Physics Test - 47

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

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

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

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

\( =2090 kJ\)

The equation of magnetic field vector of an electromagnetic field is

\(B=\frac{B_0}{\sqrt{2}}(\hat{i}+\hat{j}) \cos (k z-\omega t) T\)

Electric charges

\(\mathrm{q}_1\) at \(\mathrm{P}\left(0,0, \frac{\pi}{\mathrm{k}}\right)=4 \pi \mathrm{C}\)

\(\mathrm{q}_2\) at \(\mathrm{Q}\left(0,0, \frac{3 \pi}{\mathrm{k}}\right)=2 \pi \mathrm{C}\)

At \(\mathrm{t}=0\) and \(\mathrm{P}\left(0,0, \frac{\pi}{\mathrm{k}}\right)\), where \(\mathrm{z}=\frac{\pi}{\mathrm{k}}\)

Magnetic field vector,

\(B  =\frac{B_0}{\sqrt{2}}(\hat{i}+\hat{j}) \cos \left(k \frac{\pi}{k}-0\right) T \)

\(=\frac{B_0}{\sqrt{2}}(\hat{i}+\hat{j}) \cos \pi T=-\frac{B_0}{\sqrt{2}}(\hat{i}+\hat{j}) T\)

and at \(\mathrm{t}=0\) and \(\mathrm{Q}\left(0,0, \frac{3 \pi}{\mathrm{k}}\right)\), where \(\mathrm{z}=\frac{3 \pi}{\mathrm{k}}\). Magnetic field vector,

\(B  =\frac{B_0}{\sqrt{2}}(\hat{i}+\hat{j}) \cos \left(k \frac{3 \pi}{k}-0\right) \)

\(=-\frac{B_0}{\sqrt{2}}(\hat{i}+\hat{j}) T[\because \text { Using } \cos \pi=\cos 3 \pi=-1]\)

Velocity of charges \(q_1\) and \(q_2, v=0.5 \hat{c i}\)

We know that, force on charge in magnetic field is given by

\(\mathrm{F}=\mathrm{q}(\mathrm{v} \times \mathrm{B})\)

\(\therefore\) For \(\mathrm{q}_1, \mathrm{~F}_1=4 \pi\left[0.5 \mathrm{c} \hat{\mathrm{i}} \times\left\{-\frac{\mathrm{B}_0}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}})\right\}\right] \mathrm{N}\)

\(\mathrm{F}_1=-\frac{2 \pi \mathrm{cB}}{\sqrt{2}} \hat{\mathrm{k}} \mathrm{N}\)

Similarly, for \(q_2\)

\(\Rightarrow \mathrm{F}_2=2 \pi\left[0.5 \mathrm{c} \hat{\mathrm{i}} \times\left\{-\frac{\mathrm{B}_0}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}})\right\}\right] \mathrm{N} \)

\(\Rightarrow \mathrm{F}_2=-\frac{\pi \mathrm{c} \mathrm{B}_0}{\sqrt{2}} \hat{\mathrm{kN}}\)

Hence, ratio of magnitudes of \(F_1\) to \(F_2\)

\(\frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{\frac{2 \pi \mathrm{cB}}{0}}{\frac{\pi \sqrt{2}}{\pi \mathrm{cB}_0}}=\frac{2}{1} \)

\(\Rightarrow \mathrm{F}_1: \mathrm{F}_2=2: 1\)

Given: A solenoid of inductance \(L\) and resistance \(R\) is connected to a battery. The time is taken for the magnetic energy to reach \(\frac 14\) of its maximum value is

Initial magnetic energy \(=\frac{1}{2} L i ^{2}\)

Final magnetic energy \(=\frac{1}{4}\) of initial magnetic energy

\(=\frac{1}{4}\left(\frac{1}{2} L i ^{2}\right)\)

For final energy to become \(\frac{1}{4}\) of initial magnetic energy final current have to become \(\frac 12\) of initial current.

We have a relation \(i = i _{0}\left(1- e ^{\frac{- Rt }{ L }}\right)\)

\(t =\) time, \(L=\) inductance, \(R =\) resistance

Substituting the value of current here we get

\(\frac{i_{o}}{2}=i_{0}\left(1-e^{\frac{-R t}{L}}\right)\)

\(\Rightarrow\frac{1}{2}=\left(1-e^{\frac{-R t}{L}}\right)\)

\(\Rightarrow\frac{-1}{2}=\left(- e ^{\frac{- Rt }{ L }}\right)\) (taking \(\ln\) both sides)

\(\Rightarrow\ln \left(\frac{1}{2}\right)=\frac{- Rt }{ L }\)

\(\Rightarrow t =\frac{ L }{ R } \ln ( 2 )\)

Adding two equations gives,

\(3_{1}^{2} H \longrightarrow { }_{2}^{4} He + n + p\)

Mass-Energy conversion

\((3 \times 2.014 - 4.001 -2 \times 1.008) 3 \times 10^{8^{2}}\)

Therefore, energy released by \(3\) deuterons is

\(3.73 \times 10^{-12} J\) \(1 u =1.66 \times 10^{-27}~ kg\)

Given, average power of a star is \(10^{16} W\)

Therefore, number of deuterons lost per second is

\(N \times\frac{3.73 \times 10^{-12}}{3}=10^{16}\)

\(\Rightarrow N =\frac{1}{1.24} \times 10^{28}\)

Deuteron supply will be exhausted in

\(\frac{10^{40}}{\frac{1}{1.24} \times 10^{28}}=1.24 \times 10^{12} ~s\)

Let \(\mathrm{d}(\mathrm{w})\) and \(\mathrm{d}(\mathrm{o})\) be densities of water and oil resp, then the pressure at the bottom of the tank will be

\(=\mathrm{h}(\mathrm{w}) \times \mathrm{d}(\mathrm{w}) \times \mathrm{g}+\mathrm{h}(\mathrm{o}) \times \mathrm{d}(\mathrm{o}) \times \mathrm{g}\)

Let this pressure be equivalent to pressure due to water of height h

\(\mathrm{h\times d}(\mathrm{w}) \times \mathrm{g}=\mathrm{h}(\mathrm{w}) \times \mathrm{d}(\mathrm{w}) \times \mathrm{g}+\mathrm{h}(\mathrm{o}) \times \mathrm{d}(\mathrm{o}) \times \mathrm{g}\)

\(h=h(w)+[h(o) \times \frac{d(0)} {d(w)}]\)

\(\mathrm{h}=100+[\frac{400(0.9)} {1}]\)

\(\mathrm{h}=100+360=460 \mathrm{~cm}\)

According to Toricelli's theorem

\(v=\sqrt{2 g h} v=\sqrt{2 \times 980 \times 460} v=\sqrt{920 \times 980} v=\sqrt{901600}\)

velocity of efflux \(=949.53 \frac{{~cm}}{{s}}\)

Given that a clock keeps correct time at \(25^{\circ} C\). So initial temperature \(\theta_{1}=25^{\circ} C\).

The pendulum is made of metal. We need to find the number of seconds gained per day by the clock when the temperature falls to \(0^{\circ} \mathrm{C}\).

So final temperature \(\theta_{2}=0^{\circ} \mathrm{C}\)

So, \(\theta_{1}-\theta_{2}=25-0=25^{\circ} \mathrm{C}\)

The relation connecting gain or loss in time and the change in temperature is given as:

\(\Delta t=\frac{1}{2} \times \alpha \times \Delta \theta \times t\).....(1)

Where \(\Delta t\) is the gain or loss in time.

\(\alpha\) is the coefficient of linear expansion.

\(\Delta \theta\) is the change in temperature and \(\mathrm{t}\) is the time.

\(\alpha\) is given as \(1.9 \times 10^{-5}\) per \({ }^{\circ} \mathrm{C}\)

We need to calculate the gain of time per day.

Thus \(t=24 \times 60 \times 60 \mathrm{~s}\)

Change in temperature is \(\Delta \theta=25^{\circ} \mathrm{C}\)

Now substitute all the given values in the equation (1).

\(\Delta t=\frac{1}{2} \times 1 \cdot 9 \times 10^{-5} \times 25 \times 24 \times 60 \times 60 \)

\(\therefore \Delta t=20.520\) seconds / day

• Kinetic energy:Since the body is moving in a linear direction, so it will have kinetic energy due to its velocity.
• Rotational energy:Along with linear velocity, the body is also revolving i.e. rotating, so it will have rotation energy.
• Potential energy:Since the body is at a certain height, it will have potential energy.
• The body will haveall threeKinetic energy, Rotational energy, and potential energy.

In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwise.

\(W_{A E D A}=+\text { Area of } \triangle A E D=+\frac{1}{2} P_{0} V_{0} \)

\(W_{B C E B}=-\text { Area of } \triangle B C E=-\frac{1}{2} P_{0} V_{0}\)

The net work done by the system is:

\(W_{n e t}=W_{A E D A}+W_{B C E B} \)

\(=+\frac{1}{2} P_{0} V_{0}-\frac{1}{2} P_{0} V_{0}=0\)

Given,

\(\theta=30^{\circ}\)

External magnetic field, \(B=0.25 \mathrm{~T}\)

Torque \(\tau=4.5 \times 10^{-2} \mathrm{~J}\)

The torque experienced by a bar magnet placed at an angle θ with the external magnetic field \(B\) is given by,

\(\tau=\mathrm{MB} \sin \theta\)

\(\Rightarrow 4.5 \times 10^{-2}=\mathrm{M} \times 0.25 \sin 30^{\circ}\)

\(\Rightarrow 4.5 \times 10^{-2}=\mathrm{M} \times 0.25 \times \frac 12\)

\(\Rightarrow \mathrm{M}=0.36 \mathrm{JT}^{-1}\)