Physics Test - 20

The electric field produced by the infinite line charges at a distance d having linear charge density λ is given by the relation,

\(E=\frac{\lambda}{2 \pi \in_{0} \mathrm{~d}}\)

\(\lambda=2 \pi \in_{0} \mathrm{dE}\)

Where,

\(\mathrm{d}=2 \mathrm{~cm}=0.02 \mathrm{~m}\)

\(\mathrm{E}=9 \times 10^{4} \mathrm{~N} / \mathrm{C}\)

\(\epsilon_{0}=\) Permittivity of free space.

\(\frac{1}{4 \pi \in_{0}}=9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-2}\)

\(\lambda=\frac{0.02 \times 9 \times 10^{4}}{2 \times 9 \times 10^{9}}\)

\(=10 \mu \mathrm{C} / \mathrm{m}\)

We know that,

\(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

\(\frac{1}{f}=(\mu-1)\left(\frac{1}{R}+\frac{1}{R}\right)\)

\(\frac{1}{f}=(\mu-1) \frac{2}{R}\) ....(i)

and \(\frac{1}{f^{\prime}}=(\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right)\)

\(\frac{1}{f^{\prime}} =\frac{(\mu-1)}{R}\)...(ii)

From equations (i) and (ii), we get

\(f^{\prime}=\frac{f}{ 2}\)

There are no units for the refractive index because it is merely a ratio.

When you use the formula\(n=\frac{c}{v}\) the units for c and v cancel each other out, because they are both measured in m/s, leaving no units.

c and v are both values of speed. C being the speed of light in vacuum and v being the speed of light in the material. Both of them have the same unit i.e. m/s. So they cancel each other out leaving no unit.

Gauss's law for magnetism: The net magnetic flux of the magnetic field must always be zero over any closed surface.

This means that as many field lines should seem to enter the surface as the number of lines leaving it.

Net magnetic flux \(\left(\varnothing_{\mathrm{B}}\right)=\sum \mathrm{B} \cdot \Delta \mathrm{A}=0\)

Integral form: \(\int B . n d s=0\)

Where \(B\) is the magnetic field and \(A\) is surface area.

Let work function be \({X}\), \({K}_{1}\) and \({K}_{2}\) be kinetic energies for \({f}_{1}\) and \({f}_{2}\) respectively,
when frequency is \({f}_{1}\), \({K}_{1}={h f}_{1}-{X} \ldots (1)\)
when frequency is \({f}_{2}\), \({K}_{2}={h f}_{2}-{X} \ldots(2)\)
\({equation} (1)-{equation} (2)\) \(\Rightarrow{K}_{1}-{K}_{2}={hf}_{1}-{hf}_{2}\)
but \({K}=\frac{{m V}^{2}}{2}\), \(\frac{m\left(v_{1}^{2}-v_{2}^{2}\right)}{2}=h\left(f_{1}-f_{2}\right)\)
\(v_{1}^{2}-v_{2}^{2}=\frac{2 h\left(f_{1}-f_{2}\right)}{m}\)

The image formed by a concave mirror could be real, virtual, erect, and inverted.

Thus, from the given options possibility is the only option D.

Light converges at a point when it strikes and reflects back from the reflecting surface of the concave mirror. Thus, it is also known as a converging mirror.

Given:

Two satellites of masses \(3 \mathrm{M}\) and \(\mathrm{M}\)

Circular orbits of radii \(\mathrm{r}\) and \(3 \mathrm{r}\)

\(\mathrm{v} \propto \sqrt{\frac{1}{\mathrm{r}}}\)

This means the orbital velocity depends on the orbital radius of the satellite but not on the mass of the satellite. From the relation, we thus form an equation,

\(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\sqrt{\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}}\)

In the question,

\(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{1}{3}\)

Then,

\(\frac{\mathbf{v}_{1}}{\mathbf{v}_{2}}=\frac{\sqrt{3}}{1}\)

Let the charge on the capacitor \(\mathrm{C}\) be \(\mathrm{Q}\). Charge on the combination of \(1\) and \(2 \mu \mathrm{F}\) is also \(\mathrm{Q}\).
\(\mathrm{Q}_{2}=\frac{2}{1+2} \mathrm{Q}=\frac{2}{3} \mathrm{Q}\) But, \(\mathbf{Q}=\mathrm{E}\left(\frac{\mathbf{3 C}}{3+\mathbf{C}}\right)\)
\(\therefore \mathrm{Q}_{2}=\frac{2}{3} \mathrm{E}\left(\frac{3 \mathrm{C}}{3+\mathrm{C}}\right)=\frac{2 \mathrm{EC}}{3+\mathrm{C}}\)
As we can see, since \(\mathrm{C}\) is between \(1\) and \(3, \mathrm{Q}_{2}\) will increase until \(\mathrm{C}=3\).
Slope of curve \(\frac{d Q_{2}}{d C}=\frac{(3+C) 2 E-2 E C}{(3+C)^{2}}=\frac{6 E}{(3+C)^{2}}\)
So, the slope decreases as C increases.