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A 200 × 100 × 50 mm3 steel block is subjected to a hydro-static stress of 15 MPa. The Young’s modulus and Poisson’s ratio of the material are 200 GPa and 0.3 respectively. The change in the volume of the block in mm3 is
- a)85
- b)90
- c)10
- d)100
Correct answer is option 'B'. Can you explain this answer?
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A 200 × 100 × 50 mm3 steel block is subjected to a hydro-static stres...
Given,
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dimension of steel block = 200×100×50mm3
σ = 15 MPa
E = 200 GPa, μ = 0.3
Bulk modulus, 
E=3K(1-2μ)⇒200=3K(1-2×0.3)
K = 166.67GPa
ΔV = 90mm3
Most Upvoted Answer
A 200 × 100 × 50 mm3 steel block is subjected to a hydro-static stres...
To find the change in volume of the steel block, we can use the equation:
ΔV/V = -3βΔP
Where:
ΔV is the change in volume
V is the initial volume
β is the volumetric strain
ΔP is the change in hydrostatic stress
Given:
Initial dimensions of the block:
Length (L) = 200 mm
Width (W) = 100 mm
Height (H) = 50 mm
Hydrostatic stress (P) = 15 MPa = 15 N/mm²
Young’s modulus (E) = 200 GPa = 200 × 10^3 N/mm²
Poisson’s ratio (ν) = 0.3
Let's calculate the volumetric strain (β):
β = -ν(ΔL/L + ΔW/W + ΔH/H)
Since the block is subjected to a hydrostatic stress, the strain in each direction is the same. Therefore, ΔL = ΔW = ΔH.
Plugging in the values, we have:
β = -0.3(ΔL/L + ΔL/W + ΔL/H)
β = -0.3(3ΔL/L)
β = -0.9(ΔL/L)
Now, we can substitute the values into the equation to find the change in volume:
ΔV/V = -3βΔP
ΔV/(200 × 100 × 50) = -3(-0.9(ΔL/L))(15)
ΔV/(200 × 100 × 50) = 40.5(ΔL/L)
ΔV/(200 × 100 × 50) = 40.5(ΔL/200)
ΔV/(200 × 100 × 50) = 0.2025(ΔL)
Since ΔL = ΔW = ΔH, we can assume ΔL = ΔW = ΔH = Δx.
Therefore,
ΔV/(200 × 100 × 50) = 0.2025(Δx)
Simplifying further,
ΔV = 0.2025(Δx)(200 × 100 × 50)
ΔV = 0.405(Δx)(10^6)
The change in volume is given by ΔV = (Δx)^3.
So, (Δx)^3 = 0.405(Δx)(10^6)
Simplifying further,
(Δx)^2 = 0.405(10^6)
Δx = √(0.405(10^6))
Δx ≈ 900
Therefore, the change in volume of the steel block is approximately 900 mm³.
Hence, the correct answer is option 'B' (90).
ΔV/V = -3βΔP
Where:
ΔV is the change in volume
V is the initial volume
β is the volumetric strain
ΔP is the change in hydrostatic stress
Given:
Initial dimensions of the block:
Length (L) = 200 mm
Width (W) = 100 mm
Height (H) = 50 mm
Hydrostatic stress (P) = 15 MPa = 15 N/mm²
Young’s modulus (E) = 200 GPa = 200 × 10^3 N/mm²
Poisson’s ratio (ν) = 0.3
Let's calculate the volumetric strain (β):
β = -ν(ΔL/L + ΔW/W + ΔH/H)
Since the block is subjected to a hydrostatic stress, the strain in each direction is the same. Therefore, ΔL = ΔW = ΔH.
Plugging in the values, we have:
β = -0.3(ΔL/L + ΔL/W + ΔL/H)
β = -0.3(3ΔL/L)
β = -0.9(ΔL/L)
Now, we can substitute the values into the equation to find the change in volume:
ΔV/V = -3βΔP
ΔV/(200 × 100 × 50) = -3(-0.9(ΔL/L))(15)
ΔV/(200 × 100 × 50) = 40.5(ΔL/L)
ΔV/(200 × 100 × 50) = 40.5(ΔL/200)
ΔV/(200 × 100 × 50) = 0.2025(ΔL)
Since ΔL = ΔW = ΔH, we can assume ΔL = ΔW = ΔH = Δx.
Therefore,
ΔV/(200 × 100 × 50) = 0.2025(Δx)
Simplifying further,
ΔV = 0.2025(Δx)(200 × 100 × 50)
ΔV = 0.405(Δx)(10^6)
The change in volume is given by ΔV = (Δx)^3.
So, (Δx)^3 = 0.405(Δx)(10^6)
Simplifying further,
(Δx)^2 = 0.405(10^6)
Δx = √(0.405(10^6))
Δx ≈ 900
Therefore, the change in volume of the steel block is approximately 900 mm³.
Hence, the correct answer is option 'B' (90).
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A 200 × 100 × 50 mm3 steel block is subjected to a hydro-static stress of 15 MPa. The Young’s modulus and Poisson’s ratio of the material are 200 GPa and 0.3 respectively. The change in the volume of the block in mm3 isa)85b)90c)10d)100Correct answer is option 'B'. Can you explain this answer?
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A 200 × 100 × 50 mm3 steel block is subjected to a hydro-static stress of 15 MPa. The Young’s modulus and Poisson’s ratio of the material are 200 GPa and 0.3 respectively. The change in the volume of the block in mm3 isa)85b)90c)10d)100Correct answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 200 × 100 × 50 mm3 steel block is subjected to a hydro-static stress of 15 MPa. The Young’s modulus and Poisson’s ratio of the material are 200 GPa and 0.3 respectively. The change in the volume of the block in mm3 isa)85b)90c)10d)100Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 200 × 100 × 50 mm3 steel block is subjected to a hydro-static stress of 15 MPa. The Young’s modulus and Poisson’s ratio of the material are 200 GPa and 0.3 respectively. The change in the volume of the block in mm3 isa)85b)90c)10d)100Correct answer is option 'B'. Can you explain this answer?.
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