A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is δ T

Question

A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is δ T = 0.01 seconds and he measures the depth of the well to be L = 20 meters. Take the acceleration due to gravity g = 10 m s - 2 and the velocity of sound is 300 m s - 1 . Then the fractional error in the measurement, δ L L , is closest to

TestHub · Accepted Answer

Total time taken, T = 2 L g + L c , c is the sound speed in air. Now, for an error δ L in L, We have an error δ T in T So, T + δ T = 2 L + δ L g + L + δ L c = 2 L g 1 + δ L L + L c 1 + δ L L Since, δ T T is very small, hence δ L L is also small, so taking binomial approximation T + δ T = 2 L g 1 + 1 2 δ L L + L c 1 + δ L L ⇒ T + δ T = 2 L g + 2 L g 1 2 δ L L + L c + L c δ L L ⇒ δ T = 2 L g 1 2 δ L L + L c δ L L ⇒ δ T = L 2 g + L c δ L L ⇒ 0 . 01 = 20 2 × 10 + 20 300 δ L L ⇒ 0 . 01 = 1 + 1 15 δ L L ⇒ δ L L = 0 . 15 16 % e r r o r = δ L L × 100 % = 15 16 % ≈ 1 %

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