A rod having α = 1*10^-5per °C and length of 70cm has one of its end rigidly connected to another rod of length 1m and α= 0.5*10^-5per °C. Find the new length of the two joined rods when their temperature is increased by 30℃.(a) 170.021cm(b) 169.021cm(c) 171.031cm(d) 170.031cmThis question was posed to me in semester exam.My doubt is from Thermal Expansion in division Thermal Properties of Matter of Physics – Class 11

A rod having α = 1*10^-5per °C and length of 70cm has one of its end

1.	A rod having α = 110^-5per °C and length of 70cm has one of its end rigidly connected to another rod of length 1m and α= 0.510^-5per °C. Find the new length of the two joined rods when their temperature is increased by 30℃.(a) 170.021cm(b) 169.021cm(c) 171.031cm(d) 170.031cmThis question was posed to me in semester exam.My doubt is from Thermal Expansion in division Thermal Properties of Matter of Physics – Class 11
Answer» Correct OPTION is (a) 170.021cm For explanation I would say: The NEW length of first rod: I1 = 70(1+110^-530) = 70.021cm Similarly I2 = 100(1+ 0.510^-530) = 100.015cmTherefore, new length = 70.021 + 100.015 = 170.036cm.

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