Title 8-properties-of-matter-.pdf ✅

Properties of Matter 1. (A) 2. (C) 2k(2T − TA )S l = k(TA − T)S l ⇒ TA = 5/3T k(2T − TC )S l = 2k(TC −T)S l ⇒ TC = 4/3T 3. (D) H = 2kTS l 4. (B) ε = CT ∵ Temperature is halved ∴ emissivity is halved ∴ H = dQ dt = ( 1 2 ) 5 × 160 = 5 W 5. (B) Specific heat increases with temp. For every 1 ∘C drop in temperature of the hotter block more than 1 ∘C will be gained by the colder block Final temperature will be greater than 50∘C 6. (C) The vertical component of tension of both the string should balance weight of the frame ∴ Tension will be least in (ii) 7. (B) ∵ Area of wire B is greater than wire A. For stress (F/A) to be equal for both means tension in wire having greater cross section should be more. 8. (A) T1 = mg ⇒ Δl1 = l mg/A Y T2 = mg cos θ ⇒ Δl2 = l mg/Acos θ Y ∴ Δl2 = Δl1 cos θ 9. (B) λT = constant Q ∝ AT 4 QA:QB:QC = 4πR 2 (3000) 4 : 4π(2R) 2 (4000) 4 : 4π(3R) 2 (5000) 4 10. (A) By Hooke's law, Young's modulus (Y) = Tensile stress Tensile strain So, Y = F/A ΔL/L = FL AΔL If the rod is compressed, then compressive stress and strain appear. Their ratio Y is same as that for tensile case. Given, length of a steel wire (CL) = 10 cm Temperature (t) = 100∘C As length is constant. ∴ Strain = ΔL L αΔθ [ as ΔL = (LΔθ)] Now, pressure = stress = Y × strain [Given Y = 2 × 1011 N/m2 and α = 1.1 × 10−5 K −1 ] = 2 × 1011 × 1.1 × 10−5 × 100 = 2.2 × 108 Pa 11. (A) Lvap = 1000×900 200 J/gm = 4.5 × 103 J/gm = 4.5 × 106 J/kg 12. (AD) ∵ Change in volume is zero for ideal liquid ∴ Bulk modules becomes infinite For very small shear stress ≃ 0 Shear strain is very high (shape changes) ∴ Shear modulus = 0 13. (AD) ∵ Stress = F/A {Same for both } & Strain = stress Y {Different for both } 14. (C) Then ΔQ Δt = KA(T1−T2 ) x According to thermal conductivity, we get dQ1 dt = dQ2 dt + dQ2 dt 0.92(100 − T) 46 = 0.26(T − 0) 13 + 0.12(T − 0) 12 ⇒ T = 40∘C ∴ dQ1 dt = 0.92 × 4(100 − 40) 46 = 4.8cal/s
15. (D) Strain is same in both (A) and (B). Energy density = 1 2 ( Stress ) × ( Strain ) = 1 2 Y( Strain ) 2 Since Y is different, so energy density is different 16. (D) As soon as water temp drops below 4 ∘C colder water goes to the top surface and will freeze first. 17. (C) Δl in upper half = ∫l/2 l mgx l dx AY = 3mgl 8AY 18. (A) T = mω 2 l 2 19. (D) T = m(l − x) l ω 2 ( l + x 2 ) = mω 2 2l (l 2 − x 2 ) ⇒ Δl = ∫0 l mω 2 2l (l 2 − x 2 ) AY dx = mω 2 2AlY (l 3 − l 3 3 ) = mω 2l 2 3AY = mgl 2AY ⇒ ω = √ 3g 2l 20. (C) The various forces acting on the cylinder are shown in the figure. Let x0 be the extension in spring At equilibrium, Upward forces = Downward forces kx0 + FB = mg where, k is spring constant, FB is force due to buoyancy which is equal to weight of liquid displaced. FB = σ ⋅ L 2 ⋅ A ⋅ g ∴ kx0 + σ 1 2 Ag = Mg ⇒ x0 = Mg − σLAg 2 k = Mg k (1 − σLA 2M ) 21. (C) According to Newton's law of cooling, the rate of loss of heat is directly proportional to the difference of temperature of the body and surroundings. dQ dt = k(T2 − T1 ) and dQ = msdT2 ∴ dQ dt = msdT2 dt From equation (i) and (ii), we get : − msdT2 dt = k(T2 − T1 ) ⇒ dT2 T2 − T1 = − k ms dt = −Kdt [∴ K = k ms] On integrating, we get : log0 (T2 − T1 ) = −kt + C or [⬚ where,C ′ = e C] 22. (B) T = mω 2l ⇒ ω = √ T ml or T2 = T1 + C ′e −kt T and T = breaking stress × area = 4.8 × 107 × 10−6 = 48 N ⇒ ω = √ 48 10(0.3) = 4rad/s 23. (C) Heat required to change the temperature of vessel by a small amount dT, −dQ = mCpdT Total heat required −Q = m∫20 4 32 ( T 400) 3 dT =