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(b) (i) Name the device which (I) helps to change the electric current in the given electric circuit and (II) breaks the circuit during short-circuiting.
(ii) State Ohm's law and write the factors on which the resistance of a given conductor depends.
(iii) Find out the length of a wire of resistance $21\ \Omega$, if the diameter of the wire is $0.02\ \text{cm}$ and resistivity of the material is $$\begin{aligned}& 66 \times 10^{-8} \\ & \Omega\text{m}\end{aligned}$$. [Given : $\pi = \frac{22}{7}$]
(ii) State Ohm's law and write the factors on which the resistance of a given conductor depends.
(iii) Find out the length of a wire of resistance $21\ \Omega$, if the diameter of the wire is $0.02\ \text{cm}$ and resistivity of the material is $$\begin{aligned}& 66 \times 10^{-8} \\ & \Omega\text{m}\end{aligned}$$. [Given : $\pi = \frac{22}{7}$]
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Power consumed per second in parallel combination
$\text{P} = \frac{\text{V}^2}{\text{R}_\text{p}} = \frac{8^2}{4/3} = 48\text{W}$ (1 Mark)
(b) (i) (I) Rheostat or variable resistance ($\frac{1}{2}$ Mark)
(II) Fuse ($\frac{1}{2}$ Mark)
(ii) • The potential difference applied across the two ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it provided that its temperature remains constant. (1 Mark)
• Resistance depends on
$\blacktriangleright$ length of conductor
$\blacktriangleright$ area of cross-section/thickness of conductor
$\blacktriangleright$ Material/Electrical Resistivity of conductor
$\blacktriangleright$ Temperature of conductor. ($\frac{1}{2}$ + $\frac{1}{2}$ Mark)
(Any two)
(iii) $\text{R} = 21 \Omega$ $\text{r} = 0.01 \times 10^{-2} \text{ m}$ $\rho = 66 \times 10^{-8} \Omega\text{m}$ ($\frac{1}{2}$ Mark)
$\text{R} = \rho \frac{\text{l}}{\text{A}} = \rho \frac{\text{l}}{\pi \text{r}^2}$ ($\frac{1}{2}$ Mark)
$21 = \frac{66 \times 10^{-8} \times \text{l}}{22 \times (0.01 \times 10^{-2})^2}$ ($\frac{1}{2}$ Mark)
$21 = \frac{3 \times 10^{-8} \times \text{l} \times 7}{0.01 \times 0.01 \times 10^{-4}}$
$\text{l} = \frac{21 \times 1 \times 10^{-8}}{21 \times 10^{-8}}$
$\text{l} = 1 \text{ m}$ (1 Mark) (5 Marks)
$\text{P} = \frac{\text{V}^2}{\text{R}_\text{p}} = \frac{8^2}{4/3} = 48\text{W}$ (1 Mark)
(b) (i) (I) Rheostat or variable resistance ($\frac{1}{2}$ Mark)
(II) Fuse ($\frac{1}{2}$ Mark)
(ii) • The potential difference applied across the two ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it provided that its temperature remains constant. (1 Mark)
• Resistance depends on
$\blacktriangleright$ length of conductor
$\blacktriangleright$ area of cross-section/thickness of conductor
$\blacktriangleright$ Material/Electrical Resistivity of conductor
$\blacktriangleright$ Temperature of conductor. ($\frac{1}{2}$ + $\frac{1}{2}$ Mark)
(Any two)
(iii) $\text{R} = 21 \Omega$ $\text{r} = 0.01 \times 10^{-2} \text{ m}$ $\rho = 66 \times 10^{-8} \Omega\text{m}$ ($\frac{1}{2}$ Mark)
$\text{R} = \rho \frac{\text{l}}{\text{A}} = \rho \frac{\text{l}}{\pi \text{r}^2}$ ($\frac{1}{2}$ Mark)
$21 = \frac{66 \times 10^{-8} \times \text{l}}{22 \times (0.01 \times 10^{-2})^2}$ ($\frac{1}{2}$ Mark)
$21 = \frac{3 \times 10^{-8} \times \text{l} \times 7}{0.01 \times 0.01 \times 10^{-4}}$
$\text{l} = \frac{21 \times 1 \times 10^{-8}}{21 \times 10^{-8}}$
$\text{l} = 1 \text{ m}$ (1 Mark) (5 Marks)