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.

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  \(u=\infty ,v=f\)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( 1.55-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ -R } \right) \)
\(=0.55\times \frac { 2 }{ R } \)
\(R=0.55\times 2\times 20=22 \ cm\) 

(i) We have, initial voltage, V₁ = V volt
and charge stored, Q₁ = 360 \(\mu\)C
Q₁ = CV₁                            ....(i)
Charged potential, V₂ = V - 120
Q₂ = 120 \(\mu\)C
\(\Rightarrow\) Q₂ = CV₂ = C(V - 120)  ....(ii)
By dividing Eq. (ii) from Eq. (i), we get
\(\frac{Q_1}{Q_2}=\frac{C V_1}{C V_2} \Rightarrow \frac{360}{120}=\frac{V}{V-120}\)
\(\Rightarrow\)  V = 180 V
\(\therefore \quad C=\frac{Q_1}{V_1}=\frac{360 \times 10^{-6}}{180}=2 \times 10^{-6} \mathrm{~F}\)
= 2 \(\mu\)F
Hence, the potential, V = 180 V and unknown capacitance is 2 \(\mu\)F.
(ii) If the voltage applied had increased by 120 V, then V₃ = 180 + 120 = 300 V
Hence, charge stored in the capacitor,
Q₃ = CV₃ = 2 \(\times\)10^-6 \(\times\)300 = 600 \(\mu\)C

(i) Magnetic field due to C at D is
\(B=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2I }{ r } ={ 10 }^{ -7 }\times \frac { 2\times 3 }{ 0.10 } =6\times { 10 }^{ -6 }T\) 
(ii) Force on D, F = BI₁ lsin\(\theta \) = (6 x 10^-6) x 4 x (5 x 10^-2) x sin 90^o
 = 1.2 x 10^-6 N

(a) E = 1.25V

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).

When a semi conductor is heated more & more electrons get enough energy to jump across the forbidden energy gap from valence band to the conduction band, where they are free to conduct electricity. Thereby increasing the conductivity of a semi conductor.

(i) 10^-12 m - 10^-8 m = 0.01 \(\overset{\circ}{A}\) - 100 \(\overset{\circ}{A}\) \(\rightarrow\) X-rays.
These are used in crystallography.
(ii) 10^-3 m - 10^-1 m = 0.1 cm - 10 cm \(\rightarrow\) Radio waves.
These are used in radio communication.

Force on upper horizontal side
\(=\frac {Il\mu_0 I_1}{2 \pi l}=\frac {\mu_0 {II}_1}{2 \pi }=\frac {\mu_0I I_1}{2 \pi }(attractive)\)
Force on lower horizontal side
\(=\frac {Il\mu_0 I_1}{2 \pi (2l)}=\frac {I\mu_0 {I}_1}{4 \pi }(repulsive)\)
The direction of these forces being opposite to each other therefore net force.
\(F=\frac {\mu_0I I_1}{2\pi}(attractive)\)
(the net force on the two vertical sides is zero)

(i) Reactance of the capacitor will decrease, resulting in increase of the current in the circuit, Therefore the bulb will glow brighter.
(ii) The bulb will go dimmer.

The amount of light energy / photon energy, incident per metre square per second is called intensity of joule/Sec-meter² radiation.
SI Unit: W/m² or Joule/Sec-meter² 

A conductor will become a superconductor if its temperature becomes equals to or less than its critical temperature.

9:1

Then,R = R₁, S = R₂, l =  \(\frac { 1 }{ 3 } \)m and (100 - l) = 1 - \(\frac { 1 }{ 3 } \) = \(\frac { 1 }{ 3 } \)m
\(\therefore \frac { R1 }{ R2 } =\frac { l }{ (1oo-l) } =\frac { 1/3 }{ 2/3 } =\frac { 1 }{ 2 }\) or R₂ = 2 R₁
So, R₁ < R₂
As 6 \(\Omega \) coil is connected in series with the smaller resistance R₁, so resistance of the left gap becomes, R₁ = R₁ + 6. Resistance of right gap is R₂ .
Now, \(l'=\frac { 2 }{ 3 } m\)
and (100 - l) = \(1-\frac { 2 }{ 3 } =\frac { 1 }{ 3 } m\)
\(\therefore \frac { R1' }{ R2 } =\frac { l' }{ (100-l') } or\frac { R1+6 }{ R2 } =\frac { 2/3 }{ 1/3 } =2\) 
or R₁+ 6 = 2 R₂ = 2 x 2R₁ = 4 R₁
and  R₂ = 2 R1 = 2 x 2 = 4\(\Omega \)
R₁ = 2\(\Omega \)

\(\tau =nIBA \ sin\alpha \ ; \ Here, \ n=200, \ I=2.5 \ A;\)
\( B=0.25 \ T \ ; \ \alpha ={ 90 }^{ o }-{ 60 }^{ o } \ ={ 30 }^{ o }\)
\(A=\pi { r }^{ 2 }=\frac { 22 }{ 7 } \times { \left( 0.05 \right) }^{ 2 }{ m }^{ 2 }\)

\(\quad \quad \quad \quad \quad \ d=2.5mm=2.5\times { 10 }^{ -3 }m\\ \quad \quad \quad \quad \quad \ A=90{ cm }^{ 2 }=90\times { 10 }^{ -4 }{ m }^{ 2 }\\ potential\quad \quad V=400\quad volt\)

(a) Electrostatic energy stored in the capacitor

        \(\quad \quad \quad \quad u=\frac { 1 }{ 2 } { CV }^{ 2 }=\frac { 1 }{ 2 } .\frac { { \varepsilon }_{ 0 }A }{ d } { V }^{ 2 }\\ or\quad \quad \quad u=\frac { 1 }{ 2 } .\frac { (8.85\times { 10 }^{ -12 })\times 90\times { 10 }^{ -4 }\times { (400) }^{ 2 } }{ 2.5\times { 10 }^{ -3 } } \\ or\quad \quad \quad u=\frac { 8.85\times 90\times 16 }{ 2\times 2.5 } \times { 10 }^{ -9 }=2.55\times { 10 }^{ -6 }J\\ (b)\quad Volume=A\times d\\ \quad \quad \quad \quad =90\times { 10 }^{ -4 }\times 2.5\times { 10 }^{ -3 }\\ \quad \quad \quad \quad =225\times { 10 }^{ -7 }{ m }^{ 3 }\)

Energy per unit volume,

   \(\quad \quad \quad \quad \quad u=\frac { 2.55\times { 10 }^{ -6 }J }{ 225\times { 10 }^{ -7 }{ m }^{ 3 } } =0.113{ Jm }^{ -3 }\\ Now\quad \quad \quad u=\frac { Energy }{ Volume } =\frac { 1 }{ 2 } \frac { { \varepsilon }_{ 0 }A.{ V }^{ 2 } }{ d\times A\times d } =\frac { 1 }{ 2 } \frac { { \varepsilon }_{ 0 }.{ V }^{ 2 } }{ { d }^{ 2 } } \\ or\quad \quad \quad \quad u=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }.{ E }^{ 2 }\)