Ionisation energy is the minimum energy required to knock out an electron from an atom.Its value will be different for different atoms. Ionisation energy will also depend on the orbit from which electron is to be removed.
When an electron in hydrogen atom is replaced by a particle 200 times heavier than electron but having the same charge, ionisation energy will not change, as it depends only on charge and not on mass of particle.

A choke coil reduces the voltage across the fluorescent tube without wastage of power.

The radius R of atomic nucleus related to mass number A of the nucleus as \(R=R_0 A^{1 / 3} \text { where } R_0=1.2 \times 10^{-15} \mathrm{~m}\) an empty constant.

No, Total internal reflection cannot occur.

Resultant amplitude = 3 + 2 = 5mm

\(\Delta x=1 \mathrm{~nm}=10^{-9} \mathrm{~m}, \Delta p=? \text { As, } \Delta x \Delta p \approx \hbar\)
\( \therefore \quad \Delta p=\frac{\hbar}{\Delta x}=\frac{h}{2 \pi \Delta x}=\frac{6.6 \times 10^{-34} \mathrm{~J}-8}{2 \times(22 / 7)\left(10^{-9}\right) \mathrm{m}} \\ =1.05 \times 10^{-25} \mathrm{~kg}-\mathrm{m} / \mathrm{s} \\ \text { Energy }(E)=\frac{p^2}{2 m}=\frac{(\Delta p)^2}{2 m} \quad[\because p \approx \Delta p] \\ =\frac{\left(1.05 \times 10^{-25}\right)^2}{2 \times 9.1 \times 10^{-31}} \mathrm{~J} \\ =\frac{\left(1.05 \times 10^{-25}\right)^2}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-10}} \mathrm{eV} \\ =3.8 \times 10^{-2} \mathrm{eV} x\)

\(K E=\frac{1}{2} \frac{p^2}{m}=\frac{1}{2} \times \frac{\left(6.63 \times 10^{-24}\right)^2}{1.67 \times 10^{-27}} J=\frac{1}{2} \frac{\left(6.63 \times 10^{-24}\right)^2}{1.67 \times 10^{-27} \times\left(1.6 \times 10^{-19}\right)} \mathrm{eV}=0.082 \mathrm{eV}\)

Self-inductance is the property of a coil by virtue of which, the coil opposes any change in the strength of the current flowing through it by including an emf in itself.
Its SI unit is henry(H). 

According to Lenz's law, the direction of induced current will oppose the cause of its production. So, the current in the loop will induce in such a way that it will support the current flowing in the wire, i.e in the same direction. So, the direction of current in the loop will be clockwise.

(i) As the electric slab is introduced between the plates of the capacitor, its capacitance. Hence, the potential drop across the capacitor will i.e \(V=\frac { Q }{ C } \)
 As a result, the potential drop across the bulb will increase as they are connected in series. Thus its brightness will increase.
(ii) As a resistance R is increased, the potential drop across the resistor will increase. As a result, the potential drop across the bulb will decrease as they are connected in series. Thus, its brightness will decrease.  

(i) The frequency of reflected and refracted light remains same as that of incident light because frequency only depends on the source of light.
(ii) Since the frequency remains same, hence there is no reduction in energy.

(i) The de-Broglie wavelength of a particle is given by
 \(\lambda =\frac { 12.27 }{ \sqrt { V_{ } } } \mathring { A } \)
[Where, V is the accelerating potential of the particle]
\(\because \quad \lambda _{ p }=\lambda _{ \alpha }\)                              [given]
\(=\frac { 12.27 }{ \sqrt { V_{ p } } } =\frac { 12.27 }{ \sqrt { V_{ \alpha } } }\)
\( \Rightarrow \frac { V_{ p } }{ V_{ \alpha } } =1\)
(ii) The de-Broglie wavelength of the particle is given by
\(\lambda =\frac { h }{ mv } \)
\( \lambda _{ p }=\frac { h }{ m_{ p }.v_{ p } } \ and \ \lambda _{ \alpha }=\frac { h }{ m_{ \alpha }v_{ \alpha } } \)
We know that, \(\\ m_{ \alpha }=4 m _{ p }\)
\( \because \lambda_{p}=\lambda_{\alpha} \)                            [given]
\( \therefore \frac{h}{m_{p} \cdot v_{p}}=\frac{h}{4 m_{p} \cdot v_{\alpha}} \Rightarrow \frac{v_{p}}{v_{\alpha}}=4 \)

A biconvex lens acts as a converging lens in air because the refractive index of air is less than that of the material of the lens. The refractive index of water is less than the refractive index of the material of the lens (1.5). So, its nature will not change, it behaves as a converging lens.

You may think that the image will now show only half of the object, but taking the laws of reflection to be true for all points of the remaining part of the mirror, the image will be that of the whole object. However, as the area of the reflecting surface has been reduced, the intensity of the image will be low (in this case, half).

\(Here,\ \frac { N }{ { N }_{ 0 } } =\frac { 1 }{ 16 } \ t=4 \ days\)
\(As\quad \frac { N }{ { N }_{ 0 } } ={ \left( \frac { 1 }{ 2 } \right) }^{ n }=\frac { 1 }{ 16 } ={ \left( \frac { 1 }{ 2 } \right) }^{ 4 }\therefore n=4\)
\(Half-life\quad T=\frac { t }{ n } =\frac { 4 }{ 4 } =1\ day\)
\(Now, \ N=4\times { 10 }^{ -3 }kg,\ t=6\ days,{ N }_{ 0 }=?\)
\(As \ \frac { N }{ { N }_{ 0 } } ={ \left( \frac { 1 }{ 2 } \right) }^{ n' }={ \left( \frac { 1 }{ 2 } \right) }^{ 6/1 }=\frac { 1 }{ 64 } \)
\( \therefore \ { N }_{ 0 }=64N=64\times 4\times { 10 }^{ -3 }=0.256\ kg\)

Here, \(f_{ 0 }=1.25\ cm,\ f_{ e }=5\ cm, d=25\ cm,\ L=?,\ M=30\)
When final image is formed at the near point of eye, the magnifying power of compound microscope is given by
\(M=\frac { L }{ f_{ 0 } } \left( 1+\frac { d }{ f_{ e } } \right) \)
\( 30=\frac { L }{ 1.25 } \left( 1+\frac { 25 }{ 5 } \right) =\frac { 6L }{ 1.25 } \)
\( L=\frac { 30\times 1.25 }{ 6 } =6.25\ cm\)

In India, the voltage and frequency of mains supply is 220 V and 50 Hz, respectively.
Since the power loss at 220 V supply is less than that at 110 V. So, due to this reason, most of the electrical devices require AC voltage.
Since the transformer requires an alternating magnetic flux to operate correctly, transformers cannot, therefore, be used to transform or supply DC voltages or currents. Also, the magnetic field must be changing to induce a voltage in the secondary winding. Even if in case, the transformer's primary winding is connected to a DC supply, the inductive reactance of the winding would be zero as DC has no frequency. So, the effective impedence of the winding will therfore be very low and equal only to the resistance of the copper used. Thus the winding will draw a very high current from the DC supply causing it to overheat and eventually burn out, because as, we know \(I=\frac { V }{ R } \). Anil is very helpful and has a great presence of mind as in such condition, he used the wooden baton to switch OFF the main supply.

(i) The teacher displays the quality of the knowledge of the phenomenon and the conditions under which it occurs.In producing the diffraction pattern again, he demonstrates with compassion, kindness towards the child and eagerness to share the knowledge.
(ii) The necessary condition for diffraction is that the slit width should be less than wavelength of light.

The first refracting ABC forms the image I₁ of the object O. The image I₁ acts as virtual object for the second refracting surface ADC, which forms the real image I as shown in the diagram
For refraction at ABC 
\(\frac { { n }_{ 2 } }{ { v }_{ 1 } } -\frac { { n }_{ 1 } }{ u } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }_{ 1 } } \)
For refraction at ADC
\(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ { v }_{ 1 } } =\frac { { n }_{ 1 }-{ n }_{ 2 } }{ { R }_{ 2 } } \)
Adding equation (i) and equation (ii)
 \(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
We know, If  \(u=\infty ,v=f\)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( 1.55-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ -R } \right) \)
\(=0.55\times \frac { 2 }{ R } \)
\(R=0.55\times 2\times 20=22 \ cm\) 

Size of the candle, h = 2.5 cm
Image size = h’
Object distance, u = -27 cm
Radius of curvature of the concave mirror, R = -36 cm
\(f=\frac { R }{ 2 } =-18\)cm
Image distance = v
The image distance can be obtained using the mirror formula:
\(\frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } \)
\(\frac { 1 }{ v } =\frac { 1 }{ f } -\frac { 1 }{ u } \)
\(\frac { 1 }{ -18 } =\frac { 1 }{ -27 } =\frac { -3+2 }{ 54 } =-\frac { 1 }{ 54 } \)
∴ v = -54 cm
Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.
The magnification of the image is given as:
\(m=\frac { { h }^{ ' } }{ h } =-\frac { v }{ u } \)
\(\therefore { h }^{ ' }=-\frac { v }{ u } \times h\)
\(=-\left( \frac { -54 }{ -27 } \right) \times 2.5=-5\)cm
The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and real.
If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.

This experiment confirms the wave nature of electron.
(b)   \(\lambda =\frac { h }{ p } =\frac { h }{ \sqrt { 2mK } }\)
\( K=K.E=eV\)
\(\lambda =\frac { h }{ \sqrt { 2meV } } \)

Given, magnification, M = 20
Magnification of eyepiece, me = 5
Least distance vision, D = 20 cm
Distance between objective and' eyepiece,
L = 14 cm
We know that,
Magnification, M = m_e x m_o
\(m_{ e }=\frac { m }{ m_{ e }= } =\frac { 20 }{ 5 } =4\Rightarrow m_{ e }=1+\frac { D }{ f_{ e } } \)
where, f_e is focal length of eyepiece.
\(\Rightarrow 5=1+\frac { 20 }{ f } \Rightarrow f_{ e }=5cm\)
Using lens formula for eyepiece,
\(\frac { 1 }{ u_{ e } } =\frac { -1 }{ 20 } =\frac { 1 }{ 5 } =\frac { -5 }{ 20 } =\frac { -1 }{ 4 } \)
U = -4 cm
(objective distance for eyepiece)
L = V_o + I u, I = Ho = L -I u, 1= 14 - 4 = 10 cm
Magnification produced by objective,
\(m_{ e }=-\frac { v_{ e } }{ u_{ 0 } } \)
Object distance for objective,
\(u_{ e }=\frac { -v_{ e } }{ m_{ 0 } } =\frac { -10 }{ 4 } =-25cm\)
Using lens formula for objective,
\(\frac { 1 }{ f_{ 0 } } =\frac { 1 }{ v_{ 0 } } -\frac { 1 }{ u_{ 0 } } =\frac { 1 }{ 10 } -\frac { 1 }{ -2.5 } =\frac { 1 }{ 10 } +\frac { 1 }{ -2.5 } \Rightarrow f_{ 0 }=2cm\)

(a) The self inductance, L, of a coil equals the magnetic flux linked with it, when a unit current flows through it.
One henry is the self inductance of a coil which the magnetic flux, linked with it, due to a current of 1A, flowing in it, equals one weber
(b) Self inductance of a coil depends on
(i) Its geometry (area and length of a coil)
(ii) Number of turns
(iii) Number of turns in each coil
(iv) Nature of medium in the intervening space
(c) Mutual inductance of a given pair of coils depends on
(i) Their geometries
(ii) Their distance of separation
(iii) Number of turns in each coil.
(iv) Nature of medium in the intervening space.

(i) Mass defect (\(\Delta M\)) of any nucleus \(^A_Z X\) is the difference in the mass of the nucleus (=M) and the sum of masses of its constituent nucleons (=M').
\(\therefore\) \(\Delta M\) = M' - M
 = [Zm_p + (A - Z) m_n] - M
where m_p and m_n denote the mass of the proton and the neutron respectively.
(ii) Binding energy is the energy required to separate a nucleus into its constituent nucleons. The relation between the two is
B.E. = (mass defect) c².
(iii) There is a release of energy, i.e. the reaction is exothermic.
Reason: Increase in B.E/nucleon implies that the more mass has been converted into energy. This would result in release of energy.

Magnification produced by any lens,
m = v/u = f / f + u
given m = ± N   ±N = f / f + u
or f + u = ± f / N or u = - f ± f / N
hence magnitude of object distances,
|u| = f ± f / N
given P = 1/f = + 2.5 D
f = 1/ 2.5 = 0.4 m = 40 cm
Also N = 4
|u| = 40 ± 40/4 = 40 ± 10 = 50 cm or 30 cm