Physics Test-1

Given,

Wavelength \((\lambda)=600 ~nm=600 \times 10^{-9} {~m}\)

Slit-width (a) \(=0.2 {~mm}=0.2 \times 10^{-3} {~m}\)

\(\operatorname{Sin} \theta=\frac{n \lambda}{a}\)

For \(\frac{\text{Principal}}{\text{central maxima}}\)

\(n=1\)

\(\operatorname{Sin} \theta=\theta=\frac{\lambda}{a}\)

Angular width \((2 \theta)=\frac{2 N}{\mathrm{a}}\)

\(=\frac{\left(2 \times 600 \times 10^{-9}\right)}{\left(0.2 \times 10^{-3}\right)}=6 \times 10^{-3} \mathrm{rad}\)

Given:

Change in an electric field, \(\frac{d E}{d t}=5 \times 10^{12} {Vm}^{-1} s^{-1}\),

Side of the plate (I) \(=2\) \({cm}=2 \times 10^{-2} {~m}\),

and \(\epsilon_{0}=8.85 \times 10^{-12} C^{2} N^{-1} {~m}^{-2}\)

The area of the plate is:

Area of square\(=(\text{side})^2\)

\( A=2 \times 10^{-2} \times 2 \times 10^{-2}=4 \times 10^{-4} {~m}\)

We know that displacement current is given as:

\( I_{d}=\epsilon_{0} A \times \frac{d E}{d t}\)

\( I_{d}=8.85 \times 10^{-12} \times 4 \times 10^{-4} \times 5 \times 10^{12}\)

\( I_{d}=177 \times 10^{-4} A\)

\(I_{d}=17.7 {~m}\)

Given,

\(\mathrm{B}_{\mathrm{A}}=\mathrm{B}_{\mathrm{B}}=\mathrm{B}\)

\(\mathrm{A}_{\mathrm{A}}=\mathrm{A}_{\mathrm{B}}=\mathrm{A}\)

\(\theta_{\mathrm{A}}=\theta_{\mathrm{B}}\)

\(\frac{N_{A}}{N_{B}}=\frac{1}{2}\)

\(\frac{I_{A}}{I_{B}}=\frac{2}{1}\)

For coil \(\mathrm{A}\):

The torque on the coil \(A\) is given as,

\(\Rightarrow \mathrm{T}_{\mathrm{A}}=\mathrm{N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}} \mathrm{A}_{\mathrm{A}} \mathrm{B}_{\mathrm{A}} \sin \theta_{\mathrm{A}}\)

\(\Rightarrow \mathrm{T}_{\mathrm{A}}=\mathrm{N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}} \mathrm{AB} \sin \theta...(1)\)

For coil \(\mathrm{B}\):

The torque on the coil B is given as,

\(\Rightarrow T_{A}=N_{B} I_{B} A_{B} B_{B} \sin \theta_{B}\)

\(\Rightarrow T_{A}=N_{B} I_{B} A B \sin \theta...(2)\)

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

\(\Rightarrow \frac{\tau_{A}}{\tau_{B}}=\frac{N_{A} I_{A} A B \times \sin \theta}{N_{B} I_{B} A B \times \sin \theta}\)

\(\Rightarrow \frac{\tau_{A}}{\tau_{B}}=\frac{N_{A} I_{A}}{N_{B} I_{B}}\)

\(\Rightarrow \frac{\tau_{A}}{\tau_{B}}=\frac{1}{2} \times \frac{2}{1}\)

\(\Rightarrow \frac{\tau_{A}}{\tau_{B}}=\frac{1}{1}\)

\(\therefore\) The ratio of the torque experienced by the coil \(\mathrm{A}\) to coil \(\mathrm{B}\) will be \(\frac{\tau_{A}}{\tau_{B}}=\frac{1}{1}\).

When a body moves in a straight line under constant acceleration, it follows three equations of motion.

All unknowns are found by these equations.

\(\Rightarrow v=u+a t\)

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

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

Where u = initial velocity, v = final velocity, a = constant acceleration, t = time and s = displacement.

Given that the person throws balls into the air vertically upward with speed 'u'.

The height to which the ball will rise at this initial speed (at the highest point v will be zero)

Using the 3rd equation of motion:

\(v=0 ; u=u ; a=-g ; s=h\) (downward direction)

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

\(\Rightarrow 0^{2}=u^{2}-2 g h\)

\(\Rightarrow {u}^{2}=2 {gh}\)

\(\Rightarrow h=\frac{u^{2}}{2 g}\)

Given,

Initial magnetic flux \(\left(\phi_{1}\right)=5.5 \times 10^{-4} \mathrm{~Wb}\)

Final magnetic flux \(\left(\phi_{2}\right)=5 \times 10^{-5} \mathrm{~Wb}\)

Resiatnace \((\mathrm{R})=\) \(10 \Omega\)

Number of turns \((\mathrm{N})=1000\)

Change in time \((\Delta \mathrm{t})=0.1 \mathrm{sec}\)

Change in flux:

\(\Rightarrow d \phi=\left(\phi_{2}-\phi_{1}\right)=\left(5 \times 10^{-5}-5.5 \times 10^{-4}\right)\)

\(=-5 \times 10^{-4} \mathrm{wb}\)

Induced emf in coils,

\(e=-N \frac{d \phi}{d t}\)

\(=-1000 \frac{\left(-5 \times 10^{-4}\right)}{0.1}\)

\(=5 V\)

Induced current in the coil

\(i=\frac{e}{R}\)

\(=\frac{5}{10}\)

\(=0.5 \mathrm{~A}\)

Electromagnetic waves are produced by an acceleration charge.

As we know that the charged particle produces an electric field and it exerts a force on other charged particles. As the director of the electric field is from positive charge to negative charge, therefore positive charges accelerate in the direction of the field and negative charges accelerate in the direction opposite to the field.

From Hans Christian Oersted's experiment we find that a moving charged particle produces a magnetic field. If a charged particle moves in the magnetic field then they experience a force. The force on these charges is always perpendicular to the direction of their velocity and therefore only changes the direction of the velocity, not the speed.

An accelerating charged particle produces an electromagnetic (EM) wave.

Basically, electromagnetic waves or EM waves are 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.

A charged particle oscillating about an equilibrium position is an accelerating charged particle.

Given,

A rod of length \(l\) and radius \(\mathrm{r}\) is joined to a rod of length \(\frac l2\) and radius \(\frac r2\) of same material.

As we know,

Torque, \(\mathrm{τ}=\mathrm{C} . {\theta}=\frac{\pi \eta \mathrm{r}^{4} {\theta}}{2 \mathrm{~L}}=\mathrm{constant}\)

Here, \(\pi\) and \(\eta\) are constant.

Then, according to the question,

\(\frac{\pi \eta \mathrm{r}^{4}\left(\theta-\theta_{0}\right)}{2 l}=\frac{\pi \eta(\frac{\mathrm{r}}  2)^{4}\left(\theta_{0}-\theta^{\prime}\right)}{2(\frac l 2)}\)

At fixed end, \(\theta' = 0\)

\(\Rightarrow \frac{\left({\theta}-{\theta}_{0}\right)}{2}=\frac{{\theta}_{0}}{\mathrm{1 6}}\)

\(\Rightarrow \theta_{0}=\frac{8 \theta} 9\)

Given,

\(m=5~\mathrm{Kg}\)

\(g=10 \mathrm{~m}/\mathrm{s}^{2}\)

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

From the figure we get,

Horizontal component

\(T_{1} \cos 30^{\circ}=T_{2}\cos 60^{\circ} \)

\(T_{1} \cos 30^{\circ}-T_{2} \cos 60^{\circ}=0 \)

\(T_{1} \times \frac{\sqrt{3}}{2}-T_{2} \times \frac{1}{2}=0\)

\(\frac{\sqrt{3}T_{1}}{2}-\frac{T_{2} }{2}=0\)

\(\sqrt{3}T_{1}=T_{2}\)...(i)

Vertical component

\(T_{1} \sin 30^{\circ}+T_{2} \sin 60^{\circ}=50 N \)

\(\Rightarrow T_{1} \times \frac{1}{2} +T_{2} \times \frac{\sqrt{3}}{2}=50\)

\(\Rightarrow \frac{T_{1}}{2}+\frac{\sqrt{3}T_{2} }{2}=50\)

\(\Rightarrow \frac{T_{1}+\sqrt{3}T_{2}}{2}=50\)

\(\Rightarrow T_{1}+\sqrt{3}T_{2} =100\)...(ii)

By solving equation (i) and (ii), we get

\(T_{1}+\sqrt{3} \times \sqrt{3}T_{1} =100\)

\(T_{1}+ 3T_{1}=100 \)

\(4T_{1}=100 \)

\(\therefore T_{1}=25 N \)

\(T_{2}=\sqrt{3}T_{1}=\sqrt{3} \times 25=25 \sqrt{3} N\)

Given that \(v\) velocity is given to the pendulum when it is at rest.

At this moment it will have kinetic energy.

\(K E=\frac{1}{2}\left(m \times v^{2}\right)\)

Let this is the base point i.e. at this point of pendulum \({h}=0\)

So potential energy \(=mgh=0\)

At the highest point, there will be no speed. So there will be kinetic energy. So there will be only potential energy.

Since there is the only gravitational force acting on the given system which is a conservative force. So total mechanical energy will be conserved.

So initial mechanical energy will be equal to final mechanical energy.

So the kinetic energy at the lowest point will be converted into potential energy at the highest point.

The potential energy at highest point \(=\) kinetic energy at the lowest point

The potential energy at highest point \(=\frac{1}{2} m v^{2}\)

A rectifier is a circuit that converts the AC signal at its input to pulsating DC at its output.

For half-wave rectifier, the output is present is only for one half of the input signal and clipped for the other half.

Positive half wave rectifier clips the negative half of the input signal and only positive part of the input signal is present.

Negative half wave rectifier clips the positive half of the input signal and only negative part of the input signal is present.

Given,

Least count = 0.01 mm

Number of divisions = 50

The pitch of the screw guage is given as:

\(\mathrm{p}=\) least count \(\times\) number of divisions

Substitute the values, we get

\(\mathrm{p}=0.01 \mathrm{~mm} \times 50\)

\(\mathrm{p}=0.5 \mathrm{~mm}\)

Rydberg's formula is given as:

\(\frac{h c} {\lambda}=21.76 \times 10^{-19}\left[\frac{1 }{\left(n_{1}\right)^{2}}-\frac{1 }{\left(n_{2}\right)^{2}}\right]\)

Where,

\(\mathrm{h}=\) Planck's constant \(=6.6 \times 10^{-34} \mathrm{Js}\)

\(\mathrm{c}=\) Speed of light \(=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\)

\(\left(\mathrm{n}_{1}\right.\) and \(\mathrm{n}_{2}\) are integers)

The shortest wavelength present in the Paschen series of the spectral lines is given for values,

\(\mathrm{n}_{1}=3\)

\(\mathrm{n}_{2}=\infty\)

By putting the values, we get

\(\frac{hc }{ \lambda}=21.76 \times 10^{-19}\left[\frac{1}{ (3)^{2}}-\frac{1 }{(\infty)^{2}}\right]\)

\(\therefore \lambda=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8} \times 9} { 21.76 \times 10^{-19}}\)

\(=8.189 \times 10^{-7} \mathrm{~m}\)

\(=818.9 \mathrm{~nm}\)

Let us assume that the threshold frequency of the sphere is \(\lambda_{o} .\) Let the stopping potential is \(V'\) of the surface when the light of wavelength \(\lambda_{3}\) is used. Thus according to Einstein photoelectric equation.

\(\frac{h c}{\lambda_{1}}=\frac{h c}{\lambda_{o}}+e V\).....(i)

\(\frac{h c}{\lambda_{2}}=\frac{h c}{\lambda_{o}}+3 e V\).....(ii)

\(\frac{h c}{\lambda_{3}}=\frac{h c}{\lambda_{o}}+e V^{\prime}\).....(iii)

From equation (i) and (ii)

\(\frac{h c}{\lambda_{2}}=\frac{h c}{\lambda_{0}}+3\left(\frac{h c}{\lambda_{1}}-\frac{h c}{\lambda_{0}}\right)\)

\(\Rightarrow \frac{h c}{\lambda_{2}}=\frac{h c}{\lambda_{0}}+\frac{3 h c}{\lambda_{1}}-\frac{3 h c}{\lambda_{0}}\)

\(\Rightarrow \frac{h c}{\lambda_{2}}-\frac{3 h c}{\lambda_{1}}=-\frac{2 h c}{\lambda_{0}}\)\(\ldots(i v)\)

From equation (iii) and (iv)

\(\frac{h c}{\lambda_{2}}-\frac{3 h c}{\lambda_{1}}=-2\left(\frac{h c}{\lambda_{3}}-e V^{\prime}\right)\)

\(\Rightarrow \frac{h c}{\lambda_{2}}-\frac{3 h c}{\lambda_{1}}=-\frac{2 h c}{\lambda_{3}}+2 e V^{\prime}\)

\(\Rightarrow h c\left(\frac{1}{\lambda_{2}}+\frac{2}{\lambda_{3}}-\frac{3}{\lambda_{1}}\right)=2 e V^{\prime}\)

\(\Rightarrow \frac{h c}{e}\left(\frac{1}{ \lambda_{3}}+\frac{1}{2\lambda_{2}}-\frac{3}{2 \lambda_{1}}\right)=V^{\prime}\)

Kinetic Energy when body is moving with velocity \(10 \mathrm{~cm} / \mathrm{second}\)

Displacement from mean position \(x=4 \mathrm{~cm}\)

So, we can say

\(10=\omega \sqrt{\left(A^{2}-4^{2}\right)}...(1)\)

Also, when displacement is \(5 \mathrm{~cm}\), velocity is \(8 \mathrm{~cm} / \mathrm{second}\)

\(8=\omega \sqrt{\left(A^{2}-5^{2}\right)}...(2)\)

Dividing \((1)\) and \((2)\)

\(\frac{5}{4}=\frac{\omega \sqrt{A^{2}-4^{2}}}{\omega \sqrt{A^{2}-5^{2}}}\)

\(\Rightarrow \frac{5}{4}=\frac{\sqrt{A^{2}-4^{2}}}{\sqrt{A^{2}-5^{2}}}\)

\(\Rightarrow \frac{25}{16}=\frac{A^{2}-4^{2}}{A^{2}-5^{2}}\)

\(\Rightarrow 25 \mathrm{~A}^{2}-25 \times 5^{2}=16 \mathrm{~A}^{2}-16 \times 4^{2}\)

\(\Rightarrow 9 \mathrm{~A}^{2}=625-256=36\)

\(\Rightarrow \mathrm{A}^{2}=41...(3)\)

Putting \((3)\) in \((1)\) we get,

\(10=\omega \sqrt{\left(41-4^{2}\right)}\)

\(\Rightarrow 10=\omega \sqrt{(41-16)}\)

\(\Rightarrow 10=\omega \sqrt{(25)}\)

\(\Rightarrow 10=\omega 5\)

\(\Rightarrow \omega=0.5\)

\(\Rightarrow\frac{ 2 \pi}{T}=0.5\)

\(\Rightarrow T=2 \pi \times 0.5=\pi\)

\(\therefore\) The time period of this simple harmonic motion is \(\pi\).