Madhya Pradesh Board Class 12 Physics — NCERT Solutions

Using Coulomb’s law,

$F = k\dfrac{q_1q_2}{r^2}$

Given: $q_1=2\times10^{-7}\,\text{C}$, $q_2=3\times10^{-7}\,\text{C}$, $r=30\,\text{cm}=0.30\,\text{m}$, and $k=9\times10^9\,\text{N m}^2\text{/C}^2$.

$$F=9\times10^9\times\frac{(2\times10^{-7})(3\times10^{-7})}{(0.30)^2}$$

$$=(9\times10^9)\times\frac{6\times10^{-14}}{0.09}$$

$$=9\times10^9\times6.67\times10^{-13} =6.0\times10^{-3}\,\text{N}$$

So the force is repulsive because both charges are positive.

By Coulomb’s law,

$$F=k\frac{|q_1q_2|}{r^2}$$

Given $F=0.2\,\text{N}$, $q_1=0.4\,\mu\text{C}=0.4\times10^{-6}\,\text{C}$, $q_2=0.8\,\mu\text{C}=0.8\times10^{-6}\,\text{C}$.

So,
$$r^2=k\frac{|q_1q_2|}{F} =9\times10^9\times\frac{(0.4\times10^{-6})(0.8\times10^{-6})}{0.2}$$

$$=9\times10^9\times\frac{0.32\times10^{-12}}{0.2} =9\times10^9\times1.6\times10^{-12} =1.44\times10^{-2}$$

$$r=\sqrt{1.44\times10^{-2}}=0.12\,\text{m}$$

So the distance is 0.12 m.

For the force on the second sphere, by Newton’s third law, it has the same magnitude and opposite direction. Since the charges are unlike, the force is attractive.

Given:

• Size of candle (object height): $h_o = 2.5$ cm
• Object distance: $u = -27$ cm (negative by sign convention, object in front of mirror)
• Radius of curvature: $R = -36$ cm (concave mirror)
• Focal length: $f = R/2 = -18$ cm

Formula used (Mirror equation):
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

Calculation:
$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-18} - \frac{1}{-27}$$
$$\frac{1}{v} = -\frac{1}{18} + \frac{1}{27} = \frac{-3 + 2}{54} = \frac{-1}{54}$$
$$v = -54 \text{ cm}$$

The screen should be placed 54 cm in front of the mirror (on the same side as the object).

Magnification:
$$m = -\frac{v}{u} = -\frac{-54}{-27} = -2$$

Size of image:
$$h_i = m \times h_o = -2 \times 2.5 = -5.0 \text{ cm}$$

Nature of image: Real, inverted, and magnified (size = 5.0 cm, twice the object size).

Effect of moving candle closer: As the candle is moved closer to the mirror (but still beyond $f$), the image moves farther away from the mirror. So the screen must be moved farther away from the mirror to obtain a sharp image. When the object is between $f$ and the pole, no real image is formed and the screen cannot capture the image.

Given:

• Object height: $h_o = 4.5$ cm
• Object distance: $u = -12$ cm
• Focal length of convex mirror: $f = +15$ cm

Formula used:
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

Calculation:
$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - \frac{1}{-12} = \frac{1}{15} + \frac{1}{12}$$
$$\frac{1}{v} = \frac{4 + 5}{60} = \frac{9}{60} = \frac{3}{20}$$
$$v = +\frac{20}{3} \approx +6.7 \text{ cm}$$

The image is formed 6.7 cm behind the mirror (virtual image).

Magnification:
$$m = -\frac{v}{u} = -\frac{20/3}{-12} = +\frac{20}{36} = +\frac{5}{9} \approx +0.56$$

Size of image:
$$h_i = m \times h_o = \frac{5}{9} \times 4.5 = 2.5 \text{ cm}$$

Nature: Virtual, erect, and diminished.

As the needle is moved farther from the mirror: The image moves closer to the focus of the mirror (i.e., towards $f = +15$ cm behind the mirror) but never goes beyond the focus. The size of the image goes on decreasing. As $u \to \infty$, $v \to f = +15$ cm and the image size approaches zero.

Take the positive charge as $+5\times10^{-8}\,\text{C}$ and the negative charge as $-3\times10^{-8}\,\text{C}$, separated by $16\,\text{cm}$. For potential to be zero,

$$\frac{5}{r_1}=\frac{3}{r_2}$$

because the potentials must cancel.

There are two possible points on the line:

1. Between the charges: if the point is $x$ cm from the positive charge, then it is $(16-x)$ cm from the negative charge.

$$\frac{5}{x}=\frac{3}{16-x}$$
$$5(16-x)=3x$$
$$80-5x=3x$$
$$8x=80 \Rightarrow x=10\,\text{cm}$$

2. On the extension beyond the negative charge: if the point is $x$ cm from the positive charge, then its distance from the negative charge is $(x-16)$.

$$\frac{5}{x}=\frac{3}{x-16}$$
$$5(x-16)=3x$$
$$5x-80=3x$$
$$2x=80 \Rightarrow x=40\,\text{cm}$$

So the points are 10 cm from the positive charge between the charges, and 40 cm from the positive charge on the side of the negative charge.

If the printed options do not include these exact values, the computed answer stands.

For a regular hexagon, the centre is at a distance equal to the side length from each vertex. So each charge is at distance

$$r=10\,\text{cm}=0.10\,\text{m}$$

from the centre.

Potential due to one charge:

$$V_1=\frac{1}{4\pi\varepsilon_0}\frac{q}{r} =9\times10^9\times\frac{5\times10^{-6}}{0.10}$$

$$V_1=9\times10^9\times5\times10^{-5}=4.5\times10^5\,\text{V}$$

There are 6 identical charges, so total potential is

$$V=6V_1=6\times4.5\times10^5=2.7\times10^6\,\text{V}$$

Given:

• Wavelength of incident light: $\lambda = 589 \text{ nm} = 589 \times 10^{-9} \text{ m}$
• Refractive index of water: $\mu = 1.33$
• Speed of light in vacuum/air: $c = 3.0 \times 10^8 \text{ m s}^{-1}$

Frequency of incident light:
$$\nu = \frac{c}{\lambda} = \frac{3.0 \times 10^8}{589 \times 10^{-9}} \approx 5.09 \times 10^{14} \text{ Hz}$$

(a) Reflected Light:

Reflection occurs in the same medium (air). The frequency, wavelength, and speed do not change upon reflection.

• Wavelength: $\lambda_{\text{reflected}} = 589 \text{ nm}$
• Frequency: $\nu_{\text{reflected}} = 5.09 \times 10^{14} \text{ Hz}$
• Speed: $v_{\text{reflected}} = 3.0 \times 10^8 \text{ m s}^{-1}$

(b) Refracted Light:

When light enters a denser medium, its frequency remains unchanged but its speed and wavelength change.

• Frequency: Frequency does not change on refraction.

$$\nu_{\text{refracted}} = 5.09 \times 10^{14} \text{ Hz}$$

• Speed: Using $\mu = \dfrac{c}{v}$:

$$v = \frac{c}{\mu} = \frac{3.0 \times 10^8}{1.33} \approx 2.26 \times 10^8 \text{ m s}^{-1}$$

• Wavelength: Using $\lambda_{\text{water}} = \dfrac{v}{\nu}$:

$$\lambda_{\text{water}} = \frac{2.26 \times 10^8}{5.09 \times 10^{14}} \approx 444 \text{ nm}$$

Alternatively: $\lambda_{\text{water}} = \dfrac{\lambda}{\mu} = \dfrac{589}{1.33} \approx 443 \text{ nm}$

Summary:

Given:

• EMF of battery, $\varepsilon = 12\,\mathrm{V}$
• Internal resistance, $r = 0.4\,\Omega$

Concept: The current drawn from a battery is maximum when the external resistance $R = 0$ (short circuit condition).

Formula:
$$I_{\max} = \frac{\varepsilon}{r}$$

Calculation:
$$I_{\max} = \frac{12}{0.4} = 30\,\mathrm{A}$$

Answer: The maximum current that can be drawn from the battery is $\boxed{30\,\mathrm{A}}$.

For X-rays produced by electrons accelerated through $V=30\,\text{kV}=3.0\times10^4\,\text{V}$, the maximum photon energy is

$$E_{\max}=eV=(1.6\times10^{-19})(3.0\times10^4)=4.8\times10^{-15}\,\text{J}.$$

(a) Maximum frequency:

$$\nu_{\max}=\frac{E_{\max}}{h}=\frac{4.8\times10^{-15}}{6.63\times10^{-34}}\approx7.25\times10^{18}\,\text{Hz}.$$

(b) Minimum wavelength:

$$\lambda_{\min}=\frac{c}{\nu_{\max}}=\frac{3.0\times10^8}{7.25\times10^{18}}\approx4.14\times10^{-11}\,\text{m}.$$

Given:

• Number of turns, $N = 100$
• Radius of coil, $R = 8.0\,\text{cm} = 0.08\,\text{m}$
• Current, $I = 0.40\,\text{A}$
• $\mu_0 = 4\pi \times 10^{-7}\,\text{T m A}^{-1}$

Formula used:
The magnetic field at the centre of a circular coil of $N$ turns is:
$$B = \frac{\mu_0 N I}{2R}$$

Calculation:
$$B = \frac{4\pi \times 10^{-7} \times 100 \times 0.40}{2 \times 0.08}$$
$$B = \frac{4\pi \times 10^{-7} \times 40}{0.16}$$
$$B = \frac{4\pi \times 10^{-7} \times 250}{1}$$
$$B = 4\pi \times 250 \times 10^{-7}$$
$$B = 3.14 \times 10^{-4}\,\text{T}$$

Answer: The magnitude of the magnetic field at the centre of the coil is $B \approx 3.14 \times 10^{-4}\,\text{T}$.

In Thomson's model, the positive charge is spread throughout the atom, so the atom's size is of atomic order. Rutherford's model says most mass and positive charge are concentrated in a tiny nucleus, so the atom is much larger than the nucleus. Therefore, the size of the atom in Thomson's model is much greater than the atomic size in Rutherford's model.

For a bar magnet in a uniform magnetic field, the torque is

$\tau = mB\sin\theta$.

Given:

• $\tau = 4.5\times 10^{-2}\,\text{J}$
• $B = 0.25\,\text{T}$
• $\theta = 30^\circ$, so $\sin 30^\circ = \tfrac{1}{2}$

So,

$$m = \frac{\tau}{B\sin\theta} = \frac{4.5\times 10^{-2}}{0.25\times \tfrac{1}{2}}$$

$$m = \frac{4.5\times 10^{-2}}{0.125} = 0.36\,\text{J T}^{-1}$$

Given:

• $Z = 7$ (protons), $N = A - Z = 14 - 7 = 7$ (neutrons)
• $m\left(\frac{14}{7}\mathrm{N}\right) = 14.00307\,\mathrm{u}$
• $m_p = m_H = 1.007825\,\mathrm{u}$, $m_n = 1.008665\,\mathrm{u}$

Formula:
$$\Delta M = \bigl[Z\,m_H + (A-Z)\,m_n\bigr] - M$$

Step 1: Calculate total mass of constituents.
$$\Sigma m = 7 \times 1.007825 + 7 \times 1.008665$$
$$= 7.054775 + 7.060655 = 14.115430\,\mathrm{u}$$

Step 2: Calculate mass defect.
$$\Delta M = 14.115430 - 14.00307 = 0.11236\,\mathrm{u}$$

Step 3: Convert to energy.
$$\Delta E_b = \Delta M \times 931.5\,\mathrm{MeV/u}$$
$$= 0.11236 \times 931.5 = 104.66\,\mathrm{MeV}$$

Binding energy of $\frac{14}{7}\mathrm{N}$ nucleus $\approx 104.66\,\mathrm{MeV}$.

Since the figures cannot be seen directly, the standard NCERT descriptions and answers for each part are given below, based on the well-known content of these figures.

Concept used: Lenz's Law — the induced current opposes the change in magnetic flux through the loop.

(a) A coil is moved towards a bar magnet (North pole facing the coil).

• The flux through the coil increases as it moves towards the magnet.
• To oppose the increase, the induced current must create a magnetic field opposing the magnet's field (i.e., facing the magnet with a North pole).
• Direction: The induced current flows in the coil in the anti-clockwise direction when viewed from the magnet side (i.e., along qrpq in the coil).

(b) A coil is moved away from a bar magnet (South pole facing the coil).

• The flux through the coil decreases as it moves away.
• To oppose the decrease, the induced current must attract the magnet, so it creates a South pole facing the magnet.
• Direction: The induced current flows clockwise when viewed from the magnet side (i.e., along prqp).

(c) A wire loop is placed near a solenoid carrying increasing current.

• The magnetic flux through the loop due to the solenoid increases.
• By Lenz's law, the induced current opposes the increase.
• Direction: The induced current in the loop flows in the anti-clockwise direction (as viewed from the solenoid end), i.e., along yzxy.

(d) A wire loop is placed near a solenoid carrying decreasing current.

• The magnetic flux through the loop decreases.
• By Lenz's law, the induced current opposes the decrease.
• Direction: The induced current flows clockwise (as viewed from the solenoid end), i.e., along zyxz (opposite to case (c)).

(e) A rectangular loop is moved into a region of uniform magnetic field directed into the page.

• As the loop enters the field region, the flux through it increases (into the page).
• By Lenz's law, the induced current must create flux out of the page inside the loop.
• Direction: The induced current flows anti-clockwise, i.e., along adcba.

(f) Two circular loops — current in the outer loop is increased.

• The increasing current in the outer loop increases the flux through the inner loop.
• By Lenz's law, the induced current in the inner loop opposes this increase.
• Direction: The induced current in the inner loop flows in the clockwise direction (opposite to the current in the outer loop).

Correct Option: (c) Holes are minority carriers and pentavalent atoms are the dopants.

Justification:
In an n-type semiconductor, silicon (tetravalent) is doped with pentavalent atoms (donors) such as As, Sb, or P. Each dopant atom donates one extra electron to the conduction band. Therefore:

• Electrons are the majority carriers.
• Holes are the minority carriers.
• The dopants are pentavalent atoms.

Option (a) is wrong because it states trivalent atoms are dopants (that would give p-type). Option (b) is wrong because electrons are majority carriers, not minority. Option (d) is wrong because holes are not majority carriers in n-type.

Hence, option (c) is correct.

For a pure resistor, Ohm's law applies to rms values:

$$I = \frac{V}{R} = \frac{220}{100} = 2.2\,\text{A}$$

For a resistor, the power consumed over a full cycle is

$$P = IV = (2.2)(220) = 484\,\text{W}$$

So the rms current is $2.2\,\text{A}$ and the net power consumed is $484\,\text{W}$.

Given radius $r=12\text{ cm}=0.12\text{ m}$ and separation $d=5.0\text{ cm}=0.05\text{ m}$.

(a) Capacitance and rate of change of potential difference

For a parallel-plate capacitor,
$$C=\varepsilon_0\frac{A}{d}$$
where
$$A=\pi r^2=\pi(0.12)^2=0.0452\ \text{m}^2$$
So,
$$C=8.85\times10^{-12}\times\frac{0.0452}{0.05} \approx 8.0\times10^{-12}\ \text{F}$$
Thus, the capacitance is
$$C\approx 8.0\ \text{pF}$$

The charging current is
$$i= C\frac{dV}{dt}$$
Hence,
$$\frac{dV}{dt}=\frac{i}{C}=\frac{0.15}{8.0\times10^{-12}} \approx 1.9\times10^{10}\ \text{V s}^{-1}$$

(b) Displacement current

For a charging capacitor, the displacement current equals the conduction current:
$$i_d=\varepsilon_0\frac{d\Phi_E}{dt}=i=0.15\ \text{A}$$

(c) Kirchhoff's first rule at each plate

Yes, Kirchhoff's junction rule is valid at each plate. The conduction current arriving at a plate does not disappear; it is matched by the displacement current between the plates. Thus charge conservation is satisfied when both conduction current and displacement current are included.