Physics Test - 8

Given:

Change in an electric field, \(\frac{d E}{d t}=5 \times 10^{12} {Vm}^{-1} s^{-1}\),

Side of the plate (I) \(=2\) \({cm}=2 \times 10^{-2} {~m}\),

and \(\epsilon_{0}=8.85 \times 10^{-12} C^{2} N^{-1} {~m}^{-2}\)

The area of the plate is:

Area of square\(=(\text{side})^2\)

\( A=2 \times 10^{-2} \times 2 \times 10^{-2}=4 \times 10^{-4} {~m}\)

We know that displacement current is given as:

\( I_{d}=\epsilon_{0} A \times \frac{d E}{d t}\)

\( I_{d}=8.85 \times 10^{-12} \times 4 \times 10^{-4} \times 5 \times 10^{12}\)

\( I_{d}=177 \times 10^{-4} A\)

\(I_{d}=17.7 {~m}\)

Kepler's law tells us that the square of the Time period (T) of a satellite is proportional to the cube of its distance (R) from the orbiting planet.

∴T²∝R³

The time period of earth's orbit around the sun is 1 year, and assuming the mean distance between earth and sun as R,we can write the mean distance between mars and the sun as 1.5R.

Let, T be the time period of mars' orbit around the sun.

Thus from the above proportionality,

$\frac{T^2}{R^3}$is a constant.

$\frac{{(1year)}^2}{R^3}=\frac{T^2}{{(1.5R)}^3}$

T² = (1.5)³

$T={(1.5)}^{\frac{3}{2}}$

≈1.85 years

Displacement Current:

• The idea of displacement current was firstly developed by famous physicist James Maxwell.
• The displacement current produces due to the change in electric flux (number of electric field lines through a cross-sectional area of a closed-loop) with respect to time.
• The SI unit of displacement current is Ampere.
• The magnitude of displacement current is zero in the case of steady electric fields in conducting wire.
• The idea of displacement current was introduced to the current for making ampere circuital law consistent.

Therefore, out of all the given statements, Only the statement of option (C) is incorrect.

Let, 

Effective capacitance = C 

Then:

\(\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}\)

\(=\frac{1}{8}+\frac{1}{16}+\frac{1}{24}\)

\(=\frac{6+3+2}{48}\)

\(C=\frac{48}{11} \mu F\)

Potential difference, \(V=110 {~V}\)

We know that:

Charge, \(Q=C V\)

\(Q=\frac{48}{11} \times 10^{-6} \times 110\)

\(=480 \times 10^{-6}\)

\(=48 \times 10^{-5}\)

Since the capacitors are in series each capacitor has same charge i.e., \(48 \times 10^{-5} {C}\).

The concept of displacement current was proposed by Maxwell.

• We know that, it is the current that comes into existence, in addition to the conduction current, whenever the electric field and hence the electric flux changes with time.
• To modify Ampere’s law, Maxwell followed a symmetry consideration.
• By Faraday’s law, a changing magnetic field induces an electric field, hence a changing electric field must induce a magnetic field.
• As currents are the usual sources of the magnetic field, a changing electric field must be associated with the current.
• Maxwell called that current as displacement current.

To maintain the dimensional consistency, the displacement current is added in ampere’s law: 

\( \oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0} I+\mu_{0} \epsilon_{0}\left(\frac{d \phi_{E}}{d t}\right)\)

Where, \(\epsilon_{0}\left(\frac{d \phi_{E}}{d t}\right)\) is the displacement current.

From the above, it is clear that the concept of displacement current was proposed by Maxwell.

The net force acting on a mass that is traveling in a vertical circle is composed of the force of gravity and the tension in the string.

\(\overrightarrow{F_{c}}=\overrightarrow{m g}+\overrightarrow{T} \)

\(\frac{\overrightarrow{m v^{2}}}{R}=\overrightarrow{m g}+\overrightarrow{T}\)

At the point, when the string losses its Tension and go slang, there should be enough gravitational force to make the body continue its motion in vertical plane. For this, the critical velocity should be maintained.

\(\frac{m v^{2}}{R}=m g \Rightarrow v ^{2}= Rg \)

\(\Rightarrow v =\sqrt{Rg}\)

Where \(R=\) radius of the circle in which the stone is moving but here it is the length of the string

Given,

\(R=1 m\), \(g=9.8 ms ^{-2}\)

\(\therefore v=\sqrt{1 \times 9.8}=3.13\)

So, the critical Velocity is \(3.13 m/s\).

We know that:

Time period of a simple pendulum of length \(l\), is given by:

\(T=2 \pi \sqrt{\frac{l}{g}} \ldots(\mathrm{i})\)

Where, \(g\) is acceleration due to gravity.

When \(g^{\prime}=\frac{g}{4}\)

New, time period is

\(T^{\prime}=2 \pi \sqrt{\frac{l}{\frac{g}{ 4}}} \ldots(\mathrm{ii})\)

Dividing Eq. (ii) by Eq. (i), we get

\(\frac{T^{\prime}}{T}=\sqrt{\frac{g}{\frac{g}{ 4}}}=2\)

\(\Rightarrow T^{\prime}=2 T\)

Therefore, new time period becomes twice of the original value.

The dual nature of light exhibited bydiffraction and the photoelectric effect.

• Light: It is an electromagnetic wave.
• Light shows dual nature i.e. wave and particle.