Physics Test-18

Insulators are materials in which valence electrons are not found to conduct electricity. That's why low conductivity is found in them.

Resistivity is the property of materials that opposes electric current. The resistivity of insulators is very high.

Insulators are materials that have a large band gap, or materials that have a high energy gap between the valence and conduction bands. This large energy difference makes it difficult for electrons to move into the conduction band through which they can flow and generate an electric current.

A negative coefficient of resistance of a material means that its resistance decreases with increase in temperature. Positive coefficients are found in conductors. Negative coefficient is found in insulators. So, option (D) is wrong.

We know,

Power \(=\frac{\text{Work}}{\text{Time}}\)

P \(=\frac{W}{T}\)

Time \(=10\) s

Work done \(=\) Energy consumed by the lamp

\(=1000\)J

Power \(=\frac{1000}{10}\)

\(=100\) Js\(^{-1}\)

\(=100\) W

So, the power of the lamp is \(100\) W.

Given:

Radius of the earth \((r)= 6400 \mathrm{~km}\)

Radius of the orbit \((r)=8000 \mathrm{~km}\)

\(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}=9.8 \times 10^{-3} \mathrm{~km} / \mathrm{s}^{2}\)

So,

\(V_{o}=\sqrt{\frac{g R^{2}}{r}}\)

Put all the given values in above formula:

\(=\sqrt{\frac{9.8 \times 10^{-3} \times 6400^{2}}{8000}} \)

\(=\sqrt{50.17}\)

\(=7.08 \mathrm{~km} / \mathrm{s}\)

The phenomenon of interference is shown by all the given type type of waves.

Interference of a wave is the phenomenon that usually happens when the two waves reinforce, while travelling in the same medium. The interference of waves causes the medium to take on a shape that results from the net effects of the two individual waves simultaneously acting upon the particles of the medium.

Let \(\mathrm{M}\) and \(\mathrm{R}\) be the mass and radius of the disc respectively.

Moment of inertia about \(\mathrm{AB}\) is \(\mathrm{I}\).

Moment of inertia about an axis passing through \(\mathrm{O}\) and perpendicular to the plane,

\(\mathrm{I}_{\mathrm{O}}=\frac{1}{2} \mathrm{M} \mathrm{R}^{2}\)

Moment of inertia about an axis passing through \(\mathrm{C}\) and perpendicular to the plane,

\(\mathrm{I}_{\mathrm{C}}=\mathrm{I}_{\mathrm{O}}+\mathrm{M} \mathrm{R}^{2}\)

\(\mathrm{I}_{\mathrm{C}}=\mathrm{3 I}_{\mathrm{O}}\)

Using perpendicular axis theorem,

\(\Rightarrow \mathrm{I}_{O}=2 \mathrm{I}\)

Thus \(\mathrm{I}_{\mathrm{C}}=\mathrm{6 I}\)

When an electric current flows through a conducting solution, it causes a chemical reaction. This is a chemical effect of electric current. A chemical reaction occurs when an electric current is used to conduct a solution. This results in a colour shift, which is due to the chemical effect of electric current.

Let \(r_{1}\) and \(r_{2}\) be the radii traced by the inner and outer wheels of the bus respectively.

From the above figure \(r_{1}\).Thus, \(f=f c\)
\(\Rightarrow \mu N=m r \omega^{2}\) \(\Rightarrow N=\frac{m r \omega^{2}}{\mu}\)
where \(\mu\) is coefficient of friction between the wheels and the surface.
For the inner tyre, we have \({N}_{1}=\frac{m \omega^{2}}{\mu} {r}_{1}\)
For the outer tyre, we have \({N}_{2}=\frac{m \omega^{2}}{\mu} {r}_{2}\)
Then, we get \(\frac{N_{1}}{N_{2}}=\frac{r_{1}}{r_{2}}<1 \quad\) [Since \(r_{1}\) is given to be less than \(r_{2}\) as shown in figure.]

Since both electric and magnetic fields are vectors, the direction of propagation of EM waves is obtained from the right-hand rule.

Let the electric field be denoted by \(\vec{E}\) and magnetic field be denoted by \(\vec{B}\).

According to the rule - If the fingers of the right hand are curled so that they follow a rotation from \(\vec{E}\) to \(\vec{B}\), then the thumb will point in the direction of the vector product i.e., the direction of EM waves.

Therefore, the direction of EM waves is found from the cross product of the electric field and magnetic field.

Given:

Charge, \(q_{1}=200 \times 10^{-6} C=2 \times 10^{-4} C\)

Charge, \(q_{2}=500 \times 10^{-6} C =5 \times 10^{-4} C\)

Electrostatic force, \(F=5 g f=5 \times 10^{-3} kg f\) \(=5 \times 10^{-3} \times 10 N\) \(=5 \times 10^{-2} N\)

We have to find the distance between two charges i.e., \(r\)

Using the formula:

\(F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}\)

\(5 \times 10^{-2}=\frac{9 \times 10^{9} \times 2 \times 10^{-4} \times 5 \times 10^{-4}}{r^{2}}\)

\(r=1.34 \times 10^{2} m\)