Question

The young's modulus and rigidity modulus of a metal are 40 Gpa and 20 Gpa respectively. Then bulk modulus is
(A) 13.33 Gpa
(B) 26.67 Gpa
(C) 6.67 Gpa
(D) 120 Gpa​

Answer 1

Given:

Young's modulus (Y) = 40 GPa

Rigidity/Shear modulus (G) = 20 GPa

To Find:

Bulk modulus (B)

Answer:

Relation between Young's modulus (Y), Shear modulus (G) & Bulk modulus(B):

$\boxed{\boxed{ \bf{Y = \dfrac{9BG}{3B + G}}}}$

By substituting values we get:

$\rm \leadsto 40 = \dfrac{9B \times 20}{3B + 20} \\ \\ \rm \leadsto 40(3B + 20) = 180B \\ \\ \rm \leadsto 120B + 800 = 180B \\ \\ \rm \leadsto 180B - 120B = 800 \\ \\ \rm \leadsto 180B - 120B = 800 \\ \\ \rm \leadsto 60B = 800 \\ \\ \rm \leadsto B = \dfrac{800}{60} \\ \\ \rm \leadsto B = 13.33 \: GPa$

$\therefore$ Bulk modulus (B) = 13.33 GPa

Correct Option: $\mathfrak{(A) \ 13.33 \ GPa}$

Answer 2

Explanation:

Given:

Young's modulus (Y) = 40 GPa

Rigidity/Shear modulus (G) = 20 GPa

To Find:

Bulk modulus (B)

Answer:

Relation between Young's modulus (Y), Shear modulus (G) & Bulk modulus(B):

\boxed{\boxed{ \bf{Y = \dfrac{9BG}{3B + G}}}}

Y=

3B+G

9BG

By substituting values we get:

\begin{gathered}\rm \leadsto 40 = \dfrac{9B \times 20}{3B + 20} \\ \\ \rm \leadsto 40(3B + 20) = 180B \\ \\ \rm \leadsto 120B + 800 = 180B \\ \\ \rm \leadsto 180B - 120B = 800 \\ \\ \rm \leadsto 180B - 120B = 800 \\ \\ \rm \leadsto 60B = 800 \\ \\ \rm \leadsto B = \dfrac{800}{60} \\ \\ \rm \leadsto B = 13.33 \: GPa\end{gathered}

⇝40=

3B+20

9B×20

40(3B+20)=180B

120B+800=180B

180B−120B=800

180B−120B=800

60B=800

B= 60

800

⇝B=13.33GPa

\therefore∴ Bulk modulus (B) = 13.33 GPa

Correct Option: \mathfrak{(A) \ 13.33 \ GPa}(A) 13.33 GPa