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The resistance of a wire is $8 \Omega$ at $50^{0}C$ and $10 \Omega$ at $100^{0}C.$ The resistance of the wire at $0^{0}C$ will be -
Option: 1
$4 \Omega$

Option: 2
$6 \Omega$

Option: 3
$8 \Omega$

Option: 4
$10 \Omega$

Answers (1)

best_answer

$R_t=R_0(1+\alpha t)$

$R_t=$ The resistance of wire at $t^0C.$

$R_0=$ The resistance of wire at $0^0C.$

$\alpha=$ Temperature coeff. of resistance

$\begin{aligned} & R_{50}=R_0(1+50 \alpha)--- (I) \\ & R_{100}=R_0(1+100 \alpha)---(II) \end{aligned}$

From $(I)$ $R_{50}-R_0=(50 \alpha) R_0$

From $(II)$ $R_{100}-R_0=(100 \alpha) R_0$

So $\frac{R_{100}-R_0}{R_{50}-R_0}=\frac{(100 \alpha) R_0}{(50 \alpha) R_0}$

$\begin{aligned} & \frac{10-R_0}{8-R_0}=2 \\ & 10-R_0=16-2 R_0 \\ & \quad 2 R_0-R_0=16-10 \Rightarrow R_0=6 \Omega \end{aligned}$

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