# Physics 11th Class Book Pdf

Topics: Gas, Ideal gas law, Pressure Pages: 12 (3074 words) Published: December 15, 2012

391

Chapter 9

9.1 9.2

1.8 (a) From the given graph for a stress of 150 × 106 N m-2 the strain is 0.002 (b) Approximate yield strength of the material is 3 × 108 N m-2

9.3

(a) Material A (b) Strength of a material is determined by the amount of stress required to cause fracture: material A is stronger than material B.

9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11 9.12 9.13 9.14 9.15 9.16

(a) False

(b)

True

1.5 × 10-4 m (steel); 1.3 × 10-4 m (brass) Deflection = 4 × 10–6 m 2.8 × 10-6 0.127 7.07×10 N Dcopper/Diron = 1.25 1.539 × 10-4 m 2.026 × 109 Pa 1.034 × 103 kg/m3 0.0027 0.058 cm3 2.2 × 106 N/m2 4

392

PHYSICS

9.17 9.18 9.19 9.20 9.21

Pressure at the tip of anvil is 2.5 × 1011 Pa (a) 0.7 m (b) 0.43 m from steel wire

Approximately 0.01 m 260 kN 2.51 × 10–4 m3

Chapter 10 (a) decreases (b) η of gases increases, η of liquid decreases with temperature (c) shear strain, rate of shear strain (d) conservation of mass, Bernoulli’s equation (e) greater. 6.2 × 106 Pa 10.5 m Pressure at that depth in the sea is about 3 × 107 Pa. The structure is suitable since it can withstand far greater pressure or stress. 6.92 × 105 Pa 0.800

10.3 10.5 10.6 10.7 10.8 10.9

10.10 Mercury will rise in the arm containing spirit; the difference in levels of mercury will be 0.221 cm. 10.11 No, Bernoulli’s principle applies to streamline flow only. 10.12 No, unless the atmospheric pressures at the two points where Bernoulli’s equation is applied are significantly different. 10.13 9.8 × 102 Pa (The Reynolds number is about 0.3 so the flow is laminar). 10.14 1.5 × 103 N 10.15 Fig (a) is incorrect [Reason: at a constriction (i.e. where the area of cross-section of the tube is smaller), flow speed is larger due to mass conservation. Consequently pressure there is smaller according to Bernoulli’s equation. We assume the fluid to be incompressible]. 10.16 0.64 m s-1 10.17 2.5 × 10-2 N m-1 10.18 4.5 × 10-2 N for (b) and (c), the same as in (a). 10.19 Excess pressure = 310 Pa, total pressure = 1.0131 × 105 Pa. However, since data are correct to three significant figures, we should write total pressure inside the drop as 1.01 × 105 Pa. 10.20 Excess pressure inside the soap bubble = 20.0 Pa; excess pressure inside the air bubble in soap solution = 10.0 Pa. Outside pressure for air bubble = 1.01 × 105 + 0.4 × 103 × 9.8

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× 1.2 = 1.06 × 105 Pa. The excess pressure is so small that up to three significant figures, total pressure inside the air bubble is 1.06 × 105 Pa. 10.21 55 N (Note, the base area does not affect the answer) 10.22 (a) absolute pressure = 96 cm of Hg; gauge pressure = 20 cm of Hg for (a), absolute pressure = 58 cm of Hg, gauge pressure = -18 cm of Hg for (b); (b) mercury would rise in the left limb such that the difference in its levels in the two limbs becomes19 cm. 10.23 Pressure (and therefore force) on the two equal base areas are identical. But force is exerted by water on the sides of the vessels also, which has a nonzero vertical component when the sides of the vessel are not perfectly normal to the base. This net vertical component of force by water on sides of the vessel is greater for the first vessel than the second. Hence the vessels weigh different even when the force on the base is the same in the two cases. 10.24 0.2 m 10.25 (a) The pressure drop is greater (b) More important with increasing flow velocity. 10.26 (a) 0.98 m s-1; (b) 1.24 × 10-5 m3 s-1 10.27 4393 kg 10.28 5.8 cm s-1, 3.9 × 10-10 N 10.29 5.34 mm 10.30 For the first bore, pressure difference (between the concave and convex side) = 2 × 7.3 × 10-2 / 3 × 10-3 = 48.7 Pa. Similarly for the second bore, pressure difference = 97.3 Pa. Consequently, the level difference in the two bores is [48.7 / ( 103 x 9.8 )] m = 5.0 mm. The level in the narrower bore is higher. (Note, for zero angle of contact, the radius of the meniscus equals radius of the bore. The concave side of the surface in each bore is...