Physics Test - 23

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:

Electrostatic force \(=\frac{K Q_{1} Q_{2}}{r^{2}}\)\(\quad\)......(1)

Gravitational force \(=\frac{G m_{1} m_{2}}{r^{2}}\)\(\quad\)......(2)

On dividing the two we get,

\(\frac{\text{Electrostatic Force}}{\text{Gravitational Force}}=\frac{\frac{{KQ}_{1} \mathrm{Q}_{2}}{\mathrm{r}^{2}}}{\frac{{Gm}_{1} \mathrm{~m}_{2}}{ \mathrm{r}^{2}}}\)

\(\frac{\text{Electrostatic Force}}{\text{Gravitational Force}}=\frac{9.0 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{6.67 \times 10^{-11} \times\left(9.1 \times 10^{-31}\right)^{2}}\)

Electrostatic Force \(=4 \times 10^{42}\) Gravitational Force

Therefore, electrostatic force between two electrons is greater than gravitational force by a factor of \(10^{42}\).

The phase difference between the electric field and the magnetic field in the electromagnetic wave is zero.

• The electric field and the magnetic field components of an electromagnetic wave oscillate in such a way that they have their peak at the same time and also become zero at the same time.
• Since there is no time difference between the peaks of the electric and the magnetic field, so the phase difference between the electric and the magnetic field of the electromagnetic wave is zero.

The electric field due to any charge will be continuous, if there is no other charge in the medium. It will be discontinuous if there is a charge at the point under consideration.

Since the frequency of any light wave depends on the source frequency. Frequency doesn't depend on the medium.So whenever the light goes from one medium to another or vice versa, the frequency of light and phase of light do not change.

However, the velocity of light, the wavelength of light, and intensity depend on the medium, and hence the change.So, when light waves are incident on an air-glass boundary, then the speed, direction, and brightness of the light wave changes.

Given:

\(E=10^{4}\) Newton per Coulomb

\(q=1.6 \times 10^{-19}\) C

We know that:

\(F=q E\)

Where, \(F\) is the force due to the electric field, \(q\) is the charge, and \(E\) is the electric field.

\(F=\left(1.6 \times 10^{-19}\right) \times 10^{4}\)

\(\Rightarrow F=1.6 \times 10^{-15}\) Newton

The direction of the electric field at every point must be radial (outward if \(\lambda>0\), inward if \(\lambda<0\)).

We know that the electric field intensity due to an infinitely long straight uniformly charged wire is given as:

\(E=\frac{\lambda}{2 \pi \epsilon_{o} r}\)

\(\Rightarrow E \propto \lambda \quad \ldots\)(1)

Where, \(\lambda=\) linear charge density, \(\epsilon_{0}=\) permittivity, and \(r=\) distance of the point from the wire

When the velocity of a charged particle is perpendicular to a magnetic field, it describes a circle and the radius of the circle is given by:

\(r=\frac{m v}{q B} \quad \ldots\) (1)

Where, \(m=\) mass, \(q=\) magnitude of charge, \(v=\) speed of charge, \(B=\) magnetic field

We know that \(q , m\), and \(B\) remain unchanged during the motion of a charged particle.

\(r \propto v \quad \ldots\) (2)

\(v _{1}> v _{2}\)

\(r _{1}> r _{2}\)

Magnetic field due to Straight conductor at distance r is:

\(\mathrm{B}=0.4\)

\(\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}=0.4 \text { Tesla } \quad \ldots(1)\)

Now, μ₀ and i is constant for now, as the only distance is changing to 2r.

So, we can say that:

\(\mathrm{B} \propto \frac{1}{\mathrm{r}}\)

or

\(\mathrm{B}=\mathrm{k} \frac{1}{\mathrm{r}} \quad \ldots(2)\)

\(\mathrm{k}\) is constant.

Now, if the radius is increased to \(2 \mathrm{r}\), the new field will be \(\mathrm{B}^{\prime}\)

\(\mathrm{B}^{\prime}=\mathrm{k} \frac{1}{2 \mathrm{r}} \quad \ldots(3)\)

Using (2) and (3)

\(\mathrm{B}^{\prime}=\frac{\mathrm{B}}{2} \quad \cdots(4)\)

Using (1) and (4)

\(\mathrm{B}^{\prime}=\frac{0.4}{2} \text { Tesla }\)

\( \mathrm{B}^{\prime}=0.2 \text { Tesla }\)

So, the new magnetic field strength will be 0.2 Tesla.

When the switch \({S}\) is just closed, the capacitor will start charging through resistors \({R}_{1}, {R}_{2}\) and no current will pass through resistor \({R}_{3}\).
The current through \({R}_{1}\) is \({i}_{1}={i}_{10} {e}^{-t / {C} R_{1}}\) and current through \({R}_{2}\) is \({i}_{2}={i}_{20}{e}^{-t / {CR}_{2}}\)
As \({R}_{1}={R}_{2}=2, {i}_{10}={i}_{20}=10 / 2=5\) and \({C R}_{1}={C R}_{2}=2\)
Thus the ratio \(\left({i}_{1} / {i}_{2}\right)\) is always constant.After switch closed long time the charging of capacitor will stop and the circuit will treat as shown in figure below.

\({R}_{{eq}}=\frac{2 \times 2}{2+2}=1 \Omega\)
Current through \({R}_{3}\) is \({I}=\frac{{1 0}}{{1}+{1}}={5} {A}\)
The capacitor charged maximum when potential across \({C}\) is \({V}_{{C}}\) \(={V}_{{R}_{3}}={I R}_{3}={5 V}\)
Thus, max charge on \({C}\) is \({Q}={C} {V}_{{C}}=5 \mu {C}\)

Option (D) is the correct match for the given diodes.

1: Junction diode, 2: Zener diode, 3: Photo diode, 4: LED diode

PN Junction Diode:

A unilateral device is a device that conducts only in one direction. p-n junction diodes conduct only when the p region is connected to higher voltage and the n region is connected to lower voltage. When reverse biased, it acts as an open circuit. The symbol for a diode is as shown:

Zener diodes are normal PN junction diodes operating in a reverse-biased condition. Working of the Zener diode is similar to a PN junction diode in the forward biased condition, but the uniqueness lies in the fact that it can also conduct when it is connected in reverse bias above its threshold/breakdown voltage. Zener diodes also called avalanche diodes or Breakdown diodes  are the heavily doped P – N junction diodes. It is operated in a breakdown region. The symbolic representation of the Zener diode is as shown:

Photo-diode:

It is a light-sensing device that is used to sense the intensity of light. Some of the examples are the smoke detector, a receiver in tv for converting remote signals, etc. The symbolic representation of the Photodiode is as shown:

A light-emitting diode (LED):

The device which is used to produce the different intensity of light and different color depending upon the types of mater used in making it is called LED. The symbolic representation of the LED is as shown: