Physics Test-31

Mass spectroscopy is a process in which a compound is bombarded with electrons to generate ions. These ions are then separated by accelerating them through electric and magnetic fields. This is then separated according to their specific mass-charge ratio, where the radius of trajectory is proportional to the mass of a charged particle moving in uniform electric and magnetic fields. The sample must be in the gaseous state to be ionized. Therefore, the sample has to be converted into a gaseous state before the process can begin.

The nuclei which have the same mass number are called as isobar. Further, an isobar should have different atomic number. 

A nuclide is a specific type of atom characterized by the number of protons and neutrons in the nucleus, which approximates the mass of the nuclide.

Conduction transfers heat via direct molecular collision. An area of greater kinetic energy will transfer thermal energy to an area with lower kinetic energy. Higher-speed particles will collide with slower-speed particles. The slower-speed particles will increase in kinetic energy as a result.

From above it is clear that when a current-carrying rectangular coil is placed in a magnetic field it experiences a torque.

\(\therefore\) The moment of the deflecting couple \((\tau)=n B i A\)

Where, \(n =\) number of turns, \(B =\) magnetic field, \(i =\) current and \(A =\) area of coil

When the coil deflects, the suspension wire is twisted. On account of elasticity, a restoring couple is set up in the wire. This couple is proportional to the twist.

If \(\theta=\) the angular twist, then,

The moment of the restoring couple \(=C \theta\)

Where, \(C=\) restoring couple per unit twist.

At equilibrium,

Deflecting couple \(=\) Restoring couple

So, we can write,

\(nBi A = C \theta\)

\(\Rightarrow\left(\frac{n B A}{C}\right) i=\theta\)

Here, \(n , B\), and \(A\) are constant

Therefore, the deflection produced in the moving coil galvanometer is directly proportional to the amount of current passing through it.

If the moment of inertia of a rotating body is increased then the effect will be decrease on the angular velocity because of moment of inertia is inversely proportional to the angular velocity.

Given,

\(v=0.5 Hz ; N=100, A=0.1 m ^2\) and \(B=0.01 T\)

As we know,

The instantaneous value of the emf is \(\varepsilon=N B A \omega \sin \omega t\).

\(\varepsilon_0 =N B A(2 \pi v) \)

\(=100 \times 0.01 \times 0.1 \times 2 \times 3.14 \times 0.5 \)

\(=0.314 V\)

The maximum voltage is \(0.314 V\).

No, the photodiode cannot detect the wavelength of \(6000 \mathrm{~nm}\) because of the following reason:

The energy bandgap of the given photodiode, \(E_{g}=2.8 \mathrm{eV}\)

The wavelength is given by \(\lambda=6000 \mathrm{~nm}=6000 \times 10^{-9} \mathrm{~m}\)

We can find the energy of the signal from the following relation:

\(E=\frac{h c}{\lambda}\)

In the equation, h is the Planck's constant \(=6.626 \times 10^{-34} \mathrm{~J}\) and \(\mathrm{c}\) is the speed of light \(=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\).

Substituting the values in the equation, we get

\(E=\frac{\left(6.626 \times 10^{-34} \times 3 \times 10^{8}\right)}{6000 \times 10^{-9}}=3.313 \times 10^{-20} \mathrm{~J}\)

But, \(1.6 \times 10^{-19} \mathrm{~J}=1 \mathrm{eV}\)

Therefore, \(E=3.313 \times 10^{-20} \mathrm{~J}=\frac{3.313 \times 10^{-20}}{1.6 \times 10^{-19}}=0.207 \mathrm{eV}\)

The energy of a signal of wavelength \(6000 \mathrm{~nm}\) is \(0.207 \mathrm{eV}\), which is less than \(2.8 \mathrm{eV}\) which is the energy band gap of a photodiode. Hence, the photodiode cannot detect the signal.

Given,

Mass of the first particle \((m _{1})=2 kg\)

Mass of the second particle \((m _{2})=3 kg\)

Velocity of first particle \((u _{1})=20 m/s\)

Velocity of second particle \((u _{2})=0 m/s\)

Initial \(KE =\frac{1 }{ 2 }m _{1} u ^{2}{ }_{1}\)

\(=\frac{1 }{ 2} \times 2 \times (20)^{2}\)

\(=400 J\)

Let \(V\) is the final velocity of both the particle after collision.

\(P_{1}=m_{1} u_{1}+m_{2} u_{2}=2 \times 20+0=40 \)

\(P_{2}=m_{1} v_{1}+m_{2} v_{2}=(2+3) V=5 V \)

\(P_{1}=P_{2} \)

\(\therefore 5 V=40 \)

\(\Rightarrow V=\frac{40 }{ 5}\)

\(\Rightarrow V=8 m / s\)

Final \(KE =\frac{1}{2}mV^2=\frac{1}{2}(m_1+m_2)V^2\)

\(=\frac{1 }{ 2} \times(2+3) \times 8^{2}\)

\(=160 J\)

Loss in \(KE =\) Initial \(KE -\) Final \(KE\)

\(=400-160\)

\(=240 J\)

According to question,

Initial energy of capacitor \(E _{ i }=\frac{1}{2} CV _1^2\)

Final energy of capacitor is \(E _{ f }=\frac{1}{2} CV _2^2\)

Final energy of inductor \(E _{ L }=\frac{1}{2 LI ^2}\)

The total energy in the given circuit remains conserved I.e.

\(\frac{1}{2} CV _1^2=\frac{1}{2} CV _2^2+\frac{1}{2} LI ^2\)

\(C \left( V _1^2- V _2^2\right)= LI ^2 \)

\(I =\sqrt{\frac{ C \left( V _1^2- V _2^2\right)}{ L }}\)

Given that,

\( C =2 \mu F =2 \times 10^{-6} F \)

\( L =0.6 \times 10^{-3} H\)

\(V _1=12 V \)

\(V _2=6 V\)

Put all values in the expression of I

\(I=\sqrt{\frac{2 \times 10^{-6}(144-36)}{0.6 \times 10^{-3}}} \)

\(I=\sqrt{\frac{2 \times 10^{-3} \times 108}{0.6}} \)

\(I=0.6 A\)