Physics Test-13

It can never decrease is correctfor entropy of an isolated system,

Second law of thermodynamics:

The entropy of an isolated system not in equilibrium will tend to increase over time, approaching a maximum value at equilibrium.

In other words, The second law of thermodynamics states that the total entropy of an isolated system always increases over time, or remains constant in ideal cases where the system is in a steady-state or undergoing a reversible process. This increase in entropy accounts for the irreversibility of natural processes.

The second law gives the definition of entropy term.

\(\Delta S=\frac{\Delta Q}{T}\)

Where \(\Delta S\) is the entropy change, \(\Delta Q\) is the energy change and \(T\) is the temperature.

Isolated system: The system which doesn't interact with the surrounding is called isolated system. The universe is assumed as an isolated system.

According to the second law of thermodynamics, the entropy of an isolated system increases with time It can never decrease.

Since the universe is assumed as an isolated system and the entropy of an isolated system always increases with time.

As the entropy of an isolated system increases always, so the entropy change of an isolated system can never be negative.

Given,

Mass of luggage, \((m)=15\) kg

Displacement \((s)=1.5\) m

As we know,

Work done, \((W)=F \times s=mg \times s\)

\(=15\) kg \(\times 10 \) ms\(^{-2} \times 1.5\) m

\(=225\) kg ms\(^{-2}\) m

\(=225\) Nm

\(=225\) J

Chemical effect of current is not used in electrotyping.

In glowing of a bulb, electrical energy charges to light energy. Electrolysis refers to chemical decomposition produced by passing an electric current through a liquid or solution containing ions.

Since, 1 nm = 10^-9 m

The wavelength (λ) range of X-rays: 0.01 nm to 10 nm

So λ = (0.01 × 10^-9 ) m to (10 × 10^-9 ) m = 10^-11 m to 10^-8 m

Thus, the Wavelength of the X-ray is of the order: 10^-10 m.

Zener diodeis designed to work under breakdown region.

A Zener diode is a special type of diode designed to reliably allow current to flow "backwards" when a certain set reverse voltage, known as the Zener voltage, is reached.

Zener diodes are widely used in electronic equipment of all kinds and are one of the basic building blocks of electronic circuits. They are used to generate low-power stabilized supply rails from a higher voltage and to provide reference voltages for circuits, especially stabilized power supplies. They are also used to protect circuits from overvoltage, especially electrostatic discharge (ESD).

Given: Initially force per unit length on the wires \(= f _{ ab }= f _{ ba }= f =10^{-3}\) N and \(d =2\) m

If two current carrying long wires \(A\) and \(B\) are separated by a very small distance and placed parallel to each other. Then the force per unit length on wire \(A\) and wire \(B\) is given as,

\(f_{a b}=f_{b a}=f=\frac{\mu_{o} I_{a} I_{b}}{2 \pi d}\)

Where, \(I _{ a }=\) current in the wire \(A , I _{ b }=\) current in the wire \(B\), and \(d =\) distance between the wires

If the current in both the wires is doubled,

\(I_{a}^{\prime}=2 I_{a}\)

\(\Rightarrow I_{b}^{\prime}=2 I_{b}\)

And the distance between the wires is halved, so,

\(d^{\prime}=\frac{d}{2}\)

So, when the current in both the wires is doubled and the distance between the wires is halved, the new force per unit length on wire A and wire B will be,

\(f_{a b}^{\prime}=f_{b a}^{\prime}=f^{\prime}=\frac{\mu_{o} I_{a}^{\prime} I_{b}^{\prime}}{2 \pi d^{\prime}}\)

\(\Rightarrow f^{\prime}=2 \times \frac{\mu_{o} \times 2 I_{a} \times 2 I_{b}}{2 \pi d}\)

\(\Rightarrow f^{\prime}=8 \times \frac{\mu_{o} I_{a} I_{b}}{2 \pi d}\)

\(\Rightarrow f ^{\prime}=8 f\)

\(\Rightarrow f ^{\prime}=8 \times 10^{-3}\) N

A small liquid drop has a spherical shape, as due to surface tension the liquid surface tries to have the minimum surface area and for a given volume, the sphere has a minimum surface area.

Binding Energy can be defined as the minimum energy required to be supplied to it order to free the satellite from the gravitational influence of the planet. (in order to take the satellite from the orbit to as point if infinity.)

Consider a satellite revolving around a planet with speed V at a distance r.

Kinetic Energy \(=\frac{1}{2} \mathrm{~m}_{\mathrm{s}} \mathrm{V}^{2} \)

Potential Energy \(=\frac{\mathrm{GM}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{\mathrm{r}} \)

\(\mathrm{E}=\frac{\mathrm{GM}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{\mathrm{r}^{2}}=\frac{\mathrm{msV}^{2}}{\mathrm{r}} \)

\(\Rightarrow \mathrm{m}_{3} \mathrm{~V}^{2}=\frac{4 \mathrm{M}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{\mathrm{r}} \)

Kinetic Energy \(=\frac{1}{2} \frac{\mathrm{Gm}_{\mathrm{s}} \mathrm{M}_{\mathrm{p}}}{\mathrm{r}}\)

Total Energy\(=\mathrm{K} \cdot \mathrm{E}+\mathrm{P} \cdot \mathrm{E}=\frac{1}{2} \frac{\mathrm{GM}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{\mathrm{r}}-\frac{\mathrm{GM}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{\mathrm{r}}\)

Total Energy \(=\frac{-\mathrm{GM}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{2 \mathrm{r}} \)

Binding Energy \(+\) Total Energy = 0 (at infinite TE= 0)

Binding Energy \(=\frac{\mathrm{GP}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{2 \mathrm{r}}=0\)

Binding Energy \(=\frac{\mathrm{GM}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{2 \mathrm{r}}\)

So Binding energy of satellite \(\mathrm{G}\) of man \(\mathrm{m}_{\mathrm{s}}\) revolving around a planet of \(\operatorname{man} \mathrm{M}_{\mathrm{p}}\) in a radius \(\mathrm{r}\) \(\text { is }=\frac{\mathrm{GM}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{2 \mathrm{r}}\)

We know that:The equation for SHM is:

\(\mathrm{y}=\operatorname{Asin}(\omega+\phi)\)

As the displacement is half of the amplitudes so:\(\left(\mathrm{y}=\frac{\mathrm{A}}{2}\right)\)

or \(\frac{\mathrm{A}}{2}=\mathrm{A} \sin (\omega \mathrm{t}+\)\(\phi) \)

\((\omega t+\phi)=\frac{1}{2} \)

\(\therefore \omega \mathrm{t}+\phi=30^{\circ} \text { or } 150^{\circ}\)

Since the two particles are going in opposite directions, the phase of one is \(30^{\circ}\) and that of the other \(150^{\circ} .\)

So the phase difference between the two particles:

\(=150^{\circ}-30^{\circ}\)

\(=120^{\circ}\)

\(=\frac{2 \pi}{3}\)