Physics Test - 21

Consider an infinitesimal strip of capacitor of thickness \(\mathrm{dx}\) at a distance \(x\) as shown.

Capacitance of parallel plate capacitor of area \(\mathrm{A}\) is given by

\(\mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{t}}\)

[Here \(t=\) seperation between plates]

So, capacitance of thickness \(\mathrm{dx}\) will be

\(\therefore \mathrm{dC}=\frac{\varepsilon_0 \mathrm{ad} \mathrm{x}}{\mathrm{d}+\mathrm{x} \tan \alpha}\)

Total capacitance of system can be obtained by integrating with

limits \(x=0\) to \(x=a\)

\(\therefore \mathrm{C}_{\text {eq }}=\int \mathrm{d} C=a \varepsilon_0 \int_{\mathrm{x}=0}^{x=\mathrm{a}} \frac{\mathrm{dx}}{\mathrm{xtan} \alpha+\mathrm{d}}\) [By Binomial expansion]

\(\Rightarrow C_{e q}=\frac{\mathrm{a} \varepsilon_0}{\mathrm{~d}} \int_0^{\mathrm{a}}\left(1-\frac{\mathrm{x} \tan \alpha}{\mathrm{d}}\right) \mathrm{dx}=\frac{\mathrm{a} \varepsilon_0}{\mathrm{~d}}\left(\mathrm{x}-\frac{\mathrm{x}^2 \tan \alpha}{2 \mathrm{~d}}\right)_0^2\)

\(\Rightarrow C_{e q}=\frac{a^2 \varepsilon_0}{d}=\left(1-\frac{a \tan \alpha}{2 d}\right)=\frac{\varepsilon_0 a^2}{d}\left(1-\frac{\alpha a}{2 d}\right)\)

Given,

The Charge on the \(1^{\text {st }}\) sphere and \(2^{\text {nd }}\) sphere is \(q_1=2\times10^{-7} \mathrm{C}\) and \(\mathrm{q}_2=3 \times 10^{-7} \mathrm{C}\)

The distance between two charges is given by \(r=30 \mathrm{~cm}=0.3 \mathrm{~m}\)

The electrostatic force between the spheres is given by the relation:

\(\mathrm{F}=\frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_2}{r^2}\)

Here,

\(\epsilon_o=\) permittivity of free space and,

\(\frac{1}{4 \pi \epsilon_o}=9 \times 10^9\mathrm{~Nm}^2 \mathrm{C}^{-2}\)

Force, \(\mathrm{F}=\frac{9 \times 10^9 \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^2}=6 \times 10^{-3} \mathrm{~N}\).

The force between the charges will be repulsive as they have the same nature.

Free body diagram of the system,

As the acceleration along normal to surface in contact shall be same.

\(a \cos \alpha_{1}=a_{B} \cos \alpha_{2}\)

\(a_{B}=\frac{a \cos \alpha_{1}}{\cos \alpha_{2}}\)

Let the force F is applied at an angle θ with the horizontal.

For horizontal equilibrium,

\(\mathrm{F} \cos \theta=\mu \mathrm{R} \text {.............(i)}\)

For vertical equilibrium,

\(R+F \sin \theta=m g\)

or, \(R=m g-F \sin \theta\text {.............(ii)}\)

Substituting this value of \(R\) in eq. (i), we get

\(F \cos \theta=\mu(m g-F \sin \theta) \)

\(=\mu m g-\mu F \sin \theta\)

or, \(\mathrm{F}(\cos \theta+\mu \sin \theta)=\mu \mathrm{mg}\)

or, \(\mathrm{F}=\frac{\mu \mathrm{mg}}{\cos \theta+\mu \sin \theta}\)

For \(F\) to be minimum, the denominator \((\cos \theta+\mu \sin \theta)\) should be maximum.

\(\therefore \frac{\mathrm{d}}{\mathrm{d} \theta}(\cos \theta+\mu \sin \theta)=0\)

or, \(-\sin \theta+\mu \cos \theta=0\)

or, \(\tan \theta=\mu\)

or, \(\theta=\tan ^{-1}(\mu)\)

Then, \(\sin \theta=\frac{\mu}{\sqrt{1+\mu^2}}\) and

\(\cos \theta=\frac{1}{\sqrt{1+\mu^2}}\)

Hence, \(\mathrm{F}_{\min }=\frac{\mu \mathrm{w}}{\frac{1}{\sqrt{1+\mu^2}}+\frac{\mu^2}{\sqrt{1+\mu^2}}}=\frac{\mu \mathrm{w}}{\sqrt{1+\mu^2}}\)

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 \(933.38 J\).

As,

\(h=u t+\frac{1}{2} g t^{2}\)

Here, \(u=0, g=10 m s^{-2}, t=4 s\)

\(\therefore h=0 \times 4+\frac{1}{2} \times 10 \times 4^{2}=80\)m

Given,

Emf of the cell, \({E}_{1}=1.5 {V}\)

Balance point of the potentiometer, \(l_{1}=42 {~cm}\)

The cell is replaced by another cell of emf \({E}_{2}\).

New balance point of the potentiometer, \(l_{2}=63 {~cm}\)

The balance condition is given by the relation, 

\(\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}\)

\({E}_{2}={E}_{1} \times \frac{{l}_{2}}{l_{1}}=1.5 \times \frac{63}{42}=2.25 {~V}\)

Therefore, emf of the second cell is \(2.25 {~V}\).

When the widths of the two slits are increased, the fringes become brighter. However, the width of each slit should be considerably smaller than the separation between the slits. When the slits become so wide that this condition is not satisfied, the interference pattern disappears i.e., The fringes become less distinct.

Electric field produced by the infinite line charges at a distance \(d\) having linear charge density \(\lambda\) is given by the relation,

\(E=\frac{\lambda}{2 \pi \epsilon_0 d}\)

\(\lambda=2 \pi \epsilon_0 d E\)

Here,

\(d=2 \mathrm{~cm}=0.02 \mathrm{~m} \)

\(E=9 \times 10^4 \mathrm{~N} / \mathrm{C}\)

\(\varepsilon_0=\) Permittivity of free space and \(\frac{1}{4 \pi \epsilon_0}=9 \times10^9 \mathrm{~Nm}^2 \mathrm{C}^{-2}\)

So,

\(\lambda=\frac{0.02 \times 9 \times 10^4}{2 \times 9 \times 10^9}\)

\(\lambda=10 ~\mu \mathrm{C / m}\)

Magnetic permeability is maximum for the ferromagnetic substances.

Magnetic permeability of a material is its ability to acquire the magnetism when it is exposed to a magnetic field. It is maximum for the ferromagnetic substance.