JEE Advance - Physics (2017 - Paper 2 Offline - No. 5)
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
$$0.2\% $$
$$1\% $$
3%
5%
Explanation
Let the time taken by stone to reach bottom of will be t1 and time taken by sound to reach the observer be t2 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
vs 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
$$T + \delta T = \sqrt {{{2(L + \delta L)} \over g}} + {{(L + \delta L)} \over {{v_s}}}$$
$$T + \delta T = \sqrt {{{2L} \over g}\left( {1 + {{\delta L} \over L}} \right)} + {L \over {{v_s}}}\left( {1 + {{\delta L} \over L}} \right)$$
expanding $${\left( {1 + {{\delta L} \over L}} \right)^{1/2}}$$ using binomial approximation
$$T + \delta T = \sqrt {{{2L} \over g}} \left( {1 + {1 \over 2}{{\delta L} \over L}} \right) + {L \over {{V_s}}}\left( {1 + {{\delta L} \over L}} \right)$$
$$ = \sqrt {{{2L} \over g}} + \sqrt {{{2L} \over g}} \left( {{1 \over 2}{{\delta L} \over L}} \right) + {L \over {{v_s}}} + {L \over {{v_s}}}{{\delta L} \over L}$$
$$ = \sqrt {{{2L} \over g}} + {L \over {{v_s}}} + \left( {{1 \over 2}\sqrt {{{2L} \over g}} + {L \over {{v_s}}}} \right){{\delta L} \over L}$$
given, L = 20 m, g = 10 m/s, vs = 300 m/s and using equation (3) : $$\sqrt {{{2L} \over g}} + {L \over {{v_s}}} = T$$, above equation becomes
$$T + \delta T = T + \left( {{1 \over 2}\sqrt {{{2 \times 20} \over {10}}} + {{20} \over {300}}} \right){{\delta L} \over L}$$
$$ \Rightarrow \delta T = \left( {{1 \over 2}\sqrt 4 + {1 \over {15}}} \right){{\delta L} \over L} = \left( {1 + {1 \over {15}}} \right){{\delta L} \over L} = \left( {{{16} \over {15}}} \right){{\delta L} \over L}$$
$$ \Rightarrow {{\delta L} \over L} = \delta T\left( {{{15} \over {16}}} \right)$$
It is given that $$\delta$$T = 0.01 sec, therefore,
$$\left( {{{\delta L} \over L}} \right) \times 100\% = {{15} \over {16}} \times {1 \over {100}} \times 100\% = {{15} \over {16}}\% $$
error $$ \simeq 1\% $$
Answer (B)
$${t_1} = \sqrt {{{2L} \over g}} $$ .... (1_)
and $${t_2} = {L \over {{v_s}}}$$ .... (2)
Where L is depth of well
g is acceleration due to gravity
vs 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
$$T + \delta T = \sqrt {{{2(L + \delta L)} \over g}} + {{(L + \delta L)} \over {{v_s}}}$$
$$T + \delta T = \sqrt {{{2L} \over g}\left( {1 + {{\delta L} \over L}} \right)} + {L \over {{v_s}}}\left( {1 + {{\delta L} \over L}} \right)$$
expanding $${\left( {1 + {{\delta L} \over L}} \right)^{1/2}}$$ using binomial approximation
$$T + \delta T = \sqrt {{{2L} \over g}} \left( {1 + {1 \over 2}{{\delta L} \over L}} \right) + {L \over {{V_s}}}\left( {1 + {{\delta L} \over L}} \right)$$
$$ = \sqrt {{{2L} \over g}} + \sqrt {{{2L} \over g}} \left( {{1 \over 2}{{\delta L} \over L}} \right) + {L \over {{v_s}}} + {L \over {{v_s}}}{{\delta L} \over L}$$
$$ = \sqrt {{{2L} \over g}} + {L \over {{v_s}}} + \left( {{1 \over 2}\sqrt {{{2L} \over g}} + {L \over {{v_s}}}} \right){{\delta L} \over L}$$
given, L = 20 m, g = 10 m/s, vs = 300 m/s and using equation (3) : $$\sqrt {{{2L} \over g}} + {L \over {{v_s}}} = T$$, above equation becomes
$$T + \delta T = T + \left( {{1 \over 2}\sqrt {{{2 \times 20} \over {10}}} + {{20} \over {300}}} \right){{\delta L} \over L}$$
$$ \Rightarrow \delta T = \left( {{1 \over 2}\sqrt 4 + {1 \over {15}}} \right){{\delta L} \over L} = \left( {1 + {1 \over {15}}} \right){{\delta L} \over L} = \left( {{{16} \over {15}}} \right){{\delta L} \over L}$$
$$ \Rightarrow {{\delta L} \over L} = \delta T\left( {{{15} \over {16}}} \right)$$
It is given that $$\delta$$T = 0.01 sec, therefore,
$$\left( {{{\delta L} \over L}} \right) \times 100\% = {{15} \over {16}} \times {1 \over {100}} \times 100\% = {{15} \over {16}}\% $$
error $$ \simeq 1\% $$
Answer (B)
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