Two compressible solids are formed into spheres of the same size.The bulk modulus of sphere two is twice as large as the bulk modulus of sphere one.You now increase the pressure on both spheres by the same amount.As a result of the increased pressure,how is the change in volume of sphere two related to the change in volume of sphere one ? A)ΔV₂=2ΔV₁ B)ΔV₂= 4ΔV₁ C)ΔV₂=ΔV₁ D)ΔV₂= 1/2ΔV₁ E)ΔV₂= 1/4ΔV₁

Question

Quizplus · Accepted Answer

The relationship between pressure, volume, and bulk modulus is given by the formula ΔV/V = -ΔP/B. Since the spheres are of equal size, their initial volumes are the same, thus the initial volume fraction (ΔV/V) would be the same for both spheres. Using the formula above, we can write the following relationship for the change in volume ΔV of sphere 1 and sphere 2: ΔV2/V = -ΔP/B2 (1) ΔV1/V = -ΔP/B1 (2) Since we are increasing the pressure on both spheres by the same amount, we can equate the two changes in pressure: ΔP/B2 = ΔP/B1 We can rearrange this equation to get: ΔP = (B1/B2) * ΔP Substituting this equation into (1) and (2), we get: ΔV2/V = -[(B2/B1) * ΔP] ΔV1/V = -ΔP We want to find the relationship between ΔV2 and ΔV1. To do this, we can divide the first equation by the second: ΔV2/ΔV1 = -(B2/B1) Substituting the given ratio of bulk moduli (B2 = 2B1), we get: ΔV2/ΔV1 = -(2/1) ΔV2 = -2ΔV1 Since both spheres are being subjected to the same increase in pressure, their initial volumes are the same, hence ΔV1/V1 = ΔV2/V2. Therefore, we can write: ΔV1 = -V1*ΔP/B1 ΔV2 = -V2*ΔP/B2 Dividing the two equations, we get: ΔV2/ΔV1 = (B1/B2) * (V2/V1) Substituting the given values, we get: ΔV2/ΔV1 = (1/2) * 1 ΔV2/ΔV1 = 1/2 Since ΔV2 = -2ΔV1, we can substitute this into the equation above: -2ΔV1/ΔV1 = 1/2 Simplifying this, we get: ΔV2/ΔV1 = -2 = -2/1 Therefore, the change in volume of sphere two is twice the change in volume of sphere one: ΔV2 = -2ΔV1 which is equivalent to: ΔV₂ = 1/2ΔV₁. Hence, the correct answer is D.

Two Compressible Solids Are Formed into Spheres of the Same