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\(C_T = \frac{2 \times (2 + 2)}{2 + (2 + 2)}\) = \(\frac{8}{6}\) = 1.33 \(\mu\)F

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

studying the charges of the Gamma, Alpha, Beta and X-rays, you find that both the x-rays and theGamma rays possess zero-charge hence they could pass the charged plate undetected while the beta rays are negatively charged hence will deflect towards the positive charge of the plate. Alpha rays, on the other hand, are poisitively charged hence, they deflect towards the negative charge.

(a) The statement: The linear expansivity of a solid is \(1.0 \times 10^{-5}K^{-1}\) means that, A unit length of the solid will expand in length by a fraction \(1.0 \times 10^{-5}\) of the original per kelvin rise in temperature. (Note: \(1.0 \times 10-5 = \frac{1}{100,000}\), a fraction). (b) Experiment to determine the linear expansivity of a steel rod. The apparatus is set up as shown above. The length of the steel rod used is first measured with a meter rule and it is then placed in the jacket with one end fixed. The micrometer screw gauge is turned until the end of the screw just touches the end of the rod, and the gauge reading is noted. It is then turned back enough to make sure that the end of the rod can expand freely. The reading on the thermometer is noted and steam is now passed through the jacket for several minutes. The micrometer is used to note the new reading. More steam is passed again and the reading is taken until a constant reading is obtained. The reading is noted and the temperature is noted from the thermometer. The formula \(\alpha = \frac{l_{f} - l_{i}}{l_{i} (\theta_{f} - \theta_{i})}\) is used to determine the linear expansivity. \(l_{f} - \text{final length} ; l_{i} - \text{initial length} ; \theta_{f} - \text{final temperature} ; \theta_{i} - \text{initial temperature}\) (b)(ii) \(l = 3m ; \theta_{i} = 29^\circC ; \theta_{f} = 41^\circC\) \(\alpha = 1.0 \times 10^{-5} K^{-1}\) Safety gap = change in length. \(\Delta l = \alpha l \Delta \theta\) = \(1 \times 10^{-5} \times 3.0 \times (41 - 29)\) = \(36 \times 10^{-5} = 0.00036m\) (c) Advantages: (1) Used for marking bimetallic strip which is used in thermostat. (2) Removal of tight glass stopper (3) Red-hot rivets in ship building. (4) Expansion of metals is used in bimetallic thermometer. Disadvantages: (1) Expansion of metals or concrete bridges. (2) Cracking of glass cup when hot water is poured into the glass cup. (3) Expansion of balance wheel or wrist-watch. (4) Sagging of overhead wire. (5) Expansion of railway lines.

(a) A wave motion is succession of pulses propagated with time from one region of space to another. Energy is transmitted in the process. A is the amplitude. It is the maximum displacement from equilibrium position of the wave particle. \(\lambda\) - the wavelength, is the distance between two successive points on the wave that is in phase. (b) Experiment to determine the frequency of a note emitted by a source of sound Assuming that the source of sound is a turning fork. The vibrating tube turning fork is held close to the top of a moveable tube as shown. The tube is deeped in a cylinder that is 3/4 filled with water. Starting with the air column of length, I, with the turning fork sounding gently close to the top, the length I is gradually increased; The, sound coming from the air column is heard to increase. When the sound is loudest the air has been set into resonance by the fork. The length of I of the air column is measured and recorded. The length of the air column can be shown to be \(\frac{\lambda}{4}\) neglecting end correction. Thus using the equation: \(f = \frac{V}{\lambda} = \frac{V}{4l}\) where V is the velocity of sound in air, f the frequency of the turning-fork the source of sound, the frequency is then determined. (ii) Using \(f = \frac{V}{\lambda} = \frac{V}{4l}\) = \(\frac{340}{4 \times 1} = 85 Hz\) (c) A sound wave : (i) requires material medium for propagation. (ii) is a longitudinal wave motion. While radio wave (i) does not require material medium for propagation. (ii) is a transverse wave motion.

(a) The laws of electromagnetic induction states that (1) whenever there is a change in the magnetic flux linked with a circuit, an e.m.f. is induced, the strength of which is proportional to the rate of change of the flux linked with the circuit. The above law is sometimes called faraday's law of electromagnetic induction. (2) The direction of the induced current is always such as to oppose the change producing it. This is called Lenz's law. b(i) Experiment to show how an induced e.m.f. can be produced. A coil with many turns is connected to centre-scale zero galvanometer G. As the bar magnet is moved rapidly close and through the turns, an induced e.m.f. is produced. The galvanometer pointer is deflected showing a flow of current. The same effects could be observed if the coil is moving towards the magnet instead. Or as long as there is relative motion between the two. (ii) The magnitude of the induced e.m.f. depends on: (1) The magnitude of flux linking the coil. (2) The relative speed between the magnet and the coil. (c) The root-mean-square (r.m.s.) value of an alternating current is the steady or direct current which produces the same heating effect per secoridin a given resistor. Its relationship with the peak value of alternating current \(I_{o}\) is given by \(I_{r.m.s} = \frac{1}{\sqrt{2}} I_{o}\) d(i) \(I\) is the instantaneous current \(I_{o}\) is the peak value of the current \(\alpha\) is the angular frequency \(\alpha t\) is the phase or the current (since it is wave-like). (ii) \(I = I_{o} \sin \alpha t\) = \(I_{r.m.s} \sqrt{2} \sin 30^\circ\) = \(\sqrt{2} \times 15 \times \frac{1}{2} = 10.6A\)

a(i) Photoelectric emission occurs when electrons are emitted from the surface of metal plate when it is illuminated by light of sufficient high frequency. Threshold frequency is the minimum frequency of the illuminating light which will just be sufficient to cause photoelectric emission. Below this frequency, electrons will not be emitted. (ii) In the Einstein's photoelectric equation \(E = hf - hf_{0}\), E - is the maximum kinetic energy of the emitted electron ; hf - is the energy of the illuminating light; \(hf_{0}\) - is the work function. It represents the minimum energy required to liberate electrons from the surface of the metal \(f_{0}\) - is the threshold frequency. (b) Using \(E = hf - hf_{0}\) \(f = \frac{E + hf_{0}}{h}\) (d) The duality of matter means that matter can exhibit wave properties as well as have particle properties. This double nature of matter is general in nature e.g. particles like electrons can have wave properties. Waves such as X-rays can have particle properties. This is equally illustrated by photoelectric effect. De Broglie equation for the waves carried with particle is \(\lambda = \frac{h}{mv} = \frac{h}{p}\) mv = p is momentum of particle of mass m; h - planks constant.

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.

First, we need to write a balanced equation for the chemical reaction: H\(_2\)SO\(_4\) + Na\(_2\)CO\(_3\) → Na\(_2\)SO\(_4\) + H\(_2\)O + CO\(_2\) Stoichiometrically, N\(_A\) = 1 , N\(_B\) = 1 Given: V\(_B\) = 20cm\(^3\), V\(_A\) = 25cm\(^3\), C\(_A\) = ? C\(_B\) = 0.53 g is dissolved in 100 cm\(^3\) of water then Xg will be dissolved in 1000 cm\(^3\) of water (i.e 1dm\(^3\) = 1 Litre of water) Xg = \(\frac{{0.53}\times{1000}}{100}\) Xg = 5.3gdm\(^{-3}\) → This is the concentration of B in gdm\(^{-3}\). Hence we must have to convert it to moldm\(^{-3}\) before calculating the concentration of the acid. To convert to molar concentration, we use the relationship; Molar concentration (moldm\(^{-3}\)) = \(\frac{Mass concentration (gdm^{-3}}{Molar mass}\) But Molar mass of Na\(_2\)CO\(_3\) = (23 x 2) + 12 + (16 x 3) = 106g/mol Molar concentration (moldm\(^{-3}\)) = \(\frac{5.3}{106}\) C\(_B\) = 0.05 moldm\(^{-3}\) Using \(\frac{{C_A}{V_A}}{{C_B}{V_B}}\) = \(\frac{N_A}{N_B}\) \(\frac{C_A}{V_A}{N_B}\)= \(\frac{C_B}{V_B}{N_A}\) C\(_A\) = \(\frac{{C_B}{V_B}{N_A}}{{V_A}{N_B}}\) C\(_A\) = \(\frac{{0.05}\times{20}\times{1}}{{25}\times{1}}\) C\(_A\) = 0.04 moldm\(^{-3}\)

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 C.

\(\frac{\text{Element X}}{1s^22s^21p^4}\) Valency of X is 2 \(\frac{\text{Element Y}}{1s^22s^22p^63s^23P^1}\) valency of Y is 3 There is an exchange of valency when a compound is being formed. Hence, we have Y\(_2\)X\(_3\)

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