Physics Test - 13

The volume and temperature are the same for the two gases.

Given the ratio of partial pressure \(\frac{n_{1}}{n_{2}}=\frac{5}{2}\)

We know that:

\(P _{1} V = n _{1} R T \text { and } P _{2} V = n _{2} RT\)

Where, \(P\) is pressure, \(V\) is volume, \(T\) is temperature, \(n\) is no. of mole and \(R\) is the universal gas constant

\(\frac{P_{1}}{ P _{2}}=\frac{n_{1}}{n_{2}}=\frac{5}{2}=2.5\)

If \(N _{1}\) and \(N _{2}\) are the numbers of molecules of and \(N\) is the Avogadro's number, then:

\(\frac{n_{1}}{n_{2}}=\frac{\frac{N_{1}}{N}}{\frac{N_{2}}{N}}=\frac{5}{2}=2.5\)

Thus, the ratio of the number of molecules of helium and Neon is \(2.5\).

The circuit consists of two NOT gates connected to each input of a NOR gate.

The correct truth table of the combination will be:

\(R =\frac{ u ^{2} \sin 2 \theta}{ g }\)

\(\frac{ Rg }{ u ^{2}}=\sin 2 \theta<1\)

Therefore there should be two values of \(\theta\)

\(\theta_{1}\) and \(\left(90^{\circ}-\theta_{1}\right)\) for same range.

It is obvious that both these angles are complementary.

\(T _{1}=\frac{2 u \sin \theta_{1}}{ g }\)

and \(T _{2}=\frac{2 u \sin \left(90^{\circ}-\theta_{1}\right)}{ g }\)

\(\therefore T_{1} \times T_{2}=\frac{2 u ^{2} \sin 2 \theta_{1}}{ g ^{2}}\)

\(\frac{2 R}{g}\)

So The product of the possible times of flight from \(O\) to \(A\) is \(\frac{2 R }{ g }\)

Power of transmitter,

\(\mathrm{P}=\mathrm{1 0} \mathrm{k} \mathrm{W}\)

\(=\mathrm{1 0} \times \mathrm{10}^{3} \mathrm{W}\)

Frequency of the transmitter,

\(v=880 \mathrm{kHz}\)

\(=880 \times 10^{3} \mathrm{~Hz}\)

Number of photons emitted per second

\(\mathrm{N}=\frac{P}{h v}=\frac{10 \times 10^{3}}{6.6 \times 10^{-34} \times 880 \times 10^{3}}\)

\(\mathrm{N}=1.71 \times 10^{31}\) photons emitted per second.

Given,

The final state of transition, \(n_{1}=2\)

The initial state of transition, \(n_{2}=5\)

The Rydberg constant of hydrogen, \(R_{H}=10973731.6 {~m}^{-1}\)

The Rydberg equation for hydrogen atom is given by,

\(\frac{1}{\lambda}=R_{H}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\)...(i)

Substitute the given values of \(n_{1}, n_{2}\) and \(R_{H}\) in the equation (i),

\(\frac{1}{\lambda}=10973731.6 m^{-1}\left[\frac{1}{2^{2}}-\frac{1}{5^{2}}\right] \)

\(\frac{1}{\lambda}=10973731.6\left[\frac{1}{4}-\frac{1}{25}\right] \)

\(\frac{1}{\lambda}=10973731.6[0.25-0.04] \)

\(\frac{1}{\lambda}=10973731.6[0.21] \)

\(\frac{1}{\lambda}=2304483.636 \mathrm{~m}^{-1}\)

Taking reciprocal on both sides,

\(\lambda =\frac{1}{2304483.636 m^{-1}} \)

\(\lambda =4.3393 \times 10^{-7} m\)

Convert unit from \(m\) to \(n m\),

\(1 m=10^{9} n m\)

So,

\(\lambda=4.3393 \times 10^{-7} {~m} \times \frac{10^{9} {~nm}}{1 {~m}} \)

\(\lambda=4.3393 \times 10^{2} {~nm} \)

\(\lambda=433.93 {~nm} \)

\(\lambda \simeq 434 {~nm}\)

So, the wavelength of photon emitted during transition, \(\lambda=434 {~nm}\)

Given,

Mass of the body, \(m=3.0 \) kg

\(t=25 \) seconds

Initial velocity, \(u=2.0 \) m/s

Final velocity, \(v=3.5 \) m/s

From the first equation of motion,

\(v=u+a t\)

\(a=\frac{(3.5-2.0)}{25 }\)

\(=0.06\) m/s\(^2\)

The magnitude of the force,\({F}={ma}\)

\(=3.0 \times 0.06=0.18\)N

\(=0.18\) N

The phase difference between the electric field and the magnetic field in the electromagnetic wave is zero.

The electric field and the magnetic field components of an electromagnetic wave oscillate in such a way that they have their peak at the same time and also become zero at the same time.Since there is no time difference between the peaks of the electric and the magnetic field, so the phase difference between the electric and the magnetic field of the electromagnetic wave is zero.

The work to be done in the process is given by \(d W =P d V \)

\( =1 \times 10^5 Pa \times(1.671-0.001) m ^3 \)

\( =1.670 \times 10^5 J\)

The change in heat energy during the vaporisation process can be calculated as follows-

\(\Delta Q_{\text {supplied }} =2257 \times 1 \times 10^3 J \)

\( =22.57 \times 10^5 J\)

Hence, the change in internal energy in the process is given by

\(\Delta U =\Delta Q_{\text {supplied }}-\Delta W \)

\( =(22.57-1.67) \times 10^5 J \)

\( =20.9 \times 10^5 J \)

\( =2090 kJ\)