2024 NYJC H2 PHY 0749 P2 Answers

NYJC 2024 9749/02/J2Prelim/24 [Turn over NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME CLASS TUTOR’S NAME CENTRE NUMBER S INDEX NUMBER PHYSICS 9749/02 Paper 2 Structured Questions 10 September 2024 2 hours Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name, class, Centre number and index number in the spaces at the top of this page. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams, graphs. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Answer all questions. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 / 5 2 / 7 3 / 8 4 / 8 5 / 11 6 / 11 7 / 10 8 / 20 Total / 80 This document consists of 19 printed pages. H

2 NYJC 2024 9749/02/J2Prelim/24 Data speed of light in free space c = 3.00  108 m s−1 permeability of free space 0 = 4  10−7 H m−1 permittivity of free space 0 = 8.85  10−12 F m−1 = (1/ (36 ))  10−9 F m−1 elementary charge e = 1.60  10−19 C the Planck constant h = 6.63  10−34 J s unified atomic mass constant u = 1.66  10−27 kg rest mass of electron me = 9.11  10−31 kg rest mass of proton mp = 1.67  10−27 kg molar gas constant R = 8.31 J K−1 mol−1 the Avogadro constant NA = 6.02  1023 mol−1 the Boltzmann constant k = 1.38  10−23 J K−1 gravitational constant G = 6.67  10−11 N m2 kg−2 acceleration of free fall g = 9.81 m s−2 Formulae uniformly accelerated motion s = 21 2ut at+ v2 = u2 + 2as work done on / by gas W = pV hydrostatic pressure p = gh gravitational potential  = Gm r− temperature T / K = T/°C + 273.15 pressure of an ideal gas p = 21 3 Nm cV  mean translational kinetic energy of an ideal gas molecule E = 3 2 kT displacement of particle in s.h.m. x = 0 sinxt  velocity of particle in s.h.m. v = 0 cosvt  = 22 0xx− electric current I = Anvq resistors in series R = R1 + R2 + … resistors in parallel 1/R = 1/R1 + 1/R2 + … electric potential V = 04 Q r alternating current/voltage x = 0 sinxt  magnetic flux density due to a long straight wire B = 0 2 d   I magnetic flux density due to a flat circular coil B = 0 2r  NI magnetic flux density due to a long solenoid B = 0 nI radioactive decay x = 0 exp( )−xt decay constant  = 1 2 ln2 t

3 NYJC 2024 9749/02/J2Prelim/24 [Turn over 1 (a) A beaker in air contains a liquid. The base area of the beaker is A, as shown in Fig. 1.1. The liquid has density  and fills the beaker to a height h. Fig. 1.1 (i) Show that the pressure P due to the liquid at the base of the beaker is given by P = gh where g is the acceleration of free fall. [1] (ii) Explain why the equation in (i) does not give the total pressure at the base of the beaker. [1] (iii) Fig. 1.2 shows the variation of the total pressure inside the liquid with depth x below the surface. Fig. 1.2 both relations used appropriate algebra leading to ans and [B1 ] FmP A Ah mgP A Ah g A gh    == = = = Total pressure at the base includes atmospheric / air pressure above the liquid [B1] h beaker base area A liquid 0 1 2 3 4 5 6 7 8 pressure / 104 Pa x / cm 9.65 9.64 9.63 9.62 9.61 9.60 9.66

4 NYJC 2024 9749/02/J2Prelim/24 Use Fig. 1.2 to determine the density of the liquid. density = kg m–3 [1] (b) A spherical buoy of density 220 kg m–3 floats in equilibrium on the surface of sea water of density 1050 kg m–3, as shown in Fig. 1.3. Fig. 1.3 (not to scale) Determine the percentage of the volume of the buoy that is submerged in water. percentage = % [2] [Total: 5] ( ) ( )( ) 2 3 4 3 9.66 9.60 10 9.81 8.0 10 765 k [A1] (accept 760 to 770 kg g m m ) p g h   − − −  =  −  =  = ( )( ) ( )( )WS VV V g V g VV V V sub sub sub sub At equilibrium during floating, upthrust weight Let submerged volume be and volume of s phere be . 1050 220 0.21 percentage submerged 21% [C1] [A1]  = = = = = 760 to 770 21 sea water spherical buoy submerged portion ( )( ) p gh ptotal at 4 3 m 4 3 9.66 10 9.81 0.080 9.60 10 7 [A1] (accept65 kg 760 m to 770 kg m )    −− =+  = +  =

5 NYJC 2024 9749/02/J2Prelim/24 [Turn over 2 (a) A body travelling at a constant speed in a circular path experience centripetal acceleration. Using Newton’s laws of motion explain why there is acceleration although the speed is constant. [2] (b) A car of mass 1500 kg travels in a horizontal circ ular path of radius 50.0 m on a banked road with speed of 15.0 m s–1 without any frictional force acting on the tyres along the slope. Fig. 2.1 (i) Calculate the angle  at which the road is banked.  = o [3] (ii) Explain how friction force enables the car to travel in the same horizontal circular path at a lower speed. [2] [Total: 7] The velocity of the body changes along a circular path. By N1L this require an external resultant force to act on the body [B1]. Since the centripetal acceleration is pointing to the centre of circle and perpendicular to the instantaneous velocity, by N2L, it has no component along the path, hence speed is constant.[B1] The horizontal component of static friction acts away from the direction of centripetal force [B1], resulting in a smaller magnitude of centripetal force [B1]. A smaller centripetal force permits the car to move on the banked surface in uniform circular motion with a slower speed. 2 cos sin [M1 both eqn] N mg mvN r = =   22 15tan [M1] 50.0 9.81 v rg==   = 24.6o [A1] N W

6 NYJC 2024 9749/02/J2Prelim/24 3 (a) (i) The kinetic theory for an ideal gas of volume V at pressure p leads to the equation 21 3pV Nm c= , where the other symbols refer to their usual meanings. Use the equation of state for an ideal gas to show that the average translational kinetic energy EK of a molecule of ideal gas is given by 3 2 KE kT= . [1] (ii) One helium atom has a mass of 6.68 × 10–27 kg. Helium may be considered as an ideal gas. Show that the total kinetic energy of the helium atoms in 1.00 mol of helium gas at 25 oC is 3720 J. [1] (iii) State the value of the internal energy of 1.00 mol of helium gas at 25 oC. Explain your answer. [2] Internal energy is the sum of the random distribution of the microscopic kinetic energy (KE) and microscopic potential energy (PE) of the gas molecules [B1] For an ideal gas, there are no intermolecular forces between molecules so PE = 0, so internal energy = KE = 3720 J [B1] ( )( ) ( ) K, total 23 23 3 2 3 6.02 10 1.38 10 25 273.15 2 3715 3720 J [C1] [ (shown) 0] A E NkT − = =   + = = 2 22 1[B1] for 1 3 13 leading to seen222 pV pV Nk Nk T Nm c NkT m c kT T m c = = = =

7 NYJC 2024 9749/02/J2Prelim/24 [Turn over (iv) The helium gas is gradually cooled from 25 oC to –150 oC at which the internal energy is 1540 J. On Fig. 3.1, plot points and dra