Physics Test-12

For a geostationary satellite, its time period \(=24 \mathrm{hours}\).

Therefore, angular speed: \(\omega=\frac{\text { angular distance }}{\text { time }}\)

\(=\frac{360^{\circ}}{24}=\frac{2 \pi}{24}\)

\(=\frac{\pi}{12} \mathrm{rad} / \mathrm{hr}\)

We know that the Earth rotates from the west to east direction.

So, if a geostationary satellite orbits in the east-west direction, relative angular velocity \(\omega_{r}=\omega^{\prime}+\omega\)

\(\Rightarrow \omega_{\mathrm{r}}=\omega^{\prime}+\omega \)

\(\Rightarrow \omega_{r}=\frac{\pi}{12}+\frac{\pi}{12}\)

\(=\frac{\pi}{6} \mathrm{rad} / \mathrm{hr}\)

Relative time period, \(T_{r}=\frac{2 \pi}{\omega_{r}}=\frac{2 \pi}{\frac{\pi}{6}}\)

\(=12 \mathrm{hours}\)

Therefore, the time interval between its successive passing about a point on the equator is 12 hours.

Given:

length \(=L\) and volume \(=V\) and Force \(=F\)

Since in the stress formula we have to use area \(A\) that is not given in the question, So

Volume \(\mathrm{V}=\) Area \(\mathrm{A} \times\) length \(\mathrm{L}\)

\(\mathrm{A}=\frac{\mathrm{V} }{ \mathrm{L}}\)

From Hooke's law

stress \(=k \times\) strain

\(\frac{F}{A}=k \times \frac{\Delta L}{L}\)

\(\frac{F}{\frac{V }{ L}}=k \times \frac{\Delta L}{L} \)

\(\frac{F L}{V}=k \times \frac{\Delta L}{L} \)

\(\frac{F L^{2}}{k V}=\Delta L\)

So extension \(\Delta L \propto L^{2}\)

The extension produced in this wire by a constant force \(F\) is proportional to \(L^{2}\).

Given:

\(\mathrm{Q}_{1}=3000 \mathrm{kcal}=3 \times 10^{6} \mathrm{ca} \)

\(\mathrm{T}_{1}=627+273=900 \mathrm{~K} \)

\(\mathrm{~T}_{2}=25+273=298 \mathrm{~K}\)

Now the efficiency will be

\(\Rightarrow \eta=1-\frac{T_{2}}{T_{1}}\)

Where \(T_{1}=\) temperature of the sink and \(T_{2}=\) temperature of the source.

\(\eta=1-\frac{298}{900}=0.668\)

Now the work done will be:

\(W=\eta \times Q_{1} ; \)

\(\Rightarrow W=0.668 \times 3 \times 10^{6} \mathrm{cal} \)

\(\Rightarrow W=2 \times 10^{6} \mathrm{cal}\)

For converting it to joules, \(1 \mathrm{kcal}=4.18\);Joule

\(\Rightarrow W=2006.67 \times 4.18\)Joule

\(\Rightarrow W=8387.867 \)Joule

\(\Rightarrow W \approx 8.4 \times 10^{6}\)Joule

There are no attractive forces between the gas molecules of an ideal gas. Hence, the internal energy of the gas will only be kinetic energy i.e. \(U=\mathrm{KE}\).

The kinetic energy of an ideal gas is dependent on its temperature as:

\(K E=\frac{3}{2} k_{B} T \)

\(\Rightarrow \mathrm{KE} \propto \mathrm{T} \)

\(\Rightarrow \mathrm{U} \propto \mathrm{T}\)

Therefore, the plot of internal energy vs time will be a straight line through the origin as shown in figure.

Given:

Angular Frequency of two Simple Harmonic Motions \(\omega_{1}=100\) and \(\omega_{2}=1000\)

Amplitude displacement of both SHM is the Same (A).

Acceleration of SHM -1 given as \(a_{1}=-\omega_{1}^{2} A-(1)\)

Acceleration of SHM -2 is given as \(a_{2}=-\omega_{2}{ }^{2} A-(2)\)

Ratio of Equation (1) and Equation (2) is \(\frac{a_{1}}{a_{2}}=\frac{-\omega_{1}^{2} A}{-\omega_{2}^{2} A}\)

Putting the Values of \(\omega_{1}\) and \(\omega_{2}\) we get

\(\frac{a_{1}}{a_{2}}=\frac{-\omega_{1}^{2} A}{-\omega_{2}^{2} A} \)

\(\frac{a_{1}}{a_{2}}=\frac{100^{2}}{1000^{2}}=\frac{1}{10^{2}}\)

So, the Required Ratio is \(1: 10^{2}\)

Given:

\(\mathrm{C}_{1}=6 \mu \mathrm{F}, \mathrm{C}_{2}=4 \mu \mathrm{F}\) and \(\mathrm{V}=120 \mathrm{~V}\)

- In a series combination of the capacitor,

\(\Rightarrow \frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}} \)

\(\Rightarrow \frac{1}{C}=\frac{1}{6}+\frac{1}{4}\)

\(\Rightarrow \mathrm{C}=2.4 \mu \mathrm{F}\)

We know that for the series arrangement of the capacitor,

\(\Rightarrow \mathrm{Q}=\mathrm{CV}=\mathrm{C}_{1} \mathrm{~V}_{1}=\mathrm{C}_{2} \mathrm{~V}_{2} \quad \cdots(2)\)

By equation 1 and equation 2 ,

\(\Rightarrow V_{2}=\frac{C V}{C_{2}}\)

\(\Rightarrow V_{2}=\frac{2.4 \times 120}{4} \)

\(\Rightarrow V_{2}=72 \mathrm{~V}\)

Since the resistivity of the conductor/metal is given by:

\(\rho=\rho_{0}[1+\alpha(T-T_{0})]\).

Here \(\rho_{0}\), a and \(T_{0}\) are constant terms.

So the resistivity is linearly proportional to temperature. But in the case of metal, there are free electrons.

When the temperature of the metals increases then the randomness of the free electrons will increase and the resistivity will vary quite randomly (not a straight line).

Resistance will be increasing with the temperature but the graph will be quite parabolic (It is seen experimentally -No mathematical relation is there).

Since there is a constant addition term with \(\mathrm{T}\) in the right side that's why the graph will not start from the origin. So option 1 is correct.

Given:

Magnetisation \((\mathrm{M})=72.8 \mathrm{Am}^{-1}\), earth's radius \((\mathrm{R})=6.4 \times 10^{6} \mathrm{~m}\)

Mathematically magnetisation is written as,

\(\Rightarrow M=\frac{m_{\text {net }}}{V} \)

\(\Rightarrow m_{\text {net }}=M \times V\)

\(=M \times \frac{4}{3} \pi R^{3}\)

\(=72.8 \times 6.4 \times 10^{6}\)

\(=8.0 \times 10^{22} A m^{2}\)

Given:

Self-inductance of first coil \(\left(L_{1}\right)=108 \mathrm{mH}\),

Radius of both coils are same i.e., \(r_{1}=r_{2}=r_{1}\),

Number of turns of the first coil \(\left(N_{1}\right)=200\), and

Number of turns of the second coil \(\left(N_{2}\right)=500\)

The Self-induction for the first coil is

\(\Rightarrow L_{1}=\frac{\mu_{0} N_{1}^{2} \pi r^{2}}{2}\)

The Self-induction for the second coil is

\(\Rightarrow L_{2}=\frac{\mu_{0} N_{2}^{2} \pi r^{2}}{2} \quad \cdots \text { (2) }\)

On dividing equation 1 and 2 , we get

\(\Rightarrow \frac{L_{2}}{L_{1}}=\left(\frac{N_{2}}{N_{1}}\right)^{2} \)

\(\Rightarrow L_{2}=L_{1}\left(\frac{N_{2}}{N_{1}}\right)^{2}\)

\(\Rightarrow L_{2}=108 \times\left(\frac{500}{200}\right)^{2}\)

\(=108 \times 6.25=675 \mathrm{mH}\)

In the visible spectrum, the colour having the shortest wavelength is Violet.

Electromagnetic spectrum:

• It is a collection of a range of different waves in sequential order from radio to gamma electromagnetic waves
• Visible Light:
• A portion in the spectrum of electromagnetic waves that is visible to the human eye, ranging roughly between \(400 \mathrm{~nm}\) to \(700 \mathrm{~nm}\).
• Red has the longest wavelength and lowest frequency in this visible spectrum.
• Violet has the smallest wavelength and highest frequency in the visible spectrum.
• They are arranged in wavelength increasing order of VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, and Red)]

From above it is clear that violet has smallest wavelength in the visible spectrum. Therefore option 4 is correct.

As due to mirrors there is reflection of light from one mirror to other and finally fall on the eye of the observer.

When we replace it with the lenses then there must be a refraction of light so that then can go to other lens and finally on observer. The phenomenon of reflection of light by a lens is called as total internal reflection. Total internal reflection is a complete reflection of a ray of light within a medium such as water or glass from the surrounding surfaces back into the medium. 

The energy differences between levels in the Bohr model are given by the Rydberg formula.

Thus the wavelengths of emitted/absorbed photons are calculated by

\(\frac{1}{\lambda}=Z^{2} R_{\infty}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)\)

When atoms are excited they emit light of certain wavelengths that correspond to different colors due to the electron making transitions between two energy levels in an atom.

The light emission can be seen as a series of colored lines, known as atomic spectra.

\(\beta=80, R_{L}=5 \mathrm{k} \Omega=5000 \Omega\) and \(R_{1}=400 \Omega\)

We know that when a transistor is used in the common-emitter mode as an amplifier, then the voltage gain is given as,

\(\Rightarrow A_{V}=\beta \frac{R_{L}}{R_{I}} \)

\(\Rightarrow A_{V}=80 \times \frac{5000}{400} \)

\(\Rightarrow A_{V}=1000\)

Transistor as an amplifier (CE configuration):

• In this configuration, the emitter is common for both input and output.
• The input signal is connected in series with the voltage applied to the base-emitter junction.
• For proper functioning of a transistor, the emitter-base junction must be forward-biased and the collector-base junction must be reverse-biased.

Given:

\(l=100 \mathrm{~m}\)

We know that the minimum size of the antenna required to send the electromagnetic wave signal is given as,

\(\Rightarrow l=\frac{\lambda}{4} \quad \cdots(1)\)

Where \(\mid=\) minimum size of the antenna and \(\lambda=\) wavelength

By equation 1 the maximum wavelength of the signal that can be transmitted by an antenna of length \(100 \mathrm{~m}\) is given as,

\(\Rightarrow \lambda=41\)

\(\Rightarrow \lambda=4 \times 100\)

\(\Rightarrow \lambda=400 \mathrm{~m}\)

So we can say that the antenna of length \(100 \mathrm{~m}\) can transmit the signal of wavelength \(400 \mathrm{~m}\) or less.

We know that when a transistor is used in the common-emitter mode as an amplifier, the input signal is connected in series with the voltage applied to the base-emitter junction.

Amplifier:

• An amplifier is a device by which the amplitude of the input ac signal (voltage/current/power) can be increased.
• Transistors can act as amplifiers while they are functioning in the active region or when it is correctly biased.

Transistor as an amplifier (CE configuration):

• In this configuration, the emitter is common for both input and output.
• The input signal is connected in series with the voltage applied to the base-emitter junction.
• For proper functioning of a transistor, the emitter-base junction must be forward-biased and the collector-base junction must be reverse-biased.
• The current gain is given as,

\(\Rightarrow \beta=\frac{\Delta I_{C}}{\Delta I_{B}}\)

Where \(\Delta \mathrm{I}_{\mathrm{C}}=\) change in collector current and \(\Delta \mathrm{I}_{\mathrm{B}}=\) change in base current.