Physics Test - 46

The initial relative velocity of the express train w.r.t the passenger train,

\({u}_{{ep}}={u}_{e}-{u}_{p}\)

\(=30-5\)

\(=25 {~ms}^{-1}\)

The final relative velocity of the express train w.r.t the passenger train,

\({v}_{{ep}}=0\) (because the express train comes to rest relative to the passenger train)

From the first equation of motion,

\(v_{e q}=u_{e p}-a t\)

\(\Rightarrow 0=25-4 t\)

\(\Rightarrow 4 t=25\)

\(\Rightarrow t=6.25 {~s}\)

Given,

\(KE _{1}=50 J\)

\(KE _{2}=150 J\)

\(t =10 s\)

As we know,

The work-energy theorem states that the net work done by the forces on an object is equal to the change in its kinetic energy.

\( W =\Delta KE\)

Where \(W=\) work done and \(\Delta K E=\) change in kinetic energy

By the work-energy theorem,

\( W=\Delta KE \)

\(\therefore W=K E_{2}-K E_{1} \)

\(\Rightarrow W=150-50 \)

\(\Rightarrow W=100 J\)

So, the power of the body \( (P)=\frac{W}{t}\)

Where \(W=\) work and \(t=\) time

\(\therefore P=\frac{100}{10} \)

\(\Rightarrow P=10 W\)

Incident Energy is increased by \(20 \%\).

Initial kinetic energy \((KE_{1})=0.5 \mathrm{eV}\) 

Final kinetic energy \((K E_{2})=0.8 \mathrm{eV}\) 

\(E_{1}=E=\) initial incident energy

According to the question.

\(E_{2}=E_{1}+E_{1} \times 20 \%=E_{1}+0.2 E_{1}=1.2 \mathrm{E}_{1}=1.2 \mathrm{E}\)

Let \(\phi\) is the work function of metal. 

According to Einstein equation:

\( E=\phi+K E \)

\(E_{1} =\phi+K E_{1}=E \)

Put the value of \(K E_{1}\) in above equation,

\(E_{1} =\phi+0.5=E \cdots(i)\)

\(E_{2} =\phi+K E_{2}=1.2 E \)

Put the value of \(K E_{2}\) in above equation,

\( E_{2}=\phi+0.8=1.2 E \cdots(ii)\)

Divide equation (i) by equation (ii),

\(\frac{E_{1}}{E_{2}}= \frac{\phi+0.5}{\phi+0.8}=\frac{E}{1.2E}=\frac{1}{1.2}\)

\(\Rightarrow 1.2 \phi +0.6=\phi+0.8 \)

\(\Rightarrow 0.2 \phi =0.2 \Rightarrow \phi =\frac{0.2}{0.2}=1 \mathrm{eV}\)

Work function of metal is \(\phi=1 \mathrm{ev}\).

From the first law of thermodynamics,

\(d U=q-W\)

In adibatic process no change of heat takes place.

So, \(dU = -W\)

Since, the gas is expanding, the work is done by the gas.

\(dU =- W \text {. }\)

Hence internal energy of the system decreases.

Again, \(dU = C _{ v }\left( T _{2}- T _{1}\right)= W\)

Since here \(W\) is negative, so \(T_{1}>T_{2}\).

So temperature will also decrease.

The situation is shown of figure. After time \(t\) let separation between them is \(x\), then 

\(x^{2}=\left(l_{1}-v_{1} t\right)^{2}+\left(l_{2}-v_{2} t\right)^{2}\)......(i)

\(x\) to be minimum, \(\frac{d x}{d t}=0\) 

Differentiating equation (i) w.r.t time we have 

\(\frac{d x^{2}}{d t}=\frac{d}{d t}\left[\left(l_{1}-v_{1} t\right)^{2}+\left(l_{2}-v_{2} r\right)^{2}\right]\)

\(2 x \frac{d x}{d t}=2\left(l_{1}-v_{1} t\right) \times\left(-v_{1}\right)+2\left(l_{1}-v_{2} t\right) \times\left(-v_{2}\right)\)

or \(0=\left(l_{1}-v_{1} t\right)\left(v_{1}+\left(l_{2}-v_{2} t\right) v_{2}=\left(l_{1} v_{1}-l_{2} v_{2}\right)-t\left(v_{1}^{2}+v_{2}^{2}\right)\right.\)

which gives \(t=\left(\frac{l_{1} v_{1}+l_{2} v_{2}}{v_{1}^{2}+v_{2}^{2}}\right)\)

Now, \(x_{\min }=\sqrt{\left(l_{1}-v_{1} t\right)^{2}+\left(l_{2}-v_{2} t\right)^{2}}\).....(ii)

Substituting value of \(t\) in equation (ii), we get 

\(x_{\min }=\sqrt{\left[l_{1}-v_{1}\left(\frac{l_{1} v_{1}+l_{2} v_{2}}{v_{1}^{2}+v_{2}^{2}}\right)\right]^{2}+\left[l_{1}-v_{2}\left(\frac{l_{1} v_{1}-l_{2} v_{2}}{v_{1}^{2}+v_{2}^{2}}\right)\right]^{2}}\)

After solving, we get \(x_{\min }=\left(\frac{l_{1} v_{2}-l_{2} v_{1}}{v_{1}^{2}+v_{2}^{2}}\right)\)

We know that,

Potential gradient \(x=\frac{V}{L}=\frac{i R}{L}\)

According to the question

\(x=\frac{2}{(15+5)} \times \frac{15}{1000}\)

\(=\frac{3}{2000}\; V/ cm\)

When the new mass is hanged to the wire, the force exerted on the wire is:

\(\mathrm{F}=\mathrm{Mg}\)

The initial length of the wire is \(\mathrm{L}\) and the new length is \(\mathrm{L}_{1}\).

We know that the Young's modulus is given as:

\(\mathrm{Y}=\frac{\mathrm{FL}_{0}}{\mathrm{~A} \Delta \mathrm{l}}\)

Here, \(L_{\circ}\) is the initial length of wire and \(\Delta l\) is the change in length of wire.

Substitute the values:

\(\mathrm{Y}=\frac{\mathrm{Mg} \times \mathrm{L}}{\mathrm{A} \times\left(\mathrm{L}_{1}-\mathrm{L}\right)}\)

So, \(\mathrm{Y}=\frac{\mathrm{MgL}}{\mathrm{A}\left(\mathrm{L}_{1}-\mathrm{L}\right)}\)

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 \(2.5\).

\(\frac{\mathrm{P}}{40}=\frac{\mathrm{Q}}{60} \ldots . .(1) \)

\(\frac{\mathrm{P}+\mathrm{x}}{80}=\frac{\mathrm{Q}}{20} \ldots .(2) \)

\(\frac{\mathrm{P}}{\mathrm{P}+\mathrm{x}} \times \frac{80}{40}=\frac{20}{60} \)

\(\frac{4}{4+\mathrm{x}} \times 2=\frac{1}{3} \)

\(24=4+\mathrm{x} \)

\(\mathrm{x}=20\)

We know the relation

\(I \propto\left(a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \phi\right)\)

For intentisity to be maximum

\(\cos \phi=+1 \)

\(\text { or } \quad \phi=2 n \pi \)

\(\text { where } \quad n=0,1,2, \ldots \ldots . . \)

\(\therefore \text { Path difference } \)\((\Delta)=\frac{\lambda}{2 \pi} \times \text { phase difference }( \phi)\)

\(\Delta=\frac{\lambda}{2 \pi} \times \phi \)

Put all the given values in above equation:

\(\Delta=\frac{\lambda}{2 \pi} \times 2 n \pi\)

\(\Delta=n \lambda\)