Question

A metal cube has an edge length of 90 cm. If $2 \times 10^9$ N m$^{-2}$ of pressure (or stress) is required to reduce the edge length to 89.5 cm, then the bulk modulus of the metal is:

• A. $1 \times 10^{11}$ N m$^{-2}$
• B. $2.5 \times 10^{10}$ N m$^{-2}$
• C. $9 \times 10^{11}$ N m$^{-2}$
• D. $1.2 \times 10^{11}$ N m$^{-2}$

Hint:

For solids, volumetric strain is three times linear strain: $\Delta V/V = 3\Delta L/L$.

Answer 1

The Correct Option is D

Step 1: Work with actual volumes instead of the linear-strain shortcut.
For a cube of edge length $L$, the volume is $V = L^3$. Rather than using $\Delta V/V = 3\Delta L/L$, compute the initial and final volumes directly and take their difference.
Step 2: Compute the initial and final volumes.
\[ V_1 = (90)^3 = 729000 \ \text{cm}^3 \] \[ V_2 = (89.5)^3 = 716917.4 \ \text{cm}^3 \]
Step 3: Find the change in volume and the volumetric strain.
\[ \Delta V = V_1 - V_2 = 729000 - 716917.4 = 12082.6 \ \text{cm}^3 \] \[ \frac{\Delta V}{V_1} = \frac{12082.6}{729000} = 0.01657 \]
Step 4: Apply the definition of bulk modulus.
\[ B = \frac{\text{stress}}{\Delta V/V} = \frac{2\times10^{9}}{0.01657} \]
Step 5: Compute the value.
\[ B \approx 1.207\times10^{11} \ \text{N m}^{-2} \approx 1.2\times10^{11} \ \text{N m}^{-2} \] Computing the volumes directly instead of using the small-strain approximation gives essentially the same answer here because the strain is small, which is itself a check that the linear approximation is valid.
Final Answer:
\[ \boxed{1.2\times10^{11} \ \text{N m}^{-2}} \]