Physics Test - 6

Electromotive force is not actually a force. It is basically a voltage. It is the voltage developed by any source of electrical energy. Electromotive force is defined as the electric potential produced by either electrochemical cell or by changing the magnetic field. EMF is the commonly used acronym for electromotive force.

\(E_{A}=\frac{k_{q A}}{x^{2}}\) Rightward

And, \(E_{B}=\frac{k_{q B}}{(1-x)^{2}}\) Leftward

Since, there is no field at \(P\).

\(E=E_{A}-E_{B}=0\)

\(\Rightarrow k q\left(\frac{q_{A}}{x^{2}}-\frac{q_{B}}{(1-x)^{2}}\right)=0\)

\(\Rightarrow\left(\frac{1-x}{x}\right)^{2}=\frac{q_{B}}{q_{A}}\)

\(\Rightarrow \frac{1-x}{x}=\sqrt{\frac{40}{10}}=2\)

\(\Rightarrow 1-x=2 x\)

\(\Rightarrow x=\frac{1}{3}=\frac{90}{3}\) cm

\(\Rightarrow x=30 \) cm

When light is incident on a metal surface the maximum kinetic energy of emitted electrons vary with frequency of light.

Photoelectric effect:

\({K} {E}_{\max }={h} {v}-{W}_{{o}}\)

Where, \(W_{o}\) is the work function of the metal.

Clearly \({K} {E}\) varies linearly with the the frequency of the photons.

We know that the Magnetic field is independent of the distance for the current sheet. So, Magnetic field at point \(P\),

\(B_{1}=\frac{1}{2} \mu_{0} K\)

\(=3 \mu_{0}\)

The direction will be \(-j\). This can be found by applying the right-hand rule to the few current elements in the sheet Now Magnetic field due to current wire is given by \(B_{2}=\frac{\mu_{0} i}{2 \pi d}\)

\(B_{2}=\frac{\mu_{0}i}{2 \pi 2.5}\)

So \(B_{1}=B_{2}\)

or \(i=47.1 \mathrm{~A}\)

The direction of the current is the wire should be towards \(i\) to cancel the magnetic field of the current sheet.

Let \(R\) be the resistance (in ohm) of transmission cables.

Power of electric station,

\({P}=100 {MW}=100 \times 10^{6} {W}\)

\(=10^{8} {W}\)

(i) \({V}_{1}=20000 {V}\)

Current \({I}_{1}=\frac{{P}}{{V}_{1}}=\frac{10^{8}}{20000}=5000 {A}\)

Rate of heat dissipation at \(20000 {V}\) is:

\(P_{1}=I^{2}, R=(5000)^{2} R\)

\(=25 \times 10^{6} {Rwa}\)

(ii) \({V}_{2}=200 {V}\)

Current \({I}_{2}=\frac{{P}}{{V}^{2}}=\frac{10^{8}}{200}=5 \times 10^{5} {A}\)

Rate of heat dissipation at \(200 {V}\) is:

\(P_{2}=I_{2}^{2} R=\left(5 \times 10^{5}\right)^{2} \times R\)

\(=25 \times 10^{10} {Rwa}\)

\(\frac{{P}_{2}}{{P}_{1}}=\frac{25 \times 10^{10}}{25 \times 10^{6}}\)

\(=10^{4}\)

The weight of an astronaut in an artificial satellite revolving around the earth is zero.

The apparent weight of an astronaut in artificial satellite of earth is zero because weight of the astronaut is balanced by the centrifugal force, acted on the astronaut when the satellite revolve around the earth.

Given,

The mass of the particle is \({m}\) 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, \({u}=0\) (particle is initially at rest)

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

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

Given:

\(\mathrm{I}=\mathrm{M}\left(\frac{\mathrm{R}^{2}}{4}+\frac{\mathrm{L}^{2}}{12}\right)\)   ......(1)

As mass is constant 

\(\Rightarrow \mathrm{M}=\rho \mathrm{V}=\) constant

\(\Rightarrow\mathrm{V}=\) constant

\(\Rightarrow\pi^{2} \mathrm{R} l=\) constant 

\(\Rightarrow \mathrm{R}^{2} \mathrm{~L}=\) constant

On differentating \(R^{2}L\),

\(2 \mathrm{RL}+\mathrm{R}^{2} \frac{\mathrm{dL}}{\mathrm{dR}}=0\)

\(\frac{dL}{dR}=\frac{-2L}{R}\).....(2)

Differentating equation (1)

\(\frac{\mathrm{dI}}{\mathrm{dR}}=\mathrm{M}\left(\frac{2 \mathrm{R}}{4}+\frac{2 \mathrm{~L}}{12} \times \frac{\mathrm{dL}}{\mathrm{dr}}\right)=0\)

\(\frac{\mathrm{R}}{2}+\frac{\mathrm{L}}{6} \frac{\mathrm{dL}}{\mathrm{dR}}=0\)

Substituting value of \(\frac{\mathrm{dL}}{\mathrm{dR}}\) from eqution (2)

\(\frac{\mathrm{R}}{2}+\frac{\mathrm{L}}{6}\left(\frac{-2 \mathrm{~L}}{\mathrm{R}}\right)=0\)

\(\frac{\mathrm{R}}{2}=\frac{\mathrm{L}^{2}}{3 \mathrm{R}}\) 

\(\Rightarrow \frac{\mathrm{L}}{\mathrm{R}}=\sqrt{\frac{3}{2}}\)

Given, 

m = 10 Kg

g = 10 m/s²

First draw the free body diagram of the given figure,

Now resolve all forces into a horizontal and a vertical component.

​Horizontal Component \(= 100 \cos45^\circ\)

\(=100\times \frac{1}{\sqrt{2}}\)

\(= 50\sqrt{2} N\)

Vertical Component \(= 100\sin45^\circ\)

\(=100\times \frac{1}{\sqrt{2}}\)

\(= 50\sqrt{2} N\)

For the system to be in equilibrium \(T_{1}\) must be equal to the horizontal component but the direction must be opposite.

Here the direction of \(T_{1}\) is reversed for balancing the horizontal component.

\( T _{2}=50 \sqrt{2 } N\)

\( T _{1}=-50 \sqrt{2} N\)

So, the value of \(T_{1}\) is \(-50 \sqrt{2} N\).

Negative sign only show the direction of \(T_{1}\).

Repeater is a combination of transmitter, amplifier, and receiver.

The receiver receives the signal, which is then amplified using the amplifier. This amplified signal is then transmitted using a transmitter.

A number of repeaters are used in the transmission medium to increase the range of communication. A signal during transmission gets attenuated so repeaters are necessary for effective communication.

A communication satellite is essentially a repeater station in space. A repeater is a combination of a receiver and a transmitter. A repeater picks up the signal from the transmitter, then amplifies and retransmits it to the receiver sometimes with a change in carrier frequency.