Physics Test - 39

Since the radius of the circular path of charge particle of in the magnetic field\(r=\frac{m v}{q B}=\frac{p}{q B}\)

Now, the radius of the circular path of charge particle of given momentum p and magnetic field B is given by

\(r \propto \frac{1}{q}\)

But charge on both charge particles protons and deuterons, is same. Therefore,

\(\frac{r_{p}}{r_{D}}=\frac{q_{D}}{q_{p}}=\frac{1}{1}\)

Suppose \(\mathrm{OR}\) represents the direction of flow of the river. \(\mathrm{O}\) is the starting point and \(\mathrm{P}\) is the point where boatman wants to reach. Vector \(\mathrm{PQ}\) represents the direction in which boatman must row to reach the point \(\mathrm{P}\). Vector \(\mathrm{OP}\) shows the direction of the flow of the river.

Then, \(\mathrm{OQ} =10 \mathrm{kmh}^{-1}\)

\(\mathrm{QP}=5 \mathrm{~km} \mathrm{~h}^{-1}\)

\(\sin \theta=\frac{\mathrm{QP}}{\mathrm{OR}}=\frac{5}{10}=\frac{1}{2}=\sin 30^{\circ}\)

Or, \(\theta=30^{\circ}\) or \(\angle \mathrm{POQ}=30^{\circ}\)

Thus, \(\angle \mathrm{QOP}=+\angle \mathrm{POQ}+\angle \mathrm{POR}\)

\(=30^{\circ}+90^{\circ}=120^{\circ}\)

Thus, the boatman must row making an angle of \(120^{\circ}\) with the direction of flow of the river.

We know that:

The distance travelled by a gas molecule between two successive collisions is known as a free path. It is given as:

\(\lambda=\frac{\text { Total distance travelled by a gas molecule between successive collisions }}{\text { Total number of collisions }}\)

During two successive collisions, a molecule of gas moves in a straight line with constant velocity, and the mean free path of a gas molecule is given by:

\(\lambda=\frac{1}{\sqrt{2} \pi n d^{2}} \quad \dots\) (1)

Where, \(n=\) number of molecules per unit volume and \(d=\) diameter of a molecule

From equation (1) we get,

\( \lambda \propto d^{-2}\)

So, \(n ^{\text {th }}\) power of diameter of molecules \(= -2\)

The equation of the two superimposing waves can be give by:

\(\mathrm{y}_{1}=\mathrm{a}_{1} \sin (\omega \mathrm{t})\)

\(\mathrm{y}_{2}=\mathrm{a}_{2} \sin (\omega \mathrm{t}+ \frac{\pi}{ 2})=\mathrm{a}_{2} \cos (\omega \mathrm{t})\)

So, the equation of the resultant wave is:

\(\mathrm{y}=\mathrm{y}_{1}+\mathrm{y}_{2}\)

\(=\mathrm{a}_{1} \sin (\omega \mathrm{t})+\mathrm{a}_{2} \cos (\omega \mathrm{t})\).....(1)

Assuming \(\omega t=\phi, \mathrm{a}_{1}=\mathrm{a} \cos (\phi)\) and \(\mathrm{a}_{2}=\mathrm{a} \sin (\phi)\),

Put all values in (1).

\(\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}=\mathrm{a}^{2}\left(\sin ^{2}(\phi)+\cos ^{2}(\phi)\right)=\mathrm{a}^{2}\)

\(\Rightarrow \mathrm{a}=\sqrt{\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}}\)

Let \(E_{0}\) be the potential difference applied across the total length \(l(=10 {~cm})\) of potentiometer wire,

Potential gradient in the first case \(=\frac{E_{0}}{l}\)

As per question, \(E=\frac{l}{3}\left(\frac{E_{0}}{l}\right)=\frac{E_{0}}{3} \ldots .\) (i)

Potential gradient in second case \(=\frac{E_{0}}{\frac{3 l}{ 2}}=\frac{2 E_{0}}{3 l}\)

If \(x\) is the desired length of potentiometer to balance the emf \(E\) of the cell, then

\(E=x \times \frac{2 E_{0}}{3 l} \ldots \ldots\) (ii)

From (i) and (ii), we have

\(\frac{E_{0}}{3}=x \times \frac{2 E_{0}}{3 l}\)

or \(x=\frac{l}{2}=\frac{10}{2}=5 {~cm}\)

\(\mathrm{U}=\frac{K \mathrm{e}^{2}}{3 \mathrm{r}^{3}}\)

\(\Rightarrow \mathrm{F}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\frac{-\mathrm{dU}}{\mathrm{dr}} \propto \frac{1}{\mathrm{r}^{4}}\)

\(\Longrightarrow \mathrm{V} \propto \mathrm{r}^{-3 / 2}\)

From Bohr' quantization, \(\mathrm{mvr}=\frac{nh}{2 \pi}\)

\(\Rightarrow \mathrm{n} \propto \mathrm{r}^{-3 / 2} \mathrm{r}=\mathrm{r}^{-1 / 2} \ldots \ldots \ldots \ldots \ldots \ldots 1\)

Energy in nth orbit \(=\frac{1}{2} \mathrm{mv}^{2}+\mathrm{U}=\frac{\mathrm{Ke}^{2}}{2 \mathrm{r}^{3}}+\frac{\mathrm{Ke}^{2}}{3 \mathrm{r}^{3}}=\frac{5 \mathrm{Ke}^{2}}{6 \mathrm{r}^{3}} \propto \mathrm{r}^{-3} \propto \mathrm{n}^{6}\)

The energy in the \(n^{\text {th }}\) orbit of such a hypothetical atom will be proportional to\(n^{2}\).

The solids which have the negative temperature coefficient of resistance are insulators and semiconductors.

The negative temperature coefficient of the resistance is only present in the insulators or the semiconductors. In these, the resistance decreases with increase in temperature.

Given:

Change in an electric field, \(\frac{d E}{d t}=5 \times 10^{12} {Vm}^{-1} s^{-1}\),

Side of the plate (I) \(=2\) \({cm}=2 \times 10^{-2} {~m}\),

and \(\epsilon_{0}=8.85 \times 10^{-12} C^{2} N^{-1} {~m}^{-2}\)

The area of the plate is:

Area of square\(=(\text{side})^2\)

\( A=2 \times 10^{-2} \times 2 \times 10^{-2}=4 \times 10^{-4} {~m}\)

We know that displacement current is given as:

\( I_{d}=\epsilon_{0} A \times \frac{d E}{d t}\)

\( I_{d}=8.85 \times 10^{-12} \times 4 \times 10^{-4} \times 5 \times 10^{12}\)

\( I_{d}=177 \times 10^{-4} A\)

\(I_{d}=17.7 {~m}\)

Viscosity isa measure of resistance to flowwhich arisesdue to the internal frictionbetween layers of fluid.

Most of thefluids offer some resistance to motion.

• Thisresistance to fluid motionis likeinternal frictionanalogous tofriction when a solid moves on a surface.
• Strong intermolecular forcesbetweenmolecules hold them togetherandresist the movement of layerspast one another.
• Hence,to deliver a large volume of oil in lesser time,viscosity must be as lowas possible.
• TheSI unitofviscosityispoiseiulle (Pl).
• Itsother unitsareN s m^-2or Pa s.
• 1 poise =1 g cm^–1s^–1= 10^–1kg m^–1s^–1.

The energy equivalent of the particular mass is given as:

\(\mathrm{E}=\mathrm{mc}^{2}\)

mass is given as \(=0.5 \mathrm{~g}=0.5 \times 10^{-3} \mathrm{~kg}\)

and the speed of light \((\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\).

So, \(\mathrm{E}=0.5 \times 10^{-3} \times\left(3 \times 10^{8}\right)^{2}\)

\(E=4.5 \times 10^{13} \mathrm{~J}\)