Physics 1st Ch 07 Structural Properties of Matter

Physics - First Paper Chapter 7: Structural Properties of Matter Selected Questions and Answers (HSC English Version) Chapter 7: Structural Properties of Matter B. Creative Questions 1. Water A B In the figure radius of the tube A is 0.4 mm and radius of the tube B is 0.2 mm. Contact angle of water is 2◦. Surface tension of water is T = 7.2 × 10−2 N/m and density of water is ρ = 1000 kg/m3. (a) What is capillarity? (b) Explain why the contact angle of pure water and clean glass is zero. (c) Determine the height of liquid risen in tube B. (d) Between the tubes A and B in which tube the quantity of water is more? Analyse mathe- matically. Answer: (a) Capillarity is the phenomenon of spontaneous rise or depression of a liquid inside a narrow- bore tube (capillary tube) when dipped vertically into the liquid. (b) The contact angle depends on the relative magnitudes of adhesive force (between water and glass molecules) and cohesive force (between water molecules). For clean glass and pure water, the adhesive force is much stronger than the cohesive force. The water molecules spread completely over the glass surface, forming a hemispherical concave meniscus tangent to the glass wall. Thu

(d) The mass of liquid in a capillary tube is m = ρV = ρ(πr2h). Since h = 2T cos θ rρg , the mass is: m = ρ · πr2 · 2T cos θ rρg = 2πT cos θ g · r Notice that m ∝r. For tube A (rA = 0.4 mm) and tube B (rB = 0.2 mm): mA mB = rA rB = 0.4 0.2 = 2 Therefore, the mass (quantity) of water risen in tube A is twice as much as in tube B. 2. A liquid of co-efficient of viscosity 0.004 kg/(m s) and density 800 kg/m3 is flowing through a tube of radius 0.02 m with a velocity of 0.15 m/s. An iron sphere of radius 2 × 10−3 m and density 7800 kg/m3 is allowed to fall through this liquid. Take g = 9.8 m/s2. (a) What is terminal velocity? (b) Why raindrops fall with a constant speed instead of accelerating continuously? (c) Calculate the terminal velocity attained by the iron sphere in the liquid. (d) Determine whether the flow of the liquid in the tube is streamline or turbulent. Answer: (a) Terminal velocity is the maximum, constant velocity acquired by a body moving freely through a viscous fluid when the downward gravitational force is exactly balanced by the sum of the upward buoyant force and viscous drag. (b) As a raindrop falls, its speed increases, which causes the upward viscous drag forc

(a) State Hooke’s law. (b) Explain Poisson’s ratio. Can its value be greater than 0.5? (c) Calculate the work done in stretching the wire in the first case. (d) If the load is doubled, calculate the change in diameter of the wire. Answer: (a) Hooke’s law states that within the elastic limit, the stress developed in a deformed body is directly proportional to the corresponding strain: Stress ∝Strain. (b) Poisson’s ratio (σ) is defined as the ratio of lateral strain to longitudinal strain within the elastic limit: σ = −∆d/d ∆L/L For ordinary isotropic materials, when a body is stretched longitudinally, its volume cannot decrease; therefore, the theoretical upper limit for Poisson’s ratio is 0.5. Practical engineering materials typically have σ between 0.2 and 0.4. (c) Given: Original length L = 2 m. Diameter d = 1 mm =⇒r = 0.5 × 10−3 m. Cross-sectional area A = πr2 = π × (0.5 × 10−3)2 ≈7.854 × 10−7 m2. Elongation l = 1.2 mm = 1.2 × 10−3 m. Applied load F = mg = 10 × 9.8 = 98 N. Work done in stretching the wire: W = 1 2Fl = 1 2 × 98 × 1.2 × 10−3 = 0.0588 J W = 0.0588 J (d) When load is doubled, F ′ = 2F = 2 × 98 = 196 N. Since l ∝F, the new elongation is l′ = 2l = 2 × 1.2 × 10−3 = 2.4