Question

A glass marble that has a diameter of 1.00 cm is droppped into a graduated cylinder that contains 16.5 cm of mercury. The bulk modulus of glass is 50,0 x 10 N/m? What is the change AV in the marble's volume while at the bottom of the mercury column, where the pressure is 2.20 x 10' N/m greater than at the surface? AV = m I What is the corresponding change Ar in radius associated with the compression? Ar m

Answer 1

The change in the marble's volume (ΔV) due to increased pressure at the bottom of a mercury column can be calculated using the formula ΔV = -ΔP·V0/B. The original volume of the marble (V0) is found using its diameter, and the bulk modulus of glass (B) is given. The corresponding change in radius (Δr) can be determined after finding the new volume and using the volume-to-radius relationship.

Step-by-step explanation:

To determine the change in the marble's volume (ΔV) while at the bottom of the mercury column due to the increased pressure, we use the formula relating volume change to pressure change: ΔV = -ΔP·V0/B. Here, ΔP represents the pressure increase, V0 is the original volume of the marble, and B is the bulk modulus of the marble's material.

First, we calculate the original volume V0 of the marble using its diameter:

• V0 = (4/3) π (d/2)^3

For a diameter d = 1.00 cm,

• V0 = (4/3) π (0.5 cm)^3 ≈ 0.52 cm3

Substitute the known values into the change in volume equation:

• ΔV = -ΔP·V0/B
• ΔV = -(2.20 x 105 N/m2) · (0.52 cm3)/(50.0 x 1010 N/m2)

Performing the calculations, we find the change in volume ΔV of the marble.

Next, to find the corresponding change in radius (Δr), we can relate the new volume V1 to the new radius r1:

• V1 = (4/3) π r1^3
• r1 = (3V1/(4π))^(1/3)

We determine the new volume V1 by subtracting ΔV from V0 and then calculate the new radius r1. The change in radius Δr is the difference between the original radius r0 and the new radius r1.