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.

(i) Power = +2 dioptre.
(ii) Here, we have f = +12 cm, R₁ = +10 cm, R₂ = -15 cm.
Refractive index of air is taken as unity.
We use the lens formula. The sign convention has to be applied for f, R₁ and R₂.
Substituting the values, we have
\(\frac { 1 }{ 12 } =(n-1)\left( \frac { 1 }{ 10 } -\frac { 1 }{ 15 } \right) \)
This gives n = 1.5.
(iii) For a glass lens in air, n₂ = 1.5, n₁ = 1, f = +20 cm. Hence, the lens formula gives
\(\frac { 1 }{ 20 } =0.5\left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
For the same glass lens in water, n₂ = 1.5, n₁ = 1.33. Therefore \(\frac { 1.33 }{ f } =(1.5-1.33)\left[ \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right] \)
Combining these two equations, we find f = + 78.2 cm.

The focal length f = -15/2 cm = -7.5 cm
(i) The object distance u = -10 cm. Then Eq gives
\(\frac { 1 }{ v } +\frac { 1 }{ 10 } =\frac { 1 }{ -7.5 } \)
or \(v=\frac { 10\times 7.5 }{ -2.5 } =-30\) cm
The image is 30 cm from the mirror on the same side as the object
Also, magnification m = \(\frac { v }{ u } =-\frac { (-30) }{ (-10) } =-3\)
The image is magnified, real and inverted.
(ii) The object distance u = -5 cm. Then from Eq
\(\frac { 1 }{ v } +\frac { 1 }{ -5 } =\frac { 1 }{ -7.5 } \)
or \(v=\frac { 5\times 7.5 }{ (7.5-5) } =15\) cm
This image is formed at 15 cm behind the mirror. It is a virtual image.
Magnification m = 15 \(-\frac { v }{ u } =-\frac { 15 }{ (-5) } =3\)
The image is magnified, virtual and erect.

Since drift velocity \({ v }_{ d }\) and current, I flowing in a conductor are related by the relation:
\({ v }_{ d }=\frac { I }{ neA } ......(1)\)
Also drift velocity in terms of average relaxation time T is given by
\({ v }_{ d }=\frac { eE\tau }{ m } .......(2)\)
From (1) and (2), we have
\(\frac { eE\tau }{ m } =\frac { I }{ neA } \)
\(or\quad \frac { E }{ I } =\frac { m }{ n{ e }^{ 2 }A\tau }\)
\(or\quad \frac { V }{ lI } =\frac { m }{ n{ e }^{ 2 }A\tau }\)
\(or\quad \frac { V }{ I } =\frac { ml }{ n{ e }^{ 2 }A\tau } ........(3)\)
The R.H.S. of Eq(3) is constant
\( \frac { V }{ I } = \ Constant\)
This is Ohm′s law
\((But \ \frac { V }{ I } =R,\)the resistance of the conductor)
So Equation (3) becomes
\(R=\frac { ml }{ n{ e }^{ 2 }A\tau } \)
\(But\quad R=\rho \frac { l }{ A }\)
\(\therefore \quad \rho \frac { l }{ A } =\frac { ml }{ n{ e }^{ 2 }A\tau }\)
\(\rho =\frac { m }{ n{ e }^{ 2 }A\tau } )\)

Charge located at the origin, q = 8 mC = 8 x 10^-3 C
The magnitude of the charge taken from the point P to R and then to Q, q₁ = 2 x 10^-9 C
Here OP= d₁= 3 cm = 3 x 10^-2 m
OQ = d₂= 4 cm = 4 x 10^-2 m
Potential at the point P, V₁ \(=\frac{q}{4 \pi \epsilon_{0} d_{1}}\) 

Potential at the point Q, V₁ \(=\frac{q}{4 \pi \epsilon_{0} d_{1}}\)
The work done (W) is independent of the path
Therefore, W = q₁[V₁ – V₂]
\(=V_{2}=q_{1}\left[\frac{q}{4 \pi \epsilon_{0} d_{2}}-\frac{q}{4 \pi \epsilon_{0} d_{1}}\right]\)
\(=V_{2}=\frac{q q_{1}}{4 \pi \epsilon_{0}}\left[\frac{1}{d_{2}}-\frac{1}{d_{1}}\right]\)
Where, \(\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}\)
Therefore,
\(W=9 \times 10^{9} \times 8 \times 10^{-3} \times\left(-2 \times 10^{-9}\right)\left[\frac{1}{4 \times 10^{-2}}-\frac{1}{3 \times 10^{-2}}\right]\)
= -144 x 10^-3 x (-100/12)
= 1.2  Joule
Therefore, the work done during the process is 1.2 J

(i) In n-type semiconductor, the semiconductor is doped with pentavalent impurity. The electrons are majority carriers and holes are minority carriers or n_e >> n_h , (n_e = number density of electrons, n_h = number density of holes).
In energy band diagram of n-type semiconductor, the donor energy level E_D is slightly below the bottom of conduction band E_c and thus. the electron can move to conduction band, even with small supply of energy.
  
(ii) In p-type semiconductor, the semiconductor is doped with trivalent impurity. The holes are the majority carriers and electrons are the minority carriers, i.e. n_b >> n_e . In energy band diagram of p-type, the acceptor energy level E_A is slightly above the top of valence band E_V .
Thus, even with small supply of energy, electron from valence band can jump to level E₄ and ionise the acceptor, negatively.

The area of cross-section of the meter bridge wire should be uniform otherwise the resistance per unit length of bridge wire will be different over different length of meter bridge.

Let two coherent sources of light, S₁ and S₂ (narrow slits) are derived from a source S. The, two slits, S₁ and S ₂ are equidistant from source, S. Now, suppose S₁ and S₂ are separated by distance d. The slits and screen are distance D apart.

Considering any arbitrary point P on the screen at a distance Yn from the centre O. The path difference between interfering waves is given by S₂P - S₁P
i.e. Path difference = S₂P - S₁P = S₂M
S₂P - S = d sin \(\theta\)
where, S1 M \(\bot\) S2 P
[ \(\because\) \(\angle\) S₂S₂ M = \(\angle\) OCP ( by geometry) 
\(\Rightarrow\)  S₁P = PM \(\Rightarrow\) S₂P = S₂ M ]
If \(\theta\) is small then sin \(\theta \approx \theta\approx tan\ \theta\)
\(\therefore\) Path difference,
S₂P - S₁ P = S₂ M = dsin \(\theta\) d tan \(\theta\)
Path difference = d \(\left( {{{y}_{n}}\over{D}} \right)\)                              ...(i)
 [ \(\because\) In \(\triangle\)PCO, tan \(\theta={{OP}\over{CO}}={{{y}_{n}}\over{D}}\) ]
For constructive interference
Path difference = n\(\lambda\) where, n = 0, 1, 2,....
[from Eq. (i)]
\(\Rightarrow\)   \({y}_{n}={{Dn\lambda}\over{d}}\)
\(\Rightarrow\)  \({y}_{n+1}={{D(n+1)\lambda}\over{d}}\)
\(\therefore\)   Fringe width of dark fringe = Y_n+1 - Y_n
[ \(\because\) Dark fringe exist between two bright fringes]
\(\beta={{D\lambda}\over{d}}(n+1)-{{Dn\lambda}\over{d}}\)
= \({{D\lambda}\over{d}}(n+1-n)={{D\lambda}\over{d}}\)
Fringe width of dark fringe, \(\beta={{D\lambda}\over{d}}\)       ....(ii)
For destructive interference
Path difference = (2n -1)\({{\lambda}\over{2}},\) where n = 1, 2, 3, .....
\(\Rightarrow\)   \({{{y'}_{n}d}\over{D}}=(2n-1){{}\lambda{}}\)      [ from Eq.(i)]
\(\Rightarrow\)   \({y'}_{n}={{(2n-1)D\lambda}\over{2d }}\)
where, y'_n is the separation of nth order dark fringe from central fringe.
\(\therefore\)     \({y'}_{n+1}=(2n+1){{D\lambda}\over{}2d}\)
\(\therefore\)  Fringe width of bright fringe = Separation between (n + l)th and nth order dark fringe from centered fringe.
\(\Rightarrow\)  \(\beta = {y'}_{n+1}-{y'}_{n}\)
or   \(\beta={{(2n+1)D\lambda}\over{2d}}-{{(2n-1)D\lambda}\over{2d}}\)
= \({{D\lambda}\over{2d}}[2n+1-2+1]={{D\lambda}\over{2d}}\)
= \({{D\lambda}\over{2d}}[2n+1-2n+1]={{D\lambda}\over{2d}}\)
Fringe width of bright fringe, \(\beta={{D\lambda}\over{d}}\)   ...(iii)
From Eqs. (ii) and (iii), we can see that, fringe width of dark fringe = fringe width bright fringe \(\beta={{D\lambda}\over{d}}\) 

Given, mutal inductance, M = 10 H
Change in current, dl = 3 A
Change in time, dt = 1 millisecond = 10^-3s
Emf induced, e = ?
Emf induced in secondary coil is given by e = \(\frac { M\ dl }{ dt } \)
Emf induced = \(\frac { 10 \times 3 }{ 10^{ -3 } } \) = 3 \( \times \) 10⁴ V

The masses of three isotopes are 19.99u, 20.99u, 21.99u
Their relative abundances are 90.51%, 0.27% and 9.22% 
\(\therefore average \ atomic \ mass \ of \ Neon \ is\)
\(m=\frac { 90.51\times 19.99+0.27\times 20.99+9.22\times 21.99 }{ (90.51+0.27+9.22) }\)
\(=\frac { 1809.29+5.67+202.75 }{ 100 } \)
\( =\frac { 2017.7 }{ 100 } \)
\(=20.17u\)