I_c=95% of I_e = (95 / 100 ) I_e
I_e = (100 / 95) x 5 mA = 5.26mA
I_e= I_c+ I_b
I_b = 0.25 mA

For faster action NPN Transistor is used. In an NPN transistor, current conduction is mainly by free electron, whereas in PNP type transistor. It is mainly holes Since electron are more mobile than holes we prefer NPN for faster action as well as high conduction current.

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

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

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.

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

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

(a) There are three factors:
(i) Wider Bandwidth: High frequency provide wider bandwidth.
(ii) Distance: High-frequency wave used to carry baseband signals (low-frequency signal) over a distance of several thousand kilometres.
(iii) Cost: High frequency transmits with low cost.

Magnetic field at O due to current through one infinitely long wire is
\(B=\frac { \mu _{ 0 } }{ 4\pi } \times\frac { 2I }{ r } \hat{k}\)(i.e. along + Z-direction)
So, magnetic field at O due to all the wires is
\(B=\frac { \mu _{ 0 } }{ 4\pi } \times2\times2[\frac { 1 }{ 1 } -\frac { 1 }{ 2 } +\frac { 1 }{ 4 } -\frac { 1 }{ 8 } +\frac { 1 }{ 16 } +...] \hat{ k } \\ =\frac { { \mu }_{ 0 } }{ \pi } [(1+\frac { 1 }{ 4 } +\frac { 1 }{ 16 } +...)\hat{ k } -(\frac { 1 }{ 2 } +\frac { 1 }{ 8 } +...)\hat{ k } ]\\ =\frac { { \mu }_{ 0 } }{ \pi } [(\frac { 1 }{ 1-\frac { 1 }{ 4 } } )\hat{ k } \frac { 1 }{ 2 } (\frac { 1 }{ 1-\frac { 1 }{ 4 } } )\hat{ k } ]\\ \Rightarrow B=\frac { { \mu }_{ 0 } }{ \pi } \times\frac { 2 }{ 3 } \hat{ k } =\frac { { \mu }_{ 0 } }{ 4\pi } \times\frac { { { 8 }\hat{k} } }{ 3 } \)​​​​​​

X \(\rightarrow\) Intermediate Frequency (IF) stage
Y \(\rightarrow\) Power amplifier
The IF stage change the high-frequency carrier wave into lower frequency wave and amplifier enhances the strength of the signals.

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

When we are considering a point of sound wavw. The disturbance due to the source propagates in spherical symmetry, that is, in all directions.
       
The formation of wavefront is in accordance with Huygen's principle. So, Huygen's principle is valid for longitudinal sound waves also.

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

\(Here, \ E=648 \ J,I=9,e=?,\)
\(dI=9-0=9A,dt=0.45s\)
\(From \ E=\frac { 1 }{ 2 } { LI }^{ 2 }\)
\(648=\frac { 1 }{ 2 } { L(9) }^{ 2 }; \ L=\frac { 648\times2 }{ 9\times9 } =16H\)
\(As \ e=\frac { LdI }{ dt } \therefore e=\frac { 16(9) }{ 0.45 } =320V\)

If Changed particle was Initially at rest, it will move along the direction of electric field. If initial velocity of the charged particle makes a certain angle with electric field, then the charged particle will no move along the line of force.