Class 12 - Unit # 21 : Physics of Solids -Solved Numericals
1. Young's Modulus (Change in Length)
The 'lead' in pencils is a graphite composition with a Young's modulus of 1.0 × 10⁹ N/m². Calculate the change in length of the lead in an automatic pencil if you tap it straight into the pencil with a force of 4.0 N. The lead is 0.50 mm in diameter and 60 mm long.
Data:
Young's modulus (Y) = 1.0 × 10⁹ N/m²
Force (F) = 4.0 N
Diameter (d) = 0.50 mm = 0.50 × 10⁻³ m (Radius r = 0.25 × 10⁻³ m)
Length (L) = 60 mm = 60 × 10⁻³ m
Change in length (ΔL) = ?
Solution:
According to the definition of Young’s Modulus: Y = (F × L) / (A × ΔL)
Rearranging for ΔL: ΔL = (F × L) / (A × Y) ----(i)
Result: The change in length of the lead is 1.22 mm.
2. Stress, Strain, and Young's Modulus
A wire of 2.2 m long and 2.25 mm in diameter, when stretched by a weight of 8.8 kg, its length has been increased by 0.25 mm. Find the stress, strain, and Young's modulus of the material of the wire. Given g = 9.8 m/s².
Data:
Length (L) = 2.2 m
Diameter (d) = 2.25 mm = 2.25 × 10⁻³ m (Radius r = 1.125 × 10⁻³ m)
Result: The stress is 2.17 × 10⁷ N/m², strain is 1.14 × 10⁻⁴, and Young's modulus is 1.90 × 10¹¹ N/m².
3. Bulk Modulus & Normal Force
A farmer making juice fills a glass bottle to the brim and caps it tightly. The juice expands more than the glass when it warms up, in such a way that the volume increases by 0.2% (ΔV/V = 2 × 10⁻³) relative to the space available. Calculate the normal force exerted by the juice per square centimeter, if its bulk modulus is 1.8 × 10⁹ N/m². Assuming that the bottle does not break.
Data:
Fractional change in volume (ΔV/V) = 2 × 10⁻³
Bulk Modulus (B) = 1.8 × 10⁹ N/m²
Normal Force per cm² = ?
Solution:
According to the definition of Bulk Modulus: B = ΔP / (ΔV/V)
Per cm² conversion: Since 1 m² = 10⁴ cm², then Area = 1 cm² = 10⁻⁴ m²
Force = Pressure × Area = 3.6 × 10⁶ × 10⁻⁴ = 360 N
Result: The normal force exerted by the juice per square centimeter is 360 N.
4. Elastic Limit & Wire Diameter
The elastic limit of copper is 1.5 × 10⁸ N/m². It is to be stretched by a load of 10 kg. Find the diameter of the wire if the elastic limit is not to be exceeded. Given g = 9.8 m/s².
Data:
Maximum allowable Stress = 1.5 × 10⁸ N/m²
Mass (m) = 10 kg (Force F = 10 × 9.8 = 98 N)
Diameter (d) = ?
Solution:
According to definition of Stress: Stress = F / A ----(i)
The Area of cross section is given by: A = πd² / 4
Putting this into eq(i): Stress = (4 × F) / (π × d²)
Rearranging for d²: d² = (4 × F) / (π × Stress)
d² = (4 × 98) / (3.1416 × 1.5 × 10⁸)
d² = 392 / (4.7124 × 10⁸) ≈ 8.318 × 10⁻⁷ m²
Taking square root on both sides: d ≈ 9.12 × 10⁻⁴ m
Result: The minimum diameter of the wire is 0.912 mm.
5. Critical Hanging Length of Wire
What would be the greatest length of a steel wire which is fixed at one end, and can it be hanged freely without breaking? The breaking stress of steel is 7.8 × 10⁸ N/m², and the density of steel is 7800 kg/m³. Given g = 9.8 m/s².
Data:
Breaking Stress = 7.8 × 10⁸ N/m²
Density (ρ) = 7800 kg/m³
Greatest Length (L) = ?
Solution:
According to definition of Stress: Stress = Weight / Area = (m × g) / A ----(i)
Since density ρ = m / V, then m = ρ × V
For a uniform wire, Volume V = A × L, so m = ρ × A × L
Putting mass into eq(i): Stress = (ρ × A × L × g) / A = ρ × L × g
Rearranging for L: L = Stress / (ρ × g)
L = (7.8 × 10⁸) / (7800 × 9.8)
L = (7.8 × 10⁸) / 76440 ≈ 10204.08 m
Result: The greatest length of a steel wire without breaking is approximately 10.2 km.
6. Stress, Strain, and Elongation of Steel Wire
A mild steel wire of radius 0.55 mm and length 3.5 m is stretched by a force of 52 N. Calculate: (a) Longitudinal stress, (b) Longitudinal strain, and (c) Elongation produced in the wire if Young's modulus is 2.1 × 10¹¹ N/m².
Data:
Radius (r) = 0.55 mm = 0.55 × 10⁻³ m
Length (L) = 3.5 m
Force (F) = 52 N
Young's modulus (Y) = 2.1 × 10¹¹ N/m²
a) Stress = ?, b) Strain = ?, c) Elongation (ΔL) = ?
Solution:
a) Area A = πr² = 3.1416 × (0.55 × 10⁻³)² ≈ 9.503 × 10⁻⁷ m²
Longitudinal Stress = F / A = 52 / (9.503 × 10⁻⁷) ≈ 5.47 × 10⁷ N/m²
b) From Young's Modulus definition: Y = Stress / Strain
c) According to definition of Strain: Strain = ΔL / L
Elongation (ΔL) = Strain × L = 2.61 × 10⁻⁴ × 3.5 ≈ 9.12 × 10⁻⁴ m
Result: (a) Longitudinal stress is 5.47 × 10⁷ N/m², (b) Longitudinal strain is 2.61 × 10⁻⁴, and (c) Elongation produced is 0.912 mm.
7. Bulk Modulus, Volume Change & Compressibility
Calculate the change in volume of a lead block of volume 1.3 m³ subjected to a pressure of 12 atm. Also, calculate the compressibility of lead. Given the bulk modulus as B = 80 × 10⁹ N/m² and 1 atm = 1.013 × 10⁵ N/m².
Change in volume (ΔV) = ?, Compressibility (k) = ?
Solution:
According to the definition of Bulk Modulus: B = (ΔP × V) / ΔV
Rearranging for ΔV: ΔV = (ΔP × V) / B
ΔV = (1.2156 × 10⁶ × 1.3) / (80 × 10⁹)
ΔV = 1.58028 × 10⁶ / (80 × 10⁹) ≈ 1.98 × 10⁻⁵ m³
The compressibility (k) is reciprocal of Bulk Modulus: k = 1 / B
k = 1 / (80 × 10⁹) = 1.25 × 10⁻¹¹ m²/N
Result: The change in volume is 1.98 × 10⁻⁵ m³ and the compressibility is 1.25 × 10⁻¹¹ m²/N.
8. Shear Strength (Force to Punch a Hole)
The thickness of a metal plate is 0.35 inches. It's drilled to have a hole of radius 0.08 inches on the plate. If the shear strength is 4 × 10⁴ lbs/in², determine the force needed to make that hole.
Data:
Thickness of Plate (h) = 0.35 inches
Radius of Hole (r) = 0.08 inches
Shear Stress limit = 4 × 10⁴ lbs/in²
Force (F) = ?
Solution:
According to definition of Stress: Stress = Force / Area -----(i)
The sheared area is the cylindrical wall of the hole: Area = 2πrh
Area = 2 × 3.1416 × 0.08 × 0.35 ≈ 0.17593 in²
Putting values into eq(i) rearranged: Force = Stress × Area
Force = 4 × 10⁴ × 0.17593
Force ≈ 7037.17 lbs
Result: The force needed to make that hole is 7037.17 lbs.
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