Physics Test - 13

Given,

Least count = 0.01 mm

Number of divisions = 50

The pitch of the screw guage is given as:

\(\mathrm{p}=\) least count \(\times\) number of divisions

Substitute the values, we get

\(\mathrm{p}=0.01 \mathrm{~mm} \times 50\)

\(\mathrm{p}=0.5 \mathrm{~mm}\)

Photoelectric effect

\({K} {E}_{\max }={h v}-{W}_{{o}}\)

Here, \(W_{o}\) is the work function of the metal

For \({KE} \geq 0\)

\({hv} \geq {W}_{{o}}\)

The minimum energy required \(={W}_{{o}}\)

The work function of the metal, a characteristic property of the metal.

The situation is given in the figure. Let \(\theta\) be the temperature at point \(\mathrm{C}\).

We know that the rate flow of heat,

\(\frac{Q}{t}=\frac{K A\left(\theta_{1}-\theta_{2}\right)}{d}\)

Where, \(\mathrm{K}=\) coefficient of thermal conductivity

\(\mathrm{A}=\) area of cross-section

Then, by the conservation of heat energy,

\(\frac{K A(100-\theta)}{40}=\frac{K A(\theta-10)}{60}\)

\(\Rightarrow \frac{100-\theta}{2}=\frac{\theta-10}{3}\)

\(\Rightarrow 300-3 \theta=2 \theta-20\)

\(\Rightarrow 5 \theta=320\)

\(\Rightarrow \theta=\frac{320}{5}\)

\(\Rightarrow \theta=64^{\circ} \mathrm{C}\)

Since the particle carrying positive charge is revolving around another charge,

Electrostatic force \(=\) Centrifugal force

\(\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}=m r \omega^{2}\)

\(\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}=\frac{4 \pi^{2} m r}{T^{2}}\)

\(T^{2}=\frac{\left(4 \pi \epsilon_{0}\right) r^{2}\left(4 \pi^{2} m r\right)}{q_{1} q_{2}}\)

\(T=4 \pi r \sqrt{\frac{\pi \epsilon_{0} m r}{q_{1} q_{2}}}\)

According to the question ratio of the mass of alpha particle and proton is given as \(\frac{m_{p}}{m_{\alpha}}=\frac{1}{4}\) and the ratio of the charge of proton and alpha particle is given as \(\frac{q_{p}}{q_{\alpha}}=\frac{1}{2}\)

Since the potential difference is generated in them, therefore, energy in the particle is given by,

\(E=q \Delta V\) Here \(\Delta V\) is the potential difference

We know that this energy will be converted into kinetic energy i.e. \(E=q \Delta V=\frac{1}{2} m v^{2}\) (equation 1 ) where v is the velocity of the particle

According to the given information uniform magnetic field is set up perpendicular to the velocities of the proton and alpha particle and since we know that whenever the velocity and the magnetic field are perpendicular to each other particle goes under a circular motion whose radii is represented as;

\(r=\frac{m v}{q B}(\) Equation 2\()\)

Here v is the velocity and B is the magnetic field and q is the charge in the particle

Finding the value of v form equation 1 and substituting in equation 2

\(\frac{2 q \Delta V}{m}=v^{2}\)

\(v=\sqrt{\frac{2 q \Delta V}{m}}\)

Substituting the value of V in the formula of radius of a circular path due to uniform magnetic field acting perpendicular to the velocities of particles.

\(r=\frac{m v}{q B} \Rightarrow r=\frac{m}{q B} \sqrt{\frac{2 q \Delta V}{m}}\)

\(\Rightarrow r=\frac{1}{B} \sqrt{\frac{2 m \Delta V}{q}}\) (Equation 3)

Let use the equation 3 to find the relation of radii

For proton

\(r_{p}=\frac{1}{B} \sqrt{\frac{2 m_{p} \Delta V}{q_{p}}}\)

For alpha particle

\(r_{\alpha}=\frac{1}{B} \sqrt{\frac{2 m_{\alpha} \Delta V}{q_{\alpha}}}\)

By the equation 3 , we can say that \(r \propto \sqrt{\frac{m}{q}}\)

\(\Rightarrow \frac{r_{p}}{r_{\alpha}}=\sqrt{\frac{1}{4}} \times \sqrt{\frac{2}{1}}\)

Hence, \(\frac{r_{p}}{r_{\alpha}}=\frac{1}{\sqrt{2}}\)

Given that:

\(\phi=3 t^{2}+4 t+9\)

Number of turns is not given, so we will take is \(N=1\)

Time \((t)=2 \mathrm{sec}\)

We know that,

emf induced \((V)=-N \frac{d \varphi}{d t}\)

Where N is the number of turns, \(\phi\) is flux linked and t is time

Here negative sign shows the direction of the induced emf.

Now,

\((V)=-N \frac{d \varphi}{d t}\)

\(=-1 \times \frac{d\left(3 t^{2}+4 t+9\right)}{d t}\)

\(=-(6 t+4+0)\)

\(=-(6 t+4)\)

\(=-(6 \times 2+4)\)

\(=-16 V\)

Thus magnitude of emf induced =16V

l-V characteristics of the tunnel diode is:

So, V_P ≤ V_D < V_V

Nicol prism is used for producing a polarized beam of light from an un-polarized beam. It is based on the principle of action which involves refraction as it passes into the lower half of the prism. It leaves the prism as polarized light after undergoing another refraction as it exits the far right side of the prism. Thus its action is based on double refraction.

Double refraction is an optical property in which a single ray of unpolarized light entering an anisotropic medium is split into two rays, each traveling in a different direction.

According to Charle's law if the pressure remaining constant, the volume of the given mass of a gas is directly proportional to its absolute temperature.

i.e. \(V \propto T\)

or \(\frac{\mathrm{V}}{\mathrm{T}}=\) constant

\(\Rightarrow \frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)

For a given mass of an ideal gas at a constant temperature, the volume of a gas is inversely proportional to its pressure. This is the statement of Boyle's law.

Faraday law is related to electromagnetic induction.

The volume remaining constant, the pressure of a given mass of a gas is directly proportional to its absolute temperature. This is the statement of Gay-Lussac's law or pressure law.

Hence, the correct option is(B).