Correct Answer - A
`(dp)/(dt) = (p-400)/(2)`
`rArr (dp)/(p-400)=1/2dt`
Integrating, we get
`"ln "|p-400|=1/2t+c`
When `t=0, p=100`, we have ln 300=c
`therefore "ln"|(p-400)/(300)|=t/2`
`rArr |p-400|=300e^(t//2)`
`rArr 400-p=300e^(t//2)` (as `p lt 400)`
`rArr p=400-300e^(t//2)`