Physics Test - 12

Given,

Charge on an alpha particle, \(q_{1}=q_{2}=+2 {e}\)

Distance between the particles, \(r=3.2 \times 10^{-15} {~m}\)

We know that:

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

and,\(\frac{1}{4\pi\epsilon_0}=9 \times 10^{9}\)

Now, using coulomb's law, we get,

Force acting on the particles is given by,

\(\begin{aligned} F &=\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} \\ F &=\frac{9 \times 10^{9} \times 2 \times 1.6 \times 10^{-19} \times 2 \times 1.6 \times 10^{-19}}{3.2 \times 10^{-15} \times 3.2 \times 10^{-15}} {~N} \end{aligned}\)

\(F=90 {~N}\)

Given,

\(m =10 kg , F _{1}=4 N , F _{2}=3 N , \theta=90^{\circ}\) and \(t =10 s\)

Since the two forces are acting perpendicular to each other, so the resultant force is given as,

\( F=\sqrt{F_{1}^{2}+F_{2}^{2}} \)

\(\Rightarrow F=\sqrt{4^{2}+3^{2}} \)

\(\Rightarrow F=5 N\)...(1)

By Newton's second law of motion, the force acting on the body is given as,

\( F = ma\)

Where \(m=\) mass and \(a=\) acceleration

\( a=\frac{F}{m} \)

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

\(\Rightarrow a=\frac{1}{2} m / s^{2}\)...(2)

Kinetic energy \(KE=\frac{1}{2} mv^{2}=\frac{1}{2}m (at)^{2}\)

\(\therefore KE =\frac{1}{2} \times 10 \times (\frac{1}{2}\times 10)^{2}\)

\(=\frac{1}{2} \times 10 \times(5)^{2}\)

\(= 125J\)

So, the kinetic energy of the body at the end of \(10 s\) is \(125J\).

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:

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

\(=\sqrt{50.17}\)

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

We know that when a particle enters the magnetic field in a direction perpendicular to the direction of the field, it undergoes circular motion.

The magnetic force is then equivalent to the centripetal force and is given as:

\(F=\frac{m v^{2}}{r}=q v B\)

\(\Rightarrow r=\frac{m v}{q B}\)

So, the radius of curvature \(r=\frac{m v}{q B}\)

Where, \(m=\) mass, \(q=\) magnitude of charge, \(v=\) speed of charge, \(B=\) magnetic field

It is given that the particles have the same charge (q) and initial velocity (v). Also, the magnetic field (B) is uniform.

Thus, r ∝ m

​Therefore, the radius of curvature is greater for the particle that has more mass.

Option (C) is correct regarding the energy band of electrons.

Each electron has a unique position inside a crystal and no two electrons see exactly the same pattern of surrounding charges. That is why no two electrons will have the same energy level. Each electron will have a different energy level.

Energy Band: The different energy levels with continuous energy variation of electrons in a material are called energy bands.

Valance Band: The energy band which includes the energy levels of the valence electrons is called the valence band.

Conduction Band: The energy bands above the valence bands are called the conduction bands. The conduction band holds those electrons that are responsible for conduction.

Given:

Charge on dipole is \(\pm 5 \mu {C}=\pm 5 \times 10^{-6} {C}\)

Distance between the charges \(=1 {~mm}=10^{-3} {~m}\)

Dipole moment is given by:

\(P={q}(2 {a})\)

\(=5 \times 10^{-6} \times 2 \times 10^{-3}\)

\(=10^{-8} {Cm}\)

[Capacitance A]\(=\left[M^{-1} L 6-2 T^{2} Q^{2}\right]\)

[Magnetic induction C] \(=\left[M T^{-1} Q^{-1}\right]\)

\([C]^{2}=\left[M^{2} T^{-2} Q^{-2}\right]\)

Given, \(A=3 B C^{2}\) or \(B=\frac{A}{3 C^{2}} \Rightarrow [B]=\frac{[A]}{[C]^{2}}\)

\(\therefore [B]=\frac{\left[M^{-1} L^{-2} T^{2} Q^{2}\right]}{\left[M^{2} T^{-1} Q^{-2}\right]}=\left[M^{-3} L^{-2} T^{4} Q^{4}\right]\)

Biot-Savart Law: The law that gives the magnetic field generated by a constant electric current is Biot-savart law.

Let us take a current-carrying wire of current \(I\) and we need to find the magnetic field at a distance \(r\) from the wire then it is given by:

\(d B=\frac{\mu_{0} I}{4 \pi}\left(\frac{\overrightarrow{d l} \times \hat{r}}{r^{2}}\right)\)

Where, \(\mu_{0}=4 \pi \times 10^{-7}\) T.m/A is the permeability of free space/vacuum, \(dl =\) small element of wire and \(\hat{r}\) is the unit position vector of the point where we need to find the magnetic field.

From the above expression of the Biot-savart law, the magnetic field is:

• Directly proportional to the lengthof the wire. So statement (i)is correct.
• Directly proportional to the electric current. So statement (iii) is correct.
• Biot-savart law gives the magnetic field, not the electric field. So statement (ii) is wrong.

Given,

Radius, \(r=1000 m\)

Banking angle, \(\theta=45^{\circ}\)

Coefficient of friction, \(\mu=0.5\)

Acceleration due to gravity, \(g=9.8 m/s^2\)

As we know,

Maximum safe velocity is calculated by,

\(v=\sqrt{\frac{r g(\tan \theta+\mu)}{1-\mu \tan \theta}} \)

\(\therefore v=\sqrt{\frac{1000 \times 9.8\left(\tan 45^{\circ}+0.5\right)}{1-0.5 \times \tan 45^{\circ}}} \)

\(\Rightarrow v=\sqrt{\frac{1000 \times 9.8 (1+0.5)}{1-0.5 \times 1}}\)

\(\Rightarrow v=\sqrt{\frac{1000 \times 9.8 \times 1.5}{1-0.5}}\)

\(\Rightarrow v=\sqrt{\frac{1000 \times 9.8 \times 1.5}{0.5}}\)

\(\Rightarrow v=\sqrt{29400}\)

\(\therefore v =172 m / s\)

In a series circuit, the total resistance in the circuit is equal to the sum of all the resistances in the circuit.

• A lamp is nothing but resistance, hence with the addition of a lamp the total resistance of the circuit will increase.
• Now when the voltage is kept constant the current in the circuit will decrease with the increase of resistance in accordance with ohm’s Law (I = V/R) (with an increase in denominator value of fraction decreases).
• Thus, on adding more lamps into a series circuit, the overall current in the circuit will decrease if the voltage is kept constant.
• The opposite will happen in a circuit where lamps/resistors will be connected parallel.

As we know that:

Electric dipole moment is \(= P =\Sigma p _{ i }=\Sigma q \cdot dx =\) charge \(\times\) distance

If it is zero then we have to test its quadrupole moment for the electric field.

Given: two charges \(+q\) and \(-q\) placed at distance \(L\).

Electric monopole moment is \(=m=-q+q=0\)

Here electric monopole moment becomes zero so we have to calculate its electric dipole moment.

We are free to consider origin anywhere because the monopole moment is zero.

Electric dipole moment is \(= P =\) charge \(\times\) distance (with directions)

The electric field is towards for negative charge and away from the positive charge.

Dipole moment \(=P=q(\frac{L} { 2})(\hat{i})+(-q)(\frac{L} { 2})(-\hat{i})=q L \hat {(i)}\)

so here, the dipole moment is non-zero. we know that for dipole moment electric field varies as \(E \propto \frac{1} { R^3}\)

As we know,

NAND Gate: The logic Gate is obtained after adding NOT Gate after AND Gate.

\(Y=\overline{A . B}\)

From above, we can see that NAND gate is one of the logic gates, which is opposite of AND and output of of different inputs (A & B) are as shown below:

Threshold frequency is given as:

\( \mathrm{v}_0=\frac{W_0}{h}\)

Here \(W_0=3.3 \times 10^{-19} \mathrm{~J}\)

\(h=6.6 \times 10^{-34} \mathrm{Js}\)

\(\mathrm{v}_0=\frac{\left(3.3 \times 10^{-19}\right)}{\left(6.6 \times 10^{-34}\right)}\)

\(\mathrm{v}_0=5 \times 10^{14} \mathrm{~Hz}\)