A man is standing in a lift moving upwards with a constant speed of $10{\text{ms}}^{-1}$. The man drops a coin from a height of $4.9\text{m}$. If $g=9.8{\text{ms}}^{-2}$, the coin reaches the floor of the lift after a time

The correct option is B $1\text{s}$
Let us solve the problem in terms of relative initial velocity, relative acceleration and relative displacement of the coin with respect to the floor of the lift.
Taking downward direction as positive,
Since, the man is moving with coin in the lift with initial speed of $10\text{m/s}$ upwards, hence the intial relative velocity of coin w.r.t. lift $\left(u_{rel}\right)$ is zero.
$u_{rel}=-10-(-10)=0{\text{ms}}^{-1},a_{rel}=9.8{\text{ms}}^{-2}$
$S_{rel}=4.9\text{m}$
Using second equation of motion, we get
$S_{rel}=u_{rel}t+\frac{1}{2}{a}_{rel}{t}^2$
$4.9=0\times t+\frac{1}{2}\times 9.8\times t^2$
or $4.9t^2=4.9$
$t=1\text{s}$