Physics Test-32

We know that:

Electrostatic force \(=\frac{K Q_{1} Q_{2}}{\tau^{2}}\)

Gravitational force \(=\frac{G m_{1} m_{2}}{r^{2}}\)

On dividing the two we get,

\(\frac{\text { Electrostatic Force }}{\text { Gravitational Force }}=\frac{\frac{K Q_{1} Q_{2}}{r^{2}}}{\frac{G m_{1} m_{2}}{r^{2}}}\)

\(\frac{\text { Electrostatic Force }}{\text { Gravitational Force }}=\frac{9.0 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{6.67 \times 10^{-11} \times\left(9.1 \times 10^{-31}\right)^{2}}\)

Electrostatic Force \(=4 \times 10^{42}\) Gravitational Force

Therefore, the electrostatic force between two electrons is greater than the gravitational force by a factor of \(10^{42}\).

Sodiumis not used for electroplating metal articles.

Sodium is an alkali metal and it forms hydrogen gas when it is used as electrolyte. This makes sodium unsuitable for electroplating.

The ratio of mass defect and nucleon is called aspacking fraction.

\(Packing~fraction =\frac{\text { Mass defect }(\Delta m)}{\text { Nucleon }(A)}\)

Packing fraction depends on the manner of packing of the nucleons within the nucleus.

A negative packing fraction generally indicates the stability of the nucleus. Generally, the lower the packing fraction, the higher the binding energy per nucleon, which makes greater stability.

Maxwell showed that Ampere's circuital law is logically inconsistent. He considered a parallel plate capacitor being charged be a battery. He showed inconsistency of the law in the regions outside the plates of capacitor and just between the plates. To resolve this inconsistency displacement current was introduced by Maxwell.

The mass of planets is measured using the gravitational method. In this method, the mass is compared to the force of gravity of a known mass with the force of gravity of a known mass. The strength of the gravitational force is directly proportional to the mass.

The instrument for the accurate measurement of the e.m.f of a cell is a potentiometer.

Both potentiometer and voltmeter are devices to measure potential difference. E M F is the terminal p.d between the electrodes of a cell in open circuit, i.e., when no current is drawn from it. Potentiometer measures the potential difference using null deflection method, where no current is drawn from the cell; whereas voltmeter needs a small current to show deflection. So, accurate measurement of p.d is done using a potentiometer.

The relation between angular frequency and displacement is given as

\(\mathrm{v}=\omega \sqrt{\mathrm{A}^{2}-\mathrm{x}^{2}}\).....(1)

Suppose

\(\mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}\)

On differentiating the above equation w.r.t. time we get

\(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{A} \omega \cos \omega \mathrm{t}\)

The maximum value of velocity will be \(v_{\max}=A \omega\)

The displacement for the time when speed is half the maximum is given as

\(\mathrm{v}=\frac{\mathrm{A} \omega}{2} \)

\(\mathrm{~A}^{2} \omega^{2}=4 \omega\left(\mathrm{A}^{2}-\mathrm{x}^{2}\right)\)

By substituting the value in \((1)\) we get the displacement as:

\(\mathrm{x}=\frac{\mathrm{A} \sqrt{3}}{2}\)

Given:

Mass of earth (M)= \(5.98 \times 10^{24} \mathrm{~kg}\)

For earth to be a black hole, the escape velocity should be equal to the speed of light.

\( \text { Escape velocity= Speed of light } \)

\( v=c \ldots(1)\)

We know escape velocity is given by,

\(v=\sqrt{\frac{2 G M}{R}} \ldots(2)\)

Where,

\(\mathrm{G}\) is the universal gravitational constant

\(M\) is the mass of the body is to escaped from \(r\) is the distance from the center of the mass

Substituting equation. (2) in equation. (1) we get,

\(\sqrt{\frac{2 G M}{R}}=c\)

Squaring both the sides we get,

\(\frac{2 G M}{R}=c^{2}\)

Rearranging the above equation we get,

\(R=\frac{2 G M}{c^{2}}\).

Substituting values in above equation we get,

\(R=\frac{2 \times 6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{\left(3 \times 10^{8}\right)^{2}} \)

\(\Rightarrow R=\frac{79.77 \times 10^{13}}{9 \times 10^{16}} \)

\(\Rightarrow R=8.86 \times 10^{-3} \mathrm{~m} \)

\(\Rightarrow R=0.886 \times 10^{-2}\)

So, Earth has to be compressed to a radius of \(10^{-2} \mathrm{~m}\) to be a black hole.

In the given circuit \(2 {~A}\) current divides equally at junction \(D\) along the paths \(DAC\) and \(DBC\) (each path carry \(1~A\) current).

Potential difference between \(D\) and \(A, V_D-V_A=1 \times 2=2~V\quad\dots(1)\)

Potential difference between \(D\) and \(B, V_D-V_B=1 \times 3=3~V\quad\dots(2)\)

On solving \((1)\) and \((2)\) \({V}_{{A}}-{V}_{{B}}=+1~V\)

A passenger in a moving train tosses a five rupee coin. If the coin falls behind him, then the train must be moving with uniform acceleration.Because the coin falls behind the passenger, the train must be moving with uniform acceleration.

As the passenger tosses the coin, it goes up and is in motion with the speed (initial) of the train but the train is accelerated, so its speed increases. But the coin is at the (initial) speed in the air, so the train slightly moves forward with the person and the coin falls behind him.