Physics Test - 9

When one joule of energy is consumed in one second, then the power used is said to be One watt.

The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.

\[P=\frac{W}{t}=V I=I^{2} R=\frac{V^{2}}{R}\]

Where, \(V=\) Potential difference, \(R=\) Resistance and \(I=\) current.

If,\(W=1\) Joule and \(t=1\) second, then \(P=1\) watt

Thus,The power of an electric circuit is said to be one watt orone ampere volt if one joule of energy is consumed in one second.

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.

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.

It is given that,

The mass of the particle is \(\mathrm{m}\) having charge \({q}\) is placed at rest in uniform electric field\(E\) and then released.

We have to find the kinetic energy attained by the particle after moving a distance \(y\).

Using the third equation of motion as:

\(v^{2}-u^{2}=2 a s\)

Here, \(\mathrm{u}=0\) (particle is initially at rest)

\(\mathrm{s}=\mathrm{y}\) (displacement)

\({v^{2}=2 a y}\) .............\((i)\)

Force on a particle in electric field is,

\(F=q E\)

Since, \(F=m a\)

So, \(a=\frac{q E}{m}\) .............\((ii)\)

Put the value of \(a\) in equation \((i)\)

\(v^{2}=2 \frac{q E y}{m}\) ...........\((iii)\)

We know that kinetic energy of a particle is given by,

\(E_{k}=\frac{1}{2} m v^{2}\)

Putting equation \((iii)\) in above equation

\(E_{k}=q E y\)

In the case of a steady electric field, the electric field does not change with time. Therefore, the electric flux density through a closed-loop will be constant.

This results in zero change in electric flux with respect to time i.e.,

\( \frac{d \phi_{E}}{d t}=0\)

The displacement current will be:

\( i_{d}=\epsilon_{0} \frac{d \phi_{E}}{d t}=0\)

Let the work function for the given metal be \(\phi\). When the light of the frequency \(n\) falls on the surface of the metal, the maximum kinetic energy of the ejected photoelectron is given by,

\(K . E_{\max }=n h-\phi\)

\(\frac{1}{2} m v^{2}=n h-\phi \quad \ldots(i)\)

When the light of the frequency 3n falls on the surface of the metal, the maximum kinetic energy of the ejected photoelectron is given by,

\(K . E_{\max }=3 n h-\phi\)

Let the maximum velocity of the electron be \(v\). 

\(\frac{1}{2} m v^{2}=3 n h-\phi \quad \ldots(i i)\)

Substituting \(n h=\frac{1}{2} m v^{2}+\phi\) from equation (i) to equation (ii) 

\(\frac{1}{2} m v^{2}=3\left(\frac{1}{2} m v^{2}+\phi\right)-\phi\)

\(\frac{1}{2} m v^{2}=\frac{3}{2} m v^{2}\)

\(\frac{1}{2} m v^{2}-\frac{3}{2} m v^{2}=0\)

The work function cannot be negative or zero.

Thus, \(\frac{1}{2} m v^{2}-\frac{3}{2} m v^{2}>0\)

\(\frac{1}{2} m v^2>\frac{3}{2} m v^2\)

\(\Rightarrow v >v \sqrt{3}\)

Given:

\(y_{1}=a \cdot \cos (\omega t)\) and \(y_{2}=2 a \cdot \cos (\omega t)\)

If the two waves undergo constructive interference, then the resultant displacement is given as,

\(y=y_{1}+y_{2}\)

\(\Rightarrow y=a \cos (\omega t)+2 a \cos (\omega t)\)

\(\Rightarrow y=3 a \cdot \cos (\omega t) \quad \ldots\) (1)

By equation (1), the amplitude 'A' of the resultant wave is given as,

A \(=3{a}\)

When an electron and a proton enter a magnetic field with the same velocity, both experience the same force.

As charges and velocities are same

\(\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})\)

will be the same, but in the opposite direction.