question_answer

A drop of water of volume\[\text{5}\times \text{1}{{0}^{-\text{8}}}\text{ }{{\text{m}}^{\text{3}}}\] falls from a height of 100m. If \[\text{g}=\text{1}0\text{m}/{{\text{s}}^{\text{2}}}\], what is the work done by the gravitational force? (density of water is\[\text{1}{{0}^{\text{3}}}\text{ kg}/{{\text{m}}^{\text{3}}}\])

Answer:

Given,   \[V=5\times {{10}^{-8}}{{m}^{3}}h=100m\]                 \[g=10m/{{s}^{2}}W=?\]                 \[d={{10}^{3}}kg/{{m}^{3}}\] We know, \[W=mgh\] Formula: \[Density=\frac{mass}{volume}\] \[\Rightarrow mass=density\times Volume\] \[\Rightarrow m={{10}^{3}}\times 5\times {{10}^{-8}}=5\times {{10}^{-5}}kg\] \[\therefore W=5\times {{10}^{-5}}\times 100=5\times {{10}^{-2}}=0.05J\]