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InterviewSolution
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A stone is dropped into a well and the sound of inpact of stone on the water is heard after 2.056 sec of the release of stone form the top. If acc. Due to gravity is `980 cm//sec^(2)` and velocity of sound in air is `350 m//s`, calculate the depth of the well. |
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Answer» If the depth of well is `h` and time taken by stone to reach the bottom `t_(1)`, then `h = (1)/(2) g t_(1)^(2)` and time taken by sound to come to the surface `t_(2) = (h)/(350)` .....(ii) But according to given problem `t_(1) + t_(2) = 2.056`....(iii) Substituting `h` from Eqn. (i) in (ii) and then `t_(2)` from Eqn. (ii) in (iii), we get `t_(1) + (g t_(1)^(2))/(700) = 2.056` or `98 t_(1)^(2) + 7000 t_(1) - 14392 = 0` or `14 t_(1)^(2) + 1000 t_(1) - 2056 = 0` or `14 t_(1)^(2) - 28 t_(1) + 1028 t_(1) - 2056 = 0` or `14 t_(1) (t_(1) - 2) + 1028 (t_(1) - 2) = 0` or `(14 t_(1) + 1028) (t_(1) - 2) = 0` So `t_(1) = 2` or `t_(1) = -(1028//14)` Now as negative time is not physically acceptable, so `t_(1) = 2s` `:.` The depth of well `h = (1)/(2) xx 9.8 xx 2^(2) = 19.6 m` |
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