A well-lagged bar of length 100cm has its ends maintained at 100oC and 40oC respectively. What is the temperature at a point 60cm from the hotter end?
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Correct Answer: Option D
Explanation:
100oC ....100cm
60oC ....60cm
40oC .... 0cm
\(\frac{\theta - 40}{100 - 40}\) = \(\frac{60 - 0}{100 - 0}\)
\(\theta\) = 76oC
100oC ....100cm
60oC ....60cm
40oC .... 0cm
\(\frac{\theta - 40}{100 - 40}\) = \(\frac{60 - 0}{100 - 0}\)
\(\theta\) = 76oC