JEE MAIN - Physics (2006 - No. 22)
The resistance of bulb filmanet is $$100\Omega $$ at a temperature of $${100^ \circ }C.$$ If its temperature coefficient of resistance be $$0.005$$ per $$^ \circ C$$, its resistance will become $$200\,\Omega $$ at a temperature of
$${300^ \circ }C$$
$${400^ \circ }C$$
$${500^ \circ }C$$
$${200^ \circ }C$$
Explanation
$$R{}_1 = {R_0}\left[ {1 + \alpha \times 100} \right] = 100\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
$${R_2} = {R_0}\left[ {1 + \alpha \times T} \right] = 200\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
On dividing we get
$${{200} \over {100}} = {{1 + \alpha T} \over {1 + 100\alpha }}$$
$$ \Rightarrow 2 = {{1 + 0.005T} \over {1 + 100 \times 0.005}}$$
$$ \Rightarrow T = {400^ \circ }C$$
NOTE : We may use this expression as an approximation because the difference in the answers is appreciable. For accurate results one should use $$R = {R_0}{e^{\alpha \Delta T}}$$
$${R_2} = {R_0}\left[ {1 + \alpha \times T} \right] = 200\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
On dividing we get
$${{200} \over {100}} = {{1 + \alpha T} \over {1 + 100\alpha }}$$
$$ \Rightarrow 2 = {{1 + 0.005T} \over {1 + 100 \times 0.005}}$$
$$ \Rightarrow T = {400^ \circ }C$$
NOTE : We may use this expression as an approximation because the difference in the answers is appreciable. For accurate results one should use $$R = {R_0}{e^{\alpha \Delta T}}$$
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