Physics Test - 15

Friction is not a conservative force.Work done by friction force is lost in the form of sound energy, heat energy, etc.In the question, it is given that the block stops after some time. Here kinetic energy is lost.We can say that the kinetic energy of the block is ‘lost’ due to the frictional force.This kinetic energy loss makes a slight increase in its temperatures.The work done by friction is not lost but is transferred as heat energy. Due to this way kinetic energy of the block is converted into heat energy.This is the reason ablock is moving with some speed on the horizontal rough ground, after some time it stopped.

Concept:

Thermal conductivity is the rate of flow of heat at a given temperature difference.

The flow of heat is directly proportional to the temperature difference \(\left(T_{C}-T_{0}\right)\) and area of the cross-section \((A)\).

The flow of heat is inversely proportional to length \((L)\).

\(H=\frac{K\left(T_{C}-T_{0}\right) A}{L} \quad ... (1)\)

Calculation:

Thermal resistance per unit length is uniform.

\(R _{ T }= \frac{L}{KA}\)

Substitute it in equation \( (1)\)

\(H=\frac{\left(T_{C}-T_{0}\right)}{R_{T}}\)

\(H R_{T}=\delta T\)

Differentiate with respect to \(x\)

\(H \frac{d R_{T}}{d x}=\frac{d T}{d x}\)

As \(\frac{d R_{T}}{d x}=\) uniform

In options (A), (B) and (C) slope is uniform.

We know that if the distance is increasing temperature will decrease.

For monoatomic gases:

\(C_{V}=\frac{3}{2} R, C_{P}=\frac{5}{2} R\)

For diatomic gases:

\(C_{V}=\frac{5}{2} R, C_{P}=\frac{7}{2} R\)

then, for the mixture of both gases

\({C}_{{V}}=\frac{\frac{3}{2} {RT}+\frac{5}{2}}{2} {RT}\)

\({C}_{{V}}=\frac{8}{4} {RT}\)

\({C}_{{V}}=2 {RT}\)

\(C_{P}=\frac{\frac{5}{2} R T+\frac{7}{2}}{2} R T\)

\(C_{P}=\frac{12}{4} R T\)

\(C_{P}=3 R T\)

We know that,

\({r}=\frac{{C}_{{P}}}{{C}_{{V}}}\)

\(=\frac{3 {RT}}{2 {RT}}=1.5\)

RMS value of applied voltage \(=200 \mathrm{V}\)

Impedance of a capacitor is given by:

\(\mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}\)

So, rms current through it is:

\(\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{C}}}\)

\(\mathrm{I}_{\mathrm{rms}}=200 \times 2 \times \pi \times 50 \times 40 \times 10^{-6}\)

\(\mathrm{I}_{\mathrm{rms}}=2.5 \mathrm{~A}\)

Given,

\(N_{p}=4 N_{o}\)

\(N_{Q}=N_{o}\)

Also,

\(T_{P}=1 \mathrm{~min}\)

\(T_{Q}=2 \min\)

Now,

Amount left after time

\(N_{P_{t}}=4 N_{O}\left(\frac{1}{2}\right)^{\frac{t}{1}}\)

And

\(N_{Q_{t}}=N_{o}\left(\frac{1}{2}\right)^{\frac{t}{2}}\)

Now,

According to question,

\(N_{P_{t}}=N_{Q_{t}}\)

Thus,

\(4 N_{o}\left(\frac{1}{2}\right)^{t}=N_{o}\left(\frac{1}{2}\right)^{\frac{t}{2}}\)

Then, we get

\(4=\left(\frac{1}{2}\right)^{\frac{-t}{2}}\)

\(\Rightarrow 4=2^{\frac{t}{2}}\)

Further

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

Thus,

\(t=4 \mathrm{~min}\)

Thus,

For \(R\)

\(N_{R}=\left(N_{o}-N_{P_{t}}\right)+\left(N_{o}-N_{Q_{t}}\right)\)

Then,

\(N_{R}=\left(N_{o}-\frac{N_{o}}{4}\right)+\left(N_{o}-\frac{N_{o}}{4}\right)\)

Then, we get

\(N_{R}=\frac{15 N_{o}}{4}+\frac{3 N_{o}}{4}\)

Then,

\(N_{R}=\frac{9 N_{o}}{2}\)

Radiation is the only medium in which the heat transfer takes place without any medium. Medium is required in conduction and convection.

The distance between earth and sun has no medium.Still, the earth is heated by the sun.This happens by the method of radiation.

Radiation:

• When electromagnetic radiation is emitted or absorbed to transfer the heat or When heat is transferred in the form of electromagnetic waves is known as radiation.
• Electromagnetic radiation includes microwaves, radio waves, infrared radiation, visible light, ultraviolet radiation, gamma rays, and X-rays.
• The method does not require any medium so heat from the sun reaches us by this method.

Escape velocity is also defined as the speed at which the magnitude kinetic energy of the object is equal to the gravitational potential energy of the body. The body which achieves this escape velocity in nowhere close to the orbit of the body. This velocity of the object keeps decreasing as it moves away from the body, it goes close to zero but doesn’t become zero.

It is given by \(v_{e}=\sqrt{\frac{2 G M}{r}}\) where \(G\) is the gravitational constant, \(M\) is the mass of the body and \(r\) is the distance of the object from the center of the body.

Thus escape velocity is a function of the mass of the body and the distance of the object from the center of the body. Here, since the question says, the escape velocity of a body depends upon its mass as, i.e., the mass of the body which is projected. But we know that the escape velocity is independent of the mass of the body.

We know that:

Electrostatic force \(=\frac{K Q_{1} Q_{2}}{\tau^{2}}\)

Where \(K=\) Coulomb's constant , \(Q_{1}\) and \(Q_{2}\) are the charges of the electrons and \(r=\) distance between the charges

Gravitational force \(=\frac{G m_{1} m_{2}}{\tau^{2}}\)

Where, \(G=\) Gravitational Constant, \(m_{1}\) and \(m_{2}\) are the masses of the object and \(r=\) distance between the centre of the object

On dividing both forces we get,

\(\frac{\text { Electrostatic Force }}{\text { Gravitational Force }}=\frac{\frac{K Q_{1} Q_{2}}{ r ^{2}}}{\frac{6 m_{1} m_{2}}{r^{2}}}\)

We know that,

Charge on an electron \( = −1.6\times 10^{-19}C\),

Mass of an electron ,\(= 9.1\times 10^{-31}\) kg,

So,

\(\frac{\text { Electrostatic Force }}{\text { Gravitational Force }}=\frac{9.0 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{6.67 \times 10^{-11} \times\left(9.1 \times 10^{-31}\right)^{2}}\)

Electrostatic Force \(=4.17 \times 10^{42}\) Gravitational Force \(\approx\) \(4 \times 10^{42}\) Gravitational Force

Therefore, electrostatic force between two electrons is greater than gravitational force by a factor of \(=4 \times 10^{42}\).

In 1820, Hans Christian Oersted discovered electric current creates a magnetic field. Prior to that, scientists thought that electricity and magnetism were unrelated. Oersted also used a compass to find the direction of the magnetic field around a wire-carrying current.

The process of retrieval of information from the carrier wave at the receiver is termed demodulation.

This is the reverse process of modulation.Demodulation is a key process in the reception of any amplitude modulated signals whether used for broadcast or two way radio communication systems.

Demodulation is the process by which the original information bearing signal, i.e. the modulation is extracted from the incoming overall received signal.

The process of demodulation for signals using amplitude modulation can be achieved in a number of different techniques, each of which has its own advantage.