Physics Test - 10

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

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

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

\(R =5000-40 \Omega\)

\(R =4960 \Omega\)

Given:

\({u}=-40 {~cm} \)

\({f}=40 {~cm} \)

\(\frac{1}{v}-\frac{1}{40}=\frac{1}{40} \)

\({v}=20 {~cm}\)

So, the image is formed at \(20 {~cm}\) behind the mirror.

The purpose of most transmitters is radio communication of information over a distance. The transmitter combines the information signal to be carried with the radio frequency signal which generates the radio waves, which is called the carrier signal.

A transmitter is an electronic device used in telecommunications to produce radio waves in order to transmit or send data with the aid of an antenna. The transmitter is able to generate a radio frequency alternating current that is then applied to the antenna, which, in turn, radiates this as radio waves.

The purpose of the transmitter is to convert the message signal produced by the source of information into a suitable form for transmission through the channel.

As shown in diagram, at diametrically opposite points A and B, the magnitude of the velocity is same (=V) but the directions of the velocity are opposite. Hence the change is momentum is MV - (-MV) = 2 MV.

For any spherical conductor, if it has any charge, the charge will reside on the surface of the conductor.

Given, radius of conductor is \(10 {~cm}\) and we have to find electric field at distance of \(15 {~cm}\).

So, we can assume the sphere as point charge to calculate the electric field at distance of \(15 {~cm}\).

\(Q=3.2 \times 10^{-7} {C}\)

\(r=15 \mathrm{~cm}=0.15 {~m}\)

\(|E|=\frac{1}{4 \pi \epsilon_{0}} \times Q \times \frac{1}{r^{2}}\)

\(|E|=9 \times 10^{9} \times 3.2 \times 10^{-7} \times \frac{1}{0.15^{2}}\)

\(|E|=1.28 \times 10^{5} \mathrm{~N} / \mathrm{C}\)

Given,

Magnetic field, \(B=0.025 \mathrm{~T}\)

Radius of the loop,r \(=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}\)

Constant rate at which radius of the loop shrinks,

\(\frac{d r}{d t}=1 \times 10^{-3} \mathrm{~ms}^{-1}\)

Magnetic flux linked with the loop is

\(\phi=B A \cos \theta=B\left(\pi r^{2}\right) \cos 0^{\circ}=B \pi r^{2}\)

The magnitude of the induced emf is

\(|\varepsilon|=\frac{d \phi}{d t}=\frac{d}{d t}\left(B \pi r^{2}\right)=B \pi 2 r \frac{d r}{d t}\)

\(=0.025 \times \pi \times 2 \times 2 \times 10^{-2} \times 1 \times 10^{-3}\)

\(=\pi \times 10^{-6} V=\pi \mu V\)

Given,

Corner of a side of a square \(=93 \space mm\)

\(\therefore a= 93 \times10^{-3}\)

Potential at a point \(P\) = Potential at a point \(Q_1 +\) Potential at a point \(Q_2 +\) Potential at a point \(Q_3.\)

\(\begin{aligned} V &=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{Q_{1}}{r_{1}}+\frac{Q_{2}}{r_{2}}+\frac{Q_{3}}{r_{3}}\right) \\ &=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{33 \times 10^{-9}}{93 \times 10^{-3}}-\frac{51 \times 10^{-9}}{\sqrt{2} \times 93 \times 10^{-3}}+\frac{47 \times 10^{-9}}{93 \times 10^{-3}}\right) \\ &=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{10^{-9}}{93 \times 10^{-3}}\left(33-\frac{51}{\sqrt{2}}+47\right) \\ & \approx 4 \times 1000 \mathrm{~V} \\ &=4 \mathrm{kV} \end{aligned}\)

Temperature: It is the measure of the degree of hotness and coldness of a body.

The SI unit of temperature is Kelvin (K).

Fahrenheit and Celsius are measurements of temperature and are related to each other as follows:

\(C=(F-32) \times \frac{5}{9}\)

Where C is the temperature in Celsius and F is the temperature in Fahrenheit.

Since we need temperature in fahrenheit and celsius to be equal:

Now, let \(C=F=x\)

So, \(x=(x-32) \times \frac{5}{9}\)

or, \(9 x=5 x-160\)

or, \(4 x=-160\)

or, \({x}=-40\)

Therefore, \(-40^{\circ} \mathrm{C}=-40^{\circ} \mathrm{F}\)

We know that,

\(B=\frac{\mu_{0} i}{2 \pi r}\) or \(B \propto \frac{1}{r}\)

\(\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{2 \mathrm{r}}{\mathrm{r}}\)

\(\frac{0.4}{\mathrm{~B}_{2}}=2\)

\(\mathrm{~B}_{2}=0.2 \mathrm{~T}\)

When \(\mathrm r\) is doubled, the magnetic field becomes halved.

Now the magnetic field will be \(0.2\)tesla.

The theorem of 'equipartition of energy' states that at the thermal equilibrium of any dynamical system, the energy is equally distributed amongst its various degrees of freedom, and the energy associated with each degree freedom per molecule =\(\frac{1}{2}\mathrm{k_{B}}\mathrm{T}\)

And, The kinetic energy of a molecule having \(\mathrm{f}\) degrees of freedom is given by

\(\mathrm{E}=\frac{\mathrm{f}}{2} \mathrm{k}_{\mathrm{B}}\mathrm{T}\)

Where \(\mathrm{k_{B}}=\) Boltzmann constant, and \(\mathrm{T}=\) temperature, \(\mathrm{f}\)=degrees of freedom

The energy of one gram mole of the gas will be given by

\(\Rightarrow \mathrm{E}=\frac{\mathrm{f}}{2}\mathrm{k_{B}}\mathrm{T}\times \mathrm{N_{A}}\)

We know \(\mathrm{R}=\mathrm{N_{A}} \times \mathrm{k_{B}}\) applying this in the above equation

\(\Rightarrow \mathrm{E}=\frac{\mathrm{f}}{2}\mathrm{R}\mathrm{T}\)

The total internal energy of one gram mole of the monoatomic gas is

\(\Rightarrow \mathrm{E}=\frac{3}{2}\mathrm{R}\mathrm{T}\) (As degree of freedom of monoatomic gas is \(3\))