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Assume that a tunnel is dug across the e...

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 released from a height R above the tunnel (c ) it is thrown vertically upward along the length of tunnel with a speed of `sqrt(gR).`

Text Solution

Verified by Experts

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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