Bohr model of an atom states that only those orbits are allowed where angular momentum of electrons are integral multiple of nh/2π. These orbits have quantized energy and angular momentum associated with electron.

The model can be applied to hydrogen or hydrogen-like atoms to explain their line emission spectrum.

Given: A solenoid of inductance \(L\) and resistance \(R\) is connected to a battery. The time is taken for the magnetic energy to reach \(\frac 14\) of its maximum value is

Initial magnetic energy \(=\frac{1}{2} L i ^{2}\)

Final magnetic energy \(=\frac{1}{4}\) of initial magnetic energy

\(=\frac{1}{4}\left(\frac{1}{2} L i ^{2}\right)\)

For final energy to become \(\frac{1}{4}\) of initial magnetic energy final current have to become \(\frac 12\) of initial current.

We have a relation \(i = i _{0}\left(1- e ^{\frac{- Rt }{ L }}\right)\)

\(t =\) time, \(L=\) inductance, \(R =\) resistance

Substituting the value of current here we get

\(\frac{i_{o}}{2}=i_{0}\left(1-e^{\frac{-R t}{L}}\right)\)

\(\Rightarrow\frac{1}{2}=\left(1-e^{\frac{-R t}{L}}\right)\)

\(\Rightarrow\frac{-1}{2}=\left(- e ^{\frac{- Rt }{ L }}\right)\) (taking \(\ln\) both sides)

\(\Rightarrow\ln \left(\frac{1}{2}\right)=\frac{- Rt }{ L }\)

\(\Rightarrow t =\frac{ L }{ R } \ln ( 2 )\)

In a circuit ammeter is always connected in series and voltmeter is always connected in parallel. So the option A is the correct setups for verifying Ohm's law.

Given,

Thickness of the dielectric slab, \({t}=1 {~cm}=10^{-2} {~m}\)

Dielectric constant, \(\varepsilon_{{r}}={K}=5\)

Area of the plates of the capacitor, \(A=0.01 {~m}^{2}=10^{-2} {~m}^{2}\)

Distance between parallel plates of the capacitor, \({d}=2 {~cm}=2 \times 10^{-2} {~m}\)

We know that:

Capacity with air in between the plates,

\(C_{0}=\frac{\epsilon_{0} A}{d}\)

where,\(\epsilon_{0}=8.854 \times 10^{-12} \)

\(=\frac{8.85 \times 10^{-12} \times 10^{-2}}{2 \times 10^{-2}}\)

\({C}_{0}=4.425 \times 10^{-12}\) Farad

Capacity with dielectric slab in between the plates,

\({C}=\frac{\epsilon_{0} {~A}}{{~d}-{t}\left(1-\frac{1}{{~K}}\right)}\)

\(=\frac{8.85 \times 10^{-12} \times 10^{-2}}{\left(2 \times 10^{-2}\right)-10^{-2}\left(1-\frac{1}{5}\right)}\)

\(C=7.375 \times 10^{-12}\) Farad

Increase in capacity on introduction of dielectric:

\({C}-{C}_{0}=\left(7.375 \times 10^{-12}\right)-\left(4.425 \times 10^{-12}\right)\)

\(=2.95 \times 10^{-12}\) Farad

For the net gravitational force on a particle.

\(F=\frac{G M^{2}}{a^{2}} \).....(1)

\(F_{1}=\frac{G M^{2}}{(a \sqrt{2})^{2}}\)

\(=\frac{G M^{2}}{2 a^{2}}\).....(2)

According to fig.:

For net force, \(F=\sqrt{F^{2}+F^{2}}+F_{1}\)

\(F_{net}=F \sqrt{2}+F_{1}\).....(3)

Put values from (1) and (2) in (3).

\(F_{net}=\frac{G M^{2}}{a^{2}} \sqrt{2}+\frac{G M^{2}}{2 a^{2}}\)

This force will act as centripetal force. Distance of particle from centre of circle is \(\frac{\mathrm{a}}{\sqrt{2}}\).

\(\mathrm{~F}_{\mathrm{C}}=\frac{\mathrm{Mv}^{2}}{\mathrm{r}} \)

\(\mathrm{r}=\frac{\mathrm{a}}{\sqrt{2}}\)

\(F_{C}=F_{net}\)

\(\frac{\mathrm{Mv}^{2}}{\frac{\mathrm{a}}{\sqrt{2}}}=\frac{\mathrm{GM}^{2}}{\mathrm{a}^{2}}\left(\frac{1}{2}+\sqrt{2}\right) \)

\(\Rightarrow \mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{a}}\left(\frac{1}{2 \sqrt{2}}+1\right) \)

\(\Rightarrow \mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{a}}(1.35) \)

\(\Rightarrow \mathrm{v}=1.16 \sqrt{\frac{\mathrm{GM}}{\mathrm{a}}}\)

\(\begin{aligned} & \therefore \sin 37^{\circ}=\frac{B_v}{B_{\text {net }}} \\ & \Rightarrow B_v=B_{\text {net }} \sin 37^{\circ} \\ & \Rightarrow B_{\text {net }}=\frac{B_v}{\sin 37^{\circ}} \\ & =\frac{6 \times 10^{-5}}{\frac{3}{5}} \\ & =10 \times 10^{-5} \mathrm{~T} \\ & =1 \times 10^{-4} \mathrm{~T}\end{aligned}\)

Total path length is a scalar quantity, whereas displacement is a vector quantity. Therefore, the total path length is always greater than the magnitude of displacement. It becomes equal to the magnitude of displacement only when a particle is moving in a straight line.

According to dimension analysis the dimension of \(f, u\), and \(v\) are \([L]\).

From option C,

\(\frac{5}{f}=\frac{v^{2}-u^{2}}{2 u^{3}} \)

\(\Rightarrow {\left[L^{-1}\right]=\frac{\left[L^{2}\right]}{\left[L^{3}\right]}} \)

\(\therefore {\left[L^{-1}\right]=\left[L^{-1}\right]}\)

Given:

Two photons of same frequency are produced due to the annhiliation of a proton and antiproton.

We know that:

\(c=3 \times 10^{8}\)

Mass of a proton = \(1.67 \times 10^{-27}\)

\(E=m c^{2}\)

Put the values in above formula.

\(E=\left(2 \times 1.67 \times 10^{-27}\right) \times\left(3 \times 10^{8}\right)^{2} \mathrm{~J}\)

\(=3.006 \times 10^{-10} \mathrm{~J}\)

Also We know that:

\(2 h \nu=E\) or \(2 h \frac{c}{\lambda}=E\)

\(\therefore \lambda=\frac{2 h c}{E}\)

Put the values in above formula.

\(=\frac{2 \times 6.62 \times 10^{-34} \times 3 \times 10^{8}}{3.006 \times 10^{-10}} \mathrm{~m}\)

\(=1.323 \times 10^{-15} \mathrm{~m}\)