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No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option B.

No expanded working has been attached for this question yet. The verified answer is Option A.

No expanded working has been attached for this question yet. The verified answer is Option D.

No expanded working has been attached for this question yet. The verified answer is Option E.

(a) The specific latent heat of fusion of ice is \(3.4 \times 10^{5}Jkg^{-1}\) means that \(3.4 \times 10^{5} J\) of hoat energy is required to change 1kg mass of ice from solid state to the liquid state without a change in temperature. Weigh a clean, dry copper calorimeter empty, together with stirer. Half fill the calorimeter with water previously warmed to about 8^\circC above room temperature and reweigh it to find the mass of water. Record the temperature of the water when it is say about 5^\circC above room temperature and then immediately add small pieces of ice, to the water. Stir the water until all the ice is melted before adding another piece. Continue until final temperature is 5^\circC below the room temperature. Record this temperature. Finally, reweigh the calorimeter to find the mass of ice melted. Calculations: Heat gained by ice in forming water = Heat lost by water and calorimeter \(M_{1} L + M_{1} C_{w} \theta_{1} = (M_{w} C_{w} + M_{c} C_{c}) \Delta \theta\) \(\therefore L = \frac{(M_{w} C_{w} + M_{c} C_{c}) \Delta \theta - M_{1} C_{w} \theta_{f}}{M_{1}}\) where \(M_{1} - \text{mass of ice}\) \(M_{c} - \text{mass of calorimeter}\) \(C_{w} - \text{specific heat capacity of water}\) \(C_{c} - \text{specific heat capacity of calorimeter}\) \(\theta_{f} - \text{final temperature}\) \(\Delta \theta - \text{change in temperature of calorimeter and water}\) (c) In a solid, such as ice, the molecules, which are held together by inter molecular forces, vibrate about a mean position. As ice absorbs more heat, the vibration of the molecules have enough energy to overcome the force binding them together. They are now free to move about in a random fashion as in a liquid.

(a) Resonance occurs when one vibrating body sets other body into vibration at its own natural frequency as a result of impulse received from the first body. (ii) Simple experiment to illustrate resonance. The apparatus is arranged as shown. A cylinder is 3/4 filled with water and a tube is dipped into the water so that the length I can be varied. A tuning fork f is set vibrating just above the tube but not touching it. As the fork is set vibrating the length I is gradually increased. A point is reached when a large sound is heard. At this point resonance has occurred because the vibrating tuning fork has set the air in the tube vibrating at its own frequency. (b) Parallel light rays are reflected by a concave mirror on a piece of carbon paper. The converging of the rays on the carbon paper will cause it to ignite. The object O is kept on the centre of curvature of a concave mirror. An exact but inverted image I is formed at the centre of curvature where the screen should be placed.

(a) The step-down transformer has more turns in the primary turns than in the secondary turns. It works by electromagnetic induction, the magnetic flux from the primary coil in the iron core links the secondary coil. When an alternating e.m.f \(E_{p}\) is connected to the primary coil, an alternating current flows. This produces corresponding variation of magnetic flux in the iron core which links the secondary coil. An induced e.m.f \(E_{s}\) of the same frequency as \(E_{p}\) is obtained. From Faraday's law, the e.m.f induced in a coil proportional to the rate of change of the magnetic flux linking the coil. So, \(\frac{E_{s}}{E_{p}} = \frac{N_{s}}{N_{p}}\). \(N_{s}\) - Number of turns in secondary coil; \(N_{p}\) - number of turns in primary coil. (b) Three ways by which energy is lost and how they can be minimised are: (1) Energy is lost through Eddy current produced in the soft — iron. This is reduced by laminating the core in thin strips. (2) Heat generation in the copper coils. This is reduced by using copper coils with low resistance. (3). Loss of energy due to reversal of iroes of force. This is reduced by using soft magnetic materials for the core. (c)(i) Using \(\frac{E_{p}}{E_{s}} = \frac{N_{p}}{N_{s}}\) \(\frac{4400}{220} = \frac{N_{p}}{N_{s}}\) The ratio of the primary turns to the secondary turns is 20 : 1. (ii) \(Efficiency = \frac{\text{power output}}{\text{power input}}\) \(\frac{95}{100} = \frac{60}{input}\) \(\text{Power input} = \frac{6000}{95} = 63.2W\) \(\therefore Current I = \frac{P}{V} = \frac{63.2}{4400}\) \(I = 0.0144A = 14.4 mA\)

(a) When ultraviolet rays fall on zinc plate, electrons are liberated from the zinc plate. This phenomenon is called photoelectric emission. In general the emission of electrons from matter as a result of electromagnetic wave falling on the matter is referred to as photoelectric effect.Emission of electrons depends only on the threshold frequency. But the rate of emission depends on the intensity of light. The threshold frequency is not the same for all metals. The energy of emitted electrons varies from zero to a maximum value. (b) photo cell When light photons fall on the caesium metal, it emits electrons which move to the collector to generate current flow. As the intensity of Iight varies, the rate of emission of electrons varies and thus, the current flow varies. (c) Applications of photoelectric emission. (1) Production of television cameras (2) Burglary/ alarms (3) Automatic switches for putting on lights at dusk. (4) Industrial controls and counting operations. (d) The energy of the light is absorbed by the photocell and are used to overcome the force that brids the electrons but may not be sufficient to remove the electron until light with sufficient threshold frequency. One factor that may affect the number of emitted electrons is the intensity of the light.

No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option D.

No expanded working has been attached for this question yet. The verified answer is Option D.

100 cm3 of O2 diffuse 50 sec ∴ 100 cm3 of H2 → (100) / (100) x (50) /(1) t1 50 sec molecular mass H+ → 2 x 1 = 2 molecular mass of O2 16 x 2 = 32 T2 = t1 √M2/M1 = t2 = 50 √2/32 50√0.625 = 50 x 0.25 = 12.5 secs

No expanded working has been attached for this question yet. The verified answer is Option D.

The significant difference between a metal's melting and boiling points stems from the strength of the metallic bonds and the energy required to break those bonds during phase transitions. Melting involves loosening the attractive forces, while boiling requires complete breakage of those forces, demanding more energy.