Physics Test-3

The net force acting on a mass that is traveling in a vertical circle is composed of the force of gravity and the tension in the string.

\(\overrightarrow{F_{c}}=\overrightarrow{m g}+\vec{T}\)

\(\frac{\overrightarrow{m v^{2}}}{R}=\overrightarrow{m g}+\vec{T}\)

At the point, when the String losses its Tension and go slang, There should be Enough Gravitational Force to make the body continue its motion in Vertical Plane.

For this the Critical Velocity should be maintained.

\(\frac{m v^{2}}{R}=m g \)

\(\Rightarrow \mathrm{v}^{2}=\mathrm{Rg} \)

\(\Rightarrow \mathrm{v}=\sqrt{(\mathrm{Rg})}\)

\(\mathrm{R}\) is the radius of the Circle in which the stone is moving. Here it is the length of the string.

\(R=1 \mathrm{~m}\)

\(g=9.8 \mathrm{~ms}^{-2} . \)

\(v=\sqrt{\left(1 \times 9.8 \mathrm{~ms}^{-2}\right)}=3.16 \mathrm{~m} / \mathrm{s}\)

So, \(3.16 \mathrm{~m} / \mathrm{s}\) is Critical Velocity.

Given,

\(l=4.5 \mathrm{~m}\)

\(\mathrm{D}=6 \mathrm{~mm}=6 \times 10^{-3} \mathrm{~m}\)

So, radius \(\mathrm{r}=\frac{\mathrm{D}}{2}=3\times 10^{-3} \mathrm{~m}\)

Area \(\mathrm{A}=\pi\mathrm{r^2}\)

\(\Rightarrow A=3.14\times(3\times 10^{-3})^2\)

\(\mathrm{F}=100 \mathrm{~N}\)

\( \mathrm{Y}=4.8 \times 10^{11} \mathrm{~N}\mathrm{~m}^{-2}\)

\( \sigma=0.2\)

As, \(\mathrm{Y}=\frac{\mathrm{F}}{\mathrm{A}} \frac{l}{\Delta l}\)

\(\therefore \Delta l=\frac{\mathrm{F}}{\mathrm{A}} \frac{l}{\mathrm{Y}}\)

\(=\frac{100 \times 4.5}{3.14 \times\left(3 \times 10^{-3}\right)^{2} \times 4.8 \times 10^{11}}\)

\(=3.32 \times 10^{-5} \mathrm{~m}\)

\(=0.332 \mu \mathrm{m}\)

Electromagnetic waves or EM waves: The waves that are formed as a result of vibrations between an electric field and a magnetic field and they are perpendicular to each other and to the direction of the wave is called an electromagnetic wave. The accelerating charged particle produces an electromagnetic (EM) wave. A charged particle oscillating about an equilibrium position is an accelerating charged particle. Electromagnetic waves do not require any matter to propagate from one place to another as it consists of photons. They can move in a vacuum.

Properties of electromagnetic waves:

• Not have any charge or we can say that they are neutral.
• Propagate as a transverse wave.
• They move with the velocity the same as that of light i.e \(3 \times 10^{8} \mathrm{~m} / \mathrm{s}\).
• It contains energy and they also contain momentum.
• They can travel in a vacuum also.

From above it is clear that Electromagnetic waves do not require any matter to propagate from one place to another as it consists of photons. Therefore option (A) is correct.

In an electromagnetic wave, the electric field and magnetic field vary continuously with maxima and minima at the same place and same time. Therefore option (B) is correct.

The energy in an electromagnetic wave is divided equally between electric and magnetic fields. Therefore option (C) is correct.

An electromagnetic wave is a perpendicular variation in both the electric field(E) and Magnetic field(B). Therefore option (D) is incorrect.

Given,

\(KE _{2}=9 KE _{1}\)

We know that the kinetic energy of a body is given as,

\( K E=\frac{1}{2} m v^{2}\)...(1)

The linear momentum of a body is given as,

\( P = mv\)...(2)

By equation (1) and equation (2) the relation between the kinetic energy and the momentum is given as,

\( K E=\frac{P^{2}}{2 m}\)...(3)

By equation (3) for the initial position,

\( K E_{1}=\frac{P_{1}^{2}}{2 m}\)...(4)

By equation (3) for the final position,

\( K E_{2}=\frac{P_{2}^{2}}{2 m} \)

\(\Rightarrow 9 \times K E_{1}=\frac{P_{2}^{2}}{2 m}\)...(5)

By equation (4) and equation (5),

\(\frac{P_{2}^{2}}{2 m} \times \frac{2 m}{P_{1}^{2}}=\frac{9 \times K E_{1}}{K E_{1}}\)

\(\Rightarrow \frac{P_{2}^{2}}{P_{1}^{2}}=9 \)

\(\Rightarrow \frac{P_{2}}{P_{1}}=3 \)

\(\Rightarrow P_{2}=3 P_{1}\)

So, if the kinetic energy of a body is increased nine times then the momentum of the body will be increased by three times.

Given,

\(A=0.7 \mathrm{~m}^{2}\)

\(\mathrm{~N}=100\) turns\)

\(f=60 \mathrm{rev} / \mathrm{minute}=1 \mathrm{rev} / \mathrm{sec}\)

\(\mathrm{B}=0.02 \mathrm{~T}\)

Where \(\mathrm{A}=\) area of the coil, \(\mathrm{N}=\) number of turns, \(f=\) frequency of rotation, \(\omega=2 \pi f=\) angular speed of coil, and B = magnetic field

We know that the maximum emf induced in the AC generator is given as,

\(\Rightarrow \epsilon_{\max }=\text { NBAw } \)

\(\Rightarrow \epsilon_{\max }=100 \times 0.02 \times 0.7 \times 2 \pi \times 1 \)

\(\Rightarrow \epsilon_{\max }=\frac{100 \times 0.02 \times 0.7 \times 2 \times 22}{7} \)

\(\Rightarrow \epsilon_{\max }=8.8 \text { volt }\)

\(\therefore\) The maximum emf generated in the coil is \(8.8 ~\mathrm{volt}\).

Centripetal Force: Centripetal force is that force that is required to move a body in a circular path with uniform speed. The force acts on the body along the radius and towards the centre.

\(F=\frac{m v^{2}}{r^{2}}\)

Where \(F=\) Centripetal force, \(m=\) mass of the body, \(v=\) velocity of the body and \(r=\) radius

In a vertical circular motion, centripetal force remains the same at all points on a circular path and is always directed towards the centre of the circular path.

Escape velocity \(\left(\mathrm{V}_{\mathrm{e}}\right)\): The minimum velocity that an object must attain to escape the gravitational pull of the Earth is called the escape velocity.

It is given as:

\(V_{e}=\sqrt{2} g R\)

From the formula of escape velocity, it is clear that it only depends upon the acceleration due to gravity \((g)\) and the radius of the Earth(R).

The escape velocity of a body does not depend upon its mass.

According to the laws of motion,

the distance covered by a body is given as,\(s=u t+\frac{1}{2} a t^{2}\)

Given:

s = \(450\) m in t = \(5\) second and further s = \(700\) m in t = \(10\)second

where s = distance travelled and t = time taken for both case saperately.

When car covers \(450\) m

\(s=u t+\frac{1}{2} a t^{2}\)

\(450=5 \mathrm{u}+\frac{1}{2} \times \mathrm{a} \times(5)^{2}\) .....(1)

Now,

When the car covers the remaining \(700\) m.

\(700+450=10 u+\frac{1}{2} \times a \times(10)^{2}\) .......(2)

Solving equaiton (1) and (2)

\(\mathrm{u}=65 \mathrm{~m} / \mathrm{sec}\)

\(\therefore \mathrm{a}=10 \mathrm{~m} / \mathrm{sec}^{2}\)

Given,

\(v=\frac{A}{t}+B t^{2}+C t^{3}\)

Where,

v = Velocity

t = Time

A, B and C = Constants

\(\mathrm{As}, {v}=\frac{\mathrm{A}}{\mathrm{t}}+\mathrm{B} \mathrm{t}^{2}+\mathrm{C} \mathrm{t}^{3}\)

So, we can write the dimensional equation as:

\(\operatorname{dim}(\mathrm{v})=\operatorname{dim}\left(\mathrm{B} \mathrm{t}^{2}\right)\)

\(\therefore \operatorname{dim}(\mathrm{B})=\frac{\operatorname{dim}(\mathrm{v})}{\operatorname{dim}\left(\mathrm{t}^{2}\right)}\)

\(=\frac{\left[\mathrm{LT}^{-1}\right]}{\left[\mathrm{T}^{2}\right]}\)

\(=\left[\mathrm{LT}^{-3}\right]\)

\(\Rightarrow \operatorname{dim}(\mathrm{B})=\left[\mathrm{M}^{0} \mathrm{L} \mathrm{T}^{-3}\right]\)

Meter bridge is instruments based on the principle of Wheatstone bridge and is used to measure an unknown resistance.

In the case of a meter bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, the bridge is balanced.

If in the balanced position of bridge AB = l, BC (100 – l) so that

\(\frac{Q}{P}=\frac{(100-l)}{l}\)

Also, \(\frac{P}{Q}=\frac{R}{S} \)

\(\Rightarrow S=\frac{(100-l)}{l} R\)

Given: Balancing length \(\neq 0.25 \mathrm{~m}\)

By using the equation:

\(\frac{P}{Q}=\frac{l}{(100-l)}\)

For balancing length \(=0.25 \mathrm{~m}, \mathrm{I}=25 \mathrm{~cm},(100-1)=(100-25)=75 \mathrm{~cm}\)

\(\frac{P}{Q}=\frac{25}{75}=\frac{1}{3}\)

n options \(1 \Omega, 3 \Omega , \frac{7}{ 3} \Omega, 7 \Omega\) and \(25 \Omega, 75 \Omega\) have a ratio of \(\frac{1}{3}\).

noption \(2 \Omega\) and \(3 \Omega\), the ratio is \(\frac{2}{3}\).

So, for \(2 \Omega\) and \(3 \Omega\) pair has the balancing length is not \(0.25 \mathrm{~m}\) in a meter bridge.

Given:

Charge \(\mathrm{Q}\) at \(\mathrm{O}(0,0)\) is \(2 \mu \mathrm{C}\)

Charge \(\mathrm{Q}_{1}\) at \(\mathrm{A}(1,0)\) is \(1 \mu \mathrm{C}\)

Charge \(\mathrm{Q}_{2}\) at \(\mathrm{B}(-1,0)\) is \(-1 \mu \mathrm{C}\)

The potential energy of the system is due to the interaction between \(Q\) and \(Q_{1}\)

\( U_{1}=V q\)

\(=\frac{Q Q_{1}}{4 \pi \epsilon_{0} R_{O A}}\)

\(=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times 1 \times 10^{-6}}{1}\)

\(=18 \times 10^{-3} \mathrm{~J}\)

The potential energy of the system is due to the interaction between \({Q}\) and \({Q}_{2}\)

\(U_{2}=V q\)

\(=\frac{Q Q_{2}}{4 \pi \epsilon_{0} R_{O B}}\)

\(=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times-1 \times 10^{-6}}{1}\)

\(=-18 \times 10^{-3} \mathrm{~J}\)

The potential energy of the system is due to the interaction between \(Q_{1}\) and \(Q_{2}\)

\(U_{3}=V q=\frac{Q_{1} Q_{2}}{4 \pi \epsilon_{0} R_{A B}}\)

\(=\frac{9 \times 10^{9} \times 1 \times 10^{-6} \times(-1) \times 10^{-6}}{2}\)

\(=-4.5 \times 10^{-3} \mathrm{~J}\)

Given,

\(V_{A}=V_{B}=V=220 \mathrm{~V}\)

\(f_{A}=f_{B}=f\)

\(I_{B}=2 I_{A}\)

We know that the inductive reactance is given as,

\( X_{L}=2 \pi f L \quad \cdots(1)\)

When an AC voltage is applied to an inductor, the current in the circuit is given as,

\( I=\frac{V}{X_{L}} \quad \cdots \text {(2) }\)

By equation \((1)\) and equation \((2)\),

\( I=\frac{V}{2 \pi f L} \quad \cdots(3)\)

By equation \((3)\) , the current in inductor \(A\) is given as,

\( I_{A}=\frac{V}{2 \pi f L_{A}} \quad \cdots(4)\)

By equation 3 , the current in inductor B is given as,

\( I_{B}=\frac{V}{2 \pi f L_{B}} \)

\(\because \mathrm{I}_{\mathrm{B}}=2 \mathrm{I}_{\mathrm{A}}\)

By equation \((4)\), equation \((5)\) , and equation \((6)\) ,

\(\Rightarrow \frac{V}{2 \pi f L_{B}}=2 \times \frac{V}{2 \pi f L_{A}} \)

\(\Rightarrow \frac{L_{A}}{L_{B}}=2\)

\(\therefore\) The ratio of the inductive reactance of the inductor \(A\) to inductor \(B\) is\(2: 1\).

The de-Broglie wavelength of a particle of mass \(\mathrm{m}\) and moving with velocity \(\mathrm{v}\) is given by,

\(\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad(\because \mathrm{p}=\mathrm{mv})\)

de-Broglie wavelength of a proton of mass \(\mathrm{m}_{1}\) and kinetic energy \(\mathrm{k}\) is given by,

\(\lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{k}}} \quad(\because \mathrm{p}=\sqrt{2 \mathrm{mk}})\)

\(\Rightarrow \lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{qV}}} \ldots . .\) (i) \(\quad[\because \mathrm{k}=\mathrm{qV}]\)

For an alpha particle mass \(\mathrm{m}_{2}\) carrying charge \(\mathrm{q}_{0}\) is accelerated through potential \(\mathrm{V}\), then,

\(\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{2} \mathrm{q}_{0} \mathrm{~V}}}\)

\(\because\) For \(\alpha-\) particle \(\left({ }_{2}^{4} \mathrm{He}\right): \)

\(\mathrm{q}_{0}=2 \mathrm{q}\) and \(\mathrm{m}_{2}=4 \mathrm{~m}_{1}\)

\(\therefore \lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m}_{1} \times 2 \mathrm{q} \times \mathrm{V}}}.....\) (ii)

The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get,

\(\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{q} \mathrm{V}}} \times \frac{\sqrt{2 \times \mathrm{m}_{1} \times 4 \times 2 \mathrm{q} \mathrm{V}}}{\mathrm{h}}\)

\(=\frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\)

\(\Rightarrow \frac{\lambda_{1}}{\lambda_{2}}=2 \sqrt{2}\)

Given: f_m = 20 kHz and f_c = 3 MHz = 3000 kHz

Where, f_c = frequency of the carrier wave and f_m = frequency of the modulating signal

We know that that (f_c + f_m) and (f_c - f_m) are the sidebands of a modulated signal.

Therefore the sidebands is given as,

S₁ = (f_c + f_m)

⇒ S₁ = (3000 + 20)

⇒ S₁ = 3020 kHz -----(1)

And

S₁ = (f_c - f_m)

⇒ S₁ = (3000 - 20)

⇒ S₁ = 2980 kHz

Given,

\(n_{1}: n_{2}=1: 3\) amp

\(R_{1}: R_{2}=3: 1\)

\(I_{1}=I_{2}=1\)

The magnetic field produced at the centre of the current-carrying coil is given as,

\(B=\frac{\mu_{o} n I}{2 R}...(1)\)

Where \(R=\) radius of the coil, \(I=\) current, and \(\mu_{0}=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\)

By equation \((1)\) the magnetic field at the center of the first coil is given as,

\(\Rightarrow B_{1}=\frac{\mu_{o} n_{1} I_{1}}{2 R_{1}}\)

\(\Rightarrow B_{1}=\frac{\mu_{o} n_{1} I}{2 R_{1}}\)

By equation \((1)\) the magnetic field at the center of the second coil is given as,

\(\Rightarrow B_{2}=\frac{\mu_{o} n_{2} I_{2}}{2 R_{2}} \)

\(\Rightarrow B_{2}=\frac{\mu_{o} n_{2} I}{2 R_{2}}\)

By equation \((2)\) and equation \((3)\) ,

\(\Rightarrow \frac{B_{1}}{B_{2}}=\frac{\mu_{o} n_{1} I}{2 R_{1}} \times \frac{2 R_{2}}{\mu_{o} n_{2} I} \)

\(\Rightarrow \frac{B_{1}}{B_{2}}=\frac{n_{1}}{n_{2}} \times \frac{R_{2}}{R_{1}} \)

\(\Rightarrow \frac{B_{1}}{B_{2}}=\frac{1}{3} \times \frac{1}{3} \)

\(\Rightarrow \frac{B_{1}}{B_{2}}=\frac{1}{9}\)

\(\therefore\) The ratio of the magnetic field at the centre of the coils is \(1: 9\).