Physics Test-9

$x=A \cos t$ 

$PE=\frac{k{x}^2}{2}$ 

$PE=\frac{k{A}^2{\left(\cos t\right)}^2}{2}=\frac{k{A}^2\left(1+\cos 2t\right)}{2}$ 

At $t=0, PE$ is maximum and at $t=\mathrm{\pi } \mathrm{PE}$ is minimum.

The Graph I is for $PE$ and when $x=0 PE=0$ 

and when $x=$ maximum or $x=A, PE=\frac{k{A}^2}{2}$

therefore, Graph III is for $PE$

Given,

\(r_{1}=(2 \hat{i}-3 \hat{j}-4 k) \mathrm{m}\)

\(r_{2}=(3 \hat{i}-4 \hat{j}+5 \hat{k}) \mathrm{m}\)

Net displacement, \(r=r_{2}-r_{1}=\hat{i}-\hat{j}+9 \hat{k}\)

\(\mathrm{F}=4 \hat{\mathrm{i}}+\hat{\mathrm{j}}+6 \hat{\mathrm{k}}\)

By using vector dot product.

Work done, \(\mathrm{W}=F \cdot r\)

\(\Rightarrow \mathrm{W}=(4 \hat{\mathrm{i}}+\hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}-\hat{\mathrm{j}}+9 \hat{\mathrm{k}})\)

\(\Rightarrow \mathrm{W}=4-1+54\)

\(\Rightarrow \mathrm{W}=57 \mathrm{~J}\)

According to Kirchhoff's law, the ratio of emissive power to absorptive power is the same for all surfaces at the same temperature and is equal to the emissive power of a perfectly black body at that temperature.

\(\frac{e_{1}}{a_{1}}=\frac{e_{2}}{a_{2}}=\ldots\left[\frac{E}{A}\right]_{\text {perfectly black body }}\)

Thus, since \((E \lambda)_{\text {black }}\) is constant at a given temperature, this implies that good absorber is a good emitter (or radiator).

In the question, we are given two smooth spheres that are identical. The first sphere A moves with angular speed \(\omega\) and centre of mass velocity \(v\). The second sphere \(B\) is at rest and head-on, an elastic collision takes place there. As the given spheres are smooth, angular momentum will not be transferred between the spheres. Since this is a head on collision, sphere A transfers its velocity completely to sphere B. Further, sphere B begins to move whereas sphere A occupies the position of \(B\) and remains at rest. Hence the friction is zero, torque about their centre of mass is also having zero value. Thus the angular velocity doesn't change. Therefore, we can say that \(\omega_{A}=\omega, \omega_{B}=0\)

Given,

Acceleration, \(a=6 \mathrm{~ms}^{-2}\)

By the Newton's second law,

F = ma

Force applied by engine \(=6 m\)

When two cars are pulle,

\((m+m) a=6 m\)

\(\Rightarrow 2 m a=6 m\)

\(\Rightarrow a=3 \mathrm{~ms}^{-2}\)

\(\Rightarrow 2 m a=6 m\)

\(\Rightarrow a=3 \mathrm{~m} / \mathrm{s}^{2}\)

As we know,

Change in velocity \(=\)Area under the acceleration-time graph

Area under the given graph \(=\frac{1}{2} \times 10 \times 11\)

\(=55 \mathrm{~ms}^{-1}\) = Change in velocity

According to the question, initial velocity is zero, the final velocity is \(55 \mathrm{~ms}^{-1}\).

Given,

\(\mathrm{G}=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\) (Universal gravitational constant)

\(\mathrm{m}_{1}=\mathrm{m}_{2}=1 \mathrm{~kg}\)

Distance between two spheres, \(r=3 \mathrm{~m}\)

The force of attraction between two spheres,

\(\mathbf{F}^{\prime}=\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\)

\(\Rightarrow \mathrm{F}=\frac{6.67 \times 10^{-11} \times 1 \times 1}{3^{2}}\)

\(\Rightarrow \mathrm{F}=7.4 \times 10^{-12} \mathrm{~N}\)

From Maxwell's Electromagnetic theory, the Electromagnetic Wave propagation contains electric and magnetic fields vibrating perpendicularly to each other. Thus, changing of electric field gives rise to magnetic field.

Then the Maxwell's equation,

\(\nabla \times \mathrm{B}=\mu_{0}\left(\mathrm{~J}+\epsilon_{0} \frac{\mathrm{dE}}{\mathrm{dt}}\right)\)

Using this equation of maxwell we can say changing electric field\(\frac{\mathrm{dE}}{\mathrm{dt}}\) induces magnetic field.

Given,

The speed of light in vaccum, \(c=3 \times 10^{8} \mathrm{~ms}^{-1}\)

Planck constant, \(h=6.62 \times 10^{-34} \mathrm{Js}\)

Charge on electron, \(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\)

Wavelength \(=5000 \mathring{A}\)

Work function, \(\phi_{0}=3.31 \times 10^{-19} \mathrm{~J}\)

\(=\frac{3.31 \times 10^{-19}}{1.6 \times 10^{-19}}\)

\(=2.07 \mathrm{eV}\)

Energy of Photon,

\(E_{P}=\frac{h c}{\lambda}\)

\(=\frac{12400}{5000}\)

\(\Rightarrow E_{P}=2.48 \mathrm{eV}\)

From Einstein's Equation:

\(E_{P}=\phi_{0}+K_{\max }\)

Thus, maximum kinetic energy can be obtained as:

\(K_{\max }=E_{P}-\phi_{0}\)

\(=(2.48-2.07) e V\)

\(=0.41 \mathrm{eV}\)

Given,

Radius of curvature, \(R=+3.00 \mathrm{~m}\)

Object-distance, \(u=-5.00 \mathrm{~m}\)

Image-distance, \(v=\) ?

Focal length, \(f=\frac{R}{2}\)

\(=+\frac{3.00 \mathrm{~m}}{2}\)

\(=+1.50 \mathrm{~m}\)

By lense formula,

\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

\(\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)

\(\Rightarrow \frac{1}{v}=+\frac{1}{1.50}-\frac{1}{(-5.00)}\)

\(\Rightarrow \frac{1}{v}=\frac{1}{1.50}+\frac{1}{5.00}\)

\(\Rightarrow \frac{1}{v}=\frac{5.00+1.50}{7.50}\)

\(\Rightarrow v=\frac{+7.50}{6.50}=+1.15 \mathrm{~m}\)

The image is \(1.15 \mathrm{~m}\) at the back of the mirror.

A bar magnet is a magnetic dipole. It is analogous to an electric dipole which is a system of two equal and opposite equal charges. Isolated magnetic charges or magnetic monopoles do not exist in nature. Even if a bar magnet is broken down to the atomic level, the north and south poles can never be separated.

Use the formula of force between two bar magnets.

\(F=\frac{\mu_{0}}{4 \pi} \frac{6 M_{1} M_{2}}{r^{4}}\)

Here,

\(\mathrm{F}\) is force between two bar magnets.

\(M_1\) and \(M_2\) are magnetic moments.

\(r\) is the distance between bar magnets.

\(\mu_{0}\) is the permeability of free space.

If the length of the bar magnets is negligible compared then \(r=d\)

Thus, the force acting between them will be proportion to \(\frac{1}{d^{4}}\).

For mutual inductance, we take two coils. One with \(N_{1}\) turns and the other with \(N_{2}\) turns. When we pass current through coil 1 , the magnetic field created by the coil increases in magnitude and the magnetic field lines will also go through coil 2 and create an increasing magnetic flux. The increase in magnetic flux in one direction causes the induced current to create a magnetic field going in the opposite direction to reduce the increasing magnetic flux.

Therefore, as we increase the current in coil 1, the magnitude of the magnetic field goes up and the flux in one direction increases creating an induced EMF across the second coil.

The induced EMF in coil \(2, \varepsilon_{2}\) is given as:

\(\varepsilon_{2}=M_{12} \frac{d i_{1}}{d t}\)

Where,

\(M_{12}=\) mutual inductance coefficient.

\(\frac{d i}{d t}=\) change in current per unit time in coil \(1 .\)

The unit of mutual inductance is Henry.

This mutual inductance is expressed as:

\(M=\frac{\mu_{0} N_{1} N_{2} A}{l}\)

Where,

\(N_{1}=\) number of turns in coil \(1\)

\(N_{2}=\) number of turns in coil \(2\).

\(A=\) area of coils

\(l=\) length of coil

Here, we have

\(N_{1}=300, N_{2}=400, A=10 \mathrm{~cm}^{2}=10 \times 10^{-4} \mathrm{~m}^{2}\) and

\(l=20 \mathrm{~cm}=0.2 \mathrm{~m}\)

Also given that \(\mu_{0}=4 \pi \times 10^{-7} \mathrm{TmA}{ }^{-1}\).

Substituting these values in formula of mutual inductance, we get

\(\Rightarrow M=\frac{\left(4 \pi \times 10^{-7}\right) 300.400\left(10^{-3}\right)}{0.2}\)

On simplifying, we get

\(\Rightarrow M=4 \pi(6) 10^{-5}\)

\(\Rightarrow M=24 \pi \times 10^{-5}\)

\(\therefore M=2.4 \pi \times 10^{-4} H\)

Thus, the mutual inductance is equal to \(2.4 \pi \times 10^{-4} \mathrm{H}\).

A neon sign does not produce an absorption spectrum. Neon sign tube is filled with Neon gas which is put on a high voltage causing the electrons to get excited. When these electrons come back to the ground state they release energy in form of packets or simply photons which causes the glow. Or in other words, atoms are present there which will cause the discontinuation of a continuous spectrum. Thus a line spectrum is eminent in this case. Now, these electrons emit photons when coming back to the ground state from excited state. Hence, they will produce an emission spectrum. These electrons in the Neon gas are excited due to high voltage. The atoms do not absorb energy. Hence, there is no absorption spectrum in this case.