Physics Test-7

Given:

\(\lambda_{1}=600 \mathring{A}\) and \(\lambda_{2}=400 \mathring{A}\)

The Photon energy formula is given by,

\(\Rightarrow E=\frac{h c}{\lambda} \quad........ (1)\)

Where \(E=\) energy, \(c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\),

\( \lambda=\) wavelength and \(\mathrm{h}=\) planck's constant \(=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\)

By equation 1 ,

\(\Rightarrow E_{1}=\frac{h c}{\lambda_{1}}=\frac{h c}{600}\quad........ (2) \)

\(\Rightarrow E_{2}=\frac{h c}{\lambda_{2}}=\frac{h c}{400}\quad........ (3)\)

By equation (2) and equation (3) ,

\(\Rightarrow \frac{E_{1}}{E_{2}}=\frac{2}{3}\)

We can find out this work done by calculating the potential energy of each case. Potential energy is the energy possessed by its state of position. The energy stored in an object as the result of its height is known as the gravitational potential energy. The work done can be written as the change in gravitational potential energy.

\(\Rightarrow U_{1}=m g h_{1}........(i)\)

The Gravitational potential energy of the block at position 2 is given as

\(\Rightarrow U_{2}=m g h_{2}............ (ii)\)

Thus, work done by the external agent,

\(\Rightarrow W=\Delta U=\left(U_{2}-U_{1}\right)=m g h_{2}-m g h_{1}\)

Where \(h_{2}=\frac{b}{2}\) and \(h_{1}=\frac{a}{2}\) as it is a regular-shaped object.

\(\Rightarrow W=m g\left(\frac{b}{2}-\frac{a}{2}\right)=\frac{m g(b-a)}{2}\)

Initially, the hoop is only rotating which means it has angular momentum due to rotation only. Now it is kept on the horizontal rough surface. The word rough denotes that there is friction between the ground and the hoop and that gives some translational velocity. So after some time, there will be both translational and angular velocity.

In the initial case, the angular momentum will be due to the rotational velocity whereas later the angular momentum will be due to both rotational and translational velocity.

The diagram below depicts the initial and final condition of the hoop.

With respect to point of contact we can conserve the angular momentum.

Initial angular momentum is,

\(L_i=I \omega\)

\(\Rightarrow L_{i}=m r^{2} \omega_{0}\)

Where ' \(r^{\prime}\) is the radius and \(\omega_{0}\) is the initial angular velocity and ' \(m\) ' is the mass of hoop.

Final angular momentum is,

\(L_f=I \omega+m v r\)

\(\Rightarrow L_f=m r^{2} \frac{v}{r}+m v r\quad\) [\(\because v=r \omega\)]

\(\Rightarrow L_{f}=2 m v r\)

Where ' \(v\) ' is the translational velocity attained.

Equate the initial and final angular momentum. Then we will get,

\(L_{i}=L_{f}\)

\(\Rightarrow m r^{2} \omega_{0}=2 m v r\)

\(\Rightarrow \frac{r \omega_{0}}{2}=v\)

\(\Rightarrow v=\frac{r \omega_{0}}{2}\)

We know that

Tension at top point \(\mathrm{T}+\mathrm{mg}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\)

\(\therefore \mathrm{T}=\frac{1 \times 4^{2}}{1}-1 \times 10=6 \mathrm{N}\)

When the stone is at the bottom position forces acting on stone are:

Upward force tension \((T)\), Downward forces are weight \((mg)\) and also \(\frac{m v^{2}}{r}\) centrifugal force.

We get:

\(T= mg+\frac {mv^2}{r}\)

\(\begin{aligned} T &=10+16 \\ &=26 \mathrm{~N} \end{aligned}\)

And minimum tension is obtained at highest point which will be:

\(T= mg-\frac {mv^2}{r}\)

\( T =16-10 \)

\(=6 \mathrm{~N} \)

Therefore, minimum tension is obtained at the top of the circle.

\(k_{1}=\frac{Y{\pi(2 R)}^{2}}{L}, k_{2}=\frac{Y{\pi}(R)^{2}}{L}\)

Equivalent \(\frac{1}{k_{1}}+\frac{1}{k_{2}}=\frac{L}{4 Y{\pi R^{2}}}+\frac{L}{Y{\pi R^{2}}}\)

Since, \(k_{1} x_{1}=k_{2} x_{2}=w\)

The elastic potential energy of the system

\(U=\frac{1}{2} k_{1} x_{1}^{2}+\frac{1}{2} k_{2} x_{2}^{2}\)

\( U =\frac{1}{2} k_{1}\left(\frac{w}{k_{1}}\right)^{2}+\frac{1}{2} k_{2}\left(\frac{w}{k_{2}}\right)^{2}\)

\(=\frac{1}{2} w^{2}\left\{\frac{1}{k_{1}}+\frac{1}{k_{2}}\right\}\)

\(=\frac{1}{2} w^{2}\frac{5 L}{4 Y \pi R^{2}}\)

\(U=\frac{5 w^{2} L}{8 \pi Y R^{2}}\)

We are given that a particle of mass m is located in a one dimensional potential field and the potential energy is given by:

\(\mathrm{V}(\mathrm{x})=\mathrm{A}(1-\cos \mathrm{px})\)

So, we can find the force experienced by the particle as,

\(\mathrm{F=-\frac{d V}{d x}=-A p \sin p x}\)

For small oscillations, we have,

\(\mathrm{F} \approx-\mathrm{Ap}^{2} \mathrm{x}\)

The acceleration would be given by,

\(a=\frac{\mathrm{F}}{\mathrm{m}}=-\frac{\mathrm{Ap}^{2}}{\mathrm{~m}} \mathrm{x}\)....(i)

Also we know that,

\(a=\frac{\mathrm{F}}{\mathrm{m}}=-\omega^{2} \mathrm{x}\)....(ii)

So from (i) and (ii),

\(\omega=\sqrt{\frac{\mathrm{Ap}^{2}}{\mathrm{~m}}}\)

We know that,

Period, \(\mathrm{T}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{Ap}^{2}}}\)

Given :

\(T_{1}=627^{\circ} \mathrm{C}=900 \mathrm{~K}, T_{2}=27^{\circ} \mathrm{C}=300 \mathrm{~K}\)

The efficiency is given by

\(\Rightarrow \eta=1-\frac{T_{2}}{T_{1}} \)

\(\Rightarrow \eta=1-\frac{T_{2}}{T_{1}}=1-\frac{300}{900}=1-\frac{1}{3}\)

\(\Rightarrow \eta=\frac{2}{3}\)

Given that,

Bit rate \(=\mathrm{M} k b p s\)

Number of levels \(=\mathrm{N}=16\)

\(\therefore(B W)_{\min }=\frac{2 R_{b}}{\log _{2} N} H z=\frac{2 M}{\log _{2} 16 k H z}\)

\(=\frac{2 M}{\log _{2} 2^{4} k H z} \)

\(=\frac{2 M}{4 \times \log _{2} 2k H z} \)

\((B W)_{\min }=\frac{M}{2 k H z}\)

So. The minimum bandwidth will be \(\frac{\mathrm{M} }{ 2 \mathrm{kHz}}\), for ideal transmission.

Given:

\(g\) at the surface of the earth is \(10 \mathrm{~m} / \mathrm{s}^{2}\)

At height \(h=R_{e}\)

Acceleration due to gravity at height \(\left(g^{\prime}\right)\)

\(=\frac{g}{\left(1+\frac{h}{R_{e}}\right)^{2}}=\frac{g}{\left(1+\frac{R_{e}}{R_{e}}\right)^{2}}\)

\(g^{\prime}=\frac{g }{(1+1)^{2}}=\frac{g }{ 4}\)

\(=\frac{10 }{ 4}=2.5 \mathrm{~m} / \mathrm{s}^{2}\)

When a metal rod of length l is placed normal to a uniform magnetic field 'B' and moved with a velocity 'v' perpendicular to the field, the induced emf (called motional emf -electromotive forces) across its ends is \(\mathrm{Blv}\).

CONCEPT:

• Whenever a rod is moving in a magnetic field that is perpendicular to the plane of motion then an emf is induced in the rod.
• Faraday’s law of magnetic induction: When magnetic flux is changing in a closed-loop then an emf is induced in the coil. This emf is called induced emf.

When a rod is moving in a magnetic field then an emf is generated across its ends. The emf is given by:

Emf generated = BVl

Where B is the magnetic field perpendicular to the velocity of the rod, V is the velocity of the rod, and l is the length of the rod.

For lens in air,

\(\frac{1}{f}=(1.5-1)\left[\frac{1}{R}-\frac{1}{(-R)}\right]\)

\(\frac{1}{f}=\frac{1}{2} \times \frac{2}{R}=\frac{1}{R}\)

\(R=f\)

For calculating the focal length of the lens kept on the surface of the pond, place on object at \(\approx\) its image should be at effective focal length of system:

For surface 1, by refraction formula,

\(\frac{\mu_{9}}{\mathrm{~V}_{1}}-\frac{\mu_{\mathrm{a}}}{\infty}=\frac{\mu_{\mathrm{a}}-\mu_{\mathrm{a}}}{\mathrm{R}}\)

\(\frac{1.5}{\mathrm{~V}_{1}}=\frac{1.5-1}{\mathrm{R}} \)

\(\frac{3}{2 \mathrm{v}_{1}}=\frac{1}{2 \mathrm{R}}\)

\( \Rightarrow \mathrm{V}_{1}=3 \mathrm{R}\)

Image formed by surface 1 acts as object for surface 2 ,

For surface 2.

\(v_{1}=u=3 R \)

\(\frac{\mu_{w}}{v_{2}}-\frac{\mu_{g}}{u}=\frac{\mu_{w}-\mu_{g}}{-R} \)

\(\frac{4}{3 \times v_{2}}-\frac{3}{2 \times 3 R}=\frac{\left(\frac{4}{3}-\frac{3}{2}\right)}{-R} \)

\(\frac{4}{3 v_{2}}-\frac{1}{2 R}=\frac{1}{6 R} \)

\(\Rightarrow \frac{4}{3 v_{2}}=\frac{2}{3 R} \)

\(v_{2}=2 R \)

So, focal length of lens at water surface is \(2 R\).

For effective lens;

\(\mathrm{U}_{e}=2 \mathrm{f} ; \mathrm{F}=\mathrm{V}_{2}=2 \mathrm{R}=2 \mathrm{f}\)

\(\frac{1}{\mathrm{v}_{\mathrm{e}}}-\frac{1}{\mathrm{u}_{e}}=\frac{1}{\mathrm{~F}}\)

\( \Rightarrow \frac{1}{\mathrm{~V}_{\mathrm{e}}}-\frac{1}{(-2 \mathrm{f})}=\frac{1}{2 f}\)

\(\Rightarrow \frac{1}{\mathrm{~V}_{\mathrm{e}}}+\frac{1}{2 f}=\frac{1}{2 f}\)

\(\frac{1}{\mathrm{~V}_{\mathrm{e}}}=0\)

\(\Rightarrow \mathrm{V}_{e} \rightarrow \infty\)

The distance between the lens and the image is greater than \(2 f\).

Modulation index,

\(m=\frac{A_{m}}{A_{c}} \)

\(m=\frac{0.5 V}{10 V}\)

\(=0.05\)

The side bands frequencies are \(v_{S B}\)

\(=v_{c} \pm v_{m}\)

\(=1 \pm 0.010 \mathrm{MHz}\)

The Zener breakdown occurs in the junctions which are heavily doped and the avalanche breakdown occurs in the junctions, which are lightly doped. This statements is true with respect to semiconductor breakdown.

The avalanche breakdown is a phenomenon in which there is an increase in the number of free electrons beyond the rated capacity of the diode; This results in the flow of heavy current through the diode in reverse biased condition

Zener breakdown mainly occurs because of a high electric field; When the high electric field is applied across the PN junction diode, then the electrons start flowing across the PN-junction; Consequently, develops little current in the reverse bias The Zener breakdown occurs in heavily doped diodes.