The density of 400cm\(^3\) of palm oil was 0.9gcm-3 before frying. If the density of the oil was 0.6gcm-3 after frying, assuming no loss of oil due to spilling, its new volume was?
A
1360cm\(^3\)
B
600cm\(^3\)
C
240cm\(^3\)
D
8000m\(^3\)
Verified Correct Answer: B.
Physical Principles & Numerical Solution
\(Density= \frac{mass}{volume}\)
⇒ \(0.9gcm^{-3} = \frac{mass}{400cm^{3}} = 360g\)
without any loss in the frying process, we have
\(0.6gcm^{-3} = \frac{360g}{v}\)
\(v = \frac{360g}{0.6gcm^{-3}} = 600cm^{3}\)
Source: Myschool Past Questions Archive (Record #47376)