Question

12. 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 δT=0.01 s and he measures the depth of the well to be L=20m. Take the acceleration due to

gravity g=10ms-2 and the velocity of sound is 300ms-1

.Then the fractional error in the measurement,

/L


is closest to: a) 1% b) 5% c) 3% d) 0.2%.
Please explain the answer. And please don't spam.​

Answer 1

Given that,

Time error = 0.01 5 s

Depth of well = 20 m

Sound velocity = 300 m/s

Acceleration due to gravity = 10 m/s²

We need to calculate the fractional error in the measurement

Using formula of time period

$T=\sqrt{\dfrac{2L}{g}}+\dfrac{L}{v}$

With error limits

$T+\delta T=\sqrt{\dfrac{2(L+\delta L)}{g}}+\dfrac{L+\delta L}{v}$

$T+\delta T=\sqrt{\dfrac{2L}{g}(1+\dfrac{\delta L}{L})}+\dfrac{L}{v}(1+\dfrac{\delta L}{L})$

$T+\delta T=\sqrt{\dfrac{2L}{g}\times(1+\dfrac{\delta L}{2L})+\dfrac{L}{v}(1+\dfrac{\delta L}{L})$

$T+\delta T=\sqrt{\dfrac{2L}{g}}+\sqrt{\dfrac{2L}{g}}\times\dfrac{\delta L}{2L}+\dfrac{L}{v}+\dfrac{L}{v}\times\dfrac{\delta L}{L}$

$T+\delta T=T+\sqrt{\dfrac{2L}{g}}\times\dfrac{\delta L}{2L}+\dfrac{L}{v}\times\dfrac{\delta L}{L}$

$\delta T=\dfrac{\delta L}{L}(\dfrac{1}{2}\sqrt{\dfrac{2L}{g}}+\dfrac{L}{v})$..(I)

Put the value in the euation (I)

$0.015=\dfrac{\delta L}{L}(\dfrac{1}{2}\times\sqrt{\dfrac{2\times20}{10}}+\dfrac{20}{300})$

$\dfrac{\delta L}{L}\times100=\dfrac{0.015}{1.067}\times100$

$\dfrac{\delta L}{L}=1.4\%$

According to options

$\dfrac{\delta L}{L}=1\%$

Hence, The fractional error in the measurement is 1%.

(a) is correct option.