QJEEAdv17Phac63: JEE Advanced 2017 Physics - Class 11 JEEAdv17Phac63 Solution: A person measures the depth of a well by measuring the time…

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

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

$T + \delta T = \sqrt {{{2(L + \delta L)} \over g}} + {{(L + \delta L)} \over {{v_s}}}$

$T + \delta T = \sqrt {{{2L} \over g}\left( {1 + {{\delta L} \over L}} \right)} + {L \over {{v_s}}}\left( {1 + {{\delta L} \over L}} \right)$

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, v_s = 300 m/s and using equation (3) : $\sqrt {{{2L} \over g}} + {L \over {{v_s}}} = T$ , above equation becomes

$T + \delta T = T + \left( {{1 \over 2}\sqrt {{{2 \times 20} \over {10}}} + {{20} \over {300}}} \right){{\delta L} \over L}$

$\Rightarrow \delta T = \left( {{1 \over 2}\sqrt 4 + {1 \over {15}}} \right){{\delta L} \over L} = \left( {1 + {1 \over {15}}} \right){{\delta L} \over L} = \left( {{{16} \over {15}}} \right){{\delta L} \over L}$

$\Rightarrow {{\delta L} \over L} = \delta T\left( {{{15} \over {16}}} \right)$

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)