Physics Test - 30

We know that,

\(E=\frac{d \phi}{d t}=\frac{B d A}{d t}\)

When it is just about to move out:

\(d A=2 R(V d t)\)

\(\frac{d A}{d t}=2 R V\)

So, \(E=2BRV\)

Given,

Electric field, \(E = 10^{4}\mathrm N/\mathrm C\)

Torque,

\(T=9 \times 10^{-26} \mathrm N\mathrm m \)

\(\theta=30^{\circ}\)

When electric dipole is placed at an angle \(\theta\) with the direction of the electric field, torque acting on the dipole is given by,

\(T=p E \sin \theta\)

\(\therefore\) The electric moment of the dipole,

\(p =\frac{T}{E \sin \theta} \)

\(=\frac{9 \times 10^{-26}}{10^{4} \times \sin 30^{\circ}} \)

\(=\frac{9 \times 10^{-26}}{10^{4} \times \frac{1}{2}} \)

\(=\frac{9 \times 10^{-26}}{10^{4} \times 0.5} \)

\(=1.8 \times 10^{-19}\mathrm C \mathrm m \)

Alnico has a higher retentivity of magnetism. The value of the intensity of magnetization of a material when the magnetizing field is reduced to zero is called retentivity. The materials that are suitable for electromagnets should have high retention.

Given:

Velocity of sound waves in air (V) = \(330 \mathrm{~m} / \mathrm{s}\)

Path difference = \(40 \mathrm{~cm}\)

Phase difference = \(1.6 \pi\)

Velocity of sound waves \(=330 \mathrm{~m} / \mathrm{s}\)

\(\Delta \phi=\mathrm{K} \Delta \mathrm{x} \).....(1)

\(\Delta \phi=\) phase difference

\(\mathrm{K}=\frac{2 \pi}{\lambda}\)

\(\Delta \mathrm{x}=\) Path difference

Putting the values given in question in (1)

\(1.6 \pi=\frac{2 \pi}{\lambda}\left(\frac{40}{100}\right) \)

\(\Rightarrow \lambda=\frac{2 \pi}{1.6 \pi} \times \frac{40 \times 10}{100}\)

\(=\frac{1}{2}\)

Now,

\(\mathrm{V}=\mathrm{f} \lambda\)

\(330=\mathrm{f}\left(\frac{1}{2}\right) \)

\(\mathrm{f}=660 \mathrm{~Hz}\)

Let \(E_{1}\) be the energy required to raise it to a height \(2 R\) from the centre of the earth while \(E_{2}\) be the energy to put it into the orbit.

Then

\(E_{1}=\frac{-G M m}{2 R}+\frac{G M m}{R}=\frac{G M m}{2 R} \)

\(E_{2}=\frac{-G M m}{2(2 R)}+\frac{G M m}{R}=\frac{3 G M m}{4 R} \)

\(\frac{E_{1}}{E_{2}}=\frac{2}{3}\)

Given,

Mass of the body, \(m=20 \mathrm{~g}= 0.02 kg\)

Acceleration, \(a=5.0 \mathrm{~cm} / \mathrm{s}^{2} = 0.05 m/s^2\)

As we know,

\(F=ma\)

\(= 0.02 × 0.05\)

\(=1 \times 10^{-3} \mathrm{~N}\)

Let \(v\) be the velocity of sound.

Closed organ pipe \({P}_{1}\) of length \({L}_{1}\):

Frequency of different modes of vibration \(v_{n}^{\prime}=\frac{(2 n-1) v}{4 L_{1}}\)

First harmonic i.e., \({n}={1}, {v}_{1}^{\prime}=\frac{{v}}{4 {L}_{1}}\)

Open organ pipe \(P_{2}\) of length \(L_{2}\):

Frequency of \(m^{th}\) harmonic \( v_{{m}}=\frac{{mv}}{4 L_{2}}\)

For third harmonic i.e., \({m}={3} {v}_{3}=\frac{3 {v}}{{2} {L}_{2}}\)

But \( v_{1}^{\prime}=v_{3}\)

\(\frac{{v}}{4 L_{1}}=\frac{3 {v}}{2 L_{2}}\)

\( \Rightarrow \frac{{L}_{1}}{{~L}_{2}}=\frac{1}{6}\)

Given,

Measurement ofthe diameter ofhydrogen atom,

\(d=1.06 \times 10^{-10}\)

Least count, \(\Delta d=0.01 \times 10^{-10}\)

Accuracy \(=\frac{\text { Least count }}{\text { Orginal measurement }}\)

\(=\frac{\Delta d}{d}\)

\(=\frac{0.01 \times 10^{-10}}{1.06 \times 10^{-10}}\)

\(=\frac{1}{106}\)

If we consider this, the output frequency is certainly twice that of the input frequency.

So, if the input frequency is \(50\) Hz, the output frequency will be \(100\) Hz.

It is visible from the signal waveform shown in the image using color.

What happens is, the negative side of the signal appears at the positive side after the rectification. Since the signal to be rectified is usually symmetric in nature, the frequency of the signal at output is doubled.

Also, rectification is a process of converting alternating current into the unidirectional current. It is a part of the DC power generation unit. In a direct current, we ideally require a zero frequency signal. Thus, circuits like filter circuit and voltage regulation circuit further reduce this variable voltage to a fixed voltage by stabilizing it and removing all frequency components.