Here
I_b = 100\(\mu \)A = 0.1 mA
I_c = 3mA
a) \(\beta\) = I_c/I_b = 30
\(\beta\) = \(\alpha\) /(1-\(\alpha\)) 
\(\alpha\) = 0.97
\(\alpha\) = I_c/I_e
Ie = 3.1 mA
b) \(\triangle\)I_b = 20 \(\mu \)A 0.02 mA
\(\beta\)a.c = \(\triangle\) I_c/\(\triangle\) I_b
\(\beta\) a.c = 25

Ps = 1/2 ma² Pc
1.246 = 1 + \(\frac { ma^{ 2 } }{ 2 } \)
ma²/2 = 0.246
m_a = (2 x 0.246)^1/2 = 0.701 = 70.1%

The drift velocity of a charge carrier is proportional to electric E. Therefore V = eET/m ie. V \(\alpha\) E. But V cannot be increased indefinitely by increasing E. At high-speed relaxation time (T) begins to decrease due to increase in collision frequency. S: so drift velocity saturates at thermal velocity (10 ms^-1). An electric field of 106 V/m causes saturation of drift velocity. Hence semi- conduction obey ohm’s law for low electrical field and above this field ( E < 10⁶ V/m ) current becomes independent of potential.

(i) 6Li3 + 1no \(\rightarrow\)3H1 + 4He2 +Q (energy)
(ii) Q = \(\triangle\)mx931 MeV
Where \(\triangle\)m = 6.01512+1.0086654-4.0026044-3.0100000

The resultant intensity at a point where phase difference is \(\Phi \)is
I_R = I₁ +I₂ + 2\(\surd \)1₁1₂Cos \(\Phi \)
As I₁ = I and I₂ = 4I Therefore
IR = I + 4I + 2\(\surd \)1.41 Cos \(\Phi \) = 5I+4I cos \(\Phi \)
(i) when \(\Phi \) =0 , I_R = 5I +4I cos 0 = 9 I
(ii) when \(\Phi \) =\(\pi \)/2 , I_R = 5I +4I cos \(\pi \)/2 = 5 I
(iii) when \(\Phi \) =\(\pi \) , I_R = 5I +4I cos \(\pi \) = I 

A wavefront is a surface obtained by joining all points vibrating in the same phase. A ray is a line drawn perpendicular to the wavefront in the direction of propagation of light wave.
The wavefronts of light emerging from a point source are spherical, as shown in figure. When a point source is placed at the focus of a convex lens, the emerging light has the plane wavefronts, as shown in figure

Let the number of turns be = n
Radius = r
Length = l
Length of the wire = circumference of n turns of coil
L = n x 2\(\Pi \) r
r = L/2\(\Pi \) r
Maximum torque = nIBA = nIB \(\Pi \) r²
= nIB\(\Pi \)  (2/2\(\Pi \) n)²
= 1/n
For maximum torque n should be minimum
i.e n = 1

q \(\rightarrow\)_______Q
1/2 mv2 = kQq/r
Or, v2 a1/r
Or, r a 1/v²
Or, r' = r/4

de-Broglie wavelength is given by
\(\lambda =\frac { h }{ \sqrt { 2mK } } =\frac { h }{ \sqrt { 2mqV } }\)
\(\lambda \propto \frac { 1 }{ \sqrt { mqV } } \)
m = mass of charge particle, q = charge and V = potential difference
Raio of de-Broglie wavelengths of proton and a-particle is given by
\(\frac { \lambda _{ \alpha } }{ \lambda _{ p } } =\sqrt { \frac { m_{ \alpha }q_{ \alpha }V_{ \alpha } }{ m_{ p }q_{ p }V_{ p } } } =\sqrt { \left( \frac { m_{ \alpha } }{ m_{ p } } \right) \left( \frac { q_{ \alpha } }{ q_{ p } } \right) \left( \frac { V_{ \alpha } }{ V_{ p } } \right) } \)
\(\frac { m_{ \alpha } }{ m_{ p } } =4,\frac { q_{ \alpha } }{ q_{ p } } =2,\frac { V_{ \alpha } }{ V_{ p } } =\frac { 1 }{ 2 } \)
a-particle is 4 toimes heavier than proton and it has double the charge than that of proton
\(\frac { \lambda _{ p } }{ \lambda _{ \alpha } } =\sqrt { 4\times 2\times \frac { 1 }{ 2 } } =2\quad \Rightarrow \lambda _{ p }:\lambda _{ \alpha }=2:1\)

Angle of dip, \(\delta ={ 60 }^{ 0 }=\frac { \pi }{ 3 } \)
Horizontal component of the earth's magnetic field,
H = 0.4 gauss
Earth magnetic field, \({ B }_{ e }\)= ?
We know that,   
\(\therefore \) Horizontal component of the earth's magnetic field,
\(\Rightarrow \quad { B }_{ e }=\frac { H }{ cos\delta } =\frac { 0.4 }{ cos{ 60 }^{ 0 } } =\frac { 0.4 }{ (\frac { 1 }{ 2 } ) } =0.8\)
\({ B }_{ e }=0.8\) gauss

Applying Kirchhoff's loop rule for loop ABEFA
6 + 3 + R₁ \(\times\) 0 - 3I₁ + 1 - 2I₁ = 0
or  10 - 5I₁ = 0
or I₁ = 2A
For loop BCDEB,
4 - I₁ . R + R₁  \(\times\) 0 - 3 = 0
or  1 - 2R = 0
\(\therefore \quad R=\frac{1}{2} \Omega\)

Potential difference between A and D through path
ABCD is 6 + 4 - I₁ R = V_AD \(\Rightarrow \quad 10-2 \times \frac{1}{2}=V_{A D}\)
\(\therefore\) V_AD = 9 V

No, they will not have the same mass. B^2- has more mass as, it has gained two electrons, whereas A²⁺ has lost two electrons.

A charge moving with a circular orbit can produce electromagnetic waves because a circular motion is an accelerated motion and accelerated charges produce e.m.waves

No, image will be formed at the same position. This follows from lens maker's formula:
\(\frac { 1 }{ f } =(u-1)\left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
When we interchange R₁ and R₂, the value of f does not change except for the sign. Hence the image will be formed at the same position.