Physics Test - 14

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

Work done, \((W)=\Delta K=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}\)

Where \(m\) is the mass of the object, \(v\) is the final velocity of the object and \(u\) is the initial velocity of the object.

Work-energy theorem for a variable force:

Kinetic energy, \(K=\frac{1}{2} m v^{2}\)

\( \frac{d K}{d t}=\frac{d\left(\frac{1 }{ 2} m v^{2}\right)}{d t} \)

\(\Rightarrow \frac{d K}{d t}=m \frac{d v}{d t} v\)

\(\Rightarrow \frac{d K}{d t}=m a v\)

\(\Rightarrow \frac{d K}{d t}=F v \)

\(\Rightarrow \frac{d K}{d t}=F \frac{d x}{d t} \)

\(\Rightarrow d K=F d x\)

On integrating, we get

\( \int_{K_{i}}^{K_{f}} d K=\int_{x_{i}}^{x_{f}} F d x \)

\(\Rightarrow \Delta K=\int_{x_{i}}^{x_{f}} F d x\)

From the work-energy theorem, a change in kinetic energy equals the work done.

\( \Delta K=\int_{x_{i}}^{x_{f}} F d x \)

\(\Rightarrow \Delta K=\int_{-a}^{a}\left(P y^{2}+Q y+R\right) d y \)

\(\Rightarrow \Delta K=\left[\frac{P y^{3}}{3}+\frac{Q y^{2}}{2}+R y\right]_{-a}^{a} \)

\(\Rightarrow \Delta K=\left[\left(\frac{P a^{3}}{3}+\frac{Q a^{2}}{2}+R a\right)-\left(\frac{-P a^{3}}{3}+\frac{Q a^{2}}{2}-R a\right)\right] \)

\(\Rightarrow \Delta K=\frac{2 P a^{3}}{3}+2 R a\)

As we know that the momentum of a body is given as,

P = m × v...(1)

Where m = mass, and v = velocity of the body

By equation (1) it is clear that if the momentum of a body is constant, then its velocity will also be constant and the acceleration will be zero.

An object is in translational equilibrium if the velocity of its translational motion is constant. To be in translational equilibrium, the net force acting on the object must be zero.

Therefore we can say that if the momentum of a body is constant, then it will be in translational equilibrium.

Basically, a coaxial cable consists of a hollow (outer) cylindrical conductor surrounding a single (inner) conductor along is the axis. The two conductors are well insulated from each other. The electric field (E) and magnetic field (H) at the cross-sections are shown by solid lines and dotted lines, respectively. The outer conductor acts as the shield and minimizes interference.

Different kinds of dielectric materials, such as Teflon and polythene are covered over copper wire, it acts as a spacer. In the transmission of power through coaxial cable, the dielectric medium separating the inner conductor from outer one plays a vital role. These dielectric materials are good insulators only at low frequencies. As the frequency increase, the energy loss becomes significant That is why a coaxial cable can be used effectively for transmission up to a frequency of 20 MHz.

A steady signal flowing in a wire uniformly distributes itself throughout the cross-section of the wire. A high-frequency signal, on the other hand, distributes itself uniformly, there being a concentration of current on the outer surface of the conductor. If the frequency of the current is very high, the current is almost wholly confined to the surface layers. This is called 'Skin effect".

Given:

Mass of earth (M)= \(5.98 \times 10^{24} \mathrm{~kg}\)

For earth to be a black hole, the escape velocity should be equal to the speed of light.

Escape velocity \(=\) Speed of light

\( v=c \ldots(1)\)

We know escape velocity is given by,

\(v=\sqrt{\frac{2 G M}{R}} \ldots(2)\)

Where,

\(\mathrm{G}\) is the universal gravitational constant

\(M\) is the mass of the body is to escaped from \(r\) is the distance from the center of the mass

Substituting equation. (2) in equation. (1) we get,

\(\sqrt{\frac{2 G M}{R}}=c\)

Squaring both the sides we get,

\(\frac{2 G M}{R}=c^{2}\)

Rearranging the above equation we get,

\(R=\frac{2 G M}{c^{2}}\).

Substituting values in above equation we get,

\(R=\frac{2 \times 6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{\left(3 \times 10^{8}\right)^{2}} \)

\(\Rightarrow R=\frac{79.77 \times 10^{13}}{9 \times 10^{16}} \)

\(\Rightarrow R=8.86 \times 10^{-3} \mathrm{~m} \)

\(\Rightarrow R=0.886 \times 10^{-2}\)

So, Earth has to be compressed to a radius of \(10^{-2} \mathrm{~m}\) to be a black hole.

Given,

Applied force (F) = 250 N

Mass (m) = 40 kg

Normal (N) = mg = 40× 10 = 400 N

Friction force (f) = μ N = 400μ

For moving the block, Applied (F) ≥ Friction force

250≥ 400μ

As the givenforce is minimum force, so

400μ = 250

∴ Coefficient of friction (μ) = \(\frac{250}{400}\) = 0.625

It is given that resistance of junction diode when forward biased,

\(R_{f}=25 \Omega\)

When resistors are connected in series the total resistance is given by the sum of resistance offered by the resistors.

Thus, the total resistance in the given arrangement can be calculated as:

\(R_{\text {Total }}=R_{f}+R\)

Where \(R\) is the resistance of the resistor. We can see aresistance of \(10 \Omega\) is connected in series to the diode.

Now substitute the values

\(R_{\text {Total }}=25 \Omega+10 \Omega=35 \Omega\)

From ohm's law, we know that voltage is directly proportional to current. The relation between voltage and current is given as:

\(V=I R\)

Thus, the current can be calculated by dividing voltage by resistance.

\(I=\frac{V}{R}\)

Voltage,\(V=5 V-0 V=5 V\)

Substitute the values we get,

\(I=\frac{V}{R}\)

\(\Rightarrow I=\frac{5 V}{35 \Omega}\)

\(\therefore I=\frac{1}{7} A\)

Therefore, the current in the diode is \(\frac{1}{7} A\).

Given,

Length of the wire, \(l = 3\) cm \(= 0.03\) m

Current flowing in the wire, \(I = 10\) A

Magnetic field, \(B = 0.27\) T

Angle between the current and magnetic field, \(θ = 90°\)

Magnetic force exerted on the wire is given as:

\(F = BIl sinθ\)

\(= (0.27 × 10 × 0.03 sin90°)\)

\(= 8.1 \times 10^{-2}\)N

Hence, the magnetic force on the wire is \(8.1 \times 10^{-2}\)N. The direction of the force can be obtained from Fleming’s left-hand rule.

Given:

The angle of incidence \(={i}\)

The angle of the prism \(=A\)

The angle of emergence \(={e}\)

The refractive index of the material of the prism is \(\mu\).

For a small angle prism, angle of deviation, \(\delta=(\mu-1) {A}\).......(i)

Since, the ray emerges normally,

\(\therefore e=0\)

By the relation,

\({A}+\delta={i}+{e}\)

We have, \({i}={A}+\delta\)

or \(\delta=(i-A)\) ......(ii)

From Eqs. (i) and (ii)

\((i-A)=(\mu-1) A\)

\(\Rightarrow {i}=\mu {A}\)

The property of matter which is responsible for electromagnetic phenomenon is called charge. The space or region around the current carrying wire/moving electric charge or around the magnetic material in which force of magnetism can be experienced by other magnetic material is called as magnetic field/magnetic induction by that material/current. The space or region around the electric charge in which electrostatic force can be experienced by other charge particle is called as electric field by that electric charge.

• When a charge particle is at rest then it only creates electric field around it.
• When a charge particle is moving with a constant velocity then it equivalent to an electric current which only creates a magnetic field around it.
• When a charge particle is moving with an acceleration then it produced an electromagnetic wave around it.