Physics Test - 7

Here,Magnetic field, \(B=0.025 \mathrm{~T}\),Radius of the loop,r \(=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}\)
Constant rate at which radius of the loop shrinks,\(\frac{d r}{d t}=1 \times 10^{-3} m s^{-1}\)
Magnetic flux linked with the loop is\(\phi=B A \cos \theta=B\left(\pi r^{2}\right) \cos 0^{\circ}=B \pi r^{2}\)
The magnitude of the induced emf is\(|\varepsilon|=\frac{d \phi}{d t}=\frac{d}{d t}\left(B \pi r^{2}\right)=B \pi 2 r \frac{d r}{d t} \)
\(=0.025 \times \pi \times 2 \times 2 \times 10^{-2} \times 1 \times 10^{-3} \)\(=\pi \times 10^{-6} V=\pi \mu V\)

Since the particle carrying positive charge is revolving around another charge,

Electrostatic force \(=\) Centripetal force

\(\Rightarrow \frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}=m r \omega^{2} \)

\(\Rightarrow \frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{\tau^{2}}=\frac{4 \pi^{2} m r}{T^{2}} \)

\(\Rightarrow T^{2}=\frac{\left(4 \pi \epsilon_{0}\right) r^{2}\left(4 \pi^{2} m r\right)}{q_{1} q_{2}} \)

\(\Rightarrow T=4 \pi r \sqrt{\frac{\pi \epsilon_{0} m r}{q_{1} q_{2}}}\)

Given,

Dip angle \(\theta=22^{\circ}\)

The horizontal component of the earth's magnetic field \(H_e=0.35 \mathrm{G}=0.35 \times 10^{-4} ~T\)

As we know,

\(\mathrm{H}_e=\mathrm{B}_{\mathrm{e}} \cos \theta\)

\(\mathrm{B}_{\mathrm{e}}=\frac{0.35 \times 10^{-4}}{0.99}\)

\(=0.36 \times 10^{-4}\)

\(=0.36 \mathrm{G}\)

There are two types of waves Transverse and Longitudinal waves.

Transverse waves: The waves that produce vibrations in the medium in a direction perpendicular to their propagation (eg. Light).

Longitudinal waves: In these waves, the vibrations produced in a medium and their direction of propagation is the same.

The transverse waves can travel in a vacuum but the longitudinal waves require a medium to travel through. Sound waves are longitudinal in nature and therefore cannot travel in a vacuum.

From the above discussion, it is clear that the sound waves cannot travel in a vacuum.

Satellite going round the earth in a circular orbit loses some energy due to collision. Its speed is \(\mathrm{v}\) and distance from the earth is \(\mathrm{d}\) will decrease, \(v\) will increase.

Given that, the particle lost some energy due to a collision. So, it can no longer continue in that orbit as the earth's gravitational force is more than centripetal force. Due to this, the distance \(\mathrm{d}\) decreases gradually and the particle moves towards earth at certain accelaration gaining speed. Thus distance \(\mathrm{d}\) decreases and speed \(v\) increases.

Given,

Retarding force,\(F=-50 \mathrm{~N}\)

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

Initial velocity, \(u=15 \mathrm{~ms}^{-1}\)

Final velocity, \(v=0\)

Acceleration of the body,\(a=\frac{F}{m}\)

\(=\frac{-50}{20 }=-2.5 \mathrm{~ms}^{-2}\)

From the first equation of motion

\(v=u+at\)

\(0=15+(-2.5) \times {t}\)

\({t}=\left(\frac{15}{2.5}\right)\)

\({t}=6\) s

Given: \(\left(\frac{{T}_{1}^{2}} {{T}_{2}^{2}}\right)=\left(\frac{{R}_{1}^{3}} {{R}_{2}^{3}}\right)\)

\(\Rightarrow {T}_{2}^{2}=\left(\frac{{R}_{2}}{{R}_{1}}\right) {T}_{1}^{2}\) 

\(\Rightarrow {T}_{2}=\left(\frac{{R}_{2}}{{R}_{1}}\right)^{\frac{3 }{ 2}} {~T}_{1}\)

For the geostationary satellite, \({T}_{1}=1\) day \((24\) hrs. \()\) 

\(\therefore {T}_{2}=\left(\frac{9 {R}}{{R}}\right)^{\frac{3}{2}}=27\) days

So it will complete one revolution in 27 days.

Now, \(\omega=\frac{2 \pi }{{T}_{2}}=\frac{2 \pi}{27 \times 24 \times 3600} \frac{rad}{s}\)

\(=2.693 \times 10^{-6} \frac{rad} {sec}\)

The distance of the planet from the earth is measured by parallax method. Parallax is a displacement or difference in the apparent position of an object viewed along two different lines of sight, and is measured by the angle or semi-angle of inclination between those two lines. Parallax Method is used for large distances.

Law of conservation of charge:

The total charge of an isolated system remains constant. The electric charges can neither be created nor destroyed, they can only be transferred from one body to another. The law of conservation of charge is obeyed both in large-scale and microscopic processes. In fact, charge conservation is a global phenomenon i.e., the total charge of the entire universe remains constant.

From the above, it is clear that charge can neither be created nor be destroyed, but it can be transferred from one object to another by using some methods like induction and conduction. Therefore option (A) is incorrect and options (B) and (D) is correct.

If the charges are distributed in a system, then the net charge of the system remains constant. Therefore option (C) is correct.