alleryStart learning →Stress is defined as:
- Volume × pressure
- Force × Area
- Force per unit area: F/A
- Force/length
Show answer and solution
Correct answer: C. Force per unit area: F/A
Stress σ = F/A. SI unit: pascal (Pa = N/m²).
Tensile, compressive, shear types.
Assam CEE Mechanical Properties of Solids and Fluids — practice questions
Explore 35 free Mechanical Properties of Solids and Fluids questions in Assam CEE Physics. Open each question or exercise to read its content, reference answers and available explanation.
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Stress is defined as:Read the answer and worked solution →Strain (longitudinal) is:Read the answer and worked solution →Young's modulus Y is ratio of:Read the answer and worked solution →Hydraulic pressure at depth h below liquid surface (density ρ):Read the answer and worked solution →Pascal's law states pressure applied to enclosed fluid is:Read the answer and worked solution →Bulk modulus B for fluid relates pressure change to:Read the answer and worked solution →Poisson's ratio σ for a stretched rod relates:Read the answer and worked solution →Bernoulli's equation: along streamline of ideal fluid:Read the answer and worked solution →Equation of continuity for incompressible fluid:Read the answer and worked solution →Stokes' law: viscous drag on sphere of radius r moving at v through fluid of viscosity η:Read the answer and worked solution →Terminal velocity of sphere (radius r, density ρ_s) falling in viscous fluid (density ρ_f, viscosity η):Read the answer and worked solution →Surface tension of liquid γ has SI unit:Read the answer and worked solution →Excess pressure inside spherical drop of radius r (surface tension γ):Read the answer and worked solution →Capillary rise h for liquid in tube of radius r:Read the answer and worked solution →Reynolds number R_e for tube flow:Read the answer and worked solution →Two wires of same material, same length, but cross-sections 1 mm² and 2 mm². Same load: ratio of extensions:Read the answer and worked solution →Energy stored per unit volume in stretched wire:Read the answer and worked solution →Two wires of same material (Y, ρ) and same length, one twice the diameter. Ratio of extensions under same load:Read the answer and worked solution →For elastic wire, breaking stress depends on:Read the answer and worked solution →For a metallic rod heated through ΔT with Y and α (thermal expansion) and rigidly held:Read the answer and worked solution →Density of mercury 13.6 × 10³ kg/m³. Atmospheric pressure equivalent in mercury column:Read the answer and worked solution →Hydraulic press: small piston area 5 cm², large piston area 50 cm². Force on small piston 20 N. Force on large:Read the answer and worked solution →Velocity of efflux from small hole at depth h below free surface (Torricelli's law):Read the answer and worked solution →Water flows through pipe area A₁ = 4 cm² at 2 m/s, narrows to A₂ = 1 cm². Velocity at narrow section:Read the answer and worked solution →Drop of radius R splits into 8 drops of equal size. Each new radius:Read the answer and worked solution →Excess pressure inside soap bubble of radius 2 cm (γ = 0.025 N/m):Read the answer and worked solution →Water (γ = 0.072 N/m, ρ = 1000 kg/m³) rises in capillary 1 mm diameter (assume θ = 0):Read the answer and worked solution →For elastic potential energy in stretched wire: F × ΔL / 2. Also equals:Read the answer and worked solution →For ideal fluid, viscosity:Read the answer and worked solution →Pressure at depth h in liquid column under atmospheric pressure P₀:Read the answer and worked solution →
Worked practice questions
Strain (longitudinal) is:
- Δl / l (dimensionless)
- F/l
- Δl × l
- l × A
Show answer and solution
Correct answer: A. Δl / l (dimensionless)
Longitudinal strain = ΔL/L₀: fractional change in length. Dimensionless.
**Volume strain:** ΔV/V. Shear strain: tan θ ≈ θ.
Young's modulus Y is ratio of:
- Force to area
- Longitudinal stress to longitudinal strain
- Strain to stress
- Stress to volume
Show answer and solution
Correct answer: B. Longitudinal stress to longitudinal strain
We have Y = (F/A)/(ΔL/L).
So Y = FL/(A × ΔL).
For steel ≈ 2 × 10¹¹ Pa.
Larger Y means stiffer material.
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