Question 268: Units , Dimensions and Error Analysis

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 $$\delta T = 0.01$$ seconds and he measures the depth of the well to be $$L=20$$ meters. Take the acceleration due to gravity $$g = 10m{s^{ - 2}}$$ and the velocity of sound is $$300$$ $$m{s^{ - 1}}$$. Then the fractional error in the measurement, $$\delta L/L,$$ is closest to

Accepted Answer

Let the time taken by stone to reach bottom of will be t
1
and time taken by sound to reach the observer be t
2
then
$${t_1} = \sqrt {{{2L} \over g}} $$ .... (1_)
and $${t_2} = {L \over {{v_s}}}$$ .... (2)
Where L is depth of well
g is acceleration due to gravity
v
s
is velocity of sound
Total time taken, $$T = {t_1} + {t_2} = {{\sqrt {2L} } \over g} + {L \over {{v_s}}}$$ ..... (3)
If $$\delta$$L is error in depth of well measured and $$\delta$$T is error is time measured then
$$T + \delta T = \sqrt {{{2(L + \delta L)} \over g}}  + {{(L + \delta L)} \over {{v_s}}}$$
$$T + \delta T = \sqrt {{{2L} \over g}\left( {1 + {{\delta L} \over L}} ight)}  + {L \over {{v_s}}}\left( {1 + {{\delta L} \over L}} ight)$$
expanding $${\left( {1 + {{\delta L} \over L}} ight)^{1/2}}$$ using binomial approximation
$$T + \delta T = \sqrt {{{2L} \over g}} \left( {1 + {1 \over 2}{{\delta L} \over L}} ight) + {L \over {{V_s}}}\left( {1 + {{\delta L} \over L}} ight)$$
$$ = \sqrt {{{2L} \over g}}  + \sqrt {{{2L} \over g}} \left( {{1 \over 2}{{\delta L} \over L}} ight) + {L \over {{v_s}}} + {L \over {{v_s}}}{{\delta L} \over L}$$
$$ = \sqrt {{{2L} \over g}}  + {L \over {{v_s}}} + \left( {{1 \over 2}\sqrt {{{2L} \over g}}  + {L \over {{v_s}}}} ight){{\delta L} \over L}$$
given, L = 20 m, g = 10 m/s, v
s
= 300 m/s and using equation (3) : $$\sqrt {{{2L} \over g}}  + {L \over {{v_s}}} = T$$, above equation becomes
$$T + \delta T = T + \left( {{1 \over 2}\sqrt {{{2 	imes 20} \over {10}}}  + {{20} \over {300}}} ight){{\delta L} \over L}$$
$$ \Rightarrow \delta T = \left( {{1 \over 2}\sqrt 4  + {1 \over {15}}} ight){{\delta L} \over L} = \left( {1 + {1 \over {15}}} ight){{\delta L} \over L} = \left( {{{16} \over {15}}} ight){{\delta L} \over L}$$
$$ \Rightarrow {{\delta L} \over L} = \delta T\left( {{{15} \over {16}}} ight)$$
It is given that $$\delta$$T = 0.01 sec, therefore,
$$\left( {{{\delta L} \over L}} ight) 	imes 100\%  = {{15} \over {16}} 	imes {1 \over {100}} 	imes 100\%  = {{15} \over {16}}\% $$
error $$ \simeq 1\% $$
Answer (B)

Answer

$$0.2\% $$

Answer

$$1\% $$

Answer

3%

Answer

5%

Units , Dimensions and Error Analysis

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