InterviewSolution
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InterviewSolution
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Assume that a tunnel is dug across the earth (radius=R) passing through its centre. Find the time a particle takes to cover the length of the tunnel if (a) it is projected into the tunnel with a speed of `sqrt((gR)` (b) it is relased from a height R above the tunnel (c ) it is thrown vertically upward along the length of tunnel with a speed of `sqrt(gR).` |
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Answer» Let `y = A sin (omega t + phi)` At` t = 0` `R = A sin phi` and `(dy)/(dt) = A omega cos theta (omega + phi)` At `t = 0`, the velocity of the partical is downwards direction i.e. `v = - sqrt gR` `(dy)/(dt) = - sqrtgR = A sqrt((g)/( R)) cos phi [as omega = sqrt((g)/(R))]` `A cos phi = - R` `tan phi = - 1 implies phi = (3 pi)/(4)` Squaring and adding Eqs, (i) and (ii), `A = sqrt2 R` `y = sqrt2 sin ((sqrtg)/(R) r + (3 pi)/(4))` At `y = 0, sin ((sqrtg)/(R) r + (3 pi)/(4)) = pi implies t = (pi)/(4) sqrt((R)/(g))` Hence time to sross the tunnel `= (pi)/(2) sqrt((R)/(g))` |
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