CBSE 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.

For a concave mirror, by Cartesian sign convention: $u=-27\text{ cm}$ and $R=36\text{ cm}$, so $f=-R/2=-18\text{ cm}$.\n\nUsing the mirror formula\n\n$$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$$\n\n$$\frac{1}{v}+\frac{1}{-27}=\frac{1}{-18}$$\n\n$$\frac{1}{v}=-\frac{1}{18}+\frac{1}{27}=-\frac{3-2}{54}=-\frac{1}{54}$$\n\nSo, $$v=-54\text{ cm}.$$\n\nThus the screen should be placed 54 cm in front of the mirror.\n\nMagnification:\n\n$$m=-\frac{v}{u}=-\frac{-54}{-27}=-2$$\n\nSo the image is real, inverted, and magnified 2 times. Its size is\n\n$$h' = mh = 2\times 2.5 = 5.0\text{ cm}$$\n\nIf the candle is moved closer to the mirror, the image shifts farther away; hence the screen must be moved away from the mirror to keep the image sharp.

For a convex mirror, $f=+15\text{ cm}$ and the object distance is $u=-12\text{ cm}$.\n\nUsing the mirror formula,\n\n$$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$$\n\n$$\frac{1}{v}+\frac{1}{-12}=\frac{1}{15}$$\n\n$$\frac{1}{v}=\frac{1}{15}+\frac{1}{12}=\frac{4+5}{60}=\frac{9}{60}=\frac{3}{20}$$\n\nSo,\n\n$$v=\frac{20}{3}\text{ cm}\approx 6.67\text{ cm}$$\n\nSince $v$ is positive, the image is behind the mirror.\n\nMagnification:\n\n$$m=-\frac{v}{u}=-\frac{20/3}{-12}=\frac{5}{9}\approx 0.56$$\n\nSo the image is virtual, erect, and diminished.\n\nAs the needle is moved farther from the mirror, the image moves towards the focus behind the mirror and becomes smaller.

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

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

For light reflecting from air-water boundary:

• Frequency remains unchanged on reflection and refraction.
• In air, wavelength is given as $\lambda = 589\,\text{nm}$.

(a) Reflected light

Since reflection does not change the medium,

• wavelength = $589\,\text{nm}$
• frequency = $\dfrac{c}{\lambda} = \dfrac{3.0\times10^8}{589\times10^{-9}} \approx 5.09\times10^{14}\,\text{Hz}$
• speed = $3.0\times10^8\,\text{m s}^{-1}$

(b) Refracted light in water

Using $n = \dfrac{c}{v}$,
$$v = \frac{c}{n} = \frac{3.0\times10^8}{1.33} \approx 2.26\times10^8\,\text{m s}^{-1}$$
Frequency remains the same, so
$$\nu \approx 5.09\times10^{14}\,\text{Hz}$$
Wavelength in water is
$$\lambda' = \frac{v}{\nu} = \frac{2.26\times10^8}{5.09\times10^{14}} \approx 4.44\times10^{-7}\,\text{m} = 444\,\text{nm}$$
So the refracted light has smaller wavelength and lower speed, but the same frequency.

The maximum current is obtained when the external resistance is zero, so

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

Given $\varepsilon=12\,\text{V}$ and $r=0.4\,\Omega$,

$$I_{\max}=\frac{12}{0.4}=30\,\text{A}$$

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:

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

For a circular coil, magnetic field at the centre is

$B=\dfrac{\mu_0 N I}{2R}$

Given: $N=100$, $I=0.40\,\text{A}$, $R=8.0\,\text{cm}=0.08\,\text{m}$.

$$B=\frac{4\pi\times10^{-7}\times100\times0.40}{2\times0.08}$$

$$=\frac{16\pi\times10^{-6}}{0.16} =100\pi\times10^{-6} \approx 3.14\times10^{-4}\,\text{T}$$

So the field at the centre is $3.14\times10^{-4}\,\text{T}$.

In Thomson's model, the atom is pictured as a sphere of positive charge with electrons embedded in it, so its size is much larger than the tiny nucleus in Rutherford's model. Therefore, the atom's size 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 $\approx 104.66\,\mathrm{MeV}$.

The chapter gives only the exercise prompt, not the detailed figure information needed to apply Lenz’s law case by case.

In an n-type semiconductor, the dopant is pentavalent and the majority carriers are electrons. Therefore, holes are minority carriers. So the correct statement is (c).

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 $484\,\text{W}$.

(a) Capacitance

For a parallel plate capacitor,

$$C=\varepsilon_0\frac{A}{d}$$

Here, radius $r=12\text{ cm}=0.12\text{ m}$, so

$$A=\pi r^2=\pi(0.12)^2=\pi\times 0.0144\approx 0.0452\text{ m}^2$$

and separation

$$d=5.0\text{ cm}=0.050\text{ m}$$

Thus,

$$C=8.85\times 10^{-12}\times \frac{0.0452}{0.050}$$
$$C\approx 8.0\times 10^{-12}\text{ F}$$

So the capacitance is $8.0\times10^{-12}$ F.

The charging current is constant:

$$i=0.15\text{ A}$$

Using

$$i=C\frac{dV}{dt}$$

we get

$$\frac{dV}{dt}=\frac{i}{C}=\frac{0.15}{8.0\times10^{-12}}$$

$$\frac{dV}{dt}\approx 1.9\times10^{10}\text{ V s}^{-1}$$

So the rate of change of potential difference is $1.9\times10^{10}\text{ V s}^{-1}$.

(b) Displacement current

From the chapter, the displacement current is

$$i_d=\varepsilon_0\frac{d\Phi_E}{dt}$$

For a charging capacitor, the displacement current equals the conduction current:

$$i_d=i=0.15\text{ A}$$

So the displacement current is $0.15$ A.

(c) Kirchhoff's first rule

Yes, Kirchhoff's junction rule is valid at each plate of the capacitor if we include displacement current.

At the plates, conduction current enters the capacitor plates, while between the plates there is displacement current of the same magnitude. Therefore, there is no accumulation of charge at the junction in a way that violates current continuity. The generalized Ampere-Maxwell law makes the current the same through any surface bounded by the same loop, so the junction rule remains valid.