Physics Test - 19

Given:

Base current (IB) in a transistor circuit changes from 45μA to 140μA.

Collector current (IC) changes from 0.2 mA to 4.00 mA.

Current gain of transistor is defined as:

\(\mathrm{gain}=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}\)

Put the given values in above formula,

\(\mathrm{gain}=\frac{4 \mathrm{~mA}-0.2 \mathrm{~mA}}{140 \mu \mathrm{A}-45 \mu \mathrm{A}}\)

\(=\frac{3800}{95}\)

\(=40\)

Suppose that the distance between two protons \({p}_{1}\) and \({p}_{2}\) be \({r}_{12}\) and electron \(\left({e}^{-}\right)\) is placed at \({r}_{13}\) and distance from protons \({p}_{1}\) and \({p}_{2}\) respectively.

Given, \(r_{12}=1.5 \mathring A=1.5 \times 10^{-10} {~m}\); \(r_{13}=r_{23}=1 \mathring A=10^{-10} {~m}\); \(q_{1}=q_{2}=1.6 \times 10^{-19} {C}\) (protons); \({q}_{3}=-1.6 \times 10^{-19} {C}\) (electrons)
Now, the potential energy of the system \(W=\frac{1}{4 \pi \epsilon_{0}}\left(\frac{q_{1} q_{2}}{r_{12}}+\frac{q_{1} q_{3}}{r_{13}}+\frac{q_{2} q_{3}}{r_{23}}\right)\) \(=9 \times 10^{9}\left[\frac{1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{1.5 \times 10^{-10}}+\frac{1.6 \times 10^{-19} \times\left(-1.6 \times 10^{-19}\right)}{10^{-10}}+\frac{1.6 \times 10^{-19} \times\left(1.6 \times 10^{-19}\right)}{10^{-10}}\right]\)
\(W=\frac{1.6 \times 10^{-19} \times 1.6 \times 10^{-19} \times 9 \times 10^{9}}{10^{-10}}\left[\frac{1}{1.5}-\frac{1}{1}-\frac{1}{1}\right]\) \(=1.6 \times 1.6 \times 9 \times 10^{-19-19+9+10}\left[\frac{2}{3}-2\right]\) \(=-2.56 \times 9 \times 10^{-19} \times \frac{4}{3}=-2.56 \times 12 \times 10^{-19} \mathrm{~J}\)
\(W=\frac{-2.56 \times 12 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}\)\(=-1.6 \times 12 \mathrm{eV}\) \(=-19.2 \mathrm{e} \mathrm{V}\)

The electric field of a plane electromagnetic wave is given by

\(\vec{E}=E_{0}(\hat{x}+\hat{y}) \sin (k z-\omega t)\)

Direction of propagation \(=+\hat{k}\)

\(\hat{E}=\frac{\hat{i}+\hat{j}}{\sqrt{2}}\)

\(\hat{k}=\hat{E} \times \hat{B}\)

\(\hat{k}=\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \times \hat{B}\)

\(\Rightarrow \hat{B}=\frac{-\hat{i}+\hat{j}}{\sqrt{2}}\)

\(\hat{B}=\frac{E_{0}}{c}(-\hat{x}+\hat{y}) \sin (k z-\omega t)\)

The image will not be formed, because object is placed on the side from where reflection is not possible.

To convert a galvanometer into voltmeter, the necessary value of resistance to be connected in series with the galvanometer is:

\(\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}\)

\(R =\frac{50}{10 \times 10^{-3}}-40\)

\(R =5000-40 \Omega\)

\(R =4960 \Omega\)

\(\frac{\mathrm{I}_{\mathrm{s}}}{\mathrm{I}_{\mathrm{p}}}=\frac{\mathrm{N}_{\mathrm{p}}}{\mathrm{N}_{\mathrm{s}}}=\frac{1}{4}\)

\(\mathrm{I}_{\mathrm{s}}=\frac{1}{4} \times 4=1 \mathrm{~A}\)

Infrared rays:

• It is a part of the EM spectrum and has wavelengths range from about 700 nanometers (nm) to 1 millimeter (mm).
• it has longer wavelengths than visible light.
• Infrared cameras and telescopes or night-vision goggles etc, use infrared waves to sense or measure the heat radiated by the object.
• Infrared energy radiated from the hot objects can be detected and then converted into an electronic signal to produce a visual image by cameras telescopes.

The correct expression for the amplification factor is\(\left(\frac{\Delta I_{E}}{\Delta I_{B}}\right)_{V_{C E}}\).

Amplification factor can be defined as theratio of change in emitter current (ΔI_E) to the change in base current (ΔI_B) is known as Current Amplification factor in common collector (CC) configuration. It is denoted by γ. The current gain in CC configuration is same as in CE configuration. The voltage gain in CC configuration is always less than 1.

Given:

The value of charge on the first sphere is \(2 \times 10^{-7} \mathrm{C}\).

The value of charge on the second sphere is \(3 \times 10^{-7} \mathrm{C}\).

The distance between spheres is \(30 \mathrm{~cm}=30 \mathrm{~cm} \times\left(\frac{\mathrm{m}}{100 \mathrm{~cm}}\right)=0.3 \mathrm{~m}\)

We are required to calculate the value of force between the given two spheres due to their charge.

Let us write the expression for force of attraction or repulsion between the given spheres.

\(F=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r^{2}} \quad\quad\ldots \ldots .(1)\)

Here \(q_{1}\) is the charge on the first sphere, \(q_{2}\) is the charge on the second sphere, \(r\) is the distance between spheres and \(\varepsilon_{0}\) is free space permittivity.

We know the value of free space permittivity is:

\(\varepsilon_{0}=8.854 \times 10^{-12} \mathrm{C}^{2} \mathrm{N}^{-1} \mathrm{m}^{-2}\)

Substitute \(2 \times 10^{-7} \mathrm{C}\) for \(q_{1}, 3 \times 10^{-7} \mathrm{C}\) for \(q_{2}, 0.3 \mathrm{~m}\) for \(r\) and \(8.854 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}\) for \(\varepsilon_{0}\) in equation (1).

\(F=\frac{\left(2 \times 10^{-7} \mathrm{C}\right)\left(3 \times 10^{-7} \mathrm{C}\right)}{4 \pi\left(8.854 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)(0.3 \mathrm{~m})^{2}}\)

\(=\frac{\left(2 \times 10^{-7} \mathrm{C}\right)\left(3 \times 10^{-7} \mathrm{C}\right)}{4 \times 3.14\left(8.854 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)(0.3 \mathrm{~m})^{2}}\)

\(=5.994 \times 10^{-3} \mathrm{~N}\)