Physics Test - 24

A rectifier is a device that converts an alternating current into a direct current.In half-wave rectifier, there is only one diode, so during the positive half cycle diode conducts and gives the output similarly in the negative half cycle diode don’t conduct and gives no output.The output frequency (ω) for the half-wave rectifier is the same as that of ac.

When the P-N junction diode rectifies half of the ac wave, it is called a half-wave rectifier.The correct diagram of a half-wave rectifier is:

(i) During the positive half cycle

Diode → forward biased

Output signal → obtained

(ii) During a negative half cycle

Diode → reverse biased

Output signal → not obtained

Since the particle carrying positive charge is revolving around another charge,

Electrostatic force = Centrifugal force

\(\frac{1}{4 \pi \epsilon_{0}} \frac{{q}_{1} {q}_{2}}{{r}^{2}}={mr} \omega^{2}\)

\(\frac{1}{4 \pi \epsilon_{0}} \frac{{q}_{1} {q}_{2}}{{r}^{2}}=\frac{4 \pi^{2} {mr}}{{T}^{2}}\)

\({T}^{2}=\frac{\left(4 \pi \epsilon_{0}\right) {r}^{2}\left(4 \pi^{2} {mr}\right)}{{q}_{1} {q}_{2}}\)

\({T}=4 \pi {r} \sqrt{\frac{\pi \epsilon_{0} {mr}}{{q}_{1} {q}_{2}}}\)

Energy of photons \(=\mathrm{h\nu}\), Work function of surface \(=\mathrm{E_0}\), Maximum kinetic energy \(=\mathrm{K}\)

Now, frequency is doubled and the energy of photon is used in liberating the electron from metal surface and in imparting the kinetic energy to emitted photoelectron.

According to Einstein's photoelectric effect energy of photon = KE photoelectron + Work function of metal i.e.,

\(\mathrm{h\nu=\frac{1}{2} m v^{2}+E_{0}}\) 

\(\mathrm{h \nu=K_{\max }+E_{0}}\)

\(\mathrm{ K_{\max }= h \nu-E_{0}}  \cdots(i)\)

Now, we have given,

\(\nu'=2 \nu\)

Therefore, \(\mathrm{K_{\max }=h(2\nu)-E_{0}}\)

\(\mathrm{K_{\max }^{\prime}=2 h \nu-E_{0}}\)

From Equation (i) and (ii), we have

\(\mathrm{K'_{\max }=2\left(K_{\max }+E_{0}\right)-E_{0}}\)

\(=2 \mathrm{K_{\max }+E_{0}}\)

\(=\mathrm{K_{\max }+\left(K_{\max }+E_{0}\right)}\)

\(=\mathrm{K_{\max }+h \nu}\quad\) [From Equation (i)]

\(\mathrm{K_{\max }=K}\)

\(\mathrm{\therefore K'_{\max }=K+h\nu}\)

Given:

\(N_{a}=5 \times 10^{15} \mathrm{~cm}^{-3}\)

\(\mathrm{N}_{\mathrm{d}}=1 \times 10^{15} \mathrm{~cm}^{-3}\)

\(V_{T}=0.0259\)

Intrinsic carrier concentration \(\mathrm{n}_{\mathrm{i}}=1.5 \times 10^{10} \mathrm{~cm}^{-3}\).

The built-in voltage will be:

\(V_{b i}=V_{T} \ln \left(\frac{N_{a} N_{d}}{n_{i}^{2}}\right)\)

Now putting the given values in the above equation,

\(V_{b i}=0.0259 \ln \left(\frac{\left(5 \times 10^{15}\right)\left(10^{15}\right)}{\left(1.5 \times 10^{10}\right)^{2}}\right)\)

\(\Rightarrow V_{b i}=0.617 \mathrm{~V}\)

The given applied voltage is:

\(V_{a}=\frac{1}{2} V_{b i}\)

\(\Rightarrow V_{a}=\frac{1}{2} \times 0.617 V\)

\(V_{a}=0.308 \mathrm{~V}\)

Incident Energy is increased by \(20 \%\).

Initial kinetic energy \((KE_{1})=0.5 \mathrm{eV}\) 

Final kinetic energy \((K E_{2})=0.8 \mathrm{eV}\) 

\(E_{1}=E=\) initial incident energy

According to the question.

\(E_{2}=E_{1}+E_{1} \times 20 \%=E_{1}+0.2 E_{1}=1.2 \mathrm{E}_{1}=1.2 \mathrm{E}\)

Let \(\phi\) is the work function of metal. 

According to Einstein equation:

\( E=\phi+K E \)

\(E_{1} =\phi+K E_{1}=E \)

Put the value of \(K E_{1}\) in above equation,

\(E_{1} =\phi+0.5=E \cdots(i)\)

\(E_{2} =\phi+K E_{2}=1.2 E \)

Put the value of \(K E_{2}\) in above equation,

\( E_{2}=\phi+0.8=1.2 E \cdots(ii)\)

Divide equation (i) by equation (ii),

\(\frac{E_{1}}{E_{2}}= \frac{\phi+0.5}{\phi+0.8}=\frac{E}{1.2E}=\frac{1}{1.2}\)

\(\Rightarrow 1.2 \phi +0.6=\phi+0.8 \)

\(\Rightarrow 0.2 \phi =0.2 \Rightarrow \phi =\frac{0.2}{0.2}=1 \mathrm{eV}\)

Work function of metal is \(\phi=1 \mathrm{ev}\).

Given that:

Dipole Moment,\(P = 3 \times 10^{-3} {~cm}\)

Electric Field Intensity,\(E = 10^{4} {NC}^{-1}\)

We know that:

In rotating the dipole from the position of stable equilibrium by an angle θ, the amount of work done is given by,

\({W}={PE}(1-\cos \theta)\)

For unstable equilibrium, \(\theta=180^{\circ}\)

\(\therefore\) \(W=P E\left(1-\cos 180^{\circ}\right) \quad\left[\because \cos 180^{\circ}=-1\right]\)

\(=2 P E\)

\(=2 \times 3 \times 10^{-3} \times 10^{4} {~J}\)

\(=60 {~J}\)

Here,

Given that:

Main scale reading (MSR) = 0 mm

Circular scale reading (CSR) = 52 divisions

Pitch of the screw gauge, \(P=1 \mathrm{~mm}\)

Number of circular division, \(n=100\)

Thus,

Least count \(L C=\frac{P}{n}\)

\(=\frac{1}{100}\)

\(=0.01 \mathrm{~mm}\)

\(=0.001 \mathrm{~cm}\)

We know that:

Diameter of the wire = MSR + (CSR × LC)

\(=0+(52 \times 0.001)\) cm

\(=0.052 \mathrm{~cm}\)

The Einstein's equation for photoelectric effect is,

\(\mathrm{eV}_{0}=\frac{\mathrm{hc}}{\lambda}-\mathrm{W}\)

where, \(\mathrm{V}_{0}=\) stopping potential, \(\lambda=\) wavelength of incident light, \(\mathrm{W}=\) work function of metal.

\(E=4 \mathrm{NC}^{-1}, ~d=1m\)

\(\mathrm{V}_{0}=\frac{\mathrm{E} }{ \mathrm{d}}=\frac{4 }{ 1}=4\) volt

\(\lambda=200 \mathrm{~nm}=200 \times 10^{-9} \mathrm{~m}\)

Thus, \(\mathrm{W}=\frac{\mathrm{hc}}{\lambda}-\mathrm{eV}_{0}\)

\(=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{200 \times 10^{-9}}-\left(1.6 \times 10^{-19}\right) 4\)

\(=3.53 \times 10^{-19} \mathrm{~J}\) (as \(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\))

\(=\frac{3.53 \times 10^{-19} }{1.6 \times 10^{-19} }=2.2 \mathrm{eV}\)

The escape velocity on earth is given by:

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

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

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

For earth, \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\) and \(R=6.4 \times 10^{6} \mathrm{~m}\)

\( V_{e}=\sqrt{2 \times 9.8 \times 6.4 \times 10^{6}} \)

\(\Rightarrow \mathrm{V}_{\mathrm{e}}=11.2 \times 10^{3} \mathrm{~m} / \mathrm{s} \)

\(\Rightarrow \mathrm{V}_{\mathrm{s}}=11.2 \mathrm{~km} / \mathrm{s}\)

When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.

The net resistance/equivalent resistance(R) of resistances in parallel is given by:

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

Suppose two resistors \(R _{1}\) and \(R _{2}\) are connected in parallel \(\left( R _{1}> R _{2}\right)\)

As we know that the voltage across both resistors is the same in parallel combination. Therefore, the current through \(R_{1}\) is:

\(I_{1}=\frac{V}{R_{1}}\)

Current through \(R _{2}\):

\(I_{2}=\frac{V}{R_{2}}\)

Since, \(R _{1}> R _{2}\),

Therefore, \(l _{1}< I _{2}\)

The photoelectric cut-off voltage in a certain experiment is \(1.5 \mathrm{~V}\).

The kinetic energy of photoelectrons is given as

 K.E. \(=\mathrm{e}v_0\)

Where, the stopping potential is \(v_O\) and the charge on the electron is \(e\).

By substituting the values in above equation, we get 

K.E. \(=1.6 \times 10^{-19} \times 1.5=2.4 \times 10^{-19} \mathrm{~J}\)

According to keplars law: The square of time period of revolution of the planet is proportional to the cube of semi major axis of the ellipse traced by the planet.

\(T^{2} \propto\) \(a^{3}\)

This equation is modified to:

\(T^{2}=\frac{2 \pi a^{3}}{G M}\)

\(\mathrm{T}=\) Time period of the satellite

\(\mathrm{a}=\) semi-major axis

\(\mathrm{G}=\) gravitational constant

\(\mathrm{M}=\) mass of the earth (in this case)

Here, \(\frac{2 \pi}{G M}\) is the constant of proportionality.

So: \(T=\frac{\sqrt {2 \pi}}{\sqrt{GM}}a^{\frac{3}{2}}\).....(1)

Now, we know the semi-major axis of an elliptical path is the arithmetic mean of the minimum and the maximum distance covered by the planet.

Therefore, in this case:

\(r_{1}=\) Minimum distance covered

\(r_{2}=\) Maximum distance covered

So,

\(a=\frac{r_{1}+r_{2}}{2}\)

Put this value in (1).

\(T=\frac{\sqrt {2 \pi}}{\sqrt{GM}} \left( \frac{r_{1} +r_{2}}{2} \right)^{\frac{3}{2}}\)

Thus, this is the value of the time period of the satellite.

Given:

\(\phi_{E}=\left(10 t^{2}-20 t+5\right) V m\)

We know that:

The expression for displacement current is given by,

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

\( i_{d}=\epsilon_{0} \frac{d\left(10 t^{2}-20 t+5\right)}{d t}\)

\( i_{d}=\epsilon_{0}(20 t-20)\)

For t = 1 sec,

\( i_{d}=\epsilon_{0}(20 \times 1-20)\)

\( i_{d}=\epsilon_{0}(20-20)=0\)

\( i_{d}=0\)

An image is a copy of an object that is formed by reflected (or refracted) light. Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.

Diffuse reflection occurs when light reflects off a rough surface and forms a blurry image or no image at all.

​So, Light rays from an object fall on a surface and get reflected in a completely diffused manner; No image will be formed.