Physics Test-4

Following values are given in the question:

Number of silicon atoms, \(\mathrm{N}=5 \times 10^{28}\) atoms \(/ \mathrm{m}^{3}\)

Number of arsenic atoms, \(n_{\text {AS }}=5 \times 10^{22}\) atoms \(/ \mathrm{m}^{3}\)

Number of indium atoms, \(n_{\ln }=5 \times 10^{22}\) atoms \(/ \mathrm{m}^{3}\)

\(n_{i}=1.5 \times 10^{16} \text { electrons } / \mathrm{m}^{3}\)

\(n_{e}=5 \times 10^{22}-1.5 \times 10^{16}=4.99 \times 10^{22}\)

Let us consider the number of holes to be \(n_{h}\)

In the thermal equilibrium, \(n_{e} n_{h}=n_{i}^{2}\)

Calculating, we get

\(n_{\mathrm{h}}=\frac{(1.5\times10^{16})^{2}}{4.99\times10^{22}}\)

\(n_{\mathrm{h}}=4.51 \times 10^{9}\)

Here, \(n_{e}>n_{h}\), therefore the material is a n-type semiconductor.

Average velocity is the displacement of an object, divided by the time it took to cover that distance.

\(V _{\text {average }}=\frac{\Delta x }{\Delta t }\)

Displacement is the straight line distance between the starting point and ending point of an object's motion.

Velocity is referred to as a vector quantity because it has both magnitude and direction.

Here, length of plane tips, \(l=2 \times 20=40 m ; v=250 m s ^{-1}\);

Earth's horizontal component, \(H =2 \times 10^{-5}~ Wb m ^{-2}\) and \(\theta=60^{\circ}\)

Vertical component of earth's magnetic field, \(B_v= H \tan \theta\)

\(=2 \times 10^{-5} \times \tan 60^{\circ}\)

\(=2 \times 10^{-5} \times \sqrt3\)

\(=2 \sqrt{3}~ Wb m^{-2}\)

Since plane is flying parallel to the earth's surface, so it cuts the vertical component of earth's magnetic field.

So, induced e.m.f. is given by \(\varepsilon=B_v v l\)

\(=2 \sqrt{3} \times 10^{-5} \times 250 \times 40\)

\(=34641 \times 10^{-5}\)

\(=0.34641\) volt

\(=0.346\) volt

The amplitudes \(E_0\) and \(B_0\) of electric and the magnetic component of an electromagnetic wave respectively are related to the velocity \(c\) in vacuum as \(E_0=c B_0\).

Light is an electromagnetic wave which is transverse in nature where the phase of the magnetic field and the electric field will be the same for this EM wave.

Light is a form of energy which is stored in the small packets like structures called photons which travel with the same speed of light.

We have,

\(\frac{E_0}{B_0}=c\)

\(E_0=c B_0\)

Let,\(n\) be the number of bulbs, \(n=10\)

\(P\) be the power of each bulb, \(P=50 \mathrm{~W}\)

Total time to burn the bulb, \(\mathrm{t}=10 \times 30=300 \mathrm{hr}\)

Now the energy E, consumed in operating the bulbs is:

Energy consume by each built in \(30\) days \(\mathrm{E}=\mathrm{Pt}=50 \times 300\)

Energy consume by \(10\) bulbs in \(30\) days \(\mathrm{E}=\mathrm{nPt}=10 \times 50 \times 300 \mathrm{Wh}\)

\(\mathrm{E}=150 \mathrm{kWh}\)

Ratio of pressure drop for internal flow through ducts is known asFriction factor.

Friction factor (µ) is defined as the ratio between the force required to move a section of pipe and the vertical contact force applied by the pipe on the seabed in simpler terms it is the ratio of pressure drop for internal flow through ducts.

Human eyecannot detect polarization of light.

Polarization changes when plane of vibration of polarized light changes.Human eye is insensitive to change in polarization and hence, cannot detect polarization of light.

The net force acting on a mass that is traveling in a vertical circle is composed of the force of gravity and the tension in the string.

\(\overrightarrow{F_{c}}=\overrightarrow{m g}+\overrightarrow{T} \)

\(\frac{\overrightarrow{m v^{2}}}{R}=\overrightarrow{m g}+\overrightarrow{T}\)

At the point, when the string losses its Tension and go slang, there should be enough gravitational force to make the body continue its motion in vertical plane. For this, the critical velocity should be maintained.

\(\frac{m v^{2}}{R}=m g \Rightarrow v ^{2}= Rg \)

\(\Rightarrow v =\sqrt{Rg}\)

Where \(R=\) radius of the circle in which the stone is moving but here it is the length of the string

Given,

\(R=1 m\), \(g=9.8 ms ^{-2}\)

\(\therefore v=\sqrt{1 \times 9.8}=3.13\)

So, the critical Velocity is \(3.13 m/s\).

All monosaccharides whether aldoses or ketoses are reducing sugars. Disaccharides such as sucrose in which the two monosaccharide units are linked through their reducing centres i.e., aldehydic or ketonic groups are non-reducing. This is why sucrose is non-reducing sugar. The linkage between the glucose and fructose units in sucrose, which involves aldehyde and ketone groups, is responsible for the inability of sucrose to act as a reducing sugar.

For infinite long line charge,

\(E=\frac{\lambda}{2 \pi \epsilon_{0} r}=\frac{2 \lambda}{4 \pi \epsilon_{0} r}\)

\(\lambda=\) linear charge density

Distance, \(r=2 {~cm} \)

We have,

\({E}=9 \times {10}^{4} {~N} / {C}\)

Therefore,

\(9 \times 10^{4}=\frac{2 \times \lambda \times 9 \times 10^{9}}{2 \times 10^{-2}} \quad\) \((\because \frac {1}{4 \pi \epsilon_0} =9 \times 10^{9} \text { N m}^2 / \text{C}^2 ) \)

\(\lambda=\frac{10^{4}}{10^{11}}=10^{-7} {~cm}\)