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.

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) A galvanometer of range I_g and resistance G₁ can be converted into
(a) a voltmeter of range V, by connecting a high resistance R in series with galvanometer whose value is given by
\(R=\frac { V }{ { I }_{ g } } -G\)
(b) an ammeter of range I, by connecting a very low resistance (shunt) in parallel with galvanometer whose value is given by
\(S=\frac { { I }_{ g }G }{ I-{ I }_{ g } } -G\)
Thus, the nature of force is attractive.
When direction of flow of current is in opposite direction, the nature of force becomes repulsive.

\(\therefore \ Here, \ { n }_{ p }=500, \ { n }_{ s }=1000 \ { E }_{ p }=200V, \ { E }_{ S }=100\Omega , \ { I }_{ p }=?\)
\(In \ an \ ideal \ transformer, \ \frac { { E }_{ s } }{ { E }_{ p } } =\frac { { n }_{ s } }{ { n }_{ p } } =\frac { 1000 }{ 500 } =2\)
\({ E }_{ S }=2 \ { E }_{ P }=2\times 200=400V\)
\(\ Also, \ \frac { { I }_{ p } }{ { I }_{ s } } =\frac { { n }_{ s } }{ { n }_{ p } } =2, \ { I }_{ p } \ = \ 2{ I }_{ s }=2\left( \frac { { E }_{ S } }{ { E }_{ S } } \right) =\frac { 2\times 400 }{ 100 } =8A\)

\(Here, \ { \lambda }_{ 1 }=4000\overset { \circ }{ A } =4\times { 10 }^{ -7 }m;{ V }_{ 1 }=0.82V;\)
\({ \lambda }_{ 2 }=3000\overset { \circ }{ A } =3\times { 10 }^{ -7 }m;{ V }_{ 2 }=1.85V\)
\( As \ \frac { hc }{ { \lambda }_{ 1 } } ={ \phi }_{ 1 }+e{ V }_{ 1 } \ and \ \frac { hc }{ { \lambda }_{ 2 } } ={ \phi }_{ 1 }+e{ V }_{ 2 }\)
\(\therefore \ \frac { hc }{ { \lambda }_{ 2 } } -\frac { hc }{ { \lambda }_{ 1 } } =e\left( { V }_{ 2 }-{ V }_{ 1 } \right)\)
\( or \ hc\left( \frac { { \lambda }_{ 1 }-{ \lambda }_{ 2 } }{ { \lambda }_{ 1 }{ \lambda }_{ 2 } } \right) =e\left( { V }_{ 2 }-{ V }_{ 1 } \right) \)
\(or \ \ h=\frac { e\left( { V }_{ 2 }-{ V }_{ 1 } \right) { \lambda }_{ 1 }{ \lambda }_{ 2 } }{ \left( { \lambda }_{ 1 }-{ \lambda }_{ 2 } \right) c }\)
\( =\frac { \left( 1.6\times { 10 }^{ -19 } \right) \times \left( 1.85-0.82 \right) \times 4\times { 10 }^{ -7 }\times 3\times { 10 }^{ -7 } }{ \left( 4\times { 10 }^{ -7 }-3\times { 10 }^{ -7 } \right) \times 3\times { 10 }^{ 8 } }\)
\( =6.592\times { 10 }^{ -34 }Js\)

Given, amplitude of an electromagnetic wave,
E₀ = 120 N/C
Frequency of wave, v = 50 MHz = 50 \(\times\) 10⁶ Hz
(i) Speed of light in vacuum, \(c=\frac{E_0}{B_0}\)
\(\begin{aligned} B_0=\frac{E_0}{c}= & \frac{120}{3 \times 10^8}=40 \times 10^{-8} \end{aligned}\)
= 400 \(\times\) 10^-9 T = 400 nT
Angular frequency of electromagnetic wave,
\(\omega=2 \pi \nu=2 \times 3.14 \times 50 \times 10^6\)
\(\omega=3.14 \times 10^8 \mathrm{rad} / \mathrm{s}\)
Wave number of electromagnetic wave,
\(k=\frac{\omega}{c}=\frac{3.14 \times 10^8}{3 \times 10^8}=1.05 \mathrm{rad} / \mathrm{m}\)
Wavelength of electromagnetic wave,
\(\lambda=\frac{c}{v}=\frac{3 \times 10^8}{50 \times 10^6}=6.00 \mathrm{~m}\)
(ii) Expression of electric field, E = E₀ sin (kx - \(\omega\)t)
E = 120 sin (1.05x - 3.14 \(\times\)10⁸ t)
Expression of magnetic field B,
B = B₀ sin (kx - \(\omega\)t)
E = 120 sin (kx - \(\omega\)t)
B = 4 \(\times\)10^-7 sin (1.05x - 3.14 \(\times\) 10⁸ t)

A = 60⁰ ,  \(\delta \)_m = 30⁰
i = e = ¾ A = 450
as A + \(\delta \) = i + e
60 + \(\delta \) = 45 +45
or \(\delta \) = 30⁰
Refractive index,
\(\mu \) = sin a + \(\delta \)m /2/sin A/2 = sin 60⁰+30⁰/2/sin 60⁰/2
= sin 45⁰/sin30⁰ = 1\(\surd 2\) 1/2 = \(\surd 2\) = 1.414

V = V1 + V2
Q = C1V = 6 x 10^-6  x 2 = 12\(\mu \)C
As C₂ is in series same amount of charge will also flow through it now V₂ = Q/C₂ = (12 x 10^-6) /(12 x 10^-6) = 1 volt
Total Battery voltage, V = 2 + 1 = 3 Volt

In standard AM broadcast, ground wave propagation is used for transmitting a signal. Attenuation of surface wave increase very rapidly with upto a few MHz. In AM broadcast, range of frequencies are limited to 30 MHz.

Given, \({ \delta }_{ V }={ 3.32 }^{ 0 },{ \delta }_{ Y }={ 3.27 }^{ 0 },{ \delta }_{ R }=\ { 3.22 }^{ 0 }\quad \) 
Dispersive power \(=\frac { { \delta }_{ V }-{ \delta }_{ R } }{ { \delta }_{ Y } } \)
\(\frac { { 3.32 }^{ 0 }-{ 3.22 }^{ 0 } }{ { 3.27 }^{ 0 } } =\frac { { 0.10 }^{ 0 } }{ { 3.27 }^{ 0 } } ={ 0.0306 }^{ 0 }\)

According to the question,

Applying Kirchhoff's junction rule at F,
 I₃ = \({ I }_{ 1 }\) +\({ I }_{ 2 }\)                ......(i)
Applying Kirchhoff's second rule in loop ABCF,
-30I₁+ 20-20I₃ = 0
\(\Rightarrow \) 3I₁+2I₃ = 2           ....(ii)
In loop ABDE,
-30I₁+20I₂ - 80 =0
3I₁ + 2I₂ = 8          ......(iii)  (1)
From Eqs. (i) and (ii), we have
3I1+ 2I1 + 2I2 = 2
\(\Rightarrow \) 5I1 + 2I₂ = 2         .....(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
8I₁ = -6
\(\Rightarrow \)I1 = -\(\frac { 3 }{ 4 } \) A              (1)    

On introducing a thin metal oil, d is halved. Arrangement is equivalent to two condensor each of capacity 2 C in series.
\({1\over C_s}={1\over 2C}={1\over 2C}={1\over C}\)
\(C_s=C\)
Capacity is unchanged.

\( \mathrm{V}=210 \mathrm{~V}, \mathrm{P}=630 \mathrm{~W} ; \\ \text {Current } \mathrm{I}=\frac{P}{V}=\frac{630}{210}=3 \mathrm{~A} \)

Kilowatt is the unit of power. 1 kilowatt = 100 watt. Kilowatt hour is the unit of electric energy, where
1 kilowatt hour = 1000 watt x 1 hour
= 1000 watt x 60 x 60 second
= 3.6 x 10⁶ J

(i) Microwaves
(ii) radio waves
(iii) Gamma rays
(iv) x-rays
Infrared are used in weather forecasting. Ultraviolet rays are used in burglar alarm. X-rays are used in scientific research

As hydrogen ion is describing a circular path in a uniform magnetic field, hence force on hydrogen ion due to perpendicular magnetic field is providing the centripetal force. So
 \(Bqvsin{ 90 }^{ o }=\frac { m{ v }^{ 2 } }{ r } \ or\ r=\frac { mv }{ qB } \) 
Time period of revolution of hydrogen ion
 \(T=\frac { 2\pi r }{ v } =\frac { 2\pi }{ v } \times \frac { mv }{ qB } =\frac { 2\pi m }{ qB } \) 

Eye is most sensitive to the light of wavelength \(\lambda=5600 \dot{A}\)
\( T=\frac{1}{v}=\frac{\lambda}{c}=\frac{5600 \times 10^{-10}}{3 \times 10^8} \\ =1.87 \times 10^{-15} \mathrm{~s} \)

In a.d.c. circuit, reactance of inductor/solenoid is zero. Therefore, removal of iron core does not affect the brightness of bulb.

When an intrinsic semiconductor is doped with the impurity atoms of valence five like As, P or Sb, some additional energy levels are produced, situated in the energy gap slightly below the conduction band which are called donor energy levels. Due to it, energy gap in semiconductor decreases.

This is done to increase the light gathering capacity and hence brightness of image.

Distance between the object and the image, d = 3 m
Maximum focal length of the convex lens = f_max
For real images, the maximum focal length is given as:
f_max = \(\frac{d}{4}\)
= \(\frac{3}{4}\) = 0.75 m
Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.

Diamagnetic material
\({ \mu }_{ r }=1+{ x }_{ m }\)

Given,  W_AB = 100 J,  q = 4C,  V_A =- 10 V,  VB = V = ?
 Since,  W_AB  = q(V_B - V_A )
\(\Rightarrow \) 100 = 4(V +10)   \(\Rightarrow \) V = 15 V

\({ \left( 100000.01 \right) }_{ 2 }{ ,\left( 11000.01 \right) }_{ 2 }{ ,\left( 1000 \right) }_{ 2 }\)