Question Library

y = 0.005 sin[x(0.5X - 200t)] compare to Y = A sin(kx - wt) \(\lambda\) = \(\frac{2\pi}{k}\) and F = \(\frac{w}{2\pi}\) V= \(\lambda\)F = \(\frac{2\pi}{0.5}\) x \(\frac{200}{2\pi}\) = 400ms-1

F = EQ 1.6 x 10-19 x 108 = 1.6 x 10-11N

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

Since the system is on the verge of moving, let the resultant force on \(m_1\) and \(m_2\) be F = 0 so, \(m_2\): F = \(m_2\)g - T = 0 so that \(m_2\)g = T ---------- 1 for \(m_1\) : F = T - \mu \(m_1\) g = 0, so that T = \mu \(m_1\)g ---------- 2 equating 1 and 2 above; Therefore \(m_2\)g = \mu\(m_1\)g \mu = \(\frac{m_2}{m_1}\)

Relative density of liquid x to liquid y = \(\frac{p_x}{p_y}\) = \(\frac{h_y}{h_x}\) = \(\frac{22.5}{7.5}\) = 3( liquid x is thrice denser than liquid y)

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

(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) 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.

horizontal component = W cos(90o - \(\theta\)) = W sin

Time from tower to top, t = \(\frac{u - v}{g}\) = \(\frac{20 - 0}{10}\) = 2s distance from tower to top x = ut - \(\frac{1}{2}\)gt2 = (20 x 2) - (\(\frac{1}{2}\) x 10 x 2 x 2) = 20m Time from the top to the ground, t = 6 - 2 = 4s distance from top to ground, x = ut - \(\frac{1}{2}\)gt2 = (0 x 4) + (\(\frac{1}{2}\) x 10 x 4 x 4) = 80m therefore, height of tower = 80 - 20 = 60m

When an object moves with uniform speed in a circular path, its acceleration (centripetal acceleration) has a constant magnitude but changes direction continuously as the object moves around the circle. This acceleration is always directed towards the center of the circle.

Initial momentum = 0.1 x 10 = 1Ns final velocity = -2ms-1 Final momentum = -2 x 0.1 = -0.2Ns Change in momentum = -0.2 - 1 = -1.2Ns

P.E = \(\frac{1}{2}\)Fe = \(\frac{1}{2}\)\(\frac{F}{K}\) = \(\frac{1}{2}\) x \(\frac{75 \times 75}{1500}\) = 1.9J

Efficiency = m.A x \(\frac{1}{V.R}\) x 100% 85 = \(\frac{load}{1200}\) x \(\frac{10}{80}\) x \(\frac{100}{1}\) load = 8160N