Answer:
the cylinder with height 6 has a volume of 60/π in³ greater than the volume of the cylinder with height 10 (option B , if 60π is changed for 60/π)
Explanation:
The volume of a cylinder is
V= π*R²*H (H=height)
since the length L of the piece of paper is L=2*π*R →R=L/(2*π) (since is rolled to form the cylinder), then:
V= π*R²*H = π*L²/(2*π)²*H = L²*H/(4*π)
with L=10 in and H= 6 in we have
V₂= L²*H/(4*π)
the other way around is changing H for L
V₁= H²*L/(4*π)
the difference between the volumes will be
V₂- V₁ = L²*H/(4*π) - H²*L/(4*π) = L*H *(L-H)/(4*π)
replacing values
V₂- V₁ = L*H *(L-H)/(4*π) = 10*6*(10-6)/(4*π) = 60/π in³
then the cylinder with height 6 has a volume of 60/π in³ greater than the volume of the cylinder with height 10