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Properties of Matter question

2022 · 27 Jul · Shift 1 · Q70
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Properties of Matter question

2022 · 27 Jul · Shift 1 · Q70

JEE MainPhysicsProperties of MatterNumerical+4 / −1
A square aluminum (shear modulus is 25×109 Nm−225 \times 10^{9}\, \mathrm{Nm}^{-2}25×109Nm−2) slab of side 60 cm60 \mathrm{~cm}60 cm and thickness 15 cm15 \mathrm{~cm}15 cm is subjected to a shearing force (on its narrow face) of 18.0×104N18.0 \times 10^{4}\mathrm{N}18.0×104N. The lower edge is riveted to the floor. The displacement of the upper edge is ‾μ\underline{\hspace{2cm}}\mu​μ m.
Numerical answer
View written solutionFree

Correct answer: 48

  1. Use the definition of shear modulus

Shear modulus is

G=shear stressshear strain=F/Ax/lG = \frac{\text{shear stress}}{\text{shear strain}} = \frac{F/A}{x/l}G=shear strainshear stress​=x/lF/A​

where:

  • FFF = shearing force
  • AAA = area of the face on which force acts
  • xxx = displacement of upper edge
  • lll = thickness of slab

So,

x=FlAGx = \frac{Fl}{AG}x=AGFl​
  1. Given data
  • Side of square slab =60 cm=0.60 m= 60\text{ cm} = 0.60\text{ m}=60 cm=0.60 m
  • Thickness l=15 cm=0.15 ml = 15\text{ cm} = 0.15\text{ m}l=15 cm=0.15 m
  • Shear modulus:
G=25×109 N m−2G = 25 \times 10^9\, \text{N m}^{-2}G=25×109N m−2
  • Force:
F=18.0×104 NF = 18.0 \times 10^4\, \text{N}F=18.0×104N

Since force is applied on the narrow face, the area is

A=(60 cm)(15 cm)=(0.60)(0.15)=0.09 m2A = (60\text{ cm})(15\text{ cm}) = (0.60)(0.15) = 0.09\, \text{m}^2A=(60 cm)(15 cm)=(0.60)(0.15)=0.09m2
  1. Substitute into the formula
x=(18.0×104)(0.60)(0.09)(25×109)x = \frac{(18.0\times 10^4)(0.60)}{(0.09)(25\times 10^9)}x=(0.09)(25×109)(18.0×104)(0.60)​

Here, the relevant length in strain is the distance between lower and upper edges, i.e. side of slab =0.60=0.60=0.60 m.

Now calculate:

(18.0×104)(0.60)=1.08×105(18.0\times 10^4)(0.60)=1.08\times 10^5(18.0×104)(0.60)=1.08×105

and

(0.09)(25×109)=2.25×109(0.09)(25\times 10^9)=2.25\times 10^9(0.09)(25×109)=2.25×109

Therefore,

x=1.08×1052.25×109=4.8×10−5 mx = \frac{1.08\times 10^5}{2.25\times 10^9} = 4.8\times 10^{-5}\,\text{m}x=2.25×1091.08×105​=4.8×10−5m
  1. Convert to micrometres
1 μm=10−6 m1\,\mu\text{m} = 10^{-6}\,\text{m}1μm=10−6m

so

x=4.8×10−5 m=48×10−6 m=48 μmx = 4.8\times 10^{-5}\,\text{m} = 48\times 10^{-6}\,\text{m} = 48\,\mu\text{m}x=4.8×10−5m=48×10−6m=48μm
  1. Final answer
48\boxed{48}48​

This matches the stored correct answer.

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