Physics Test - 38

Total angular momentum of the system initially \(\mathrm{L}_{\mathrm{i}}=\mathrm{I}_{1} \mathrm{\omega}+\mathrm{I}_{2}(\mathrm{0})=\mathrm{I}_{1} \mathrm{\omega}\)

Total angular momentum of the system finally \(\mathrm{L}_{\mathrm{f}}=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \mathrm{\omega}_{2}\)

According to conservation of angular momentum i.e. \(\mathrm{L}_{\mathrm{i}}=\mathrm{L}_{\mathrm{f}}\)

\(\Rightarrow \mathrm{I}_{1} \omega=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \omega_{2}\)

\(\Rightarrow \omega_{2}=\frac{\mathrm{I}_{1} \omega}{\mathrm{I}_{1}+\mathrm{I}_{2}}\)

Given,

\(x(t)=\frac{v}{A}\left(1-e^{-A t}\right)\)

As we know that \( (1- {e}^{\text {-At }}) \) is a constant value and will have no dimension.

Thus, the dimension of \(\frac{v}{A}\) will be equal to the dimension of \(x\).

Dimension of position, \(x=\left[M^{0} L^{1} T^{0}\right]\)

Thedimension of velocity,\(v=\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-1}\right]\)

\(\Rightarrow x=\frac{v}{A}\)

\(\Rightarrow\left[M^{0} L^{1} T^{0}\right]=\frac{\left[M^{0} L^{1} T^{-1}\right]}{A}\)

\(\Rightarrow A=\left[T^{-1}\right]\)

\(\begin{aligned} & \mathrm{B}_1=\frac{\mu_0 \mathrm{i}}{2 \mathrm{R}} \\ & \mathrm{B}_2=\frac{\mu_0 \mathrm{R}^2}{2\left(\mathrm{R}^2+\mathrm{x}^2\right)^{\frac{3}{2}}} \\ & \Rightarrow \frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{1}{\mathrm{R}^3}\left(\mathrm{R}^2+\mathrm{x}^2\right)^{\frac{3}{2}} \\ & =\frac{1}{\mathrm{R}^3}\left(8 \mathrm{R}^3\right) \\ & =8\end{aligned}\)

As we know that the dimension of Planck's constant = \(\left[M L^{2} T^{-1}\right]\)

The dimension of angular momentum = \(\left[M L^{2} T^{-1}\right]\)

The dimension of Linear momentum = \(\left[M^{1} L^{1} T^{-1}\right]\)

The dimension of Torque = \(\left[M L^{2} T^{-2}\right]\)

The dimension of Velocity = \(\left[\mathrm{L} \mathrm{T}^{-1}\right]\)

Thus we can see that the dimension of Planck's constant is similar to angular momentum.

Given,

Charge on an alpha particle, \(q_{1}=q_{2}=+2 e\)

Distance between the particles, \(r=3.2 \times 10^{-15} m\)

We know that:

Charge on an electron, \(e=1.6 \times 10^{-19}\)

and, \(\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9}\)

Now, using coulomb's law, we get,

Force acting on the particles is given by,

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

\(F=\frac{9 \times 10^{9} \times 2 \times 1.6 \times 10^{-19} \times 2 \times 1.6 \times 10^{-19}}{3.2 \times 10^{-15} \times 3.2 \times 10^{-15}} \)

\(F=\frac{36 \times 10^{9} \times 2.56 \times 10^{-19}\times 10^{-19}}{3.2 \times 10^{-15} \times 3.2 \times 10^{-15}}\)

\(F= 90 N\)

The wave theory of light was first demonstrated by Thomas Young in 1801 through Young's double-slit experiment.

This experiment shows that the observed pattern of interference occurs due to the superposition of light waves which proves the wave nature of light.

In this experiment, two narrow slits that are close to each other, are illuminated by a monochromatic light source.

The two slits are responsible for the production of two different wavefronts which then superimpose on a screen forming the definite interference pattern.

For two coherent sources \(s_1\) and \(s_2\), the resultant intensity at some point p is given by:

\(I=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2} \cos \phi}\)

Thus, interference proves wave nature.

Given:

\(m_{p}=1.6 \times 10^{-27} \mathrm{~kg}, g=10 \mathrm{~ms}^{-2}\)

\(\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-27} \mathrm{Kg}\)

\(\mathrm{h}=1 \mathrm{~mm}=10^{-3} \mathrm{~m}\)

\(V=5 \mathrm{~V}\)

Time of flight of electron \(t_{e}=\sqrt{\frac{2 h}{a}}\) (ignoring gravity)

We know that \(\mathrm{F}=\mathrm{ma}\)

Similarly \(\mathrm{F=e E}\)

\(\therefore \mathrm{a}=\frac{e E}{m}\left[\right.\) also \(\left.\mathrm{E}=\mathrm{V} / \mathrm{d}=\frac{5}{10^{-3}}=5000 \mathrm{Vm}^{-1}\right]\)

\(\therefore \mathrm{t}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{hm}_{e}}{e E}}=\sqrt{\frac{2 \times 10^{-3} \times 9.1 \times 10^{-31}}{1,6 \times 10^{-19} \times 5000}}\)

\(=\left[\frac{2 \times 9.1 \times 10^{-34}}{1.6 \times 5 \times 10^{-16}}\right]^{1 / 2}=\left[2.275 \times 10^{-18}\right]^{1 / 2}\)

\(t_{e} \simeq 1.5 \times 10^{-9} \mathrm{~S}\) or \(\simeq 1.5 \mathrm{~ns}\)

Time flight for proton is

\(t_{p}=\sqrt{\frac{2 h m_{p}}{e E}}\)

\(=\sqrt{\frac{2 \times 10^{-3} \times 1.6 \times 10^{-27}}{1.6 \times 10^{-19} \times 5000}}=\left[\frac{2 \times 10^{-30}}{5 \times 10^{-16}}\right]^{1 / 2}\)

\(=\left[0.4 \times 10^{-14}\right]^{1 / 2}\)

\(=\left[4000 \times 10^{-18}\right]^{1 / 2}\)

\(=63.25 \times 10^{-9}\)

\(t_{p}=63.25 \mathrm{~ns}\)

Some gases like Hydrogen, oxygen, nitrogen behaves as an ideal gas at low Pressure and High Temperature.

An ideal gas is a gas that obeys the law of the Ideal Gas Equation.

\(PV = nRT\)

While deriving the ideal gas equation, the following assumptions should be considered:

• The size of the ideal gas molecules is negligibly small.
• There must be no force of attraction among the molecules of the gases.
• Ideal gas work at Low Pressure and High Temperature.

Given that, \(r=50 \mathrm{~cm}, R=100 \mathrm{~cm}\)

Mass supported on four columns, \(M=50 \times 10^3 \mathrm{~kg}\)

Mass supported on each column, \(m=\frac{M}{4}\)

\(\Rightarrow m=\frac{50 \times 10^3}{4}=12.5 \times 10^3 \mathrm{~kg}\)

Now, weight, \(w=m g=12.5 \times 9.8 \times 10^3 \mathrm{~N}=1225 \times 10^5 \mathrm{~N}\)

Area of cross-section of each column

\(A=\pi\left(R^2-r^2\right) \)

\(=3.14\left\{(100)^2-(50)^2\right\} \times 10^{-4} m^2=2.35 m^2\)

Young's modulus, \(Y=2.0 \times 10^{11} \mathrm{~Pa}\)

By using Hooke's law,

Stress \(=Y \times\) Strain

\(\therefore\) Compressive strain \(=\frac{\text { Stress }}{Y}=\frac{W}{A Y}\)

Substituting the values, we get

\(\text { Compressive strain }=\frac{1.225 \times 10^5}{2.35 \times 2.0 \times 10^{11}}=2.60 \times 10^{-7}\)