Physics Test - 22

We know that:

Charge of one electron \(=1.6 \times 10^{-19}\) Coulomb.

Therefore,

Charge of \(25 \times 10^{31}\) electrons will be:

\({Q}=25 \times 10^{31} \times 1.6 \times 10^{-19}\)

\({Q}=40 \times 10^{12} {C}\)

A communication satellite is a repeater between many transmitting stations and many receiving stations.

Communications satellite

• It is an artificial satellite that relays and amplifies radio telecommunications signals via a transponder; it creates a communication channel between a source transmitter and a receiver at different locations on Earth. We can say satellite works as a repeater between transmitting and receiving antenna. Communications satellites are used for television, telephone, radio, internet, and military applications. Based on working it has two types:
• Active Satellites
• Passive Satellites

The process in which an electron escapes from the metal surface is known as electron emission. This occurs due to various mechanisms such as heating (thermionic emission), exposure to light (photoelectric emission), or application of a strong electric field (field emission). Once freed, these electrons contribute to electrical conductivity or can be manipulated for various technological applications like vacuum tubes or semiconductor devices. 

Electric Field at a point due to a line charge distribution \(+\lambda\) at the mid distance (i.e. \(\frac{R}{2}\) ) is:

\(E_{+\lambda}=\frac{\lambda}{2 \pi \varepsilon_{O}\left(\frac{R}{2}\right)}\)

Similarly, Electric field due to line charge \(-\lambda\) is:

\(E_{-\lambda}=\frac{\lambda}{2 \pi \varepsilon_{O}\left(\frac{R}{2}\right)}\)

Both the fields have the same direction at the point A i.e. from \(+\lambda\) towards \(-\lambda\)

Thus, resultant electric field is equal to:

\(E=E_{+\lambda}+E_{-\lambda}\)

\(=\frac{\lambda}{2 \pi \varepsilon_{O}\left(\frac{R}{2}\right)}+\frac{\lambda}{2 \pi \varepsilon_{O}\left(\frac{R}{2}\right)}\)

\(=\frac{2 \lambda}{\pi \varepsilon_{o} R}\)

The electric field produced by a point charge at a point in the electric field is given as,

\(E=\frac{k Q}{r^{2}}\)

\(\Rightarrow E \propto \frac{1}{r^{2}} \quad \ldots\)(1)

Electric potential due to a point charge at a point in the electric field is given as,

\(V=\frac{k Q}{r}\)

\(\Rightarrow V \propto \frac{1}{r}\quad \ldots\)(2)

Electric field due to an electric dipole on the equatorial line is given as,

\(E=\frac{k P}{r^{3}}\)

\(\Rightarrow E \propto \frac{1}{r^{3}}\quad \ldots\)(3)

Electric field due to an electric dipole on the axial line is given as,

\(E=\frac{2 k P}{r^{3}}\)

\(\Rightarrow E \propto \frac{1}{r^{3}}\quad \ldots\)(4)

From equation (1), equation (2), equation (3), and equation (4) it is clear that only the electric field produced by a point charge is inversely proportional to the square of the distance. So it follows the inverse-square law.

The wavelength that is associated with an object in relation to its momentum and mass is known as de Broglie wavelength. A particle's de Broglie wavelength is usually inversely proportional to its force.

Momentum of a photon is given by

\(P=\frac{E}{c}=\frac{h}{\lambda}\) 

where \(E\) is the energy of the photon, \(c\) is the speed of light in vacuum, \(h\) is the Planck's constant and \(\lambda\) is the de Broglie wavelength.

According to de Broglie

\(P=\frac{h}{\lambda}\) or \(P ~\alpha \frac{1}{\lambda}\).

By this relation we can conclude that the linear momentum of a photon is inversely proportional to the de Broglie wavelength. 

The graph of \(P\) vs \(\lambda\) shall be a rectangular hyperbola.

Gravitational force is not the same for every pair of bodies, it will vary with different masses and distance between them. So option (D) is wrong.

Newton's law of gravitation:

Every particle in the universe attracts every other particle with a force, which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Visible spectrum lies betweeninfrared spectrum and ultra violet rays.

• The increasing order of frequency of EM waves are given as Radio, Microwave, Infrared, Visible, Ultraviolet, X Rays, and Gamma Rays.
• From the above, it is clear that the visible spectrum lies between infrared spectrum and ultraviolet rays.

Twinkling of stars is due to atmospheric refraction of starlight.

• The atmosphere of the earth is made of different layers.
• It is affected by winds, varying temperatures, and different densities as well.
• When light from a distant source (a star) passes through our turbulent (moving air) atmosphere, it undergoes refraction many times.
• When we finally perceive this light from a star, it appears to be twinkling.
• This is because some of the light rays reach us directly and some bends away from and toward us. It happens so fast that it gives a twinkling effect.