Physics Test - 14

If wavelength is \(\lambda\) then the speed of the electron is \(v\). 

Let us assume that when wavelength is \(\frac{3 \lambda}{4}\) then the speed of electron is \(v_{1}\).

According to the equation of photoelectric emission:

\(\frac{\mathrm{hc}}{\lambda}=\frac{1}{2} \mathrm{mv}^{2}+\phi \quad \ldots . .\)(1)

 When the wavelength \(\frac{3 \lambda}{4}\). Then the speed of electron is \(v_{1}\).

\(\frac{4 \mathrm{hc}}{3 \lambda}=\frac{1}{2} \mathrm{mv}_{1}^{2}+\phi \quad \ldots . .\)(2)

On dividing (2) by (1)

\(\frac{4}{3}=\frac{\frac{1}{2} m v_{1}^{2}+\phi}{\frac{1}{2} m v^{2}+\phi}\)

\(\Rightarrow \frac{1}{2} m v_{1}^{2}=\frac{4}{3}\left(\frac{1}{2} m v^{2}\right)+\frac{4\phi}{3}-\phi\)

\(\Rightarrow \frac{1}{2} m v_{1}^{2}=\frac{4}{3}\left(\frac{1}{2} m v^{2}\right)+\frac{\phi}{3}\)

or \(v_{1}=\) greater than \(v\left(\frac{4}{3}\right)^{\frac{1 }{ 2}}\)

The longer length of the potentiometer wire is preferred for accurate measurements. The longer the wire length, the lesser is the value of fall of potential per unit length of wire resulting in a lesser value of least count (more accuracy).

Copper wire is not used in potentiometers, this is because the temperature coefficient of resistance of the copper wire is large and its resistivity is small.

In a potentiometer, Electric field intensity,\(E=\frac{V}{l}=\) constant i.e., the potential gradient is constant. Therefore, the electric field inside the potentiometer is variable.

If the length of the wire is halved, the potential gradient along the length of the potentiometer wire increases. As a result position of zero deflection will occur at a shorter length.

The ripple factor is the ratio between the RMS value of the AC voltage (on the input side) and the DC voltage (on the output side) of the rectifier.

\( r=\frac{\sqrt{I_{r m s}^{2}-I_{d c}^{2}}}{I_{d c}}\).....(1)

Where, \(r =\) ripple factor, \(I_{r m s} =\) root mean square current and \(I_{dc}=\) dirrect current

We Know that,

\(I_{r m s}=\frac{I_{o}}{\sqrt{2}}\) and \(I_{d c}=\frac{I_{o}}{\pi}\)

A rectifier is a device that converts an alternating current into a direct current. A p-n junction can be used as a rectifier because it permits current in one direction only.

A p-n junction can be used as a rectifier because it permits current in one direction only.

Put all the given values in equation (1),

\( I_{r m s}=\frac{\sqrt{\frac{I_{o}^{2}}{2}-\frac{I_{0}^{2}}{\pi^{2}}}}{\frac{I_{0}}{\pi}}\)

\(=1.21\)

The relative permittivity of the water is highest among the other materials given.

Permittivity describes the amount of charge needed to generate one unit of electric flux in a particular medium. Accordingly, a charge will yield more electric flux in a medium with low permittivity than in a medium with high permittivity. Thus, permittivity is the measure of a material's ability to resist an electric field.

Relative permittivity is the ratio of its absolute permittivity ‘\(\epsilon \)’ to free space (empty of matter) permittivity '\(\epsilon_0 \)’ i.e.,

\(\epsilon_{r}=\frac{\epsilon}{\epsilon_{0}}\)

From the given relation, \(D=\frac{n\left(x_{2}-x_{1}\right)}{n_{2}-n_{1}}\)

Here, \([n]=\left[\frac{1}{\text { area } \times \text { time }}\right]=\frac{1}{\left[L^{2} T\right]}-\left[L^{-2} T^{-1}\right]\)

\(x_{2}-x_{1}=[L]\) and \(n_{2}-n_{1}=\left[\frac{1}{\text { volume }}\right]=\left[\frac{1}{L^{3}}\right]\)

\(=\left[L^{-3}\right]\)

So, \([D]=\frac{\left[L^{-2} T^{-1} L\right]}{\left[L^{-3}\right]}=\left[L^{2} T^{-1}\right]\)

The time period of a geostationary satellite that moves in a circular orbit around the earth in the equatorial plane is 24 hours.

Geostationary Satellite:

• The satellite which appears stationary relative to the earth is called a geostationary or geosynchronous satellite.
• A geostationary satellite always stays over the same place above the earth such a satellite is never at rest.
• Such a satellite appears stationary due to its zero relative velocity w.r.t. that place on earth.
• The orbit of a geostationary satellite is known as the parking orbit.
• It takes 24 hours to complete one revolution of the earth by satellite, which is the same time as the earth takes to rotate once on its axis.
• A geostationary orbit has an orbital period equal to the Earth's rotational period i.e., 23 hours, 56 minutes, 4 seconds, which is rounded off to 24 hours.

The microscope has both objective and eyepiece convex lenses. So, statement 1 is correct.

The focal length of the objective lens of the telescope is larger than the focal length of their eyepiece lens. So, statement 2 is correct.

Magnification of a telescope is \({M}=\frac{{f}_{0}}{{f}_{{e}}}\) so magnification increases with an increase in focal length of the objective. So, statement 3 is correct.

\(M=\frac{L}{f_{0}}\left(\frac{D}{f_{e}}\right)\) therefore, magnification of microscope decreases with an increase in focal length of the objective. So, statement 4 is not correct.

In the above figure, \(6 \Omega\) and \(6 \Omega\) are connected in parallel.

The net resistance/equivalent resistance(R) of two \(6 \Omega\) resistances connected in parallel is given by:

\(\frac{1}{R^{\prime}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\)

\(=\frac{1}{6}+\frac{1}{6}\)

\(=\frac{2}{6}\)

\(=\frac{1}{3}\)

\(R ^{\prime}=3 \Omega\)

Now \(R ^{\prime}\) and \(3 \Omega\) are in series,

\(\therefore R_{\text {net }}=R^{\prime}+3 \Omega\)

\(\Rightarrow R _{\text {net }}=3 \Omega+3 \Omega=6 \Omega\)

Given:

\(V _{ e }=12 ~km / s =12 \times 10^{3} ~m / s\)

Radius \(( R )=6000~km =6 \times\) \(10^{6} ~m\)

The escape velocity on earth is given by:

\(V_{e}=\sqrt{\frac{2 G M}{R}}\)

As we know, \(G M=g R^{2}\)

\( V_{e}=\sqrt{2 g R}\)

By squaring both sides, we get,

\(\left(V_{e}\right)^{2}=2 g R\)

\(\Rightarrow g=\frac{V_{e}^{2}}{2 R}\)

Put all the given values in above formula:

\( g=\frac{\left(12 \times 10^{3}\right)^{2}}{2 \times 6 \times 10^{6}}\)

\(=12 m / s ^{2}\)

Electromagnetic waves are generated by accelerated charge.

• An accelerating charge produces a changing electric field which in turn produces a magnetic field. These alternatively changing magnetic and electric field give rise to electromagnetic waves.
• Whereas, the static force generate only electric field and moving particle generates only magnetic field.