Title 6-properties-of-matter-c05a8a2c-9ccc-4772-be21-586113a257d2.pdf ✅

PROPERTIES OF MATTER 1. (A) Tension at point x = mw 2 r T = { m l (l − x)} ⋅ w 2 { l + x 2 } T = mw 2 2l (l 2 − x 2 ) 2. (A) Let x be latent heat of fusion w be water equivalent of apparatus 200(70 − 40) + w(70 − 40) = 50(x + 40) ... ... ... 250(40 − 10) + w(40 − 10) = 80(x + 10) ... ... ... . Solving (i) & (ii) x = 90cal/g ≃ 3.8 × 105 J/kg 3. (B) dq = msdT ⇒ q = ∫1 2 1 × aT 3dT = 15a 4 4. (B) dl l = 1% ∴ dA A = 2 × 1% = 2% 5. (C) dl = dx(αΔT) ⇒ dl = [(3x + 2) × 10−6 × 20]dx Integrating Δl = 1.208 cm ∴ length = 2.0120 m 6. (C) R1 = l K1πr 2 &R2 = l 3K2πr 2 Both rods are in parallel 1 Req = 1 R1 2K − Kx l H = −K(x)A ( dT dx) 4πr 2Keq l = πr 2K1 l + 3πr 2K2 l ∴ Keq = K1 + 3K2 4 7. (B) K(x) = 2K − Kx l H = −K(x)A ( dT dx) H = − [2K − Kx l ]A ( dT dx) Integrate and get temperature as a function of x Alternative approach Initially K is higher ∴ dT dx should be smaller in magnitude Later K decrease dT dx increases 8. (B) πR 2 l = nπr 2 l. (i) and n(2πrl) = 2(2πRl) (ii) Solve for n ∴ n = 4 9. (B) −10 4 = −K(55 − To ) (i) { Newtons low of cooling average method } −10 8 = −K(35 − To ) (ii) Solving (i)&(ii) To = 15∘C 10. (B) Let T : tension at mid point, for the outer half of the rod T = m 2 ω 2 3l 4 and T A = Y. strain ⇒ strain = 3mω 2l 8AY 11. (B) 2.4×108×3×10−4 3 − 1200 × 10 = 1200a ⇒ a = 10 m/s 2
12. (D) ms dθ dt = σA ∈ (T 4 − T0 4 ) ⇒ m1 = 3m2 ⇒ r1 = 3 1/3 r2 13. (A) Error = 1 2 αθ. 86400 = 4.32sec 14. (A) According to Newton'slaw of cooling dT dt ∝ ΔT ⇒ tan θ1 tan θ2 = 35 − 20 25 − 20 = 15 5 = 3 1 ⇒ tan2 θ1 = 9tan2 θ2 ∵ sec2 θ1 = 1 + tan2 θ1 ⇒ sec2 θ1 = 1 + 9tan2 θ2 15. (A) Y = Stress Strain for same strain (stress) AA > (stress) )B YA > YB Breaking point of B > Breaking point of A So, A is less ductile. 16. (ABC) In the case of a perfectly rigid body and incompressible liquid, volume strain is zero. 17. (BCD) W = ms or, m = W s = 4.5 0.09 = 50 g The thermal capacity and the water equivalent of a body have the same numerical value. Also, Q = 4.5 × 8 = 36cal Since, the temperature remains constant, during the process of melting, no heat is exchanged with the calorimeter and hence, Q = 15 × 80 = 1200cal. Hence, the correct choices are (B), (C) and (D). 18. (AC) (Stress) )s = F 2A , (Stress) Cu = F A Given that YS YCu = 2 1 ( strain )s ( strain )Cu = ( stress )S/Ys ( stress )Cu/Ycu = (F/2A) (F/A) × 1 2 = 1 4 So, options (A), (B) and (C) are correct. 19. (BCD) Thermal force = YAαdθ = Yπr 2αdθ r1 = r, r2 = r√2, r3 = r√3, r4 = 2r ⇒ F1:F2: F3: F3 = 1: 2: 3: 4 Thermal stress = Yαdθ As Y and α are same for all the rods, hence stress developed in each rod will be the same. As strain = αdθ, so strain will also be the same. E = Energy stored = 1 2 Y( strain ) 2 × A × L ∴ E1: E2: E3: E4 = 1: 2: 3: 4 So, option (B) and (C) are correct. 20. (BC) We can see the slope of second step (liquid heating) is less than that of first step (solid heating) hence specific heat of liquid is greater than that of solid. We can also see that the horizontal portion of first part is smaller than that of second part which implies that solid melting is taking less time than liquid vaporization hence (C) is also correct. 21. (AD) H ∝ A l (T1 − T2 ) (A) Temp. diff. quadrupled, c/s area halved, H doubles (correct) (B) Temp. diff. doubled, length quadrupled, H is halved (incorrect) (C) c/s area halved, length doubled, H becomes 1/4th (incorrect) (D) Temp. diff. doubled, Area doubled, length doubled, H is doubled (correct) 22. (AC) mA = 4mB ⇒ SA = (4) 2/3SB (mass) (surface area) HA HB = SA SB also dT dt ∝ S m 23. (AD) γHg > γFe ⇒ density of Hg decreases more on heating. Hence cube will be submerged more. If fraction submerged is f ρHgVfg = ρFeVg ⇒ f = ρFe ρHg Change of fraction = ρFe ′ ρHg ′ − ρFe ρHg ρFe ρHg = 1+γHg×25 1+γFe×25 − 1 = 1 + 180 × 10−6 × 25 1 + 35 × 10−6 × 25 − 1 = (180 − 35) × 10−6 × 25 1 = 3.6 × 10−3 = 0.36% ≃ 0.4% 24. (BD) 25. (AD) dQ dt = eAσ(T 4 − T0 4 ) = same ; dθ dt = eAσ ms (T 4 − T0 4 ) = different 26. (ACD) Considering heat flow at junction A Heat in = heat out x + 1 + 3 = 5 x = 1 x is positive so heat flows in from E to A
∴ TE > TA ∴ TC > TA > TD TB − TA = TE − TA ∴ TB = TE 27. (ABCD) When a body is heated all dimension of the material as well as the enclosed lengths, areas and volumes also increase. 28. (AC) As the two ends of the rod are at constant temperature that means the rod is in steady state of thermal conduction hence throughout the length the temperature gradient will remain constant and in steady state of thermal conduction the rate of heat flow (heat current) is given by the product of thermal conductivity, cross sectional area and the temperature gradient. 29. (BC) When temperature increases of a bimetallic strip, the one which is having more value of coefficient of expansion will expand more so that strip will bend toward the metal with low value of coefficient of expansion and it will bend toward high value of coefficient of expansion if it cooled. 30. (AD) By wein's displacement law we use λmT = constant ⇒ λmA λmB = TB TA ⇒ TA TB = λmB λmA = 800 400 = 2 By Stefan's law, we use EA EB = AATA 4 ABTB 4 = 4πrA 2 4πrB 2 ( TA TB ) 4 = 4 31-32. 31. (B) 32. (A) 33. (C) Power of Heater H = H2 − H1 = 80−0 10 − 100−80 10 = 6cal/s H2 = Lf dm dt ( rate of melting = dm dt ) ⇒ 80 − 0 10 = 80 dm dt ⇒ dm dt = 0.1gr/sec Heater Since temperature of middle container is constant, H2 is constant, hence the rate of melting of ice will be constant 34. (D) 40−36 10 = k(36 − 30); 39−x 10 = k ( 36+x 2 − 30) 4 36 − x = 8 ( x 2 − 30 + 18) ; x 2 − 12 = 72 − 2x; 5x 2 = 72 + 12; 5x 2 = 84; x = 2 × 84 5 = 168 5 = 33.6 35. (D) dθ dt = − kA ms (T − T0 ) Magnitude of slope will decrease with time. 36. (C) ∵ 40−36 10 = k (38 − 30 ⇒ k = 4 10×8 = 1 20 When the block is at 38∘C and room temperature is at 30∘C the rate of heat loss ms × dθ dt = msk(38 − 30) Total heat loss in 10 minute ⇒ dQ = msk(38 − 30) × 10 = 2 × 1 20 × 8 × 10 = 8 J Now heat gained by the object in the said 10 minutes. Q = msΔθ = 2 × 4 = 8 J Total heat required = 8 + 8 = 16 J 37. (B) Material which is most ductile is easy to expand is used for making wire. 38. (C) Material which breaks just after proportional point. 39. (D) Material which retains permanent deformation. 40. (B) V = V0 (1 + γLΔT) = (10)(100)[1 + 5 × 10−5 × 20] = 1000(1 + 0.001) = 1001 cm3 = 1001cc 41. (B) Cross-sectional area of vessel at 40∘C A = A0(1 + 2αgΔT) = 100(1 + 2 × 10−5 × 20) = 100.04 cm2 Actual height of liquid = Actual volume of liquid Cross-sectional area of vessel = 1001 100.04 = (1001)(100 + 0.04) −1 = ( 1001 100 ) (1 + 0.04 100) −1
= (1001) 100 (1 − 0.04 100) = 1 100 (1001 − 0.4) = 1000.6 100 = 10.006 cm 42. (B) ∵ TV = SR(1 + αgΔT) = (TV)(1 − αgΔT) ∴ SR = (TV)(1 + αgΔT) −1 = (10.006)(1 − 10−5 × 20) = 10.006 − 0.002 = 10.004 cm 43. [A − r; B − p; C − q;D − s] σ = Fcos φ A/cos φ = F A cos2 φ τ = Fsin φ A/cos φ = F A cos φsin φ = F 2A sin 2φ T will be max if 2φ = 90∘ ⇒ φ = 45∘ σ will be max if φ = 0 ∘ 44. [A − prs;B − prs; C − p; D − pqrs] 45. [A − q;B − r; C-s; D-p] (a) 10×A 2 (100 − T) = 20×A 1 (T − 0) ; 100 − T − 4T = 0; T = 20 (b) 10×A(100) 1 + 20×A(100) 1 3000 × 100 × 10−4 = 30 W (c) Rth = (0.2−0.1) 4π×0.2×0.1 = 40 SI units (d) dt dx = 100−0 10 = 10∘C/m 46. (1) (T) Tension at any point x = F2 + ( F1−F2 l ) x ⇒ dl = T AY dx Integrating Δl = (F1+F2 )l 2AY ⇒ Δl = 1 × 10−9 m ∴ x = 1 F1 ⟷→ F2 47. (2) Tension in ring = ρAω 2 r 2 [ A = C/s are of ring ] For breaking ρAω2r 2 A = σ ; ω 2 = σ r 2ρ ∴ ω = 2rad/s 48. (3) Let x kg of ice remaining (2)(15) × 500 + (2 − x) × 80,000 = (2.5)(25)(1000) x ≃ 1.4 kg ∴ Water = 3 kg 49. (8) ∵ Conduction rate ∝ A l H ∝ radius 2 length ∴ H(max) = K 4r 2 l ; Hmin = K r 2 2l ∴ Hmax Hmin = 8 50. (8) Let S power be lost to surrounding 16 − S = 10 m ... (i) and 32 − S = 10(3m) Solving we get S = 8 watts. 51. (3) dp = mg A K = −dp dv/v ⇒ K 3dr r = −dp ∴ dr r = − mg 3AK n = 3 52. (2) 5 = 1 2 α(T − 15)86400 10 = 1 2 α(33 − T)86400 Where T is temperature at which clock shows correct time. Solving T = 21∘C and α = 2 × 10−5 / ∘C 53. (4) ΔT Δt = −k(Tavg − T0) −10 2 = −k(70 − 30) ...(i) and −10 t = −k(50 − 30) Solving (i) and (ii) t = 4 min. 54. (3) For no heat flow in AB temperature of Junction B must be 20∘C ∵ Material and cross-section are same 80 − 20 lBD = 20 − 0 lBC Temperature gradient is also same ∴ lBD lBC = 3 55. (4) Qab = 200 × 80 + 200 × 1 × 100 + (200 × 1 × 45) Qloss = ms(540) 16000 + 20000 + 9000 = (540)ms ; 45000 540 = ms ; ms = 250 3 = 83.33 Steam left = 87.33 − 83.33 = 4 ∴