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A spherical ball of steel of diameter 200 mm goes down to a depth of 1 km in sea water. If specific weight of sea water in 10.2 kN/m3, (and bulk modulus of steel is 170 kN/mm2), the change in volume of the spherical ball is ___________mm3
- a)248.0
- b)252.0
Correct answer is option ''. Can you explain this answer?
Verified Answer
A spherical ball of steel of diameter 200 mm goes down to a depth of ...
Depth, h = 1000 m
Weight density, w = 10.2 x 1000 N/m3
Pressure, p = wh = 1000 x 10.2 x 1000 N/m2
= 10.2 x 106 N/m2 Bulk modulus,
K = 170 x 103 N/mm2
= 170 x 109 N/m2
Volumetric strain, 
=0.06 x 10-3
Original volume, V = 
=4.188 x 106mm3
Change volume,
= 0.25 x 103 mm3
= 250 mm3
Question_Type: 5
This question is part of UPSC exam. View all Mechanical Engineering courses
Most Upvoted Answer
A spherical ball of steel of diameter 200 mm goes down to a depth of ...
To calculate the change in volume of the spherical ball, we can use the equation for the bulk modulus of elasticity:
Change in volume = -(Bulk modulus of steel * Original volume * Change in pressure) / (Specific weight of sea water)
Given:
Diameter of the ball = 200 mm
Radius of the ball = 100 mm
Depth in sea water = 1 km = 1000 m
Specific weight of sea water = 10.2 kN/m3 = 10.2 * 106 N/m3
Bulk modulus of steel = 170 kN/mm2 = 170 * 109 N/m2
Converting the diameter and radius to meters:
Diameter = 200 mm = 0.2 m
Radius = 0.1 m
Calculating the original volume of the ball:
Original volume = (4/3) * π * (Radius)3
Calculating the change in pressure:
Change in pressure = Specific weight of sea water * Depth
Substituting the values into the equation:
Change in volume = -(170 * 109 N/m2 * [(4/3) * π * (0.1)3] * (10.2 * 106 N/m3 * 1000 m))
Simplifying the equation:
Change in volume = -(170 * 109 N/m2 * (4/3) * π * (0.1)3 * 10.2 * 106 N/m3 * 1000 m)
Calculating the change in volume:
Change in volume = -248 mm3
Therefore, the change in volume of the spherical ball is 248 mm3.
Change in volume = -(Bulk modulus of steel * Original volume * Change in pressure) / (Specific weight of sea water)
Given:
Diameter of the ball = 200 mm
Radius of the ball = 100 mm
Depth in sea water = 1 km = 1000 m
Specific weight of sea water = 10.2 kN/m3 = 10.2 * 106 N/m3
Bulk modulus of steel = 170 kN/mm2 = 170 * 109 N/m2
Converting the diameter and radius to meters:
Diameter = 200 mm = 0.2 m
Radius = 0.1 m
Calculating the original volume of the ball:
Original volume = (4/3) * π * (Radius)3
Calculating the change in pressure:
Change in pressure = Specific weight of sea water * Depth
Substituting the values into the equation:
Change in volume = -(170 * 109 N/m2 * [(4/3) * π * (0.1)3] * (10.2 * 106 N/m3 * 1000 m))
Simplifying the equation:
Change in volume = -(170 * 109 N/m2 * (4/3) * π * (0.1)3 * 10.2 * 106 N/m3 * 1000 m)
Calculating the change in volume:
Change in volume = -248 mm3
Therefore, the change in volume of the spherical ball is 248 mm3.
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A spherical ball of steel of diameter 200 mm goes down to a depth of 1 km in sea water. If specific weight of sea water in 10.2 kN/m3, (and bulk modulus of steel is 170 kN/mm2), the change in volume of the spherical ball is ___________mm3a) 248.0b) 252.0Correct answer is option ''. Can you explain this answer?
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A spherical ball of steel of diameter 200 mm goes down to a depth of 1 km in sea water. If specific weight of sea water in 10.2 kN/m3, (and bulk modulus of steel is 170 kN/mm2), the change in volume of the spherical ball is ___________mm3a) 248.0b) 252.0Correct answer is option ''. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A spherical ball of steel of diameter 200 mm goes down to a depth of 1 km in sea water. If specific weight of sea water in 10.2 kN/m3, (and bulk modulus of steel is 170 kN/mm2), the change in volume of the spherical ball is ___________mm3a) 248.0b) 252.0Correct answer is option ''. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A spherical ball of steel of diameter 200 mm goes down to a depth of 1 km in sea water. If specific weight of sea water in 10.2 kN/m3, (and bulk modulus of steel is 170 kN/mm2), the change in volume of the spherical ball is ___________mm3a) 248.0b) 252.0Correct answer is option ''. Can you explain this answer?.
A spherical ball of steel of diameter 200 mm goes down to a depth of 1 km in sea water. If specific weight of sea water in 10.2 kN/m3, (and bulk modulus of steel is 170 kN/mm2), the change in volume of the spherical ball is ___________mm3a) 248.0b) 252.0Correct answer is option ''. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A spherical ball of steel of diameter 200 mm goes down to a depth of 1 km in sea water. If specific weight of sea water in 10.2 kN/m3, (and bulk modulus of steel is 170 kN/mm2), the change in volume of the spherical ball is ___________mm3a) 248.0b) 252.0Correct answer is option ''. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A spherical ball of steel of diameter 200 mm goes down to a depth of 1 km in sea water. If specific weight of sea water in 10.2 kN/m3, (and bulk modulus of steel is 170 kN/mm2), the change in volume of the spherical ball is ___________mm3a) 248.0b) 252.0Correct answer is option ''. Can you explain this answer?.
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