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Standard XII

Physics

Beta Decay

A magnesium p...

Question

A magnesium plate in biaxial stress is subjected to tensile stresses $σ_{x} = 24 M P a a n d σ_{y} = 12 M P a$ as shown in figure below. The corresponding strains in the plate are $ε_{x} = 440 \times 10^{- 6} a n d ε_{y} = 80 \times 10^{- 6}$ . The poisson’s ratio of the plate material is (Rounded upto two decimal places).

0.35

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Solution

The correct option is A 0.35
$σ_{x} = 24 M P a σ_{y} = 12 M P a$

$ε_{x} = 440 \times 10^{- 6} ε_{y} = 80 \times 10^{- 6}$

We know that, $ε_{x} = \frac{σ_{x}}{ε} (-) μ \frac{σ_{y}}{ε}$

$\Rightarrow ε \times 440 \times 10^{- 6} = 24 - 12 μ$ ….(I)

Also $ε_{y} = \frac{σ_{y}}{ε} (-) μ \frac{σ_{x}}{ε}$

$\Rightarrow ε \times 80 \times 10^{- 6} = 12 - 24 μ$ ….(II)

From (I) and (II)

$μ = 0.35$

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A magnesium plate in biaxial stress is subjected to tensile stresses σx=24 MPa and σy=12 MPa as shown in figure below. The corresponding strains in the plate are εx=440×10−6 and εy=80×10−6. The poisson’s ratio of the plate material is (Rounded upto two decimal places). 0.35

The correct option is A 0.35 σx=24 MPa σy=12 MPa εx=440×10−6 εy=80×10−6 We know that, εx=σxε(−)μσyε ⇒ε×440×10−6=24−12μ ….(I) Also εy=σyε(−)μσxε ⇒ε×80×10−6=12−24μ ….(II) From (I) and (II) μ=0.35