From the given values

\(\large E_{2}=hv_{1}\)

and \(\large E_{3}=hv_{3}\)

Now \(\large hv_{2}=E_{3}-E_{2}\, \, so\, \, v_{2}=v_{3}-v_{1}\)

Hence \(\large v_{1}+v_{2}=v_{3}\)

It is given three capacitors of equal capacities are connected in parallel

And also one of the same capacity is connected in series with its combination

And the resultant capacity is 3.75 . \(\large \mu F\)

Resultant capacity \(\Large C_{eq}=\frac{3C\times C}{3C+C}\)

\(\large C_{eq}=\frac{3C}{4}=3.75\)

Hence C = 5 \(\large \mu F\)

\(\Large S=\frac{G}{8}\, \, ,\, \, S=\frac{G}{n-1} \)

\(\Large \Rightarrow \frac{G}{8}=\frac{G}{n-1} \)

So n = 8 + 1 = 9

range of galvanometer in increased nine times so \(\large S' = \frac {S}{9}\)

\(\large \iota _{x} =40\, cm,\, \,\iota _{R} =60 \, cm\)

\(\Large \frac{x}{R}=\frac{\imath _{x}}{\imath _{R}}=\frac{40}{60}=\frac{2}{3}-----(i)\)

\(\Large \frac{x+30}{R}=\frac{60}{40}\)

\(\large {2(x+30)}=3R ----- (ii)\)

From equation (i) and (ii) we can derive the value

\(\large x=24\Omega \)

Given values are \(\large a=0.2\times 10^{-3}m, \,D=2m, \, \lambda =5\times 10^{-7}m\)

\(\Large x= \frac{\lambda D}{a}=\frac{5\times 10^{-7}\times 2}{0.2\times 10^{-3}}\)

\(\large x= 5\times 10^{-3}m\)

Distance between 1st minima on either side \(\large = (5 \times 10^{-3}) + (5 \times 10^{-3})=10\times 10^{-3}\)

\(\large = 10^{-2}\)

\(\large Power=V_{RMS}.I_{RMS}\,cos \Phi \)

\(\Large V_{RMS}.\frac{V_{RMS}}{z}.\frac{R}{z}\)

\(\Large \frac{220\times 220\times 18}{33\times 33}=800\, W\)

\(\large \frac{1}{C_{1}}=\frac{1}{C}+\frac{1}{C}\Rightarrow C_{1}=\frac{C}{2}\)

\(\large \frac{1}{C_{2}}=\frac{1}{C}+\frac{1}{KC}\)

So \(\large C_{2}=\frac{C}{\left ( 1+\frac{1}{K} \right )}=\frac{CK}{K+1}\)

\(\large \Delta C=\frac{CK}{K+1}-\frac{C}{2}=C\left [ \frac{K}{K+1}-\frac{1}{2} \right ]\)

\(\large =\frac{C}{2}\left [ \frac{K-1}{K+1} \right ]\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\large tan\theta =\mu =\frac{C}{V}\)

\(\large \Rightarrow cot\theta =\frac{V}{C}\)

\(\large \Rightarrow \theta =cot^{-1}\left ( \frac{V}{C} \right ) \)

\(\large I_{R}=4I\, cos^{2}\frac{\Phi }{2}\)

Hence \(\large I_{R} \propto cos^{2}\frac{\Phi }{2}\)

\(\large \beta _{dc}=\frac{\alpha _{dc}}{1-\alpha _{dc}}\)

\(\large \Rightarrow \frac{\alpha _{dc}}{\beta _{dc}}=1-\alpha _{dc}\)

\(\Large \Rightarrow \frac{\beta_{dc}-\alpha _{dc}}{\beta_{dc}.\alpha _{dc}} =\frac{\beta_{dc}\left ( 1-\frac{\alpha _{dc}}{\beta_{dc}} \right )}{\beta_{dc}.\alpha _{dc}} \)

\(\Large = \frac{1-\frac{\alpha _{dc}}{\beta_{dc}}}{\alpha _{dc}} \)

\(\Large = \frac{1-1+\alpha _{dc}}{\alpha _{dc}}=1 \)

\(\large = \frac{N}{N_{0}}=e^{-\lambda t}\)

Now on applying values

\(\large e^{-\lambda \times 4}=\frac{2500}{10000}\)

\(\large \Rightarrow e^{\lambda 4}=4\)

\(\large \Rightarrow 4\lambda = log_{e}4\)

\(\large \Rightarrow 4\lambda = 2 log_{e}2\)

\(\Large \Rightarrow \lambda = 0.5\times log_{e}2\)

What is number of waves in glass slab ? No. of waves in glass slab is same as number of waves in water column

\(\large \mu _{g}.h _{g}=\mu _{w}.h _{w}\)

\(\large \mu _{w}=\frac{\mu _{g}.h _{g}}{h _{w}}\)

On applying values

\(\Large \mu _{w}=\frac{1.5\times 6}{7}=1.286\)

\(\Large E=\frac{13.6}{1}-\frac{13.6}{2^{2}}\)

\(\large =10.2eV (=hV)\)

\(\large hV=\Phi _{0}+eV_{s}\)

By given values

\(\large 10.2 \,eV=4.2 \,eV+eV_{s}\)

Hence \(\large V_{s}= 6V\)

\(\Large M_{0}=\frac{e}{2m_{0}}\times L_{0}\)

\(\large L_{0}\propto n\)

So \(\Large M_{0}\propto n\)

It is on theoritycal fact as

Photodiode is a device which is always operated in reverse bias

The given values are \(\large I=2\, Kg\, m^{2},\omega _{0}=60 \, rad/s, t =5\, min =60\, sec\)

\(\large \alpha =\frac{0-60}{300}=\frac{-1}{5}\, rad/s^{2}\)

The question is asking for 2 minutes i.e. 120 sec

\(\large \omega = 60-\frac{1}{5}\times 120=36\, rad/s\)

\(\large L=I\omega = 2\times 36=72\, kg\, m^{2}/s\)

\(\large y=A\, sin\left [ 2\pi \left ( \frac{t}{T}-\frac{x}{\lambda } \right ) +\frac{\pi }{4}\right ]\)

Compare with given equation in question

\(\large y=3\, sin\left [ 2\pi \left ( \frac{t}{6}-\frac{x}{10 } \right ) +\frac{\pi }{4}\right ]\)

Hence \(\large \lambda =10\)

\(\Large \frac{Q}{t}=\sigma AT^{4}\)

So \(\Large A\propto \frac{1}{T^{4}}\)

So \(\Large \frac{A_{2}}{A_{1}}=\frac{T_{1}^{4}}{T_{2}^{4}}\)

\(\Large \Rightarrow \frac{4\pi r_{2}^{2}}{4\pi r_{1}^{2}}=\frac{T_{1}^{4}}{T_{2}^{4}}\)

\(\Large \Rightarrow \frac{r_{2}}{r_{1}}=\left ( \frac{T_{1}}{T_{2}} \right )^{2}\)

What is first overtone for open pipe \(\Large v_{1}=\frac{v}{L}\)

for closed pipe \(\Large {v}'_{1}=\frac{3v}{4L}\)

\(\Large v_{1}-{v}'_{1}=\frac{v}{L}-\frac{3v}{4L}=3\)

So \(\Large \frac{v}{4L}=3 \,\,then\,\, \frac{v}{L}=12\)

When length of open pipe is made  \(\Large \frac{L}{3}\) the fundamental frequency  \(\Large \frac{v}{2\left ( \frac{L}{3} \right )}=\frac{3v}{2L}\)                             

the fundamental frequency of closed pipe when length is made 3 times \(\Large {v}'=\frac{v}{4\left ( 3L \right )}=\frac{v}{12L}\)

What Beats produced is \(\large v-{v}'\)=\(\large \frac{3v}{2L}-\frac{v}{12L}=\frac{17}{12}.\frac{v}{L}=17\)

What is the maximum tension in the rope ?

It is m (g + a)

What is the Stress in the rope

it is \(\Large \frac{m(g+a)}{\pi r^{2}}\)

So \(\Large T=\frac{m(g+a)}{\pi r^{2}}=\frac{m(g+a)}{\pi \left ( \frac{d}{2} \right )^{2}}\)

Now \(\Large T=\frac{4m(g+a)}{\pi d^{2}}\)

Then \(\Large d=\left [ \frac{4m(g+a)}{\pi T} \right ]^{\frac{1}{2}}\)

What is the K E  of rolling solid sphere ?

\(\large \frac{1}{2}mV^{2}+\frac{1}{2}I\omega ^{2}\)

\(\large \frac{1}{2}mV^{2}+\frac{1}{2}\times \frac{2}{5}mr^{2}\omega ^{2}=\frac{7}{10}mV^{2}\)

What is the P E of the spring on maximum compression x

\(\Large \frac{1}{2}kx^{2}\)

So  \(\Large \frac{1}{2}kx^{2}\)\(\Large =\frac{7}{10}mV^{2}\)

\(\Large x^{2}=\frac{14}{10}\frac{mV^{2}}{k}\)

Let's apply values to this equation

\(\large x^{2}=\frac{14}{10}\frac{2\times 6^{2}}{36}\)

\(\large x=\sqrt{2.8}\,\,m\)

\(\Large \alpha =\frac{\omega -\omega _{0}}{t}\)

On applying given values \(\Large \alpha =\frac{24}{3}=3\: rad/s^{2}\)

Total angle turned \(\Large \theta =\omega _{0}t+\frac{1}{2}\alpha t^{2}\)

\(\large \theta =0+\frac{1}{2}\times 3\times 8^{2}=96 \,rad\)

How can we represent fundamental frequency of the first wire

\(\Large f=\frac{1}{2L_1}\sqrt{\frac{T}{m}}=\frac{1}{2L_1}\sqrt{\frac{T}{\pi r_{1}^{2}\rho }}=\frac{1}{2L_{1}r_{1}}\sqrt{\frac{T}{\pi \rho }}\)

The first overtone \(\Large f_{1}=\frac{1}{L_{1}r_{1}}\sqrt{\frac{T}{\pi \rho }}\)

The second overtone \(\Large f_{2}=\frac{3}{2L_{2}r_{2}}\sqrt{\frac{T}{\pi \rho }}\)

\(\large f_{1}=f_{2}\)

So \(\Large \frac{1}{L_{1}r_{1}}\sqrt{\frac{T}{\pi \rho }}\)\(\Large =\frac{3}{2L_{2}r_{2}}\sqrt{\frac{T}{\pi \rho }}\)

\(\large 3L_{1}r_{1}=2L_{2}r_{2}\)

\(\Large \frac{L_{1}}{L_{2}}=\frac{2}{3}\frac{r_{2}}{r_{1}}\, \, Here \, \, r_{1}=2r_{2}\, \, so \, \, \frac{L_{1}}{L_{2}}=\frac{1}{3}\)

\(\large F = 2\pi rT\, \, \Rightarrow 2\pi r=\frac{F}{T}\)

Now let's put the given value

\(\Large F = 2\pi rT\, \, \Rightarrow 2\pi r=\frac{105\times 10^{-5}}{7\times 10^{-2}}=1.5\, cm\)

\(\large \frac{C_{p}}{C_{v}}=\frac{7}{5}\, \, \Rightarrow C_{v}=C_{p}\frac{5}{7} \)

Again \(\large R=C_{p}-C_{v}\)

\(\Large \Rightarrow R=C_{p}-\frac{5}{7}C_{p}=\frac{2}{7}C_{p}\)

Hence \(\Large n=\frac{2}{7}=0.2857\)

In case of  ideal gas PV = nRT

\(\Large PV=\frac{{m}'}{M}RT=\frac{{m}'}{V}.\frac{RT}{M}\)  (where m' is the mass of the gas and M molecular weight)

So \(\Large P=\frac{\rho RT}{m} ,\, \, \, \, \, \,\rho \) is the density

\(\large \rho =\frac{PM}{RT}=\frac{PM}{NKT}\, \, \, \, \, \, \, N \, \,is\, \, Avogadro \, \,number\)

\(\Large \rho =\frac{Pm}{KT}\)   (where \(\Large m = \frac {M}{N}\)= mass of each molecule)

Volume of big drop = n(Volume of small drop)

\(\large \frac{4}{3}\pi R^{3}=n.\frac{4}{3}\pi r^{3}\, \, \Rightarrow R^{3}=nr^{3}\)

\(\Large R=n^{\frac{1}{3}}.r\)

\(\large E_2\) = Surface energy of n drops =\(\large n\times 4\pi r^{2}\times T\)

\(\large E_1\)= Surface energy of big drop =\(\large 4\pi R^{2} T\)

\(\Large \frac{E_{2}}{E_{1}}=\frac{nr^{2}}{R^{2}}\)

On applying value of R in above equation \(\Large \frac{E_{2}}{E_{1}}=n^{\frac{1}{3}}\)

So the Ratio is \(\Large \sqrt[3]{n}:1\)

\(\Large E_{1}=\frac{GMm}{R}\)

\(\Large E_{2}=\frac{GMm}{2\left ( R+h \right )}\)

So \(\Large \frac{E_{1}}{E_{2}}=\frac{2\left ( R+h \right )}{R}\)

Displacement \(\large X=A \, \, sin\omega t\)

Velocity \(\Large v=\frac{dx}{dt}=A\omega cos\omega t\)

We know \(\large v=\pi \,m/s \,\, and\, \, T=16 sec \)

Now \(\Large \omega =\frac{2\pi }{T}=\frac{2\pi }{16}=\frac{\pi }{8}\, rad/s\)

So \(\large \pi =A\times \frac{\pi }{8}\times cos \frac{\pi }{8}\times 2\)

Based on value it is \(\large 1=\frac{A }{8}cos \frac{\pi }{4}\, \, \Rightarrow A=8\sqrt{2}\, m\)

\(\large {g}'=g\left ( 1-\frac{d}{R} \right )\)

\(\large {g}'=\frac{g}{n}\)

\(\large \Rightarrow \frac{g}{n} =g\left ( 1-\frac{d}{R} \right )\)

\(\large \Rightarrow \frac{1}{n} = 1-\frac{d}{R} \Rightarrow \frac{d}{R}=1-\frac{1}{n}\)

\(\large d=R\left ( \frac{n-1}{n} \right )\)

What is the fundamental frequency for closed pipe ?

\(\Large n_{c}=\frac{V}{4L}=100 \, HZ\)

What is the fundamental frequency for open pipe ?

\(\large n_{o}=\frac{V}{2L}=2n_{c}=200 \, HZ\)

So open at both the ends, the frequencies produced in Hz are 200, 400, 600, 800, …