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Physics Question on mechanical properties of solids

If Y,KY, K and η\eta are the values of Young's modulus, bulk modulus and modulus of rigidity of any material respectively. Choose the correct relation for these parameters.

A

Y=9Kη3K−ηN/m2Y =\frac{9 K \eta}{3 K -\eta} N / m ^{2}

B

η=3YK9K+YN/m2\eta=\frac{3 Y K}{9 K+Y} N / m^{2}

C

Y=9Kη2η+3KN/m2Y =\frac{9 K\eta }{2 \eta+3 K } N / m ^{2}

D

K=Yη9η−3YN/m2K =\frac{ Y\eta }{9 \eta-3 Y } N / m ^{2}

Answer

K=Yη9η−3YN/m2K =\frac{ Y\eta }{9 \eta-3 Y } N / m ^{2}

Explanation

Solution

Y- Younge modulus, K- Bulk modulus,
η\eta - modulus of rigidity
We know that
y=3k(1−2σ)y=3 k(1-2 \sigma)
σ=12(1−y3k)\sigma=\frac{1}{2}\left(1-\frac{y}{3 k}\right)...(i)
y=2η(1+σ)y=2 \eta(1+\sigma)
σ=y2η−1\sigma=\frac{y}{2 \eta}-1...(ii)
From E(i) and E (ii)
12(1−Y3k)=y2η−1\frac{1}{2}\left(1-\frac{ Y }{3 k }\right)=\frac{ y }{2 \eta}-1
1−y3k=yη−21-\frac{ y }{3 k }=\frac{ y }{\eta}-2
y3k=3−yη\frac{ y }{3 k }=3-\frac{ y }{\eta}
y3k=3η−yη\frac{ y }{3 k }=\frac{3 \eta- y }{\eta}
ηy3k=3η−y\frac{\eta y }{3 k }=3 \eta- y
k=ηy9η−3yk =\frac{\eta y }{9 \eta -3 y }