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A metallic ring of radius 2cm and cross sectional area 4cm^2 is fitted into a wodden circular disc of radius 4cm.if the Young's modulus of the material of the ring is 2×10^11N/m^2,the force with which the metal ring expands is ? . the answer for this is 8×10^7N.can u explain this ?
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A metallic ring of radius 2cm and cross sectional area 4cm^2 is fitted...
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A metallic ring of radius 2cm and cross sectional area 4cm^2 is fitted...
Given Data

Radius of the metallic ring (r1) = 2 cm
Cross-sectional area of the metallic ring (A1) = 4 cm^2
Radius of the wooden circular disc (r2) = 4 cm
Young's modulus of the material of the ring (Y) = 2 × 10^11 N/m^2

Explanation


Step 1: Calculating the Area of the Wooden Disc

The area of a circular disc is given by the formula A = πr^2, where A is the area and r is the radius of the disc.
Substituting the given radius (r2 = 4 cm) into the formula, we can find the area of the wooden disc.
A2 = π(4 cm)^2 = 16π cm^2

Step 2: Calculating the Change in Radius

When the metallic ring expands, its radius will increase. Let's assume the change in radius as Δr.
The final radius of the metallic ring after expansion (r1') = r1 + Δr

Step 3: Calculating the Change in Area

The change in area of the metallic ring is given by the formula ΔA = A1' - A1, where ΔA is the change in area, A1' is the final area of the metallic ring after expansion, and A1 is the initial area of the metallic ring.
A1' = π(r1')^2
ΔA = A1' - A1

Step 4: Calculating the Force

The force with which the metal ring expands is given by Hooke's law: F = Y * (ΔA/A2), where F is the force, Y is the Young's modulus, ΔA is the change in area, and A2 is the area of the wooden disc.

Substituting the Values

Substituting the given values into the formulas:
ΔA = π(r1')^2 - A1
= π(r1 + Δr)^2 - A1

As Δr is very small compared to r1, we can neglect the squared term in the expansion of the equation above:
ΔA ≈ 2πr1Δr

Now, substituting the values into the force formula:
F = Y * (ΔA/A2)
= Y * (2πr1Δr / 16π)

Simplifying the equation by canceling out π:
F = Y * (r1Δr / 8)

Final Calculation

Substituting the given values into the equation:
F = (2 × 10^11 N/m^2) * (2 cm × Δr / 8 cm)

As the radius of the ring doubles (from 2 cm to 4 cm), the change in radius Δr is also 2 cm.
F = (2 × 10^11 N/m^2) * (2 cm × 2 cm / 8 cm)
F = 8 × 10^11 N/m^2

Therefore, the force with which the metal ring expands is 8 × 10^
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A metallic ring of radius 2cm and cross sectional area 4cm^2 is fitted into a wodden circular disc of radius 4cm.if the Young's modulus of the material of the ring is 2×10^11N/m^2,the force with which the metal ring expands is ? . the answer for this is 8×10^7N.can u explain this ?
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A metallic ring of radius 2cm and cross sectional area 4cm^2 is fitted into a wodden circular disc of radius 4cm.if the Young's modulus of the material of the ring is 2×10^11N/m^2,the force with which the metal ring expands is ? . the answer for this is 8×10^7N.can u explain this ? 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 metallic ring of radius 2cm and cross sectional area 4cm^2 is fitted into a wodden circular disc of radius 4cm.if the Young's modulus of the material of the ring is 2×10^11N/m^2,the force with which the metal ring expands is ? . the answer for this is 8×10^7N.can u explain this ? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A metallic ring of radius 2cm and cross sectional area 4cm^2 is fitted into a wodden circular disc of radius 4cm.if the Young's modulus of the material of the ring is 2×10^11N/m^2,the force with which the metal ring expands is ? . the answer for this is 8×10^7N.can u explain this ?.
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