it was found that Higg's particles or God particles give mass to the particles like protons, neutrons etc.

Parrot fish's source of bite → mimic → Ultra durable synthetic material.
Lotus leaf surface → SEM → Self Cleaning Process
The scales on the wings of a morpho butterfly → mimic → Interaction of colours.
Peacock feathers → mimic → Glowing in different colours.

Size of the particle is between
1-100 nm - Nano.
Size of the particle is greater
than-100 nm - Bulk.

A = 1, B = 0, C = 1
y = A + B, y = (A + B).C
y = (1 + 0).1 ⇒ y = 1

\(Y_1=\overline{A+B}, y_2=\overline{A+B}=A+B, y=\overline{A+B}\)

\(y=\overline{A}=\overline{1011}=0100\)
The Boolean expression for NOT gate, \(y=\bar{A}\)

B = ∆m x c²
∆m = \(\frac{B}{c^2}\)
N M_n + Z M_p - M(N,Z) = \(\frac{B}{c^2}\)
N (N,Z) = N M_n + ZM_p - \(\frac{B}{c^2}\)

r ∝ A^1/3
Surface Area = 4πr²
Hence, Surface Area ∝ A^2/3

Electric potential in n^th orbit
\(\mathrm{V} =\mathrm{V}_0 \ln \left(\frac{\mathrm{r}_{\mathrm{n}}}{\mathrm{r}_0}\right)  \)
\(=\mathrm{V}_0\left(\ln \mathrm{r}_{\mathrm{n}}-\ln \mathrm{r}_0\right)  \)
\(=\mathrm{V}_0 \ln \mathrm{r}_{\mathrm{n}}-\mathrm{V}_0 \ln \mathrm{r}_0\)
\(\left|\mathrm{F}_\epsilon\right|=\mathrm{e} \frac{\mathrm{dv}}{\mathrm{dr}}  =\mathrm{e} \frac{\mathrm{d}}{\mathrm{dr}}\left(\mathrm{V}_b / n \mathrm{r}_B-\mathrm{V}_0 / m \mathrm{r}_0\right)  \)
\(=\mathrm{c}\left(\frac{\mathrm{V}_0}{\mathrm{r}_n}-0\right)=\frac{\mathrm{eV}}{\mathrm{r}_{\mathrm{n}}} \)
Centripetal force = coulomb force
\(\frac{m v^2}{r_n}=\mathrm{c} \frac{V_0}{r_n} \Rightarrow v=\sqrt{\frac{e V_0}{m}}=\text { constant }\)
Angular momentum,
\(\mathrm{mvr}_n=\frac{\mathrm{nh}}{2 \pi}\)
\(\mathrm{m}, \mathrm{v}, \mathrm{h}, 2 \pi\) are constants
hence, r_n ∝ n

Cathode rays are stream of negatively charged electron.

\(V_{ionisation}=\frac{13.6}{n^2}Z^2 volt\)
\(Z=\sqrt\frac{V\times n^2}{13.6}=\sqrt\frac{122.4 \times I^2}{13.6}=\sqrt{9}=3\)

\(\text {K.E }  =\frac{B^2 q^2 r^2}{2 m}  \)
\(\phi  =\frac{h c}{\lambda}-K . E  \)
\(=\frac{h c}{\lambda}-\frac{B^2 q^2 r^2}{2 m}  \)
\(\phi  =\frac{h c}{\lambda}-2 m\left(\frac{B q r}{2 m}\right)^2\)

\(\lambda_0  =\frac{h c}{\phi}  \)
\(=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.313 \times 1.6 \times 10^{-19}}  \)
\( =\frac{19.8782400}{5.3} \times 10^{-7}  \)
\(\lambda_0  =3.750 \times 10^{-7} \simeq 3750 \stackrel{o}A\)

\(P =\frac{E}{t}=\frac{n h v}{t}=\frac{n h c}{\lambda t} \Rightarrow P \propto n / t  \)
\(\frac{P_1}{P_2}  =\frac{1.38 \times 10^{15}}{460} \times \frac{520}{1.04 \times 10^{15}}=1.5\)

\(\lambda_{\mathrm{i}}  \frac{1}{\mathrm{mv}}  \)
\(\frac{\lambda_p}{\lambda_e}  =\frac{m_e v_e}{m_P v_P}  \)
\(1  =\frac{9.1 \times 10^{-31} \times 6 \times 10^6}{3 \times 10^{-9} \times v_p}  \)
\(\mathrm{v}_{\mathrm{p}}  =9.1 \times 10^{-16} \times 2  \)
\(\mathrm{v}_{\mathrm{p}}  =18.2 \times 10^{-16}  \)
\(\mathrm{v}_{\mathrm{p}}  =1.82 \times 10^{-15} \mathrm{~m} \mathrm{~s}^{-1}\)

\(\mathrm{E}_{\mathrm{p}}  =\frac{\mathrm{hc}}{\lambda_{\mathrm{p}}}  \)
\(\mathrm{E}_{\mathrm{e}}  =\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda_{\mathrm{e}}^2}  \)
\(\frac{\mathrm{hc}}{\lambda_{\mathrm{p}}}  =\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda_{\mathrm{e}}^2}  \)
\(\lambda_{\mathrm{p}}  \propto \lambda_{\mathrm{e}}^{{ }^2}\)

n = tan i_p = tan 60^o = \(\sqrt{3}\)

d' = 2d, β' = β, D' = ?
W.K.T, Fringe width
\(\beta = \frac{D\lambda}{d} \Rightarrow D' = \frac{Dd'}{d}\)
\(D' = \frac{D2d}{d}=2D\)

Refractive index for violet is more and wavelength for violet is very low comparing other colours. So, the letter which appears to be raised more is violet.

For total internal reflection,
sin i > sin c
\(n=\frac{1}{sin \ c}\)
\(sin \ c=\frac{1}{n}\)
\(sin \ i>\frac{1}{n}\)
\(n>\frac{1}{sin \ i}\)
n >\(\sqrt{2}\)
n >1.414 = 1.5

\(\frac{1}{f}  =(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)  \)
\(=(1.5-1)\left(\frac{1}{\infty}-\frac{1}{(-10)}\right)\)
(Since plano convex lens)
\(=0.5\left[\frac{1}{10}\right]=\frac{1}{20}  \)
\(\mathrm{f}_t  =20 \mathrm{~cm}\)
Formula for silvered lenses
\(\frac{1}{\mathrm{~F}}  =\frac{2}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_m}  \)
\(\frac{1}{\mathrm{~F}} =\frac{2}{20}+\frac{1}{\infty}  \)
\(\therefore \mathrm{F}  =\frac{20}{2}=10 \mathrm{~cm}\)

At end A,
\(\frac{1}{f}  =\frac{1}{u_A}+\frac{1}{v_A}  \)
\(\therefore \frac{1}{v_A}  =\frac{1}{-10}-\frac{1}{-20}  \)
\(\frac{1}{v_A}  =-\frac{1}{10}+\frac{1}{20}=\frac{-2+1}{20}=-\frac{1}{20}  \)
\(v_A  =-20 \mathrm{~cm} \)
\(\left|v_{\wedge}\right|=20 \mathrm{~cm}\)
At end B,
\(\frac{1}{f}  =\frac{1}{u_B}+\frac{1}{v_B}  \)
\(\frac{1}{v_B} =\frac{1}{f}-\frac{1}{u_B},  \)
\(u_B  =-30 \mathrm{~cm}  \)
\(\frac{1}{v_B}  =-\frac{1}{10}+\frac{1}{30}  \)
\(=\frac{-3+1}{30}=\frac{-2}{30}=\frac{-1}{15}  \)
\(v_B  =-15 \mathrm{~cm}  \)
\(\left|v_B\right|  =15 \mathrm{~cm} \)
\(\therefore \quad\left|\mathrm{v}_{\mathrm{A}}\right|-\left|\mathrm{v}_{\mathrm{B}}\right| \) is the length of the image
= 20 - 15 = 5 cm

\( { E } ={ E }_{ 0 }\hat { k } \) and \({ B } ={ B }_{ 0 }\hat { j } \)

Q = q₁ + q₂ = 4 x 10^-2C
\(q_{2f}=Q[\frac{r_2}{r_1+r_2}]\)
= 4 x 10^-2 \([\frac{3}{4}]\)
q_2f = 3 x 10^-2 C

Energy density u_E \(=\frac{U}{volume}\)
If A' = 2A     d ' = 2d
Then V ' = 2A  x 2d = 4Ad = 4V
Then volume would be increased. So, energy density will change.

If voltage is increased from V to 2 V
Then Q₁ = CV
Q₂ = C(2 V) = 2 CV
∴ Q is doubled and C remains same

Force ∝ charge 
After the separation, the magnitude of charge will be increased. So the force will be more than before.