In series combination of resistors, current I is given by
\(I=\frac{E}{R+n R}\)
whereas, in parallel combination current 10 I is given by
\(\frac{E}{R+\frac{R}{n}}=10 I \Rightarrow \frac{E}{R+\frac{R}{n}}=10\left(\frac{E}{R+n R}\right)\)
Now, according to problem,
\(\frac{1+n}{1+\frac{1}{n}}=10 \Rightarrow 10=\left(\frac{1+n}{n+1}\right) n \Rightarrow n=10\)

P is in the plane of S, needle is in North, so the declination is zero.

P is also on the magnetic equator, so the angle of dip is zero, because the value of angle of dip at equator is zero. Q is also on the magnetic equator, so the angle of dip is zero. 
As, the earth is tilted on its axis by 113°, thus the declination at Q is 113°.

Given, mass of a paisa coin, m = 0.75 g
Atomic mass of aluminium, M = 26.9815 g
Length of the diagonal of square shaped paisa coin= 17 mm
Avogadro's number, N_A = 6.023 x 10²³
\(\Rightarrow \quad n=\frac{N_A}{M} \times m=\frac{6.023 \times 10^{23}}{26.9815 \mathrm{~g}} \times 0.75 \mathrm{~g}\)
\(=1.6742 \times 10^{22}\)
Since, atomic number (Z) of AI is 13, therefore each atom of AI contains 13 protons and 13 electrons. Now, find out the magnitude of positive and negative charges present in one paisa coin.
nZe = 1.6742 x 10²² x 13 x 1.6 x 10^-19 C
= 34.8 kC
Now, write the conclusion drawn from this magnitude of charge.
34.8 kC is a very large amount of charge. This concludes that ordinary neutral matter contains an enormously large amount ot positive and negative charges.

If emitter is reverse biased and collector is forward biased, then emitter works as collector and collector is work as emitter. In such situation, transistor works but has lower amplification because the size of  emitter is smaller than collector and doping of emitter is more than collector.

On switching on the current in a galvanometer, the coil of the galvanometer does not come to rest immediately. It oscillates about its equilibrium position but the coil of a dead beat galvanometer comes to rest immediately. It is due to the reason that the eddy currents are set up in the metallic frame, over which the coil is wound and the eddy currents oppose the oscillatory motion of the coil.

Radius of curvature of objectrive mirror, 
R₁ = 220 mm
\({ f }_{ 1 }=\frac { { R }_{ 1 } }{ 2 } =\frac { 220 }{ 2 } =110 \ mm\) 
Radius of curvature of secondary mirrors, R₂ = 140 mm
\({ f }_{ 2 }=\frac { { R }_{ 2 } }{ 2 } =\frac { 140 }{ 2 } =70 \ mm\) 
Distance between two mirrors, d = 20 mm from objective mirror.
Now, for secondary mirror, u = f₁ - d = 110 - 20
= 90 mm
From mirror formula,
\(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ { f }_{ 2 } } \Longrightarrow \frac { 1 }{ v } =\frac { 1 }{ { f }_{ 2 } } -\frac { 1 }{ u } \) 
\( =\frac { 1 }{ 70 } -\frac { 1 }{ 90 } \Longrightarrow v=\frac { 630 }{ 2 } =315\ mm\)
i.i., final image will be at 31.5 cm to the right of secondary mirror.

(i)  

The direction of magnetic field is from S to N-pole of magnet.
(ii) 

The direction of magnetic field is from N to S-pole of magnet.

According to the formula, E.dr = dV, the value of dV = 0 at each point of equipotential surface.
\(\therefore \) E.dr = 0
so, the angle between electric field vector and distance vector will be 90^o . Thus, the electric field is always normal to the plane passing through AB.

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.

Intensity of light is reduced to one fourth because the light beam spreads as it approaches into a spherical region of area \(4\pi { r }^{ 2 },i.e.,I\infty 1/{ r }^{ 2 }\) But laser beam does not spread, hence its intensity remains constant. Laser beam is unidirectional, monochromatic and coherent light, whereas the light from a bulb does not posses the above properties.