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# CBSE Class 12 Physics Solved Question Paper 2018

1.

A bar magnet of magnetic moment 6 J/T is aligned at 60° with a uniform external magnetic field of 0.44 T. Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magnetic field, (ii) opposite to the magnetic field, and (b) the torque on the magnet in the final orientation in case (ii).

M = 6J/T

θ = 60o

B = 0.44T

$\mathrm{\tau }=\mathrm{mB}\mathrm{sin\theta }\phantom{\rule{0ex}{0ex}}\mathrm{\tau }=6\mathrm{x}0.44\mathrm{sin}{60}^{\mathrm{o}}\phantom{\rule{0ex}{0ex}}=6\mathrm{x}0.44\mathrm{x}\frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}=3\sqrt{3}\mathrm{x}0.44\phantom{\rule{0ex}{0ex}}=2.836\phantom{\rule{0ex}{0ex}}\mathrm{dw}{\int }_{\mathrm{\theta }={60}^{0}}^{{90}^{0}}\mathrm{\tau }.\mathrm{d\theta }\phantom{\rule{0ex}{0ex}}=-\mathrm{mB}\left[\mathrm{cos}{90}^{0}-\mathrm{cos}{60}^{0}\right]\phantom{\rule{0ex}{0ex}}=6\mathrm{x}0.44\mathrm{x}\left[-\frac{1}{2}\right]\phantom{\rule{0ex}{0ex}}=3\mathrm{x}0.44=1.32\mathrm{J}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\mathrm{dw}={\int }_{\mathrm{\theta }={60}^{0}}^{{180}^{0}}\mathrm{mB}\mathrm{sin}\mathrm{\theta }.\mathrm{d\theta }\phantom{\rule{0ex}{0ex}}=-\mathrm{mB}\left[\mathrm{cos}{180}^{0}-\mathrm{cos}{60}^{0}\right]\phantom{\rule{0ex}{0ex}}=6\mathrm{x}0.44\mathrm{x}\left[-1-\frac{1}{2}\right]\phantom{\rule{0ex}{0ex}}=-6\mathrm{x}0.44\left[-\frac{3}{2}\right]\phantom{\rule{0ex}{0ex}}=9\mathrm{x}0.44\phantom{\rule{0ex}{0ex}}=39.6\mathrm{J}\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)\stackrel{\to }{\mathrm{\tau }}=\stackrel{\to }{\mathrm{m}}\mathrm{x}\stackrel{\to }{\mathrm{B}}\phantom{\rule{0ex}{0ex}}\mathrm{\tau }=\mathrm{m}\mathrm{Bsin}\mathrm{\theta }\phantom{\rule{0ex}{0ex}}=6\mathrm{x}0.44\mathrm{x}\mathrm{sin}{180}^{0}\phantom{\rule{0ex}{0ex}}\mathrm{\tau }=0$

2.

Two electric bulbs P and Q have their resistances in the ratio of 1: 2. They are connected in series across a battery.Find the ratio of the power dissipation in these bulbs.

Let ratio be x

Rp = x (Resistance of bulb P)

Rq = 2x (Resistance of bulb Q)

We know, P = VI = V.V/R

$\frac{{\mathrm{P}}_{\mathrm{p}}}{{\mathrm{P}}_{\mathrm{q}}}=\frac{{\mathrm{V}}_{2}}{{\mathrm{R}}_{\mathrm{p}}}\mathrm{x}\frac{{\mathrm{R}}_{\mathrm{q}}}{{\mathrm{V}}_{2}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{In}\mathrm{series}\mathrm{Potential}\mathrm{will}\mathrm{be}\mathrm{same}\right)\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{P}}_{\mathrm{p}}}{{\mathrm{P}}_{\mathrm{q}}}=\frac{2\mathrm{x}}{\mathrm{x}}\phantom{\rule{0ex}{0ex}}2:1$

3.

Define the term 'conductivity' of a metallic wire. Write its SI unit.

The conductivity of a metallic wire is defined as the degree to which a specified material conduct electricity calculated as the ratio of current density in the material to the electric field which causes the flow of current,

$R=\frac{\rho \mathcal{l}}{A}\phantom{\rule{0ex}{0ex}}\rho =\frac{RA}{\mathcal{l}}\phantom{\rule{0ex}{0ex}}\frac{1}{\rho }=\frac{\mathcal{l}}{R.A}\phantom{\rule{0ex}{0ex}}\sigma =\frac{\mathcal{l}}{R.A}$

The SI unit of conductivity is ohm-1 metre-1 or mho metre-1 or siemen metre-1

4.

Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field E.

As we know that,

$\mathrm{R}=\frac{\mathrm{\rho }\mathcal{l}}{\mathrm{A}}\phantom{\rule{0ex}{0ex}}\mathrm{V}=\mathrm{IR}\phantom{\rule{0ex}{0ex}}\mathrm{V}=\frac{\mathrm{I\rho }\mathcal{l}}{\mathrm{A}}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{I}}{\mathrm{A}}=\mathrm{j}=\mathrm{current}\mathrm{density}\phantom{\rule{0ex}{0ex}}\mathrm{V}=\mathrm{j\rho }\mathcal{l}\phantom{\rule{0ex}{0ex}}\mathrm{V}=\mathrm{E}.\mathcal{l}\phantom{\rule{0ex}{0ex}}\mathrm{E}.\mathcal{l}=\mathrm{j\rho }\mathcal{l}\phantom{\rule{0ex}{0ex}}\mathrm{E}=\mathrm{j\rho }\phantom{\rule{0ex}{0ex}}\mathrm{j}=\frac{\mathrm{E}}{\mathrm{\rho }}\phantom{\rule{0ex}{0ex}}\frac{1}{\mathrm{\rho }}=\mathrm{\sigma }\phantom{\rule{0ex}{0ex}}\mathrm{j}=\mathrm{\sigma E}\phantom{\rule{0ex}{0ex}}\mathrm{V}=\mathrm{u}+\mathrm{at}\phantom{\rule{0ex}{0ex}}\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\phantom{\rule{0ex}{0ex}}\mathrm{a}=\frac{-\mathrm{eE}}{\mathrm{m}}\phantom{\rule{0ex}{0ex}}|{\mathrm{V}}_{\mathrm{d}}|=\frac{\mathrm{eE}}{\mathrm{m}}\mathrm{\tau }\phantom{\rule{0ex}{0ex}}\mathrm{I}.∆\mathrm{t}=\mathrm{neA}\left({\mathrm{V}}_{\mathrm{d}}\right)∆\mathrm{t}\phantom{\rule{0ex}{0ex}}\mathrm{j}=\mathrm{i}/\mathrm{A}\phantom{\rule{0ex}{0ex}}\mathrm{j}=\mathrm{neA}\frac{\mathrm{eE}}{\mathrm{m}}\mathrm{\tau }\phantom{\rule{0ex}{0ex}}\mathrm{j}=\frac{{\mathrm{ne}}^{2}\mathrm{\tau }}{\mathrm{m}}\mathrm{E}$

5.

A 10 V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38 Ω as shown in the figure. Find the value of current in the circuit.

The positive terminals are connected to the offsite ends of the resistor, they send the current in opposite directions.

Hence, the net emf = 200-10 = 190 V

∴ current in the circuit I = E/R

= 190V / 38 Ω

= 5 A

6.

Four point charges Q, q, Q and q are placed at the corners of a square of side ‘a’ as shown in the figure

Find the
(a) resultant electric force on a charge Q,

(b) the potential energy of this system.

${\mathrm{F}}_{1}=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}=\frac{\mathrm{qQ}}{{\mathrm{a}}^{2}}\phantom{\rule{0ex}{0ex}}{\mathrm{F}}_{2}=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{\mathrm{qQ}}{{\mathrm{a}}^{2}}\phantom{\rule{0ex}{0ex}}{\mathrm{F}}_{3}=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{\mathrm{QQ}}{\left(\sqrt{2\mathrm{a}{\right)}^{2}}}=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{{\mathrm{Q}}^{2}}{2{\mathrm{a}}^{2}}$

F1 and F2 are perpendicular to each other so their resultant will be

$\mathrm{F}\text{'}=\sqrt{{\mathrm{F}}_{1}^{2}+{\mathrm{F}}_{2}^{2}+2{\mathrm{F}}_{1}{\mathrm{F}}_{2}\mathrm{cos}{90}^{\mathrm{o}}}\phantom{\rule{0ex}{0ex}}{\mathrm{F}}_{1}={\mathrm{F}}_{2}\phantom{\rule{0ex}{0ex}}\mathrm{F}\text{'}=\sqrt{2}\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{\mathrm{qQ}}{{\mathrm{a}}^{2}}$

F3 and resultant of F1 and F2 will be in the same direction

Net force

F = F' + F3

$\mathrm{F}=\frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_{0}}\left[\frac{\sqrt{2\mathrm{q}}}{{\mathrm{a}}^{2}}+\frac{\mathrm{Q}}{2{\mathrm{a}}^{2}}\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_{0}}\left[\frac{2\sqrt{2\mathrm{q}}+\mathrm{Q}}{2{\mathrm{a}}^{2}}\right]\phantom{\rule{0ex}{0ex}}\mathrm{F}=\frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_{0}}\left[\frac{2\sqrt{2\mathrm{q}}+\mathrm{Q}}{2{\mathrm{a}}^{2}}\right]$

(b) The potential energy of the given system,

$\mathrm{W}=\frac{4\mathrm{KQq}}{\mathrm{a}}+\frac{{\mathrm{Kq}}^{2}}{\sqrt{2}\mathrm{a}}+\frac{{\mathrm{KQ}}^{2}}{\sqrt{2}\mathrm{a}}\phantom{\rule{0ex}{0ex}}\mathrm{K}=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}$

7.

An iron ring of relative permeability μr has windings of insulated copper wire of n turns per metre. When the current in the windings is I, find the expression for the magnetic field in the ring.

There will be two magnetic field one due to current and another due to magnetics.

$\mathrm{B}={\mathrm{B}}_{0}+{\mathrm{B}}_{\mathrm{m}}\phantom{\rule{0ex}{0ex}}\mathrm{B}={\mathrm{\mu }}_{0}\mathrm{nI}+{\mathrm{\mu }}_{0}\mathrm{M}\phantom{\rule{0ex}{0ex}}\mathrm{B}={\mathrm{\mu }}_{0}\mathrm{H}+{\mathrm{\mu }}_{0}\mathrm{\chi H}\phantom{\rule{0ex}{0ex}}\mathrm{B}={\mathrm{\mu }}_{0}\left(1+\mathrm{\chi }\right)\mathrm{H}\phantom{\rule{0ex}{0ex}}\mathrm{B}={\mathrm{\mu }}_{0}{\mathrm{\mu }}_{\mathrm{r}}\mathrm{H}$

8.

In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9 ohm is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.

Potentiometer at open circuitℓ1 = 350

R = 9

2 = 300

$\mathrm{r}=\mathrm{R}\left(\frac{{\mathrm{I}}_{1}}{{\mathrm{l}}_{2}}-1\right)\phantom{\rule{0ex}{0ex}}\mathrm{r}=9\left(\frac{350}{300}-1\right)\phantom{\rule{0ex}{0ex}}=1.5\mathrm{\Omega }$

9.

Three point charges q, – 4q and 2q are placed at the vertices of an equilateral triangle ABC of side ‘l’ as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge q.

(b)Find out the amount of the work done to separate the charges at infinite distance.

$|{\stackrel{\to }{\mathrm{F}}}_{1}|=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{\left(4\mathrm{q}\right)\left(\mathrm{q}\right)}{{\mathcal{l}}^{2}}\phantom{\rule{0ex}{0ex}}{\mathrm{F}}_{1}=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{\left(4{\mathrm{q}}^{2}\right)}{{\mathcal{l}}^{2}}\phantom{\rule{0ex}{0ex}}{\mathrm{F}}_{1}=\frac{1}{{\mathrm{\pi \epsilon }}_{0}}\frac{{\mathrm{q}}^{2}}{{\mathcal{l}}^{2}}\phantom{\rule{0ex}{0ex}}{\stackrel{\to }{\mathrm{F}}}_{2}|=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{\left(2\mathrm{q}\right)\left(\mathrm{q}\right)}{{\mathcal{l}}^{2}}\phantom{\rule{0ex}{0ex}}{\mathrm{F}}_{2}=\frac{1}{2{\mathrm{\pi \epsilon }}_{0}}\frac{{\mathrm{q}}^{2}}{{\mathcal{l}}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{angle}\mathrm{between}\stackrel{\to }{{\mathrm{F}}_{1}}\mathrm{and}\stackrel{\to }{{\mathrm{F}}_{2}}\mathrm{is}{120}^{\mathrm{o}}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{F}}=\sqrt{{\mathrm{F}}_{1}^{1}+{\mathrm{F}}_{2}^{2}+2{\mathrm{F}}_{1}{\mathrm{F}}_{2}\mathrm{cos}{120}^{\mathrm{o}}}\phantom{\rule{0ex}{0ex}}{\mathrm{F}}_{1}=2{\mathrm{F}}_{2}\phantom{\rule{0ex}{0ex}}\mathrm{F}=\sqrt{\left(2{\mathrm{F}}_{2}{\right)}^{2}+{\mathrm{F}}_{2}^{2}+4{\mathrm{F}}_{2}^{2}\mathrm{cos}{120}^{0}}\phantom{\rule{0ex}{0ex}}\mathrm{F}=\sqrt{4{\mathrm{F}}_{2}^{2}+{\mathrm{F}}_{2}^{2}-2{\mathrm{F}}_{2}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{F}=\sqrt{3{\mathrm{F}}_{2}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{F}=\sqrt{3{\mathrm{F}}_{2}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{F}=\sqrt{3}\frac{1}{2{\mathrm{\pi \epsilon }}_{0}}\frac{{\mathrm{q}}^{2}}{{\mathcal{l}}^{2}}$

(b) The amount of work done to separate the charges at infinity will be equal to potential energy.

$\mathrm{U}=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}\mathcal{l}}\left[\mathrm{q}\mathrm{x}\left(-4\mathrm{q}\right)+\left(\mathrm{q}\mathrm{x}2\mathrm{q}\right)+\left(-4\mathrm{q}\mathrm{x}2\mathrm{q}\right)\phantom{\rule{0ex}{0ex}}\mathrm{U}=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}\mathcal{l}}\left[-4{\mathrm{q}}^{2}+2{\mathrm{q}}^{2}-8{\mathrm{q}}^{2}\right]\phantom{\rule{0ex}{0ex}}\mathrm{U}=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}\mathcal{l}}\left[-10{\mathrm{q}}^{2}\right]\phantom{\rule{0ex}{0ex}}\mathrm{U}=-\frac{1}{4{\mathrm{\pi \epsilon }}_{0}\mathcal{l}}\left[10{\mathrm{q}}^{2}\mathrm{unit}$

10.

The susceptibility of a magnetic material is 0.9853. Identify the type of magnetic material. Draw the modification of the field pattern on keeping a piece of this material in a uniform magnetic field.

The susceptibility of this material is between 0 and 1 so its a paramagnetic material.