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A rectangular metallic block of dimensions $40 \mathrm{~mm} \times 20 \mathrm{~mm}$ when pulled with a tension of $50 \mathrm{kN}$ undergoes only elastic deformation. The strain in the block is
(The shear modulus of the material of the block is $40 \times 10^9 \mathrm{Nm}^{-2}$ )
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(The shear modulus of the material of the block is $40 \times 10^9 \mathrm{Nm}^{-2}$ )
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Verified Answer
The correct answer is:
$1.56 \times 10^{-3} \mathrm{~m}$
Tension, $\mathrm{F}=50 \times 10^3 \mathrm{~N}$
Area, $A=40 \times 20 \times 10^{-6} \mathrm{M}$
$\begin{aligned} & \text { Modulus of the material }=\frac{\text { Stress }}{\text { Strain }} \\ & \text { Strain }=\frac{\text { Force }}{\text { Area } \times \text { shear modulus }}\end{aligned}$
$=\frac{50 \times 10^3}{40 \times 20 \times 10^{-6} \times 40 \times 10^9}=1.5 \times 10^{-3} \mathrm{~m}$
Area, $A=40 \times 20 \times 10^{-6} \mathrm{M}$
$\begin{aligned} & \text { Modulus of the material }=\frac{\text { Stress }}{\text { Strain }} \\ & \text { Strain }=\frac{\text { Force }}{\text { Area } \times \text { shear modulus }}\end{aligned}$
$=\frac{50 \times 10^3}{40 \times 20 \times 10^{-6} \times 40 \times 10^9}=1.5 \times 10^{-3} \mathrm{~m}$
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