Physics Test - 18

When the block is moving on a Rough surface Kinetic Frictional forces try to resist the movement.

Here, a block is moving with constant velocity, which means that no resultant force is acting on the block.

Thus, the frictional force is equal and opposite to the applied force and therefore 40 N force is acting on the block but in opposite direction to the external force.

Friction force = external force = 40 N

The magnetic flux that links the larger loop with the smaller loop of side \(l(l<

\(\phi_{12}=B R^{2}=\frac{2 \sqrt{2} \mu_{0} I^{2}}{\pi L}\)

\(\therefore\) Mutual inductance \(M_{12}\)

\(=\frac{\phi_{12}}{I}=\frac{2 \sqrt{2} \mu_{0}}{\pi}\left(\frac{l^{2}}{L}\right)\)

i.e., \(M_{12} \propto \frac{l^{2}}{\mathrm{L}}\)

The natural frequency is given by:

\(\mathrm{f}_{\mathrm{n}}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{\mathrm{M}}}\)

\(\Rightarrow \mathrm{f}_{\mathrm{n}}=\frac{1}{2 \pi} \sqrt{\frac{4000}{5}}\)

\(=4.5 \mathrm{~Hz}\)

The damped frequency is given by:

\(f=f_{n} \sqrt{1-\delta^{2}}\)

\(=4.5 \times \sqrt{1-0.04}\)

\(=4.41 \mathrm{~Hz}\)

When heat is given to particles, they gain energy. On gaining enough energy, particles start moving randomly to acquire kinetic energy and get mixed up with the particles present in air. It is a surface phenomenon.

We know that mercury does not evaporate at room temperature. It has its various applications like it is used in thermometers to check body temperatures also. We will see that mercury has a higher boiling point that is why it does not show the evaporation process at room temperature.

Current sensitivity,

\(\mathrm{I}_{\mathrm{g}}=\frac{\mathrm{NBA}}{\mathrm{K}}\)

\(=\frac{5 \mathrm{div}}{\mathrm{mA}}\)

\(=5000 \frac{\text { div }}{\mathrm{A}}\)

Voltage sensitivity,

\(\mathrm{V}_{\mathrm{S}}=\frac{\mathrm{NBA}}{\mathrm{KR}}\)

\(=20 \frac{\operatorname{div}}{\mathrm{V}}\)

\(\Rightarrow \frac{\mathrm{I}_{\mathrm{g}}}{\mathrm{V}_{\mathrm{S}}}=\mathrm{R}\)

\(\Rightarrow R=\frac{5000}{20}=250 \Omega\)

At \(\mathrm{i}=\mathrm{i}_{\mathrm{C}}\), refracted ray grazes with the surface.

So angle of refraction is \(90^{\circ}\).

When \(L\) is removed, the phase difference between the voltage and current is:

\(\tan \phi_{1}=\frac{X_{C}}{R}\)

\(\tan \frac{\pi}{3}=\frac{X_{C}}{R}\)

or \(X_{C}=R \tan 60^{\circ}\)

or \(X_{C}=\sqrt{3} R\)

When \(\mathrm{C}\) is removed, the phase difference between the voltage and current is:

\(\tan \phi_{2}=\frac{X_{L}}{R}\)

\(\tan \frac{\pi}{3}=\frac{X_{L}}{R}\)or \(X_{L}=R \tan 60^{\circ}=\sqrt{3} R\)

As \(X_{L}=X_{C}\)

The series \(L C R\) circuit is in resonance.

Impedance of the circuit,

\(Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}=R \quad\left(\because X_{L}=X_{C}\right)\)

Power factor, \(\cos \phi=\frac{R}{Z}=\frac{K}{R}=1\)

In air angular fringe width is given by \(\theta_{0}=\frac{\beta}{D}\)

Angular fringe width in water,

\(\theta_{\mathrm{w}}=\frac{\beta}{\mu \mathrm{D}}=\frac{\theta_{0}}{\mu}\)

\(=\frac{0.2^{\circ}}{\left(\frac{4}{3}\right)}=0.15^{\circ}\)

In Fluid Mechanics, the matter of concern is the general state of motion at various points in the fluid system (as in Eulerian approach) rather than the motion of each particle (as in Lagrangian approach). Hence, the Eulerian method is extensively used in Fluid Mechanics.

Given that electron has a mass \({m}\).

De-Broglie wavelength for an electron will be given as,

\(\lambda_{e}=\frac{h}{p} \quad \ldots\)(i)

where, \({h}=\) Planck's constant, and, \(p=\) linear momentum of electron

Kinetic energy of electron is given by, 

\({E}=\frac{\mathrm{p}^{2}}{2 {~m}}\Rightarrow p=\sqrt{2 m E} \quad \ldots\) (ii)

From equation (i) and (ii), we have

\(\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \quad \cdots\)(iii)

Energy of a photon can be given as,

\(E=h v\)

\(\Rightarrow E=\frac{h c}{\lambda_{p}}\)

\(\Rightarrow \lambda_{p}=\frac{h c}{E}\quad \cdots\)(iv)

Thus, \(\lambda_{\mathrm{p}}\) is the de-Broglie wavelength of photon.

Now, dividing equation (iii) by (iv), we get

\( \frac{\lambda_{e}}{\lambda_{p}}=\frac{h}{\sqrt{2 m E}} \cdot \frac{E}{h c}\)

\(\frac{\lambda_{e}}{\lambda_{p}}=\frac{\sqrt{E}}{\sqrt{2m}c}\)

\(\Rightarrow  \frac{\lambda_{e}}{\lambda_{p}}=\frac{1}{c} \cdot \sqrt{\frac{E}{2 m}}\)