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

Modulation index ,mf = \(\frac { Maximum\ frequency\ deviation }{ minimum\ signal\ freqency } \)

(i) Pulse Amplitude Modulation : Amplitude of the pulse varies in accordance with the modulating signal.

(ii) (ii) Pulse Position Modulation. : Pulse position (ie) time of rise or fall of the pulse ) changes with the modulating signal.

The factors upon which transconductance of a transistor depend are as follows.
i) Geometry of the transistor
ii) Doping levels.
iii) Biasing of the transistors

As shown in figure the bright fringes B1 and b2 on either side of O coincide with S1 and S2 respectively. Clearly ,
\(\beta\) = d/2
As \(\beta\) = \(D\lambda /d\) d/2 = \(D\lambda /d\) or \(\lambda ={ d }^{ 2 }/2d\)

Twinkling of stars. The light from stars undergoes refraction continuously before it reaches earth. So the apparent position of the stars is slightly different than its actual position. Due to variation in atmosphere conditions, like change in temperature, density etc., and this apparent position keeps on changing. The amount of light entering our eyes from a particular star increases and decreases randomly with time. Sometimes, the star appears brighter and other times, it appears fainter. This gives rise to the twinkling effect of stars
The planets do not show twinkling effect. As the planets are much closer to the earth, the greater and the fluctuations caused in the amount of light due to atmospheric refraction are negligible as compared to the amount of light received from them.

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

a) in parallel, power dissipation \(\alpha\) 1/R 
Therefore 3\(\Omega \) wire will dissipate more power
b) In series, power dissipation \(\alpha\) R
Therefore 9\(\Omega \) wire will dissipate more power

Give, power = 1 W, Zener breakdown, V_Z = 5 V
Minimum voltage, V_min = 3 V
Maximum voltage, V_max = 7 V
Current, \({ I }_{ Zmax }=\frac { P }{ { V }_{ Z } } =\frac { 1 }{ 5 } =0.2A\)
The values of R_S for safe operation,
\({ R }_{ S }=\frac { { V }_{ max }-{ V }_{ Z } }{ { I }_{ Zmax } } =\frac { 7-5 }{ 0.2 } =\frac { 2 }{ 0.2 } =10\Omega\)

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
\(\frac { \lambda _{ p } }{ \lambda _{ \alpha } } =\sqrt { 4\times 2\times \frac { 1 }{ 2 } } =2\quad \Rightarrow \lambda _{ p }:\lambda _{ \alpha }=2:1\)

\(Given,radius\ of\ ring,r=2m,{ q }_{ 1 }={ 10 }^{ -6 }C\)
\({ q }_{ 2 }={ 10 }^{ -7 }C,m=9\times { 10 }^{ -4 }kg\)
\(l=2cm=0.02m,T=?\)
\(Force\ on\ particle\ is\ given\ by\)
\(F=\frac { { q }_{ 1 }{ q }_{ 2 }l }{ 4\pi { \varepsilon }_{ 0 }{ ({ l }^{ 2 }+{ r }^{ 2 }) }^{ 2 } } =\frac { (9\times { 10 }^{ 9 })({ 10 }^{ -6 }){ 10 }^{ -7 })l }{ ({ 4+0.0004 })^{ 3/2 } } \)
\(=\frac { 9\times { 10 }^{ -4 }l }{ 8.001 } =1.125\times { 10 }^{ -4 }l\)
\(As,acceleration,a=\frac { F }{ m } =\frac { 1.125\times { 10 }^{ -4 }l }{ 9{ \times 10 }^{ -4 } } \)
\(Thus,the\ motion\ of\ particle\ is\ SHM.\)
\(Now,\ thime\ period\ of\ oscillation\ of\ particle\ is\quad given\ by\)
\(T=2\pi \sqrt { \frac { l }{ a } } =2\pi \sqrt { \frac { l }{ 0.125l } } \Rightarrow 17.7s\)

Dielectric constant of water is much greater than that of mica because of the following reason
(i) water has a symmetrical shape as compared to mica
(ii) water has permanent dipole moment.

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.

Yes, because objective with larger diameter will collect more light and even the distant objects can be seen.