Physics Test - 20

NAND Implementation:

NOR Implementation:

NAND and NOR gates are known as universal gates. Any one of these gates can be used to implement any kind of logic gate. This kind of feasibility does not exist with other gates i.e. any other gate cannot solely implement all logic gates. For example, AND gate cannot be implemented using an OR gate and vice-versa. The implementation of NAND and NOR gates to generate other logic gates is shown above.

Given,

Massof a scooter, \(m=120\) kg

Velocity \(v=108\) km/h \(=108 \times \frac{5}{18} = 30\)m/s

Time taken to stop \(t=10\) sec

Final velocity, \(v = 0\) m/s​

From the first equation of motion,

\(v=u+a t\)

\(0=30+a \times 10\)

\(a=-\frac{30 }{ 10}=-3\)m/s\(^2\)

Since force cannot be negative we will use a positive value for acceleration.

Force, \(F = m a\)

\(= 120 × 3 = 360 \) N

We know that the photoelectric effect of the electron is the emission of the electron due to electromagnetic radiation. This photo electron gets excited and leaves the metal, commonly called photo emission.

Then the energy \(E\) of the photoelectron is given by Planck as:

\(E=h \nu=\frac{h c}{\lambda}\) 

Where, \(h\) = Planck's constant, \(\nu\) = frequency of the incident light, \(c\) = speed of light  \(\lambda\) = wavelength of the incident light

Also, the frequency of the incident light and the kinetic energy is related as,

\(E_{p}=\phi+E_{e}\),

Where \(E_{p}\) = energy of the photon, \(\phi\) = work function,  \(E_{e}\) = energy of the photoelectron.

Given that, \(\phi

According to Kepler's second law, the angular momentum of the planet is constant. The angular momentum of the planet does not vary, i.e., the aerial velocity of the planet remains constant.

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. 

Given that,

The initial angular velocity of the wheel, \(\left(\omega_{i}\right)=33\) revolution/min

The Final angular velocity of the wheel, \(\left(\omega_{f}\right)=0\)

Time taken by wheel to stop \(t=20~s\)

We know that,

The initial angular velocity in \({rad}/s, \left(\omega_{i}\right)\) is given by

\(= 33 \times \frac{2 \pi}{60}\) (\(\frac{2 \pi}{60}\) is the conversion factor from revolution/minute to \(rad/s\))

\(= 1.1 \pi~ rad / s\)

The angular retardation can be calculated by using the formula,

(Since retardation is mentioned, hence we have to take the negative value of alpha and not positive value).

\(\omega_{f}=\omega_{i}-\alpha t\) (Putting the values given in the question, we have the following)

\(\Rightarrow 0=1.1 \pi-\alpha \times 20\)

\(\Rightarrow \alpha=\frac{1.1 \pi}{20}\)

\(=\frac{11 \pi}{200} \mathrm{~rad} / \mathrm{s}^{2}\)

Force acting on \(q\) due to \(\mathrm{Q}_{1}\) and \(\mathrm{Q}_{5}\) are opposite direction, so cancel to each other. Force acting on \(q\) due to 

\(\mathrm{Q}_{3}\) is \(\mathrm{F}_{3}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q Q_{3}}{R^{2}}\)

Force acting on \(q\) due to \(Q_{2}\) and \(Q_{4}\)

Resolving in two component method:

(i) Vertical Component: \(\mathrm{Q}_{2} \operatorname{Sin} \theta\) and \(\mathrm{Q}_{4} \operatorname{Sin} \theta\) are equal and opposite direction, so they are cancel to each other.

(ii) Horizontal Component: \(\mathrm{Q}_{2} \operatorname{Sin} \theta\) and \(\mathrm{Q}_{4} \cos \theta\) are equal and same direction, so they can get added.

\(F_{24}=F_{2 q}+F_{4 q}=F_{2} \cos 45^{\circ}+F_{4} \cos 45^{\circ}\)

\(F_{24}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q Q_{2}}{R^{2}} \cos 45^{\circ}+\frac{1}{4 \pi \varepsilon_{0}} \frac{q Q_{ 4}}{R^{2}} \cos 45^{\circ}\)

Resultant net force \(\mathrm{F}\)

\(\mathrm{F}=\mathrm{F}_{3}+\mathrm{F}_{24}+\mathrm{F}_{15}\)

\(\cos 45^{\circ}=\frac{1}{\sqrt{2}}\)

\(\ \mathrm{F} =\frac{q \mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}\left[1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]\)

\(=\frac{1}{4 \pi \varepsilon_{0}} \frac{q \mathrm{Q}}{\mathrm{R}^{2}}\left[1+\frac{2}{\sqrt{2}}\right]\)

 \(\mathrm{F} =\frac{1}{4 \pi \varepsilon_{0}} \frac{q \mathrm{Q}}{\mathrm{R}^{2}}[1+\sqrt{2}] \mathrm{N} \)

Vector form:

\(\overrightarrow{\mathrm{F}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q \mathrm{Q}}{\mathrm{R}^{2}}(1+\sqrt{2}) \mathrm{N} \hat{i}\)

The increase in the width of the depletion region is due to the absence of the electrons and holes in the region. This occurs only in the case of the reverse bias only in a diode.

Electric Power: The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.,

\(P=\frac{W}{t}=V I=I^{2} R=\frac{V^{2}}{R}\)

Where V = Potential difference, R = Resistance and I = current

From above equation it is clear that all options are correct.