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 ms -2 and the velocity of sound is 300 ms -1. Then the fractional error in the measurement, δ L / L, is closest to

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 ms -2 and the velocity of sound is 300 ms -1. Then the fractional error in the measurement, δ L / L, is closest to (A) 0.2 % (B) 1 % (C) 3 % (D) 5 %

Tardigrade Admin · Accepted Answer

t=√(2 L/g)+(L/C) (d t/d L)=√(L/g) × (1/2 √L)+(1/C) dL =( dt /(1)√2 gL +(1) C ⇒ ( dL / L ) × 100=(( dt /(1)√2 gL +(1) C ) (1/ L ) × 100 =(15/16 %) ≈ 1 %

$0.2%$

$1%$

$3%$

$5%$

$t = \frac{2 L}{g} + \frac{L}{C}$
$\frac{d t}{d L} = \frac{L}{g} \times \frac{1}{2 L} + \frac{1}{C}$
$d L = \frac{d t}{\frac{1}{2 gL} + \frac{1}{C}}$
$\Rightarrow \frac{d L}{L} \times 100 = (\frac{d t}{\frac{1}{2 gL} + \frac{1}{C}}) \frac{1}{L} \times 100$
$= \frac{15}{16%} \approx 1%$

0.2%

1%

3%

5%

t=g2L​​+CL​ dLdt​=gL​​×2L​1​+C1​ dL=2gL​1​+C1​dt​ ⇒LdL​×100=(2gL​1​+C1​dt​)L1​×100 =16%15​≈1%

$0.2%$

$1%$

$3%$

$5%$

$t = \frac{2 L}{g} + \frac{L}{C}$
$\frac{d t}{d L} = \frac{L}{g} \times \frac{1}{2 L} + \frac{1}{C}$
$d L = \frac{d t}{\frac{1}{2 gL} + \frac{1}{C}}$
$\Rightarrow \frac{d L}{L} \times 100 = (\frac{d t}{\frac{1}{2 gL} + \frac{1}{C}}) \frac{1}{L} \times 100$
$= \frac{15}{16%} \approx 1%$