The resistance of a wire is 5, Omega at 50, ^ | vedclass

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(D) We know that the resistance at temperature $t$ is given by the formula:$R_{t} = R_{0}(1 + \alpha t)$where $R_{t}$ is the resistance at $t\, ^\circ

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$4$

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(D) We know that the resistance at temperature $t$ is given by the formula:$R_{t} = R_{0}(1 + \alpha t)$where $R_{t}$ is the resistance at $t\, ^\circ C$,$R_{0}$ is the resistance at $0\, ^\circ C$,and $\alpha$ is the temperature coefficient of resistance.For $t = 50\, ^\circ C$:$5 = R_{0}(1 + 50\alpha)$  ......$(i)$For $t = 100\, ^\circ C$:$6 = R_{0}(1 + 100\alpha)$ ......$(ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{5}{6} = \frac{R_{0}(1 + 50\alpha)}{R_{0}(1 + 100\alpha)}$$\frac{5}{6} = \frac{1 + 50\alpha}{1 + 100\alpha}$Cross-multiplying:$5(1 + 100\alpha) = 6(1 + 50\alpha)$$5 + 500\alpha = 6 + 300\alpha$$200\alpha = 1$$\alpha = \frac{1}{200} = 0.005\, ^\circ C^{-1}$Now,substitute $\alpha$ back into equation $(i)$ to find $R_{0}$:$5 = R_{0}(1 + 50 	imes 0.005)$$5 = R_{0}(1 + 0.25)$$5 = R_{0}(1.25)$$R_{0} = \frac{5}{1.25} = 4\, \Omega$Thus,the resistance at $0\, ^\circ C$ is $4\, \Omega$.

The resistance of a wire is $5\, \Omega$ at $50\, ^\circ C$ and $6\, \Omega$ at $100\, ^\circ C$. The resistance of the wire at $0\, ^\circ C$ will be .............. $\Omega$.

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