a berry is dropped into a cup of cooking oil. the berry comes to rest with most of its volume under the…

a berry is dropped into a cup of cooking oil. the berry comes to rest with most of its volume under the surface of the oil and a small part of it above the surface. if the oil has a density of 0.9 g/cm³, which is a correct estimate of the density of the berry? a. much less than 0.9 g/cm³ b. slightly less than 0.9 g/cm³ c. slightly more than 0.9 g/cm³ d. much more than 0.9 g/cm³

a berry is dropped into a cup of cooking oil. the berry comes to rest with most of its volume under the surface of the oil and a small part of it above the surface. if the oil has a density of 0.9 g/cm³, which is a correct estimate of the density of the berry? a. much less than 0.9 g/cm³ b. slightly less than 0.9 g/cm³ c. slightly more than 0.9 g/cm³ d. much more than 0.9 g/cm³

Answer

Explanation:

Step1: Recall Archimedes' principle

When an object floats in a fluid, the buoyant force equals the weight of the object. The buoyant force is given by $F_b=\rho_{fluid}gV_{submerged}$, and the weight of the object is $W = \rho_{object}gV_{object}$. Since the berry is floating with most of its volume submerged in the oil, $\rho_{object}gV_{object}=\rho_{oil}gV_{submerged}$. And $V_{submerged}\approx V_{object}$ (because most of the berry's volume is under - the surface).

Step2: Analyze density relationship

If $\rho_{object}V_{object}=\rho_{oil}V_{submerged}$ and $V_{submerged}\approx V_{object}$, then $\rho_{object}\approx\rho_{oil}$. Since only a small part of the berry is above the surface, the density of the berry is slightly less than the density of the oil.

Answer:

B. slightly less than 0.9 g/cm³