RI 2021 Y6 H2 Phy T3 CT Soln

© Raffles Institution [Turn over 2021 July Common Test H2 Physics Solutions 1 C The other three statements are equivalent to the statement “The molecules of the two systems have the same root -mean-square speed”. In general, root -mean-square speed is different from mean speed. 2 C In equilibrium, the two ideal gases have the same temperature: 11 1 22 2 ,pV n RT p V n RT = = Taking ratio, 11 1 1 12 2 22 22 1 2.0 0.50 1.0 1.13.0 0.30 0.90 pV n p nV pV n p nV ×=⇒= = = = × 3 D A process is adiabatic if there is no heat transfer into or from the gas. All four processes involve the expansion of an ideal gas. So, work done by the gas is positive, or the work done on the system is negative. If there is not heat absorbed into the system (as in an adiabatic process), the internal energy of the system has to decrease (ΔU = W + Q, with W negative and Q = 0, implies that ΔU must be negative). Hence, the temperature of the system has to decrease (since U ∝ T for an ideal gas). 4 A Option B: As a counter example, a system undergoing phase transitions such as melting or boiling has an increasing internal energy at a fixed temperature. Options C and D: A system with higher temperature can have less internal energy than another at a lower temperature by having less mass, and vice versa. 5 D 222 00 0 6 2 12 44 4 QqF rddπε πε πε ×= = = After contact, both charges have +2.0 μC charge. The electric force is repulsive. 1 22 0 0 2 2 16 4 344 2 FF dd πεπε ×= = =   6 B ( ) ( ) ( ) 3 6 3 6 35 3 40 30 10 405.0 10 30 10 40work done 5.0 10 12 10 8.0 10 J30 10 E VE d F qE − − − − −− − ∆= =∆ × = = ×  × = × −× = − × ×

2 © Raffles Institution 7 B 0000 0 10 4 32electric potential 22224444 2222 4321 42 2 5.09 10 Q QQQ Q Q πε πε πε πε πε −−=+++                −− + +=  = −× 8 D D 2 4 19 28 3 20 0.0021 4.5 10 1.6 104 8.02 10 m nAv q n n π −− − = = × × ×× ×× = × I 9 C There is work done when current flows through internal resistance r and a potential difference across it. Hence, the terminal p.d. decreases. 10 B The effect of the increase in the number of charge carriers is large than the effect of an increase in the frequency of collision between lattice ions and charge carriers. Hence, the ratio V I decreases and the curve gets steeper. 11 A Power dissipated, 2VP R= Since the resistors are identical, P ∝ V2 Y and Z are connected in parallel. Hence, VX : VY : VZ or VX 2 : VY 2 : VZ 2 2 : 1 : 1 or 4 : 1 : 1 Power dissipated across X = 1 24 4.0 W6 ×= 12 A Direction of electric force on electrons is upwards. Therefore, the direction of magnetic force on the electrons has to be downwards (magnetic field is into paper). Direction of electric force on alpha particles is downwards. Direction of magnetic force on alpha particles is upwards. When magnitude of B is increased, the magnetic force (Fm = B’ × 2e × v) is greater than the electric force (FE = 2e × E). Hence, the alpha particles deviate towards PQ. 13 B F = BIL sinθ FB θ∝ sin FB FB θ θ=22 2 111 sin sin ⇒ F =2 sin60 8 2sin90   ⇒ F2 = 3.46 mN

3 © Raffles Institution [Turn over 14 C 1. Change in flux linkage = initial flux linkage (final flux linkage is zero) 2. Change in flux linkage = more than initial flux linkage ( 0 2B r µ π= I ) 3. Change in flux linkage = less than initial flux linkage (final flux linkage is not zero) The change in flux linkage in ascending order is 1, 3, 2. Since the time taken of the changes are the same, the induced emf is in the same order. 15 C Induced emf, V t ∆Φ= ∆ Hence, 0NBA NBAt VV V ∆Φ −∆= = = 16 C ( ) 222 rms rm s 120 28.8 500 VVPR RP= ⇒= = = Ω Power dissipated in 240 V d.c. source 2 2 d.c. 240 2000 W28.8 V R= = = 17 B ss pp VN VN= Decreasing number of turns in primary coil will increase the voltage in the secondary coil. The voltage in the secondary coil is dependent on the ratio of the number of turns between the secondary and primary coils. 18 D 12 3 12 3 123 123 3 3 Let , and , Since , 1 11 840 nm (this is infrared radiation)350 600 hc hc hc EE E hc hc hcEEE λλ λ λλλ λλ = = = =+⇒ = + ∴ = +⇒= E1 E2 E3 2L L I 6L O A 0 L 2L 3L r B Area under the graph from L to 3L is proportional to the initial flux linkage. Area under the graph from 0 to 2L is proportional is the final flux linkage. Clearly, change in area (area B − area A) is larger than area under the graph from L to 3L. B

4 © Raffles Institution 19 A 22 2 22 2 11 2 1 22 2 (/) 1 22 1 4 2 16 16 phKE K K mm m Km m Km m KK λ λ λ λ λλ = == ⇒∝ ∴= = = ∴=          20 D 1 34 6 31 0.50 0.50 15000 75.0 m s100 100 6.63 10 9.70 10 m() 9.11 10 75.0 vv hhx p mv − − − − ∆= = × = ×∆> = = = ×∆∆ ×× 21 (a) (i) 0.905 800 724 kgmV ρ= = ×= B1 (ii) Q = cmΔT = 1050 × 724 × (400-273) = 96545400 = 9.65 × 107 J B1 (iii) Since density is inversely proportional to temperature, 273 400 400 273 ρ ρ = . 3 273 400 400 400( ) 0.905 1.326 1.33 kg m273 273ρρ −= ×= × = = B1 (iv) 3 273 273 724 546 m1.326 mV ρ= = = B1 (b) work done by air = p∆V = p(Vf − Vi) = 1.03 × 105 × (800 − 546) = 2.62 × 107 J B1 (c) ΔU = Q + W = 9.65 × 107 + (−2.62 × 107) (M1 for the negative sign) = 7.03 × 107 J ( ) ( )35 or 22U PV U PV∆= ∆ ∆= ∆ are not acceptable as air is not a monatomic or diatomic gas. It is a mixture of different gases. M1 A1 (d) 1. Heat is lost through the surface of the balloon to the surrounding air. 2. The balloon itself is heated up. 3. Air spilled out during the heating process carries away some heat (convection). 4. Air outside the balloon was also heated up. B1 B1 Markers’ Comments (b) Many students made mistakes in the sign of the work done. Instead of relying on the sign in the formula (WD = −p∆V, which is for work done ON the gas), what you can do is to remember a simple fact: if a gas expands, it does positive work on its surroundings. By this criterion, the work done in this question (BY the gas on the atmosphere) must be positive, so should your final answer for this part be. Also, remember that the work done on a gas and that by the gas differ by a negative sign. (c) The work done in the first law equation is work done ON the gas, which should be the negative of your answer in (b) (which was for work done BY the gas). To score the error-carried-forward mark here if your answer for (b) was wrong, you need the above - mentioned negative sign.

5 © Raffles Institution [Turn over (d) Many students stated the two points as “Heat loss to the surroundings” and “Heat loss to the balloon/gas spilt out/atmosphere”. These answers can score only 1 mark out of 2, since the balloon/gas spilt out/atmosphere is part of the surroundings, so the two points are essentially the same point. The question specifically mentioned that we cons ider only the air of mass m. Hence, everything else is part of the “surroundings”. To score both points, students can identify two parts of the environment, which absorb energy via different mechanisms (conduction and convection, for example). Many students stated that air is not an ideal gas, with intermolecular forces, hence part of heat provided is used to increase the distance between particles, hence more energy is needed for the same amount of temperature change (which only measures the change in th e average kinetic energy of the particles). This, although true (hence accepted), is a very small effect. Air under normal conditions is an excellent approximation to an ideal gas. The dominating effect for this question should still be the heat loss to the su