Physics Test - 13

The value of atomic specific heat of gas under constant pressure is:

\(C_{P}=C_{V}+R\)

also value of atomic specific heat of gas under constant volume is by mathematical relation \(C_{V}=\frac{f}{2} R\).

For monoatomic gases, \(f=3\), as they have only translational degrees of freedom.

\(C_{V}=\frac{f}{2} R=\frac{3}{2} R\)

The value of atomic specific heat of gas under constant pressure is \(C_{P}=C_{V}+R\)

\(C_{P}=C_{V}+R=\frac{3}{2} R+R=\frac{5}{2} R\)

Since, The value of atomic number \( N=1\) specific heat of gas under constant pressure is \(C_{P}=\frac{5}{2} R\).

The parallax method is used to determine large distances, such as the distance from Earth to a planet or a star. Parallax is the projected shift in the position of one object with respect to another when we move the point observation slant. The distance between two points of observation is called the base \((b)\). The distance of the object from the two points of view is \(D\). The angle between the two directions along which the object is viewed is the parallax angle or displacement angle \((θ)\).  

If \(S\) is the position of the object and \(AB\) is the two points of observation,

\(\theta=\frac{b}{D}\)

The triangulation method finds the angles in a triangle formed by three observation points. The other distances in a triangle are calculated using trigonometry and the measured length of just one side. It is used for geographic measurement and not astronomical distance.

The echo method uses the principle of reflection of sound or light. Knowing the speed of the wave and the time of reflection back to the source, the distance between the two points is calculated. This is not a very accurate method and therefore astronomical distances cannot be relied upon.

Nuclear radius \(=10^{-15} \mathrm{~m}\)

Atomic radius \(=10^{-10} \mathrm{~m}\)

Ratio \(=\frac{\text { Nuclear radius }}{\text { Atomic radius }}=\frac{10^{-15}}{10^{-10}}=10^{-5}\)

So, the ratio of the radius of the nucleus of the radius of the atom is of the order of \(10^{-5}\).

Coulomb’s law: When two charged particles of charges q₁ and q₂ are separated by a distance r from each other then the electrostatic force between them is directly proportional to the multiplication of charges of two particles and inversely proportional to the square of the distance between them.

Force (F) ∝ q₁ × q₂

$F\propto \frac{1}{r^2}$

$F=K\frac{q_1\times q_2}{r^2}$

Where K is a constant = 9 × 10⁹ Nm²/C²

From the above, it is clear that Columbu's force between two point charges varies with distance 'r' in relation to $\frac{1}{r^2}$.

From the graph it can be said that

\(v = v _{0}+\alpha x\)

then

\(\frac{d v}{d x}=\alpha\)

Acceleration of the particle can be written as

\(a = v \frac{d v}{d x}\)

\(=\left( v _{0}+\alpha x \right) \alpha\)

This is linear equation.

Thus acceleration varies linearly with \(x\).

The mean free path of electrons is defined as the distance travelled by the electrons before collision, here mean free path \(\lambda=4 \times 10^{-8} \mathrm{~m}\).

And the energy provided by the electric field is 2eV.

\(\mathrm{V}_{0}=2 \mathrm{eV}\)

Now, electric field \(\mathrm{E}=\frac{V_{0}}{\lambda}\)

Put the values in above formula.

\(\mathrm{E}=\frac{2}{4 \times 10^{-8}}\)

\(=0.5 \times 10^{8}\)

\(=5 \times 10^{7} \mathrm{Vm}^{-1}\)

For constructive interference between two waves of equal wavelength the path difference should be a multiple of \(\lambda\).

Let it be \(n \lambda\)

It path difference is \(n \lambda\) then phase difference is \(2 n \pi\) because \(\lambda\) corresponds to \(2 \pi\).

Thus, \(\delta=2 \mathrm{n} \pi\)

\(\cos ^{2} \frac{\delta}{2}=1\)

\(I=\frac{m\left[(20)^{2}+(10)^{2}\right]}{2}=\mathrm{mk}^{2}\)

\(\mathrm{k}=\sqrt{250}\)

\(⇒\mathrm{k}=5 \sqrt{10}\)

\(⇒\mathrm{k}=16\)