QUARTER FINAL STAGE 2026
PRESEC, Legon: 56 points
Mfantsipim School: 39 points
Presby SHS, Bompata: 24 points
QUESTION
Find Young's modulus for a linear isotropic substance whose bulk modulus is $1.8$ GPa and whose Poisson ratio is $0.30$.
ANSWER: $2.2$ GPa
SOLUTION 1:
$E = 3K(1-2\nu)$
$E = 3 \times 1.8 \times (1 - 2 \times 0.30)$
$E = 5.4 \times (1 - 0.60)$
$E = 5.4 \times 0.40$
$E = 54\times10^{-1} \times 4\times10^{-1}$
$E = 54 \times 4 \times10^{-2}$
$E = 2.16\text{ GPa}$
$E \approx 2.2\text{ GPa}$
SOLUTION 2:
$E = 3K - 6K\nu$
$E = 3(1.8) - 6(1.8)(0.30)$
$E = 5.4 - 10.8(0.30)$
$E = 5.4 - 3.24$
$E = 2.16\text{ GPa}$
$E \approx 2.2\text{ GPa}$
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PRACTICE QUESTIONS
1. QUESTION: Find Young's modulus for a linear isotropic substance whose bulk modulus is 2.4 GPa and whose Poisson ratio is 0.35.
ANSWER: 2.2 GPa
SOLUTION:
$E = 3K(1 - 2\nu)$
$E = 3 \times 2.4 \times (1 - 2 \times 0.35)$
$E = 3 \times 2.4 \times 0.30$
$E = 3 \times 24\times10^{-1} \times 3\times10^{-1}$
$E = 3 \times 24 \times 3 \times10^{-2}$
$E \approx 2.2\text{ GPa}$
2. QUESTION: Find the bulk modulus of a linear isotropic substance whose Young's modulus is 36 GPa and whose Poisson ratio is 0.10.
ANSWER: 15 GPa or $1.5\times10^{1}$ GPa
SOLUTION:
$E = 3K(1 - 2\nu)$
$K = \dfrac{E}{3(1 - 2\nu)}$
$K = \dfrac{36}{3(1 - 2 \times 0.10)}$
$K = \dfrac{36}{3 \times 0.80}$
$K = \dfrac{36}{3 \times 8\times10^{-1}}$
$K = \dfrac{9}{3 \times 2} \times10^{1}$
$K = \dfrac{3}{2} \times10^{1}$
$K = 15\text{ GPa}$
3. QUESTION: Find Young's modulus for a linear isotropic substance whose bulk modulus is 1.5 GPa and whose Poisson ratio is 0.30.
ANSWER: 1.8 GPa
SOLUTION:
$E = 3K(1 - 2\nu)$
$E = 3 \times 1.5 \times (1 - 2 \times 0.30)$
$E = 3 \times 1.5 \times 0.40$
$E = 3 \times 15\times10^{-1} \times 4\times10^{-1}$
$E = 3 \times 15 \times 4 \times10^{-2}$
$E = 6 \times 3 \times10^{-1}$
$E = 1.8\text{ GPa}$
4. QUESTION: Find Young's modulus for a linear isotropic substance whose bulk modulus is 25 GPa and whose Poisson ratio is 0.20.
ANSWER: 45 GPa or $4.5\times10^{1}$ GPa
SOLUTION:
$E = 3K(1 - 2\nu)$
$E = 3 \times 25 \times (1 - 2 \times 0.20)$
$E = 3 \times 25 \times 0.60$
$E = 3 \times 25 \times 6\times10^{-1}$
$E = 45\text{ GPa}$
5. QUESTION: Find Young's modulus for a linear isotropic substance whose bulk modulus is 75 GPa and whose Poisson ratio is 0.26.
ANSWER: 110 GPa or $1.1\times10^{2}$ GPa
SOLUTION:
$E = 3K(1 - 2\nu)$
$E = 3 \times 75 \times (1 - 2 \times 0.26)$
$E = 3 \times 75 \times 0.48$
$E = 3 \times 75 \times 48\times10^{-2}$
$E \approx 1.1\times10^{2}\text{ GPa}$
6. QUESTION: Find the bulk modulus of a linear isotropic substance whose Young's modulus is 7.2 GPa and whose Poisson ratio is 0.25.
ANSWER: 4.8 GPa
SOLUTION:
$E = 3K(1 - 2\nu)$
$K = \dfrac{E}{3(1 - 2\nu)}$
$K = \dfrac{7.2}{3(1 - 2 \times 0.25)}$
$K = \dfrac{7.2}{3 \times 0.50}$
$K = \dfrac{72}{3 \times 5}$
$K = \dfrac{24}{5}$
$K = 4.8\text{ GPa}$
7. QUESTION: Find the bulk modulus of a linear isotropic substance whose Young's modulus is 3.6 GPa and whose Poisson ratio is 0.40.
ANSWER: 6.0 GPa
SOLUTION:
$E = 3K(1 - 2\nu)$
$K = \dfrac{E}{3(1 - 2\nu)}$
$K = \dfrac{3.6}{3(1 - 2 \times 0.40)}$
$K = \dfrac{3.6}{3 \times 0.20}$
$K = \dfrac{36}{3 \times 2}$
$K = \dfrac{12}{2}$
$K = 6.0\text{ GPa}$
8. QUESTION: Find Young's modulus for a linear isotropic substance whose bulk modulus is 4.0 GPa and whose Poisson ratio is 0.35.
ANSWER: 3.6 GPa
SOLUTION:
$E = 3K(1 - 2\nu)$
$E = 3 \times 4.0 \times (1 - 2 \times 0.35)$
$E = 3 \times 4.0 \times 0.30$
$E = 3 \times 4 \times 3\times10^{-1}$
$E = 3.6\text{ GPa}$
9. QUESTION: Find Young's modulus for a linear isotropic substance whose bulk modulus is 2.5 GPa and whose Poisson ratio is 0.45.
ANSWER: 0.75 GPa or $7.5\times10^{-1}$ GPa
SOLUTION:
$E = 3K(1 - 2\nu)$
$E = 3 \times 2.5 \times (1 - 2 \times 0.45)$
$E = 3 \times 2.5 \times 0.10$
$E = 3 \times 25\times10^{-1} \times 10^{-1}$
$E = 3 \times 25 \times10^{-2}$
$E = 0.75\text{ GPa}$
10. QUESTION: Find the bulk modulus of a linear isotropic substance whose Young's modulus is 9.0 GPa and whose Poisson ratio is 0.20.
ANSWER: 5.0 GPa
SOLUTION:
$E = 3K(1 - 2\nu)$
$K = \dfrac{E}{3(1 - 2\nu)}$
$K = \dfrac{9.0}{3(1 - 2 \times 0.20)}$
$K = \dfrac{9.0}{3 \times 0.60}$
$K = \dfrac{9}{3 \times 6\times10^{-1}}$
$K = \dfrac{3}{6} \times10^{1}$
$K = \dfrac{1}{2} \times10^{1}$
$K = 5.0\text{ GPa}$
11. QUESTION: Find Young's modulus for a linear isotropic substance whose bulk modulus is 1.2 GPa and whose Poisson ratio is 0.45.
ANSWER: 0.36 GPa or $3.6\times10^{-1}$ GPa
SOLUTION:
$E = 3K(1 - 2\nu)$
$E = 3 \times 1.2 \times (1 - 2 \times 0.45)$
$E = 3 \times 1.2 \times 0.10$
$E = 3 \times 12\times10^{-1} \times 10^{-1}$
$E = 3 \times 12 \times10^{-2}$
$E = 0.36\text{ GPa}$
12. QUESTION: Find Young's modulus for a linear isotropic substance whose bulk modulus is 18 GPa and whose Poisson ratio is 0.15.
ANSWER: 38 GPa or $3.8\times10^{1}$ GPa
SOLUTION:
$E = 3K(1 - 2\nu)$
$E = 3 \times 18 \times (1 - 2 \times 0.15)$
$E = 3 \times 18 \times 0.70$
$E = 3 \times 18 \times 7\times10^{-1}$
$E \approx 38\text{ GPa}$