(i) A material whose electrical resistivity is low is a good conductor of electricity. Since the electrical resistivity of silver is less than that of the copper, so silver is a better conductor than the copper.
(ii) For making the elements of heating devices, alloy is used instead of a pure metal This is because the resistivity of an alloy is more than that of a metal and alloy does not burn (or oxidise) even at higher temperature. Out of the given substances, nichrome is an alloy, so nichrome is used in electrical heating devices.

a) Wire is cut into 5 equal parts. Since all dimensions are same, resistance of each wire is equal and has a value = \(\frac {R}{5}\)
b) Formula for finding the effective resistance when connected in parallel is
\(\frac{1}{R_{p}^{\prime}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}+\frac{1}{R_{5}}\)
Here, \(\mathrm{R}_{1}=\mathrm{R}_{2}=\mathrm{R}_{3}=\mathrm{R}_{4}=\mathrm{R}_{5}=\frac{\mathrm{R}}{5}\)
\(\frac{1}{R_{p}}=\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}\)
\(\frac{1}{R_{p}}=\frac{25}{R}  \)
\(R_{p}=\frac{R}{25} \Omega\)
c) If the resistors are connected in series, then the effective resistance will be
\(\mathrm{R}_{\mathrm{s}}=\frac{\mathrm{R}}{5}+\frac{\mathrm{R}}{5}+\frac{\mathrm{R}}{5}+\frac{\mathrm{R}}{5}+\frac{\mathrm{R}}{5} \)
\(\mathrm{R}_{\mathrm{s}}=\frac{5 \mathrm{R}}{5}=\mathrm{R}\)
Ratio of effective resistance in series connection to that of the parallel connection is
\(\frac{R_{s}}{R_{p}}=\frac{R}{R / 25}=\frac{25}{1}\Rightarrow R_s:R_p=25:1\)

(i) LED Television is one of the most important applications of Light Emitting Diodes.
(ii) An LED TV is actually an LCD TV (Liquid Crystal Display) with LED display.
(iii) An LED display uses LEDs for back light and an array of LEDs act as pixels.
(iv) LEDs emitting white light are used in monochrome (black and white) TV; Red, Green and Blue (RGB) LEDs are used in colour television.

(i) Resistor is used to fix the magnitude of the current through the circuit
(ii) Rheostat is used to fix the magnitude of the current through the circuit
(iii) Ammeter is used to measure the current.
(iv) Voltmeter is used to measure the potential difference.

(i) Kalaivani's mother took a small PVC pipe which has water in it and held in a horizontal position. Water came out drop by drop. When she raised one of the ends of the pipe, the water through the other end came out with a greater speed. She said initially when I was holding the pipe in a horizontal position, pressure at the both the ends are same. So water did not come out. The difference in pressure (Pressure difference) between the two ends is zero. But when I raised one of the ends, the top end has high pressure, and the bottom end has low pressure. Due to this pressure difference, water came out with a faster rate. So compare a conductor or a wire to this PVC pipe The wire is connected to a battery, whose one end has higher potential and the other end has lower potential. Due to this difference in potential, electrons from negative terminal (lower potential end) flow towards the positive terminal (higher potential end). Here electron flow is compared with the water flow and the electric potential is compared with the pressure. Kalaivani felt very happy, after hearing her mother's genius explanation.
(ii) When Kalaivani asked how to understand about the concept of resistance from her mother's scientific explanations, her mother continued. If the water pipe is smooth inside, then there is no charge in the speed of flow of water. i.e., zero resistance. But when the pipe has some blocks inside at different points, the water changes its speed of flow at different points, likewise, whenever electrons flow inside the conductor, they may collide with each other, due to which their speed will be resisted (blocked), which forms as electrical resistance. Her explanation made Kalaivani satisfied.
(iii) When two resistors of resistances, R₁ & R₂ are connected in series to a battery source.
\({ R }_{ s }={ R }_{ 1 }+{ R }_{ 2 }\)
When R₁ & R₂ are connected in parallel,
\(\frac { 1 }{ { R }_{ p } } +\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } \Rightarrow \frac { 1 }{ { R }_{ p } } =\frac { { R }_{ 2 }+{ R }_{ 1 } }{ { R }_{ 1 }{ R }_{ 2 } } \)
\({ R }_{ p }=\frac { { R }_{ 1 }{ R }_{ 2 } }{ { R }_{ 1 }+{ R }_{ 2 } } \)

Supply Voltage is same
Amount of heat produced is also same
Time taken to heat water t = 15 min.
\(H=\frac { { V }^{ 2 } }{ R } .t\quad [H={ I }^{ 2 }Rt\Rightarrow (V=IR)\therefore I=\frac { V }{ R } ]\)
\(\therefore \frac { { R }_{ 1 } }{ { R }_{ 2 } } =\frac { { t }_{ 1 } }{ { t }_{ 2 } } \quad ...(1)\)
\(R\alpha l\quad \quad \quad [R=\rho \frac { l }{ A } ]\)
\(\frac { { R }_{ 1 } }{ { R }_{ 2 } } =\frac { { l }_{ 1 } }{ { l }_{ 2 } } \quad \quad \quad ...(2)\)
From (i) and (2)
\(\frac { { l }_{ 1 } }{ { l }_{ 2 } } =\frac { { t }_{ 1 } }{ { t }_{ 2 } } =\frac { 15 }{ { t }_{ 2 } } \quad \quad ....(iii)\)
\({ l }_{ 2 }=\frac { 1 }{ 3 } { l }_{ 1 }\Rightarrow \frac { { l }_{ 1 } }{ { l }_{ 2 } } =3\)
Sub in (iii) \(3=\frac { 15 }{ { t }_{ 2 } } \)
\(\therefore { t }_{ 2 }=\frac { 15 }{ 3 } =5min\)
Time taken to heat \(\frac{1}{3}\) length of the wire = 5 min

Given 
For resistance in parallel, the effective resistance R = ?
\(\frac { I }{ { R }_{ p } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } +\frac { 1 }{ { R }_{ 3 } } \)
\(=\frac { 1 }{ 20 } +\frac { 1 }{ 100 } +\frac { 1 }{ 50 } \)
\(\frac { 1 }{ { R }_{ p } } =\frac { 5+1+2 }{ 100 } =\frac { 8 }{ 100 } \)
\({ R }_{ p }=\frac { 100 }{ 8 } =12.5\Omega \)
\({ R }_{ p }=12.5\Omega \)
i) Total current flowing 
through the circuit \({ I }_{ p }=\frac { { V }_{ p } }{ { R }_{ p } } \)
\({ I }_{ p }=\frac { 125 }{ 12.5 } \)
\({ I }_{ p }=10A\)
(ii) In a parallel circuit, the potential drop across each resistor is same.
∵ V₁ = V₂ = V₃ = 125 V
(iii) The current through each resistor according to ohm's law V = IR
\({ I }_{ 1 }=\frac { { V }_{ 1 } }{ { R }_{ 1 } } =\frac { 125 }{ 20 } =6.25A\)
\({ I }_{ 2 }=\frac { { V }_{ 2 } }{ { R }_{ 2 } } =\frac { 125 }{ 100 } =1.25A\)
\({ I }_{ 3 }=\frac { { V }_{ 3 } }{ { R }_{ 3 } } =\frac { 125 }{ 50 } =2.50A\)
(iv) Total current flowing the circuit
I_p= I₁+ I₂+I₃
I=6.25 + 1.25 + 2.50 = 10A
(v) The electric power of each resistor
P₁=V₁I₁= 125 x 6.25 = 781.25 W
P₂ = V₂I₂ = 125 x 1.25 = 156.25 W
P₃ = V₃I₃ = 125 x 50 = 312.50 W

Given :
Potential difference, V = 110 V
Charge, q = 5 C
To find: Work done, W = ?
Solution
\(V=\frac { W }{ q } \therefore q\times V=W\)
W= 5 x 110 = 550 J
Work, W = 550J

Current flowing through the wire, I = 6 A
Time taken, t = 2 miniatures = 2 x 60 = 120 S
To find : Amount of charge passing
through wire, q = ?
Solution
\(I=\frac { q }{ t } \Rightarrow =6\times 120=720C\)
Charge, q = 720 C