The control spring of the instrument have fol...
The control spring of the instrument have following dimensions:
Length = 250 mm, thickness of the strip = 0.005 mm, width of the strop = 0.55 mm, Young modulus = 129 gN/m2
Find out the torque exerted by spring when it turns through 90°
• a)
0
• b)
11 x 10-9 μN/m2
• c)
205 x 10-9 μN/m2
• d)
20.25 x 10-9 μN/m2
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The control spring of the instrument have following dimensions:Length ...
Given,
l = 0.25 m
t = 5 × 10-6m
θ = 90
b = 55 × 10-5
E = 129 gN/m2
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The control spring of the instrument have following dimensions:Length ...
To find the torque exerted by the spring when it turns through 90 degrees, we first need to calculate the moment of inertia of the spring.

The moment of inertia of a rectangular strip about its centroidal axis is given by the formula:

I = (1/12) * m * (h^3 + b^3)

Where I is the moment of inertia, m is the mass, h is the height (thickness), and b is the width.

First, let's calculate the mass of the strip:

Mass = density * volume

The density of the strip can be calculated using the formula:

Density = mass / volume

Given that the Young's modulus is 129 gN/m^2 and the dimensions of the strip, we can calculate the mass of the strip as follows:

Density = 129 gN/m^2 = mass / (0.005 mm * 0.55 mm * 250 mm)
mass = Density * (0.005 mm * 0.55 mm * 250 mm)
mass = 129 gN/m^2 * (0.005 mm * 0.55 mm * 250 mm)

Next, let's calculate the moment of inertia:

I = (1/12) * mass * (thickness^3 + width^3)
I = (1/12) * (mass) * (0.005 mm)^3 + (0.55 mm)^3)

Now, let's calculate the torque exerted by the spring when it turns through 90 degrees:

Torque = I * angular acceleration
Torque = I * (angular velocity / time)

Since we are given that the spring is turning through 90 degrees, which is equal to pi/2 radians, we can calculate the angular velocity:

Angular velocity = (pi/2 radians) / time

Finally, we can calculate the torque exerted by the spring:

Torque = I * angular velocity
Torque = (1/12) * (mass) * (0.005 mm)^3 + (0.55 mm)^3) * ((pi/2 radians) / time)
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The control spring of the instrument have following dimensions:Length = 250 mm, thickness of the strip = 0.005 mm, width of the strop = 0.55 mm, Young modulus = 129 gN/m2Find out the torque exerted by spring when it turns through 90°a)0b)11 x 10-9μN/m2c)205 x 10-9 μN/m2d)20.25x 10-9 μN/m2Correct answer is option 'D'. Can you explain this answer?
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The control spring of the instrument have following dimensions:Length = 250 mm, thickness of the strip = 0.005 mm, width of the strop = 0.55 mm, Young modulus = 129 gN/m2Find out the torque exerted by spring when it turns through 90°a)0b)11 x 10-9μN/m2c)205 x 10-9 μN/m2d)20.25x 10-9 μN/m2Correct answer is option 'D'. Can you explain this answer? for SSC JE 2024 is part of SSC JE preparation. The Question and answers have been prepared according to the SSC JE exam syllabus. Information about The control spring of the instrument have following dimensions:Length = 250 mm, thickness of the strip = 0.005 mm, width of the strop = 0.55 mm, Young modulus = 129 gN/m2Find out the torque exerted by spring when it turns through 90°a)0b)11 x 10-9μN/m2c)205 x 10-9 μN/m2d)20.25x 10-9 μN/m2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for SSC JE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The control spring of the instrument have following dimensions:Length = 250 mm, thickness of the strip = 0.005 mm, width of the strop = 0.55 mm, Young modulus = 129 gN/m2Find out the torque exerted by spring when it turns through 90°a)0b)11 x 10-9μN/m2c)205 x 10-9 μN/m2d)20.25x 10-9 μN/m2Correct answer is option 'D'. Can you explain this answer?.
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