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^