DEPARTMENT OF PHYSICS AND ASTRONOMY Dt Provided: A formul sheet nd tble of physicl constnts is ttched to this pper. Liner grph pper is vilble. Spring Semester 2015-2016 PHYSICS 1 HOUR Answer questions ONE, plus ONE other question, TWO questions in ll. All questions re mrked out of twenty. The brekdown on the right-hnd side of the pper is ment s guide to the mrks tht cn be obtined from ech prt. Plese clerly indicte the question numbers on which you would like to be exmined on the front cover of your nswer book. Cross through ny work tht you do not wish to be exmined. 1 TURN OVER
SECTION A 1. COMPULSORY () Stte the reltionship between temperture mesured in Celsius (θ) nd the equivlent temperture in Kelvin (K). [1] (b) Define the fixed points of the Kelvin temperture scle. [2] (c) Define the mole. [2] (d) 48 The molr mss of the titnium isotope Ti Clculte: is 47.948 g mol -1. 48 i. The mss of one tom of Ti [2] ii. The number of toms in 1 cm 3 of 48 Ti. [2] (The density of titnium is 4.51 g cm -3 ) (e) A cylinder of helium locted t se level hs temperture of 25 C nd pressure of 450 kp. The cylinder is then tken to the top of mountin t n ltitude of 6 km where the temperture is t -15 C. The cylinder is llowed to come to therml equilibrium with its surroundings. Clculte the pressure in the cylinder t the top of the mountin. [4] (f) Distinguish between elstic nd plstic behviour of solid. [4] (g) Explin wht is ment by blck body s used in the study of therml rdition. [3] (20 mrks in totl) 2 CONTINUED
2. () Define in full the Young modulus E. [2] (b) A nickel wire is stretched just tut (zero tension) t 150 C. Find the stress in the wire, if it is cooled to 20 C without being llowed to contrct. For Nickel: Young Modulus = 207 GP Expnsivity = 1.3 10 K -1 [5] (c) Sketch stress-strin grphs for ductile mteril. [3] (d) i. Define the coefficient of therml conductivity of mteril. [2] ii. A window pne is 1.50 m wide, 0.75 m tll nd 8.0 mm thick. It is mde from glss with coefficient of therml conductivity of 0.85 W m -1 K -1. Clculte the rte of het flow through the window when the temperture difference cross it is 25 K. [3] (e) The electricl element in shower hs power rting of 10.5 kw nd increses the temperture of the wter flowing through the shower by 35 C. Clculte the mss of wter tht is flowing through the shower every second. You my ssume tht ll the energy from the element is trnsferred to the wter. (Specific het cpcity of wter = 4200 J kg -1 K -1 ) [5] (20 mrks in totl) 3 TURN OVER
3. () A smll cubicl spce probe mesures 20 cm long ech side. The equipment in the probe dissiptes energy t rte of 25.0 W. Use Stefn s lw to clculte the surfce temperture, in Kelvin, of the probe when it is in deep spce, very distnt from the Sun nd plnets. [4] (b) In n experiment to mesure specific het cpcity of metl, 250 g of metl rivets t 90 C is dded to 80 g of wter t 20 C in polystyrene cup of negligible het cpcity. The mximum temperture reched by the mixture is 37.6 C. i. Clculte the specific het cpcity of the metl. (The specific het cpcity of wter is 4200 J kg -1 K -1 ) [4] ii. Stte two ssumptions mde in your clcultion of the specific het cpcity. [2] iii. Suggest two experimentl precutions to reduce sources of error in the procedure outlined. [2] (c) Stte the idel gs eqution nd define ny symbols tht you use. [4] (d) A cylinder of compressed nitrogen hs volume of 5.0 10-3 m 3 nd contins gs t pressure of 10 MP nd temperture of 15 C. i. Clculte the number of moles of nitrogen in the cylinder. [3] ii. Explin why the temperture of the cylinder decreses when the high pressure gs is rpidly relesed into the tmosphere. [1] (20 mrks in totl) END OF EXAMINATION PAPER 4 CONTINUED
PHYSICAL CONSTANTS & MATHEMATICAL FORMULAE Physicl Constnts electron chrge e = 1.60 10 19 C electron mss m e = 9.11 10 31 kg = 0.511 MeV c 2 proton mss m p = 1.673 10 27 kg = 938.3 MeV c 2 neutron mss m n = 1.675 10 27 kg = 939.6 MeV c 2 Plnck s constnt h = 6.63 10 34 J s Dirc s constnt ( = h/2π) = 1.05 10 34 J s Boltzmnn s constnt k B = 1.38 10 23 J K 1 = 8.62 10 5 ev K 1 speed of light in free spce c = 299 792 458 m s 1 3.00 10 8 m s 1 permittivity of free spce ε 0 = 8.85 10 12 F m 1 permebility of free spce µ 0 = 4π 10 7 H m 1 Avogdro s constnt N A = 6.02 10 23 mol 1 gs constnt R = 8.314 J mol 1 K 1 idel gs volume (STP) V 0 = 22.4 l mol 1 grvittionl constnt G = 6.67 10 11 N m 2 kg 2 Rydberg constnt R = 1.10 10 7 m 1 Rydberg energy of hydrogen R H = 13.6 ev Bohr rdius 0 = 0.529 10 10 m Bohr mgneton µ B = 9.27 10 24 J T 1 fine structure constnt α 1/137 Wien displcement lw constnt b = 2.898 10 3 m K Stefn s constnt σ = 5.67 10 8 W m 2 K 4 rdition density constnt = 7.55 10 16 J m 3 K 4 mss of the Sun M = 1.99 10 30 kg rdius of the Sun R = 6.96 10 8 m luminosity of the Sun L = 3.85 10 26 W mss of the Erth M = 6.0 10 24 kg rdius of the Erth R = 6.4 10 6 m Conversion Fctors 1 u (tomic mss unit) = 1.66 10 27 kg = 931.5 MeV c 2 1 Å (ngstrom) = 10 10 m 1 stronomicl unit = 1.50 10 11 m 1 g (grvity) = 9.81 m s 2 1 ev = 1.60 10 19 J 1 prsec = 3.08 10 16 m 1 tmosphere = 1.01 10 5 P 1 yer = 3.16 10 7 s
Polr Coordintes x = r cos θ y = r sin θ da = r dr dθ 2 = 1 ( r ) + 1r 2 r r r 2 θ 2 Sphericl Coordintes Clculus x = r sin θ cos φ y = r sin θ sin φ z = r cos θ dv = r 2 sin θ dr dθ dφ 2 = 1 ( r 2 ) + 1 r 2 r r r 2 sin θ ( sin θ ) + θ θ 1 r 2 sin 2 θ 2 φ 2 f(x) f (x) f(x) f (x) x n nx n 1 tn x sec 2 x e x e x sin ( ) 1 x ln x = log e x 1 x cos 1 ( x sin x cos x tn ( 1 x cos x sin x sinh ( ) 1 x cosh x sinh x cosh ( ) 1 x sinh x cosh x tnh ( ) 1 x ) ) 1 2 x 2 1 2 x 2 2 +x 2 1 x 2 + 2 1 x 2 2 2 x 2 cosec x cosec x cot x uv u v + uv sec x sec x tn x u/v u v uv v 2 Definite Integrls 0 + + x n e x dx = n! (n 0 nd > 0) n+1 π e x2 dx = π x 2 e x2 dx = 1 2 Integrtion by Prts: 3 b u(x) dv(x) dx dx = u(x)v(x) b b du(x) v(x) dx dx
Series Expnsions (x ) Tylor series: f(x) = f() + f () + 1! n Binomil expnsion: (x + y) n = (1 + x) n = 1 + nx + k=0 ( ) n x n k y k k n(n 1) x 2 + ( x < 1) 2! (x )2 f () + 2! nd (x )3 f () + 3! ( ) n n! = k (n k)!k! e x = 1+x+ x2 2! + x3 x3 +, sin x = x 3! 3! + x5 x2 nd cos x = 1 5! 2! + x4 4! ln(1 + x) = log e (1 + x) = x x2 2 + x3 3 n Geometric series: r k = 1 rn+1 1 r k=0 ( x < 1) Stirling s formul: log e N! = N log e N N or ln N! = N ln N N Trigonometry sin( ± b) = sin cos b ± cos sin b cos( ± b) = cos cos b sin sin b tn ± tn b tn( ± b) = 1 tn tn b sin 2 = 2 sin cos cos 2 = cos 2 sin 2 = 2 cos 2 1 = 1 2 sin 2 sin + sin b = 2 sin 1( + b) cos 1 ( b) 2 2 sin sin b = 2 cos 1( + b) sin 1 ( b) 2 2 cos + cos b = 2 cos 1( + b) cos 1 ( b) 2 2 cos cos b = 2 sin 1( + b) sin 1 ( b) 2 2 e iθ = cos θ + i sin θ cos θ = 1 ( e iθ + e iθ) 2 nd sin θ = 1 ( e iθ e iθ) 2i cosh θ = 1 ( e θ + e θ) 2 nd sinh θ = 1 ( e θ e θ) 2 Sphericl geometry: sin sin A = sin b sin B = sin c sin C nd cos = cos b cos c+sin b sin c cos A
Vector Clculus A B = A x B x + A y B y + A z B z = A j B j A B = (A y B z A z B y ) î + (A zb x A x B z ) ĵ + (A xb y A y B x ) ˆk = ɛ ijk A j B k A (B C) = (A C)B (A B)C A (B C) = B (C A) = C (A B) grd φ = φ = j φ = φ x î + φ y ĵ + φ z ˆk div A = A = j A j = A x x + A y y + A z z ) curl A = A = ɛ ijk j A k = ( Az y A y z φ = 2 φ = 2 φ x + 2 φ 2 y + 2 φ 2 z 2 ( φ) = 0 nd ( A) = 0 ( A) = ( A) 2 A ( Ax î + z A ) ( z Ay ĵ + x x A ) x y ˆk