A particle falls in a medium whose resistance is propotional to the square of the velocity of the particles. If the differential equation of the free fall is `(dv)/(dt) = g-kv^(2)` (k is constant) thenA. `v=2sqrt(g/k)(e^(2tsqrt(g//t))+1)/(e^(2rsqrt(g//k))-1)`B. `v=sqrt(g/k)(e^(2tsqrt(gk))-1)/(e^(2tsqrt(gk))+1`C. `v to 0` as `t to infty`D. `v to sqrt(g/k)` as `t to infty`

A particle falls in a medium whose resistance is propotional to the sq

1.	A particle falls in a medium whose resistance is propotional to the square of the velocity of the particles. If the differential equation of the free fall is `(dv)/(dt) = g-kv^(2)` (k is constant) thenA. `v=2sqrt(g/k)(e^(2tsqrt(g//t))+1)/(e^(2rsqrt(g//k))-1)`B. `v=sqrt(g/k)(e^(2tsqrt(gk))-1)/(e^(2tsqrt(gk))+1`C. `v to 0` as `t to infty`D. `v to sqrt(g/k)` as `t to infty`
Answer» Correct Answer - B::D `(dv)/(dt) = g-kv^(2)` `rArr (dv)/(g-kv^(2))` `rArr (1/k int(dv)/((g/k)-v^(2))) = int(dt+C)` `rArr 1/(2sqrt(gk)) log\|(sqrt(g/k)+v)/(sqrt(g//k)-v)\|=t+C` At t=0, v=0 `rArr C=0` `rArr v=sqrt(g/k) (1-1/(e^(2tsqrt(gk))/(1-1/(e^(2tsqrt(gk))))))` Clearly when `v to sqrt(g/k)` as `t to infty`

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