Physics Test - 9

When a body is in linear motion, its energy due to motion will be kinetic energy.

\(K =\frac{1 }{ 2} \times m \times v ^{2}\)

where \(K\) is the kinetic energy of the body, and \(m\) is the mass of the body, and \(v\) is its velocity.

Work-Energy Theorem: The net work done on an object is equal to a change in its Kinetic energy.

\( W =\Delta K \)

\( W = K _{ f }- K _{ i }\)

Where \(W\) is work done, \(K_{f}\) is final kinetic energy and \(K_{i}\) is initial kinetic energy.

Given,

Initial velocity \(= v\)

Final velocity \(=2 v\)

Let the mass of the truck is \(m\).

Initial kinetic energy \(K _{ i }=\frac{1 }{ 2} \times m \times v ^{2}\)

Final kinetic energy \(K _{ f }=\frac{1 }{ 2} \times m \times(2 v)^{2}=2 \times m \times v^{2}\)

Net work done \(W=\Delta K=K_{f}-K_{i}\)

\(\therefore W=2 \times m \times v^{2}-\frac{1 }{ 2} \times m \times v^{2}\)

\(\Rightarrow W=\frac{3 }{ 2} \times m \times v^{2} \)

\(\Rightarrow W=3 K_{i}\)

So, Net work done is three times as the initial kinetic energy of the mass or work done in accelerating it from rest to \(v\).

Given:

\(1^{\text {st }}\) orbital radius \(\left(R_{1}\right)=R\), and \(2^{\text {nd }}\) orbital radius \(\left(R_{2}\right)=4 R\)

According to the law of periods: The square of the period of revolution (T) of any planet around the sun is directly proportional to the cube of the semimajor axis of the orbit i.e., \(T^{2} \propto R^{3}\)

\( \mathrm{T}^{2} \propto \mathrm{R}^{3} \)

\(\Rightarrow\left(\frac{T_{1}}{T_{2}}\right)^{2}=\left(\frac{R_{1}}{R_{2}}\right)^{3} \)

\(\Rightarrow\left(\frac{T_{1}}{T_{2}}\right)^{2}=\left(\frac{R}{4 R}\right)^{3}\)

\(=\frac{1}{64} \)

\(\Rightarrow \frac{T_{1}}{T_{2}}=\left(\frac{1}{64}\right)^{\frac{1}{2}}\)

\(=\frac{1}{8}\)

 "Light speeds down as it leaves a water surface and enters the air"  is not correct for light rays.

• The bending of the ray of light passing from one medium to the other medium is called refraction.
• The refraction of light takes place on going from one medium to another because the speed of light is different in the two media.
• The greater the difference in the speeds of light in the two media, the greater will be the amount of refraction.
• A medium in which the speed of light is more is known as an optically rarer medium and a medium in which the speed of light is less is known as an optically denser medium.
• Light travels at almost 300 million metres per second in air.
• Light speeds up as it leaves a glass surface and enters the air.

We know that the formula for the stress is:

stress \(=\frac{\text { Force }}{\text { Area }}\)

\(=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^{2}\right]}\)

\(=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\)

Meter Bridge:It is an electrical instruments based on the principle ofWheatstone bridgeand is used tomeasurethe resistance of a resistor.

The metre bridge, also known as the slide wire bridge consists of a one metre long wire of uniform cross sectional area, fixed on a wooden block. A scale is attached to the block. Two gaps are formed on it by using thick metal strips in order to make the Wheat stones bridge.

The formula meter bridge is given below:

ρ = Lπr2X

Where, L be the length of the wire and r be its radius.

Given: \(N=30\) turns, \(x=20\) cm \(=0.2\) m, and \(I=10\) A

Where, \(N=\) number of turns in the coil, \(I=\) current in the loop, \(x=\) side of the square loop, and \(A =\) area enclosed by the loop

The area enclosed by the square loop is given as,

\(A = x ^{2}\)

\(\Rightarrow A =0.2^{2}\)

\(\Rightarrow A =0.04\) m \(^{2}\)

We know that the magnetic moment '\(m\)' of the current-carrying loop is given as,

\(m = NIA\)

\(\Rightarrow m =30 \times 10 \times 0.04\)

\(\Rightarrow m =12\) A-m\(^{2}\)

In the region (I)

• Fermi level is above the mid-gap so material is n-type.

In the region (II)

• The Fermi level is very near to conduction band.
• Thus, this region is more doped than 1.
• So n⁺ region.

In region (III)

• The fermi level is inside the valence band. So this is degenerate so p⁺⁺.

Kepler's Second Law: The line that joins any planet to the sun sweeps equal areas in equal intervals of time.

The planet \(\mathrm{P}\) moves around the sun in an elliptical orbit. The shaded area is the area \(\Delta A\) swept out in a small interval of time \(\Delta t\). i. e.,

\(\frac{\mathrm{dA}}{\mathrm{dt}}=\) constant

From above it is clear that Kepler's second law (law of areas) is nothing but a statement of conservation of angular momentum.

We know that:

Permittivity of free space is given by:

\(\epsilon_{0}=\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{4 \pi \mathrm{Fr}^{2}}\)

Dimensions of \([\mathrm{F}]=\left[\mathrm{MLT}^{-2}\right]\)

Dimensions of \([\mathrm{q}]=[\mathrm{AT}]\)

Dimensions of \([\mathrm{r}]=[\mathrm{L}]\)

Thus, dimensional formula of \(\left[\epsilon_{0}\right]=\frac{[\mathrm{AT}][\mathrm{AT}]}{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^{2}\right]}\)

\(\therefore \left[\epsilon_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]\)

Biot-Savart’s law is an equation that gives the magnetic field produced due to a current-carrying segment.Biot-Savart’s law is used to determine the magnetic field at any point due to a current-carrying conductor.This law is although for infinitesimally small conductors yet it can be used for long conductors.

According to this law, the magnetic field produced at a distance r from the current-carrying element is given as,

\(\overrightarrow{d B}=\frac{\mu_{o}}{4 \pi} \frac{\overrightarrow{d l} \times \vec{r}}{r^{3}}\)

Where, \(\overrightarrow{d B}=\) magnetic field, \(r =\) distance of a point from the element, \(dl =\) length of the element, and \(\theta=\) angle between \(d l\) and \(r\)

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 \(T_{1}\).)

Metal A has higher threshold frequency,

We know that for a given photosensitlve material,

The stopping potential varies linearly with the frequency of the incident radiation.

Also, Work function, \(\phi_{0}=\mathrm{hv}_{0}\)

Also, \(\phi_{0}=\mathrm{e} \times\) magnitude of intercept on potential axis.

Here for metal A, \(y\) - intercept is more negative.

In an n-p-n transistor, there are two junctions n-p junction and p-n junction.So the left side of the base will be an n-p junction and the right side of the base will be a p-n junction.Therefore, n-p-n transistor = opposite p-n diode and p-n diode.

Given:

Capacitance, (C) \(=1 \mu {F}\),

Rate of change in voltage, \(\frac{d V}{d t}=10^{6} {~V} / {s}\)

The expression for displacement current \({i}_{{d},}\) in the case of capacitance is given as:

\( i_{d}=\frac{d q}{d t}\)\(\quad\).....(i)

As we also know that, the charge on the capacitor is:

\( {q}={CV}\)

Where q = charge on the capacitors

On substituting the value of q = CV in equation (i), we get,

\( i_{d}=C \frac{d V}{d t}\)\(\quad\).....(ii)

On substituting the given values in equation (ii), we get,

\( i_{d}=\left(10^{-6} \times 10^{6}\right) A\)

\( {i}_{{d}}=1 {~A}\)