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A lead cube measures 6.00 cm on each side. The bottom face is held in place by very strong glue to a flat horizontal surface, while a horizontal force F is applied to the upper face parallel to one of the edges. How large must F be to cause the cube to deform by 0.250 mm? (Shear modulus of lead = 0.6 × 1010Pa)
- a)7.0 * 105 N
- b)6* 105 N
- c)6.6 * 105 N
- d)5.6 * 105 N
Correct answer is option 'C'. Can you explain this answer?
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To solve this problem, we can use Hooke's Law for shear deformation:
Shear stress (τ) = Shear modulus (G) * Shear strain (γ)
Here, the shear stress is the force per unit area applied parallel to the edge, and the shear strain is the displacement divided by the original length.
The shear strain (γ) is given by:
γ = Δx / L
Where Δx is the displacement and L is the original length of the cube.
Given:
Initial length (L) = 6.00 cm = 0.06 m
Displacement (Δx) = 0.250 mm = 0.250 * 10^-3 m
Shear modulus (G) = 0.6 x 10^9 N/m^2
We need to find the force (F) required to cause the deformation.
Using Hooke's Law, we can rearrange the equation:
F = τ * A
Where A is the cross-sectional area of the cube.
Since the force is applied parallel to one of the edges, the cross-sectional area is the area of the face of the cube.
A = L^2
Substituting the values into the equations:
Shear strain (γ) = Δx / L = (0.250 * 10^-3 m) / (0.06 m) = 4.17 x 10^-3
Using Hooke's Law:
Shear stress (τ) = Shear modulus (G) * Shear strain (γ)
τ = (0.6 x 10^9 N/m^2) * (4.17 x 10^-3) = 2.5 x 10^6 N/m^2
Using the equation for force:
F = τ * A = (2.5 x 10^6 N/m^2) * (0.06 m)^2 = 9 x 10^3 N
Therefore, the force required to cause the cube to deform by 0.250 mm is 9 x 10^3 N.
Shear stress (τ) = Shear modulus (G) * Shear strain (γ)
Here, the shear stress is the force per unit area applied parallel to the edge, and the shear strain is the displacement divided by the original length.
The shear strain (γ) is given by:
γ = Δx / L
Where Δx is the displacement and L is the original length of the cube.
Given:
Initial length (L) = 6.00 cm = 0.06 m
Displacement (Δx) = 0.250 mm = 0.250 * 10^-3 m
Shear modulus (G) = 0.6 x 10^9 N/m^2
We need to find the force (F) required to cause the deformation.
Using Hooke's Law, we can rearrange the equation:
F = τ * A
Where A is the cross-sectional area of the cube.
Since the force is applied parallel to one of the edges, the cross-sectional area is the area of the face of the cube.
A = L^2
Substituting the values into the equations:
Shear strain (γ) = Δx / L = (0.250 * 10^-3 m) / (0.06 m) = 4.17 x 10^-3
Using Hooke's Law:
Shear stress (τ) = Shear modulus (G) * Shear strain (γ)
τ = (0.6 x 10^9 N/m^2) * (4.17 x 10^-3) = 2.5 x 10^6 N/m^2
Using the equation for force:
F = τ * A = (2.5 x 10^6 N/m^2) * (0.06 m)^2 = 9 x 10^3 N
Therefore, the force required to cause the cube to deform by 0.250 mm is 9 x 10^3 N.
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A lead cube measures 6.00 cm on each side. The bottom face is held in place by very strong glue to a flat horizontal surface, while a horizontal force F is applied to the upper face parallel to one of the edges. How large must F be to cause the cube to deform by 0.250 mm? (Shear modulus of lead = 0.6 × 1010Pa)a)7.0 * 105Nb)6* 105Nc)6.6 * 105Nd)5.6 * 105NCorrect answer is option 'C'. Can you explain this answer?
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A lead cube measures 6.00 cm on each side. The bottom face is held in place by very strong glue to a flat horizontal surface, while a horizontal force F is applied to the upper face parallel to one of the edges. How large must F be to cause the cube to deform by 0.250 mm? (Shear modulus of lead = 0.6 × 1010Pa)a)7.0 * 105Nb)6* 105Nc)6.6 * 105Nd)5.6 * 105NCorrect answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A lead cube measures 6.00 cm on each side. The bottom face is held in place by very strong glue to a flat horizontal surface, while a horizontal force F is applied to the upper face parallel to one of the edges. How large must F be to cause the cube to deform by 0.250 mm? (Shear modulus of lead = 0.6 × 1010Pa)a)7.0 * 105Nb)6* 105Nc)6.6 * 105Nd)5.6 * 105NCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A lead cube measures 6.00 cm on each side. The bottom face is held in place by very strong glue to a flat horizontal surface, while a horizontal force F is applied to the upper face parallel to one of the edges. How large must F be to cause the cube to deform by 0.250 mm? (Shear modulus of lead = 0.6 × 1010Pa)a)7.0 * 105Nb)6* 105Nc)6.6 * 105Nd)5.6 * 105NCorrect answer is option 'C'. Can you explain this answer?.
A lead cube measures 6.00 cm on each side. The bottom face is held in place by very strong glue to a flat horizontal surface, while a horizontal force F is applied to the upper face parallel to one of the edges. How large must F be to cause the cube to deform by 0.250 mm? (Shear modulus of lead = 0.6 × 1010Pa)a)7.0 * 105Nb)6* 105Nc)6.6 * 105Nd)5.6 * 105NCorrect answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A lead cube measures 6.00 cm on each side. The bottom face is held in place by very strong glue to a flat horizontal surface, while a horizontal force F is applied to the upper face parallel to one of the edges. How large must F be to cause the cube to deform by 0.250 mm? (Shear modulus of lead = 0.6 × 1010Pa)a)7.0 * 105Nb)6* 105Nc)6.6 * 105Nd)5.6 * 105NCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A lead cube measures 6.00 cm on each side. The bottom face is held in place by very strong glue to a flat horizontal surface, while a horizontal force F is applied to the upper face parallel to one of the edges. How large must F be to cause the cube to deform by 0.250 mm? (Shear modulus of lead = 0.6 × 1010Pa)a)7.0 * 105Nb)6* 105Nc)6.6 * 105Nd)5.6 * 105NCorrect answer is option 'C'. Can you explain this answer?.
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ample number of questions to practice A lead cube measures 6.00 cm on each side. The bottom face is held in place by very strong glue to a flat horizontal surface, while a horizontal force F is applied to the upper face parallel to one of the edges. How large must F be to cause the cube to deform by 0.250 mm? (Shear modulus of lead = 0.6 × 1010Pa)a)7.0 * 105Nb)6* 105Nc)6.6 * 105Nd)5.6 * 105NCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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