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Suppose that a mothball loses volume by evaporation at a rateproportional to its instantaneous area. If the diameter of the ball decreasesfrom 2cm to 1cm in 3 months, how long will it take until the ball haspractically gone? |
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Answer» Let at any instance (t), the radius of moth ball be r and v be its volume. Then `v=4/3pir^(3)` or `(dv)/(dt) = 4pir^(2)(dr)/(dt)` Thus, as per the information `4pir^(2)(dr)/(dt) = -k4pir^(2)`, where `k in R^(+)` or `(dr)/(dt) = -k` or `r=-kt+c` at t=0, r=2cm, t=3 month, r=1 cm `rArr c=2, k=1/3` `rArr r=-1/3t+2` now `r to 0`, `t to6` Hence, it will take six months until the ball is practically gone. |
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