Distance between the image (screen) and the object, D = 90 cm
Distance between two locations of the convex lens, d = 20 cm
Focal length of the lens = f
Focal length is related to d and D as:
f = \(\frac { { D }^{ 2 }-{ d }^{ 2 } }{ 4D } \)
= \(\frac { { (90) }^{ 2 }-({ 20) }^{ 2 } }{ 4\times 90 } =\frac { 770 }{ 36 } =21.3\) cm
Therefore, the focal length of the convex lens is 21.39 cm.

The ray diagram for the formation of the image of the phone is shown in Fig. The image of the part which is on the plane perpendicular to principal axis will be on the same plane. It will be of the same size, i.e., B'C = BC. You can yourself realise why the image is distorted.

The 4 bonding electrons of C, Si or Ge lie, respectively, in the second, third and fourth orbit. Hence, energy required to take out an electron from these atoms (i.e., ionisation energy Eg) will be least for Ge, followed by Si and highest for C. Hence, number of free electrons for conduction in Ge and Si are significant but negligibly small for C.

63% tin + 37% lead.

De-Broglie wavelength associated with a body of mass m, moving with velocity v is given by.\(\lambda =\frac { h }{ mv } \).
Since the mass of the objects used in our daily life is very large, hence the de-Broglie wavelength associated with them is quite small and is not visible. Hence the wave nature of matter is not more apparent to our daily observations.

(i) The given circuit can be redrawn as shown in the figure.

As \(\frac{R_{1}}{R_{2}}=\frac{R_{3}}{R_{4}}\)
The circuit is a balanced wheatstone bridge.
\(\therefore \ V_{C}=V_{D} \text { and } I_{C D}=0\)
Hence, an equivalent circuit is redrawn as shown. 
Thus, \(R_{A B}=\frac{(2 R)(2 R)}{4 R}=R \Omega\)
(ii) Being a balanced wheatstone bridge, l_CD = 0

Given: \( R=2 \Omega, V_{A B}=10 \mathrm{~V} \)
\(R_{A C B}=4 \Omega \)
\(I_{A C B}=\frac{10}{4}=2.5 \mathrm{~A}\)

(a) dl and r for each element of the straight segments are parallel. Therefore, dl x r = 0. Straight segments do not contribute to |B|.
(b) For all segments of the semicircular arc, dl x r are all parallel to each other (into the plane of the paper). All such contributions add up in magnitude. Hence direction of B for a semicircular arc is given by the right-hand rule and magnitude is half that of a circular loop. Thus B is 1.9 x 10^–4 T normal to the plane of the paper going into it.
(c) Same magnitude of B but opposite in direction to that in (b).

(a) As \(V \propto \frac{1}{r}, V_{P}>V_{Q^{}}\) Thus, (V_P – V_Q) is positive. Also V_B is less negative than V_A . Thus, V_B > V_A or (V_B – V_A) is positive.
(b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive. Similarly, (P.E.)_A > (P.E.)_B and hence sign of potential energy differences is positive.
(c) In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative.
(d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive.
(e) Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going from B to A.

we know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius 5.3 × 10^–11 m is 2.2 × 10^–6 m/s. Thus, the frequency of the electron moving around the proton is
\(v=\frac{v}{2 \pi r}=\frac{2.2 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}}{2 \pi\left(5.3 \times 10^{-11} \mathrm{~m}\right)}\)
\(\approx \) 6.6 × 10¹⁵ Hz
According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of its revolution around the nucleus. Thus the initial frequency of the light emitted is 6.6 × 10¹⁵ Hz.

Let a current I₂ flow through the outer circular coil. The field at the centre of the coil is B₂ = μ₀I₂ / 2r₂. Since the other co-axially placed coil has a very small radius, B₂ may be considered constant over its cross-sectional area. Hence,
\(\Phi_{1}=\pi r_{1}^{2} B_{2}\)
\(=\frac{\mu_{0} \pi r_{1}^{2}}{2 r_{2}} I_{2}\)
= M₁₂ I₂
Thus_,
\(M_{12}=\frac{\mu_{0} \pi r_{1}^{2}}{2 r_{2}}\)
From Eq
\(M_{12}=M_{21}=\frac{\mu_{0} \pi r_{1}^{2}}{2 r_{2}}\)
Note that we calculated M₁₂ from an approximate value of Φ₁, assuming the magnetic field B₂ to be uniform over the area πr₁². However, we can accept this value because r₁ << r₂.

Density of nucleus (of water) 
\(\rho =\frac { 3m }{ 4\pi { R }_{ 0 }^{ 3 } } =\frac { 3\times 1.67\times { 10 }^{ -27 } }{ 4\times \frac { 22 }{ 7 } { \left( 1.2\times { 10 }^{ -15 } \right) }^{ 3 } }\)
\(=\frac { 7\times 3\times 1.67\times { 10 }^{ 18 } }{ 88\times 1.2\times 1.2\times 1.2 } =2.307{ \times 10 }^{ 17 }kg/{ m }^{ 3 }\)
Density of water,  \({ \rho \prime =10 }^{ 3 }kg/{ m }^{ 3 }\)
\(\therefore \quad \frac { \rho }{ \rho \prime } =\frac { 2.307\times { 10 }^{ 17 } }{ { 10 }^{ 3 } } =2.307\times { 10 }^{ 14 }\)

Given, amplitude of the magnetic field part of harmonic electromagnetic wave,
B₀ = 510 nT = 510 \(\times\)10^-9 T
Speed of light in a vacuum, c = 3 × 10⁸ m/s
Amplitude of electric field of the electromagnetic wave is given by the relation,
E = cB₀
= 3 × 10⁸ × 510 × 10⁻⁹ = 153 N/C
Therefore, the electric field part of the wave is 153 N/C.

(a) We know that P = I V cos\(\phi \) where cos\(\phi \) is the power factor. To supply a given power at a given voltage, if cos\(\phi \) is small, we have to increase current accordingly. But this will lead to large power loss (I²R) in transmission.
(b) Suppose in a circuit, current I lags the voltage by an angle \(\phi \). Then power factor \(\phi \) = R/Z.
We can improve the power factor (tending to 1) by making Z tend to R. Let us understand, with the help of a phasor diagram.

how this can be achieved. Let us resolve I into two components. I_p along the applied voltage V and I_q perpendicular to the applied voltage. I_q as you have learnt in Section 7.7, is called the wattless component since corresponding to this component of current, there is no power loss. I_P is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit.
It’s clear from this analysis that if we want to improve power factor, we must completely neutralize the lagging wattless current I_q by an equal leading wattless current I'_q. This can be done by connecting a capacitor of appropriate value in parallel so that I_q and I′_q cancel each other and P is effectively I_p V.

Focal length of the objective lens, f₁ = 2.0 cm
Focal length of the eyepiece, f₂ = 6.25 cm
Distance between the objective lens and the eyepiece, d = 15 cm
(a) Least distance of distinct vision,  d' = 25 cm
∴ Image distance for the eyepiece, v₂ = -25 cm
Object distance for the eyepiece = u₂
According to the lens formula, we have the relation
\(\frac { 1 }{ { v }_{ 2 } } -\frac { 1 }{ { u }_{ 2 } } =\frac { 1 }{ { f }_{ 2 } } \)
\(\frac { 1 }{ { u }_{ 2 } } =\frac { 1 }{ { v }_{ 2 } } -\frac { 1 }{ { f }_{ 2 } } \)
= \(\frac { 1 }{ -25 } -\frac { 1 }{ 6.25 } =\frac { -1-4 }{ 25 } =\frac { -5 }{ 25 } \)
∴ u₂ = -5 cm
Image distance for the objective lens,
v₁ = d + u₂ = 15 - 5 = 10 cm
Object distance for the objective lens = u₁
According to the lens formula, we have the relation:
\(\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { u }_{ 1 } } =\frac { 1 }{ { f }_{ 1 } } \)
\(\frac { 1 }{ { u }_{ 1 } } =\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { f }_{ 1 } } \)
= \(\frac { 1 }{ 10 } -\frac { 1 }{ 2 } =\frac { 1-5 }{ 10 } =\frac { -4 }{ 10 } \)
\(\therefore { u }_{ 1 }=\)-2.5 cm
Magnitude of the object distance, |u₁| = 2.5 cm
The magnifying power of a compound microscope is given by the relation:
\(m=\frac { { v }_{ 1 } }{ \left| { u }_{ 1 } \right| } \left( 1+\frac { { d }^{ ' } }{ { f }_{ 2 } } \right) \)
= \(\frac { 10 }{ 2.5 } \left( 1+\frac { 25 }{ 6.25 } \right) =4(1+4)=20\)
Hence, the magnifying power of the microscope is 20.
(b) The final image is formed at infinity.
∴ Image distance for the eyepiece, v₂ = ∞
Object distance for the eyepiece = u₂
According to the lens formula, we have the relation:
\(\frac { 1 }{ { u }_{ 2 } } =\frac { 1 }{ { v }_{ 2 } } -\frac { 1 }{ { f }_{ 2 } } \)
\(\frac { 1 }{ \infty } -\frac { 1 }{ { u }_{ 2 } } =\frac { 1 }{ 6.25 } \)
∴ u₂ = -6.25 cm
Image distance for the objective lens,
v₁ = d + u₂ = 15 - 6.25 = 8.75 cm
Object distance for the objective lens = u₁
According to the lens formula, we have the relation:
\(\frac { 1 }{ { u }_{ 1 } } =\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { f }_{ 1 } } \)
\(\frac { 1 }{ { u }_{ 1 } } =\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { f }_{ 1 } } \)
\(=\frac { 1 }{ 8.75 } -\frac { 1 }{ 2.0 } =\frac { 2-8.75 }{ 17.5 } \)
Magnitude of the object distance,  |u₁| = 2.59 cm
The magnifying power of a compound microscope is given by the relation:
\(m=\frac { { v }_{ 1 } }{ \left| { u }_{ 1 } \right| } \left( \frac { { d }^{ ' } }{ \left| { u }_{ 2 } \right| } \right) \)
\(=\frac { 8.75 }{ 2.59 } \times \frac { 25 }{ 6.25 } =13.51\)
Hence, the magnifying power of the microscope is 13.51.

\(V=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2} \times \frac{4 \times 10^{-7} \mathrm{C}}{0.09 \mathrm{~m}}\)
= 4 x 10⁴ V 
(b) W = qV = 2 x 10⁻⁹C x 4 x 10⁴ V
= 8 x 10^–5 J
No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along r and another perpendicular to r. The work done corresponding to the later will be zero.

(i) (b): Here, \(f_{0}=2.0, f_{e}=6.25 \mathrm{~cm}, u_{0}=?\) 
When the final image is obtained at the least distance of distinct vision:
V_e = - 25 cm
\(\text {As } \frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}} \)
\(\therefore \ \frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}=\frac{1}{-25}-\frac{1}{6.25} \)
\(=\frac{-1-4}{25}=\frac{-5}{25}=-\frac{1}{5} \)
\(\text {or } u_{e}=-5 \mathrm{~cm}\)
(ii) (b): Distance between objective and eye-piece = 15cm
\(\therefore\) Distance of the image from objective is
\(v_{0}=15-5=10 \mathrm{~cm} \)
\(\therefore \quad \frac{1}{u_{0}}=\frac{1}{v_{0}}-\frac{1}{f_{0}}=\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=-\frac{2}{5} \)
\(\text {or } \ u_{0}=-\frac{5}{2}=-2.5 \mathrm{~cm}\)
\(\therefore\) Distance of object from objective = 2.5 cm
(iii) (a): Magnifying power 
\(m=m_{0} \times m_{e}=\frac{v_{0}}{u_{0}}\left(1+\frac{D}{f_{e}}\right)=\frac{10}{2.5}\left(1+\frac{25}{6.25}\right)=20\)
(iv) (a): The intermediate image formed, by the objective of a compound microscope is real, inverted and magnified.
(v) (d)

(i) (d): \(E=\frac{V}{l}=\frac{2}{0.1}=20 \mathrm{~V} / \mathrm{m} ;\)
\(A=1.0 \mathrm{~cm}^{2}=1.0 \times 10^{-4} \mathrm{~m}^{2}\)
\(\begin{array}{l} v_{e}=\mu_{e} E=0.14 \times 20=2.8 \mathrm{~m} \mathrm{~s}^{-1} \\ I_{e}=n_{e} A e v_{e} \end{array}\)
\(\begin{array}{l} =\left(1.5 \times 10^{16}\right) \times\left(1.0 \times 10^{-4}\right) \times\left(1.6 \times 10^{-19}\right) \times 2.8 \\ =6.72 \times 10^{-7} \mathrm{~A} \end{array}\)
(ii) (c): In a pure semiconductor,
\(n_{e}=n_{h}=1.5 \times 10^{16} \mathrm{~m}^{-3}\)
\(\begin{array}{l} v_{h}=\mu_{h} \times E=0.05 \times 20=1.0 \mathrm{~ms}^{-1} \\ I_{h}=n_{h} \mathrm{Aev}_{h} \end{array}\)
\(\begin{array}{l} =\left(1.5 \times 10^{16}\right) \times\left(1.0 \times 10^{-4}\right) \times\left(1.6 \times 10^{-19}\right) \times 1.0 \\ =2.4 \times 10^{-7} \mathrm{~A} \end{array}\)
(iii) (a): \(\sigma=e n_{e} m_{e}\)
\(\text { or } n_{e}=\frac{\sigma}{e \mu_{e}}=\frac{6.4 \times 10^{2}}{\left(1.6 \times 10^{-19}\right) \times 0.14}\)
\(=3.14 \times 10^{22} \approx 3 \times 10^{22} \mathrm{~m}^{-3}\)
(iv) (c): \(n_{e}=\frac{n_{i}^{2}}{n_{h}}=\frac{\left(1.5 \times 10^{16}\right)^{2}}{4.5 \times 10^{22}}=5 \times 10^{9} \mathrm{~m}^{-3}\)
(v) (c): In case of a p-n junction diode, width of the depletion region decreases as the forward bias voltage increases.