Physics Test - 8

Whenever there is a change in magnetic flux in the closed-circuit then an emf is induced in the circuit and if resistance is there in the circuit then current will flow in the circuit. This current is called an induced current.

The device which stores magnetic energy in a magnetic field is called an inductor.

Inductive reactance \(\left(\mathrm{X}_{\mathrm{L}}\right)=w \mathrm{~L}\)

Where \(L=\) inductance of the coil and \(\omega=\) angular frequency of the AC circuit.

Current \(({I})=\frac{V}{{X}_{{L}}}\)

Where \({V}\) is potential

When an iron rod is inserted into the interior of the inductor then inductance \((L)\) of the coil increases. Here due to the iron rod, the inductance \(\left({X}_{{L}}\right)\) of the coil increases.

Current (I) \(=\frac{{V}}{{X}_{{L}}}\)

If the inductance increase then the current will decrease and the bulb will glow less brightly.

Given,

\(\frac{v_{1}}{v_{2}}=4\)

We know that when a wave passes from one medium to another medium, then its speed changes.

If \(v_{1}\) and \(v_{2}\) represent the speed of the wave in medium 1 and medium 2, respectively, and the wave is moving from medium 1 to medium 2. Then by Snell's law,

\(\mu_{21}=\frac{v_{1}}{v_{2}} \quad \text {..(1) }\)

\(\mu_{12}=\frac{v_{2}}{v_{1}} \quad \ldots(2)\)

Where \(\mu_{12}=\) refractive index of medium 1 with respect to medium 2 and \(\mu_{21}=\) refractive index of medium 2 with respect to medium 1

By equation 2, the refractive index of medium 1 with respect to medium 2 is given as,

\(\mu_{12}=\frac{1}{4}\)

\(\mu_{12}=0.25\)

The wavelength that is associated with an object in relation to its momentum and mass is known as de Broglie wavelength. A particle's de Broglie wavelength is usually inversely proportional to its force.

Momentum of a photon is given by

\(P=\frac{E}{c}=\frac{h}{\lambda}\) 

where \(E\) is the energy of the photon, \(c\) is the speed of light in vacuum, \(h\) is the Planck's constant and \(\lambda\) is the de Broglie wavelength.

According to de Broglie

\(P=\frac{h}{\lambda}\) or \(P ~\alpha \frac{1}{\lambda}\).

By this relation we can conclude that the linear momentum of a photon is inversely proportional to the de Broglie wavelength. 

The graph of \(P\) vs \(\lambda\) shall be a rectangular hyperbola.

Given,

\(m_{1}=2 {~kg}\)

\({~m}_{2}=4 {~kg}\)

\({h}=60\) feet

Initial \({PE}_{1}={m}_{1} {~g} {~h}=2 {~g} \times 60=120 {~g}\)

Initial \({PE}_{2}=m_{2} g {~h}\)

\(=4 {~g} \times 60\)

\(=240 {~g}\)

Below \(30\) feet,

Final \({PE}_{1}={m}_{1} {~g} {~h}^{\prime}\)

\(=2 {~g} \times 30\)

\(=60 {~g}\)

Final \({PE}_{2}={m}_{2} {~g} {~h}^{\prime}=4 {~g} \times 30=120 {~g}\)

So \({KE}_{1}=\) Initial \({PE}_{1}\) - Final \({PE}_{2}\)

\(=120 {~g}-60 {~g}\)

\(=60 {~g}\)

\({KE}_{2}=\) Initial \({PE}_{2}-\) Final \({PE}_{2}\)

\(=240 {~g}-120 {~g}\)

\(=120 {~g}\)

The ratio of \({KE}=\frac{{KE}_{1}}{ {KE}_{2}}=\frac{60 {~g}}{ 120 {~g}}=1: 2\)

In a step up transformer, the output voltage from the secondary transformer is increased in the ratio of number of coils in the primary and secondary.

Thus, \(\frac{V_{\text {secondary }}}{V_{\text {primary }}}=\frac{4}{1}\)

Since the power \(V I\) should remain the same,

\(V_{\text {primary }} I_{\text {primary }}=V_{\text {secondary }} I_{\text {secondary }}\)

\(\frac{I_{\text {secondary }}}{I_{\text {primary }}}=\frac{V_{\text {primary }}}{V_{\text {secondary }}}\)

\(=\frac{1}{4}\)

\(I_{\text {secondary }}=1 \mathrm{~A}\)

As we know that angular momentum of a satellite is given by \(\mathrm{L}=\mathrm{mvr}\) (where \(\mathrm{r}\) is the radius of orbital path of the satellite around earth and \(\vee\) is the orbital velocity of satellite)

\(\mathrm{L}=\mathrm{m} \sqrt{\frac{G M}{R+h}}(R+h) \)

\(\mathrm{L}=\mathrm{m} \sqrt{G M(R+h)}\)

Angular momentum is dependent on \(M, m\) and \(h\).

First, draw the free body diagram of the given figure and resolve it into a horizontal and vertical component.

As we know that the value of the horizontal component is always zero irrespective of the inclination of the body.

So, the Horizontal component will be,

\(T_{1} \cos 30^{\circ}=100 \cos 60^{\circ} \)

\(\therefore T_{1}=\frac{100 }{ \sqrt{3}} N\)

Vertical Component will be,

\(T_{1} \sin 30^{\circ}+100 \sin 60^{\circ}= W\)

By putting the value of \(T_{1}\) from above we get

\(\frac{100}{\sqrt{3}} \times \frac{1}{2}+100 \times \frac{\sqrt{3}}{2}= W \)

\( \therefore W =\frac{200 }{ \sqrt{3}} N\)

Let the mass of the rope be \(M\). If a small part of length \(x\) hangs through the hole, its weight

\(=\frac{M x}{L}(g)\)

and its acceleration is \(a=\frac{M x}{L} \frac{g}{M}=\frac{x g}{L}\)

\(\therefore\) Force on the remaining part of length \((L-x)=\) mass of part of length \((L-x) \times\) acceleration \((a)\)

\(=\frac{M}{L}(L-x) \times \frac{x g}{L}=\frac{M g}{L^{2}}\left(L x-x^{2}\right)\)

If the rope falls through a distance \(d x,\) the work done by gravity is

\(d W=\frac{M g}{L^{2}}\left(L x-x^{2}\right) d x\)

\(\therefore\) Total work done is

\(W =\int_{0}^{L} d W=\frac{M g}{L^{2}} \int_{0}^{L}\left(L x-x^{2}\right) d x\)

\(=\frac{M g}{L^{2}}\left|\frac{L x^{2}}{2}-\frac{x^{3}}{3}\right|_{0}=\frac{M g L}{6}\)

Since the centre of mass of the rope falls through a distance \(\frac{L}{2},\) decrease in \(\mathrm{PE}=\frac{M g L}{2}\).

From the principle of conservation of energy, we have

Decrease in PE - Work done (W) \(=\) Increase in KE

or

\(\frac{M g L}{2}-\frac{M g L}{6}=\frac{1}{2} M v^{2}\)

which gives \(v=\sqrt{\frac{2 g L}{3}}\)

Binding Energy can be defined as the minimum energy required to be supplied to it order to free the satellite from the gravitational influence of the planet. (in order to take the satellite from the orbit to as point if infinity.)

Consider a satellite revolving around a planet with speed V at a distance r.

Kinetic Energy \(=\frac{1}{2} \mathrm{~m}_{\mathrm{s}} \mathrm{V}^{2} \)

Potential Energy \(=\frac{\mathrm{GM}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{\mathrm{r}} \)

\(\mathrm{E}=\frac{\mathrm{GM}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{\mathrm{r}^{2}}=\frac{\mathrm{msV}^{2}}{\mathrm{r}} \)

\(\Rightarrow \mathrm{m}_{3} \mathrm{~V}^{2}=\frac{4 \mathrm{M}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{\mathrm{r}} \)

Kinetic Energy \(=\frac{1}{2} \frac{\mathrm{Gm}_{\mathrm{s}} \mathrm{M}_{\mathrm{p}}}{\mathrm{r}}\)

Total Energy\(=\mathrm{K} \cdot \mathrm{E}+\mathrm{P} \cdot \mathrm{E}=\frac{1}{2} \frac{\mathrm{GM}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{\mathrm{r}}-\frac{\mathrm{GM}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{\mathrm{r}}\)

Total Energy \(=\frac{-\mathrm{GM}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{2 \mathrm{r}} \)

Binding Energy \(+\) Total Energy = 0 (at infinite TE= 0)

Binding Energy \(=\frac{\mathrm{GP}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{2 \mathrm{r}}=0\)

Binding Energy \(=\frac{\mathrm{GM}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{2 \mathrm{r}}\)

So Binding energy of satellite \(\mathrm{G}\) of man \(\mathrm{m}_{\mathrm{s}}\) revolving around a planet of \(\operatorname{man} \mathrm{M}_{\mathrm{p}}\) in a radius \(\mathrm{r}\) \(\text { is }=\frac{\mathrm{GM}_{\mathrm{p}} \mathrm{m}_{\mathrm{s}}}{2 \mathrm{r}}\)

The formula for determining the temperature on an unknown scale is given by:

\(\frac{\text { (UFP on unknown scale) }-\text { (LFP on unknown scale) }}{\text { (UFP on known scale) }-(\text { LFP on known scale })}=\frac{t-t \text { on known scale }}{t \text { on known scale }}\quad\quad\).....(i)

Here, UFP is the upper fixed point, LFP is the lower fixed point and \(t\) is the temperature in the unknown scale corresponding to the LFP of the known scale.

The new scale is called \(W\) scale. We have given that on the new scale which is linear, the freezing and boiling points of the water are \(39^{\circ} \mathrm{W}\) and \(239^{\circ} \mathrm{W}\) respectively.

LFP on W scale \(=39^{\circ} \mathrm{W}\)

UFP on \(\mathrm{W}\) scale \(=239^{\circ} \mathrm{W}\)

We know that the freezing point and boiling point of water on the Celsius scale are and \(100^{\circ} \mathrm{C}\) respectively.

LFP on Celsius scale \(=0^{\circ} \mathrm{C}\)

UFP on Celsius scale \(=100^{\circ} \mathrm{C}\)

We have asked to calculate the temperature corresponding to \(39^{\circ} \mathrm{C}\) on the W scale. We can calculate the temperature corresponding to \(39^{\circ} \mathrm{C}\) on \(\mathrm{W}\) scale using equation (i).

Rewrite equation (i) for known Celsius scale and unknown W scale.

\(\frac{(\text { UFP on } \mathrm{W} \text { scale })-(\text { LFP on } \mathrm{W} \text { scale })}{(\mathrm{UFP}-\mathrm{LFP}) \text { on Celsius scale }}=\frac{t-t \text { on Celsius scale }}{t \text { on Celsius scale }}\)

Substitute \(239^{\circ} \mathrm{W}\) for UFP on \(\mathrm{W}\) scale, \(39^{\circ} \mathrm{W}\) for LFP on \(\mathrm{W}\) scale, \(100^{\circ} \mathrm{C}\) for UFP on Celsius scale, \(0^{\circ} \mathrm{C}\) for LFP on Celsius scale and \(39^{\circ} \mathrm{C}\) for t on Celsius scale in the above equation.

\(\frac{\left(239^{\circ} \mathrm{W}\right)-\left(39^{\circ} \mathrm{W}\right)}{\left(100^{\circ} \mathrm{C}\right)-\left(0^{\circ} \mathrm{C}\right)}=\frac{t-39^{\circ} \mathrm{C}}{39^{\circ} \mathrm{C}}\)

\(\Rightarrow \frac{\left(200^{\circ} \mathrm{W}\right) \times\left(39^{\circ} \mathrm{C}\right)}{100^{\circ} \mathrm{C}}=t-39^{\circ} \mathrm{C}\)

\(\Rightarrow 78=t-39\)

\(\Rightarrow t=78+39\)

\(\therefore t=117^{\circ} \mathrm{W}\)

Therefore, the required temperature on the \(\mathrm{W}\) scale is \(117^{\circ} \mathrm{W}\).