A) 3.316 × 10?5/ºC
B) 2.316 × 10?5/ºC
C) 4.316 × 10?5/ºC
D) None of these
Correct Answer: B
Solution :
Loss of weight at 27ºC is = 46 ? 30 = 16 = V1 × 1.24 rl × g ?(i) Loss of weight at 42ºC is = 46 ? 30.5 = 15.5 = V2 × 1.2 r l × g ?(ii) Now dividing (i) by (ii), we get \[\frac{16}{15.5}\] = \[\frac{{{V}_{1}}}{{{V}_{2}}}\times \frac{1.24}{1.2}\] But \[\frac{{{V}_{2}}}{{{V}_{1}}}\] = 1 + 3a (t2 ? t1) = \[\frac{15.5\times 1.24}{16\times 1.2}\] = 1.001042 Þ 3a (42º ? 27º) = 0.001042 Þ a = 2.316 × 10?5/ºC.You need to login to perform this action.
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