Physics Test - 19

Given,

\(m=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}\)

Initial speed \(m=1 \mathrm{~ms}^{-1}\)

Thickness, \(s=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}\)

Resistance offered by the wall,\(F=-2.5 \times 10^{-2} \mathrm{~N}\)

So, deacceleration of bullet,

\(F=ma\)

\(a=\frac{F}{m}\)

\(=\frac{-2.5 \times 10^{-2}}{20 \times 10^{-3}}\)

\(=-\frac{5}{4} \mathrm{~ms}^{-2}\)

Now, using the equation of motion,

\(v^{2}=u^{2}+2 a s\)

\(v^{2}=1+2\left(-\frac{5}{4}\right)\left(20 \times 10^{-2}\right)\)

\(v^{2}=\frac{1}{2}\)

\(v=\frac{1}{\sqrt{2}}=0.7 \mathrm{~ms}^{-1}\)

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

\(\begin{aligned}& =1 \times 10^5 \mathrm{~Pa} \times(1.671-0.001) \mathrm{m}^3 \\& =1.670 \times 10^5 \mathrm{~J}\end{aligned}\)

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

\(\begin{aligned}& \Delta Q_{\text {supplied }}=2257 \times 1 \times 10^3 \mathrm{~J} \\& =22.57 \times 10^5 \mathrm{~J}\end{aligned}\)

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

\(\begin{aligned}& \Delta U=\Delta Q_{\text {supplied }}-\Delta W \\& =(22.57-1.67) \times 10^5 J \\& =20.9 \times 10^5 J \\& =2090 \mathrm{~kJ}\end{aligned}\)

Initial flux through the coil,

\(\Phi_{\mathrm{B} \text { (initial) }}=B A \cos \theta\)

\(=3.0 \times 10^{-5} \times\left(\pi \times 10^{-2}\right) \times \cos 0^{\circ}\)

\(=3 \pi \times 10^{-7} \mathrm{~Wb}\)

Final flux after the rotation,

\(\Phi_{\mathrm{B}(\text { final) }}=3.0 \times 10^{-5} \times\left(\pi \times 10^{-2}\right) \times \cos 180^{\circ}\)

\(=-3 \pi \times 10^{-7} \mathrm{~Wb}\)

Therefore, estimated value of the induced emf is,

\(\varepsilon=N \frac{\Delta \Phi}{\Delta t}\)

\(=500 \times\left(6 \pi \times 10^{-7}\right) / 0.25\)

\(=3.8 \times 10^{-3} \mathrm{~V}\)

\(I=\frac{\varepsilon}{R}\)

\(=1.9 \times 10^{-3} \mathrm{~A}\)

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

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

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

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

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

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

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

Given that,

Mass of Nitrogen, \(m=2.0 \times 10^{-2} kg =20 g\).

Rise in temperature, \(\Delta T=45^{\circ} C\).

Molecular mass of \(N_{2}, M=28\)

Universal gas constant, \(R=8.3 J \operatorname{mol}^{-1} K^{-1}\)

Number of moles, \(n=\frac{m}{M}\)

\(n=\frac{2 \times 10^{-2} \times 10^{3}}{28}\)

\( n=0.714\)

Now, molar specific heat at constant pressure for nitrogen,

\(C_{p}=\frac{7}{2} R \)

\(C_{p}=\frac{7}{2} \times 8.3 \)

\( C_{p}=29.05 J mol ^{-1} K ^{-1}\)

The total amount of heat to be supplied is given by the relation:

\(\Delta Q=n C_{p} \Delta T \)

\(\Delta Q=0.714 \times 29.05 \times 45 \)

\( \Delta Q=933.38 J\)

Clearly, the amount of heat to be supplied is \(933.38 J\).

Steeper curve (B) is adiabatic

Adiabatic \(\Rightarrow \mathrm{PV}^v=\) const.

Or \(\mathrm{P}\left(\frac{\mathrm{T}}{\mathrm{P}}\right)^{\mathrm{v}}=\) const.

\(\frac{\mathrm{T}^{\mathrm{v}}}{\mathrm{p}^{\mathrm{v}}-1}=\) const.

Curve \((A)\) is isothermal

\(\mathrm{T}=\) const.

\(\mathrm{PV}=\) const

We know that the energy carried by the photon is given by the following equation:-

\(E=p c \Rightarrow p=\frac{E}{c}\cdots(i)\)

Where \(p\) denotes momentum, \(E\) is energy and \(c\) is speed of light.

One important fact is that the momentum incidence on the surface is equal to the momentum reflected from the surface. But reflected momentum is in the opposite direction.

Therefore, change in momentum, \(\Delta p=p_{i}-p_{r}\)

Where \(p_{i}\) and \(p_{r}\) are incident and reflected momentum respectively.

\(p_{i}=\frac{E}{c} \cdots(ii)\)

Since, \(p_{i}=p_{r}\cdots(iii)\)

So, \(p_{r}=\frac{-E}{c}\)

\(\left(p_{r}\right.\) is in opposite direction of \(p_{i}\) )

But change in momentum is given as follows:-

\(\Delta p=p_{i}-p_{r}\cdots(iv)\)

Putting value of \((i i)\) and \((i i i)\) in \((i v)\), we have

\(\Rightarrow \Delta p=\frac{E}{c}-\frac{-E}{c} \)

\(\Rightarrow \Delta p=\frac{E}{c}+\frac{E}{c}=\frac{2 E}{c}\)

\(\begin{aligned} & \text { Particle velocity }=\frac{\partial x}{\partial t} \\ & \Rightarrow \text { Maximum particle velocity }=(2 \pi n)(10) \\ & \Rightarrow(2 \pi n)(10)=(n \lambda)(4) \\ & \Rightarrow \lambda=5 \pi\end{aligned}\)

The potential difference across \(PQ\)

i.e., potential difference across the resistance of \(20 \Omega\) which is \({V}={i} \times 2 {0}\)

\(i=\frac{48}{100+100+80+20}\)

\(=0.16 \mathrm{~A}\)

\(\mathrm{V}=0.16 \times 20\)

\(=3.2 \mathrm{~V}\)

Alternating current cannot be measured by DC ammeter because average value of complete cycle is zero.

The average of all the instantaneous values of alternating currents over one complete cycle is called the mean value.The mean value of the alternating current is zero.Alternating current keeps changing its polarity periodically. In positive half cycle, the alternating current has positive polarity whereas in negative half cycle, the alternating current has negative polarity with the same amount. Therefore, the average value of the alternating current for one complete cycle is zero.