Physics Test - 6

Given,

\(V=220 \mathrm{~V}\)

\(\mathrm{f}=50 \mathrm{~Hz} \)

\( \omega=2 \pi \mathrm{f}=2 \pi \times 50=100 \pi\; \mathrm{rad} / \mathrm{s}\)

\(\mathrm{R}=20 \Omega\)

\(\mathrm{L}=0.4 \mathrm{H}\)

\( \Rightarrow \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=100 \pi \times 0.4=40 \;\Omega\)

In the given circuit, capacitor is absent.

\(\therefore \tan \theta=\frac{X_{L}-X_{C}}{R}=\frac{X_{L}}{R}=\frac{40 \pi}{20}=2 \pi\)

Therefore, \(\theta=\tan ^{-1}(2 \pi)\)

Here, \(H=1500 \mathrm{A} \mathrm{m}^{-1}, \phi=2.4 \times 10^{-5}\) weber

\(A=0.5 \mathrm{~cm}^{2}=0.5 \times 10^{-4} \mathrm{~m}^{2}\)

\(\therefore B=\frac{\phi}{A}=\frac{2.4 \times 10^{-5}}{0.5 \times 10^{-4}}=4.8 \times 10^{-1} \mathrm{~T}\)

and \(\mu=\frac{B}{H}=\frac{4.8 \times 10^{-1}}{1500}=3.2 \times 10^{-4}\)

So relative permeability,

\(\mu_{r}=\frac{\mu}{\mu_{0}}=\frac{3.2 \times 10^{-4}}{4 \pi \times 10^{-7}}=0.255 \times 10^{3}=255\)

The unit of light year, wavelength and displacement is meter.

The unit of light year is meter. A light year is the distance traveled by light in one year. and wavelength is the distance between two consecutive vertices or descents. The unit of wavelength is also the meter. The minimum distance covered by an object in a certain direction with respect to a reference point is called displacement. The unit of displacement is also the meter.

When we rub a glass rod with silk, then the charge on the glass rod will be positive. When we rub a glass rod with silk, some of the electrons from the rod are transferred to the silk cloth. Thus the rod gets positively charged and the silk gets negatively charged.

The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

Work done, \((W)=\Delta K=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}\)

Where \(m\) is the mass of the object, \(v\) is the final velocity of the object and \(u\) is the initial velocity of the object.

Work-energy theorem for a variable force:

Kinetic energy, \(K=\frac{1}{2} m v^{2}\)

\( \frac{d K}{d t}=\frac{d\left(\frac{1 }{ 2} m v^{2}\right)}{d t} \)

\(\Rightarrow \frac{d K}{d t}=m \frac{d v}{d t} v\)

\(\Rightarrow \frac{d K}{d t}=m a v\)

\(\Rightarrow \frac{d K}{d t}=F v \)

\(\Rightarrow \frac{d K}{d t}=F \frac{d x}{d t} \)

\(\Rightarrow d K=F d x\)

On integrating, we get

\( \int_{K_{i}}^{K_{f}} d K=\int_{x_{i}}^{x_{f}} F d x \)

\(\Rightarrow \Delta K=\int_{x_{i}}^{x_{f}} F d x\)

From the work-energy theorem, a change in kinetic energy equals the work done.

\( \Delta K=\int_{x_{i}}^{x_{f}} F d x \)

\(\Rightarrow \Delta K=\int_{-a}^{a}\left(P y^{2}+Q y+R\right) d y \)

\(\Rightarrow \Delta K=\left[\frac{P y^{3}}{3}+\frac{Q y^{2}}{2}+R y\right]_{-a}^{a} \)

\(\Rightarrow \Delta K=\left[\left(\frac{P a^{3}}{3}+\frac{Q a^{2}}{2}+R a\right)-\left(\frac{-P a^{3}}{3}+\frac{Q a^{2}}{2}-R a\right)\right] \)

\(\Rightarrow \Delta K=\frac{2 P a^{3}}{3}+2 R a\)

Given,

 \(n=800\)

\(\mathrm{A}=2.5 \times 10^{-4} \mathrm{~m}^{2}\)

\(\mathrm{I}=3.0 \mathrm{~A}\)

A magnetic field develops along the axis of the solenoid. Therefore current-carrying solenoid acts like a bar magnet.

Associated magnetic moment,

\(\mathrm{m}=\mathrm{nIA}\)

\(=800 \times 3 \times 2.5 \times 10^{-4} \)

\(=0.6 \mathrm{JT}^{-1}\)

Maxwell Boltzmann distribution for velocity is given by:

\(n(E) d E=\frac{2 \pi N}{V(\pi k T)^{\frac{3}{2}}} E^{\frac{1}{2}} e^{-\frac{E}{k T}} dE\)

Where, \(n=\) number of molecules, \(T=\) temperature, \(k=\) Boltzmann constant, and \(E=\) energy

From the above equation, it is clear that the fraction of gas molecules having an energy between \(E\) and \(E + d E\) is proportional to \(E^{\frac{1}{2}} \exp \left(-\frac{E}{k T}\right)\).

Given,

Mass \(( m )=5 kg\)

Radius \(( R )=1 m\)

Velocity \(( v )=300 rpm =\frac{300 }{ 60}=5 rpm\)

As we know,

The angular speed is given by,

\(\omega=2 \pi v\)

\(\therefore \omega =2 \pi \times 5=10 \pi rads ^{-1} \)

\(v=\omega R\)

\(\therefore v =10 \pi \times 1=10 \pi ms ^{-1}\)

Kinetic energy, \((KE)=\frac{1}{2} m v^{2}=\frac{1}{2} m \omega^{2} R^{2}\)

\(=\frac{1}{2} \times 5 \times(10 \pi)^{2}\)

\(=250 \pi^{2}\)

The loudness and pitch of a sound depends on intensity and frequency.

Loudness is a sensation of how strong a sound wave is at a place. It is always a relative term. It is a dimensionless quantity. Its unit is decibel (dB).

Pitch is the characteristic of sound by which an acute (or shrill) note can be distinguished from a grave or a flat note. The term 'pitch' is often used in music. It depends on the frequency of the sound wave. A note of higher frequency is at a higher pitch than a note of lower frequency. It is a qualitative term and cannot be quantified.

The following figure shows the cyclic process of gas. If an object returns to its initial position after one or more processes it went through.

\(A B C D A\) is the cycle. The pressure at the points both \(D\) and \(C\) remain the same, which is \(2 P\). The volume at the point \(D\) is \(V\) and at the point \(C\) is \(3 V\). So, the work done by the gas from point \(D\) to point \(C, W_{D C}=2 P(3 V-V)=4 P V\)

The pressures at the points \(C\) and \(B\) are \(2 P\) and \(P\) respectively. The volume at the points both \(C\) and \(B\) remain the same, which is \(3 V\). So, the work done by the gas from point \(C\) to point \(B, \)

\(W_{C B}=P(3 V-3 V)=0\)

The pressure at the points both \(B\) and \(A\) remain the same, which is \(P\). The volume at the point \(B\) is \(3 V\) and at the point \(A\) is \(V\),

So, the work done by the gas from point \(B\) to point \(A, W_{B A}=P(V-3 V)=-2 P V\)

The pressures at the points \(A\) and \(D\) are \(P\) and \(2 P\) respectively. The volume at the points both \(A\) and \(D\) remain the same, which is \(V\).

So, the work done by the gas from point \(A\) to the point \(D, W_{A D}=P(V-V)=0\)

Hence the total work done in the whole cycle,

\(W=4 P V-2 P V=2 P V\)

We know the heat rejected from the cycle is equal to the amount of total work done by the gas, so \(Q=W\)

\( Q=2 P V\)