Physics Test - 31

Given:

\(n_{1}\) = 3 and \(n_{2}\) = 4

As we know,

When an electron jumps from \(n_{2}\) orbit to the \(n_{1}\) orbit \((n_{2}>n_{1})\), the energy of the atom changes from \(En_{2}\) to \(En_{1}\). This extra energy \((En_{2} - En_{1})\) is emitted as a photon of electromagnetic radiation. This corresponding wavelength is given as,

\(\frac{1}{\lambda}=R Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)\).......(i)

Where, Rydberg constant \((\mathrm{R})=1.0973 \times 10^{7} \mathrm{~m}^{-1}\) , \(Z=2\), for \(\mathrm{He}^{+}\)ion

Put all the given values in (i),

\(\frac{1}{\lambda}=4 R\left(\frac{1}{9}-\frac{1}{16}\right)\)

\(=4 R \times \frac{7}{144}\)

\(=\frac{7}{36} R\)

\(\Rightarrow\lambda=\frac{36}{7 R}\)

\(=\frac{36}{7 \times 1.0973 \times 10^{7}}\)

\(=468 \mathrm{~nm}\)

The magnetic field inside (for \(r

\(\oint H . d l=I_{\text {enclosed }}\)

\(H=\frac{I r}{2 \pi a^{2}}\)

For \(r=a\)

\(H = \frac{I}{2\pi a}\)

Analysis:

Radius \(=R\)

Distance from axis \(=r\)

\(\because r

\(\oint H . d l=I_{\text {enclosed }}\)

\(H=\frac{I r}{2 \pi R^{2}}\)

Field emission involves the emission of electrons from a material's surface when subjected to a strong electric field. In this process, electrons are emitted from a material's surface when a strong electric field is applied. It's termed "quick" because it occurs rapidly under the influence of the electric field, without requiring significant heating or other conditions.

Given:

\(Q=1\), and \(r=\) radius of the sphere

By the Gauss law, if the total charge enclosed in a closed surface is \(Q\), then the total electric flux associated with it will be given as,

\(\Rightarrow \phi=\frac{Q}{\varepsilon_{o}} \quad \cdots(1)\)

By equation \(1\) the total flux linked with the sphere is given as,

\(\Rightarrow \phi=\frac{Q}{\varepsilon_{o}}\)

\(\Rightarrow \phi=\frac{1}{\varepsilon_{o}}\)

\(\Rightarrow \phi=\varepsilon_{o}^{-1}\)

When an inductive load is connected, the power factor is lagging.

The overall power factor is defined as the cosine of the angle between the phase voltage and phase current.In AC circuits, the power factor is also defined as the ratio of the real power flowing to the load to the apparent power in the circuit.

In a purely inductive circuit, the current lags the voltage by 90° and the power factor is zero lagging.

In a purely capacitive circuit, the current leads the voltage by 90° and the power factor is zero leading.

In a periscope, the two plane mirrors are keptparallel to each other.

A periscope is an instrument used for observing over an obstacle or object which is prevented by a direct line of sight.When light falls on one of the mirrors, it gets reflected back and the reflected rays fall on the other mirror which further gets reflected towards the observer’s eyes.In a periscope, the two plane mirrors are kept parallel to each other.The mirrors are at 45°to the horizontal plane.

If two unit masses are placed at unit distance apart the force of attraction between them is equal to universal gravitational constant.

Therefore , the force of attraction between two unit point massess separated by a unit distance is called universal gravitational constant.

Given:

\(m_{p}=1.6 \times 10^{-27} \mathrm{~kg}, g=10 \mathrm{~ms}^{-2}\)

\(\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-27} \mathrm{Kg}\)

\(\mathrm{h}=1 \mathrm{~mm}=10^{-3} \mathrm{~m}\)

\(V=5 \mathrm{~V}\)

Time of flight of electron \(t_{e}=\sqrt{\frac{2 h}{a}}\) (ignoring gravity)

We know that \(\mathrm{F}=\mathrm{ma}\)

Similarly \(\mathrm{F=e E}\)

\(\therefore \mathrm{a}=\frac{e E}{m}\left[\right.\) also \(\left.\mathrm{E}=\mathrm{V} / \mathrm{d}=\frac{5}{10^{-3}}=5000 \mathrm{Vm}^{-1}\right]\)

\(\therefore \mathrm{t}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{hm}_{e}}{e E}}=\sqrt{\frac{2 \times 10^{-3} \times 9.1 \times 10^{-31}}{1,6 \times 10^{-19} \times 5000}}\)

\(=\left[\frac{2 \times 9.1 \times 10^{-34}}{1.6 \times 5 \times 10^{-16}}\right]^{1 / 2}=\left[2.275 \times 10^{-18}\right]^{1 / 2}\)

\(t_{e} \simeq 1.5 \times 10^{-9} \mathrm{~S}\) or \(\simeq 1.5 \mathrm{~ns}\)

Time flight for proton is

\(t_{p}=\sqrt{\frac{2 h m_{p}}{e E}}\)

\(=\sqrt{\frac{2 \times 10^{-3} \times 1.6 \times 10^{-27}}{1.6 \times 10^{-19} \times 5000}}=\left[\frac{2 \times 10^{-30}}{5 \times 10^{-16}}\right]^{1 / 2}\)

\(=\left[0.4 \times 10^{-14}\right]^{1 / 2}\)

\(=\left[4000 \times 10^{-18}\right]^{1 / 2}\)

\(=63.25 \times 10^{-9}\)

\(t_{p}=63.25 \mathrm{~ns}\)

Given that,

Mass of Nitrogen, \(m=2.0 \times 10^{-2} kg =20 g\).

Rise in temperature, \(\Delta T=45^{\circ} C\).

Molecular mass of \(N_{2}, M=28\)

Universal gas constant, \(R=8.3 J \operatorname{mol}^{-1} K^{-1}\)

Number of moles, \(n=\frac{m}{M}\)

\(n=\frac{2 \times 10^{-2} \times 10^{3}}{28}\)

\( n=0.714\)

Now, molar specific heat at constant pressure for nitrogen,

\(C_{p}=\frac{7}{2} R \)

\(C_{p}=\frac{7}{2} \times 8.3 \)

\( C_{p}=29.05 J mol ^{-1} K ^{-1}\)

The total amount of heat to be supplied is given by the relation:

\(\Delta Q=n C_{p} \Delta T \)

\(\Delta Q=0.714 \times 29.05 \times 45 \)

\( \Delta Q=933.38 J\)

Clearly, the amount of heat to be supplied is \(933.38 J\).

Let 1 and 2 represent for Argon atom and Helium atom.

rms speed of Argon, \(\mathrm{v}_{\mathrm{rms}_{1}}=\sqrt{\frac{3 \mathrm{R} \mathrm{T}_{1}}{\mathrm{M}_{1}}}\)

rms speed of Helium, \(\mathbf{v}_{\mathrm{rms} _2}=\sqrt{\frac{3 \mathrm{R} \mathrm{T}_{2}}{\mathrm{M}_{2}}}\)

According to question,

\(\mathbf{v}_{\mathrm{rms}_{1}}=\mathbf{v}_{\mathrm{rms}_{2}}\)

\(\therefore \sqrt{\frac{3 \mathrm{R} \mathrm{T}_{1}}{\mathrm{M}_{1}}}=\sqrt{\frac{3 \mathrm{R} \mathrm{T}_{2}}{\mathrm{M}_{2}}}\)

or \( \frac{\mathrm{T}_{1}}{\mathrm{M}_{1}}=\frac{\mathrm{T}_{2}}{\mathrm{M}_{2}}\)

or \(\mathrm{T}_{1}=\frac{\mathrm{T}_{2}}{\mathrm{M}_{2}} \times \mathrm{M}_{1}=\frac{253}{4} \times 39.9=2.52 \times 10^{3} \mathrm{~K}\)