A right circular cylinder with radius $R$ and height $H$ contains a liquid which evaporates at a rate proportional to its surface area in contact with air (proportionality constant = $K > 0$). If $T$ is time after which cylinder will be empty, then
Step-by-Step Solution
Key Concept: The evaporation rate depends on the surface area A = πR² in contact with air, so dV/dt = -KπR². Since V = πR²h for a cylinder, this gives dh/dt = -K (independent of R). Integrating h from H to 0 yields T = H/K, making T dependent only on H and K, not R.
Given $\frac{dv}{dt} = -kA$ and $\frac{dH}{dt} = -kA$, we equate to get $\frac{dH}{dv} = 1$. Integrating from 0 to $t$ gives $\int_0^t dH = -k\int_0^t dt$, which directly relates height change to time elapsed through the constant $k$.
Correct Answer: 2,3,4