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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 D.

(a) The specific heat capacity of copper is 400 \(JKg^{-1}K^{-1}\) means that 400 J of heat energy is required to raise (or lower) the temperature of 1 kg of a piece of copper by 1 kelvin. (b) Experiment to determine the specific heat capacity of copper using a copper ball. The piece of copper ball is suspended by a thread and placed in a beaker of boiling water. While the temperature of the copper is being raised to boiling point of water, a copper calorimeter is half filled with water. The initial temperature of the water is measured and recorded and the hot copper ball is lifted by the thread and transferred to the calorimeter when the water has reached boiling point. The calorimeter cover is put on, the mixture is well stirred and the final steady temperature read from the themometer. The copper ball is weighed either before or after the experiment. The result is tabulated as shown below. Mass of copper ball - m (kg) Temperature of boiling water - \(\theta\) (K) Mass of copper calorimeter - \(m_{c}\) (kg) Mass of calorimeter + water - \(m_{x}\) (kg) Mass of water \((m_{x} - m_{c})\) - \(m_{w}\) (kg) Specific heat capacity - C \((Jkg^{-1} K^{-1})\) Initial temperature of water = \(\theta_{w}\) (K) Final temperature of water = \(\theta_{f}\) (K) Using Heat lost by copper ball = Heat gained by calorimeter + Heat gained by water \(mc(\theta - \theta_{f}) = m_{c} c (\theta_{f} - \theta_{w}) + m_{w} c_{w} (\theta_{f} - \theta_{w})\) \(c = \frac{m_{w} c_{w} (\theta_{f} - \theta_{w})}{m (\theta - \theta_{f}) - m_{c} (\theta_{f} - \theta_{w})}\) Precautions: (i) The copper ball should be transferred quickly into the calorimeter without splashing. (ii) Just before transfer, the copper should be shaken slightly to remove adhering hot water. (iii) Heat lost by copper ball = heat gained by copper calorimeter + Hoat gained by water. \(m_{c} (\theta - \theta_{f}) = m c (\theta_{f} - \theta_{w}) + m_{w} c_{w} (\theta_{f} - \theta_{w})\) \(20 \times 0.4 \times (200 - \theta_{f}) = 60 \times 0.4 \times (\theta_{f} - 30) + 50 \times 4.2 \times (\theta_{f} - 30)\) \(8(200 - \theta_{f}) = 24(\theta_{f} - 30) + 210 (\theta_{f} - 30)\) \(\theta_{f} = \frac{8620}{242} = 35.6K\)

(a) The experiment is set up as shown in the diagram. The wooden box effective length, I of the string remains constant throughout the experiment. Also the string used is of uniform thickness and must not be changed to ensure that the mass per unit length is constant throughout the experiment. The length of string is tied to the stake at one free end while a known mass is added to the other end. This creates a tension in the string which is equal to the weight of the attached mass. A tuning fork of known frequency is set vibrating and the string is tuned in unison with the fork. The frequency of the fork is recorded. The procedure is repeated by increasing the masses and using different tuning forks four more times. The corresponding tension (weights) and frequency are recorded in a table. Experiment to show that frequency is proportional to the square root of the tension, T. Mathematically, the graph of F against \(\sqrt{T}\) is a straight line through the origin. (b)(i) (ii) Second overtone (c)(i) (ii)

(a) A magnetic field is a region in which a force of magnetic origin is experienced. It is usually due to the presence of magnetic flux which is a vector quantity. (b) A straight wire carrying current is filled through a hole in a cardboard as shown. Compasses are placed round the conductor or the cardboard. The needles of the compass will be noticed to deflect in a particular direction. They all conform to a circular pattern indicating that the magnetic field around the conductor carrying current is circular or concentric in shape. When the current is reversed, the direction of the compass needle reverses as well. (c) (d) Current in the same direction The material is surrounded by a ring of iron. This shields the flux of the earth away from the material.

(a) Whenever there is a change in the magnetic flux linked with a circuit an electromotive force is induced, the strength of which is proportional to the rate of change of the flux linked with the circuit. Lenz's law states that the direction of the induced current is always such as to oppose the charge producing it. (b) Lenz's law illustrates the law of conservation of energy. In this case mechanical energy i.e kinetic energy, due to the motion of the bar magnet is changed to electrical energy in the form of induced current. (c) A simple d.c. electric motor can supply direct current (d.c.) by using a split - ring commutator.The brush (-) touches one half of the split ring X while the other brush (+) touches the other half Y. As the coil rotates it cuts through the magnetic flux linking the magnets and current is induced which flows through the coil as shown in the diagram. As the coil rotates past the vertical Y moves round to touch (-) while X moves round to touch (+). This action reverses the connections to the external resistance connected to (-) and (+). The reverse happens when the current is just zero (coil vertical) so instead of the current reversing in the external resistance, it continues to flow in the same direction. c(ii) (1) Part of the current is wasted in form of heat due to the resistance of the coil and this reduces efficiency; (2) Friction between the moveable parts of the motor makes the efficiency to reduce.

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 B.

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

1 mole of any gas has a volume of 22.4dm3 at STP, SO 1mole of propane = 44g = 22.4dm3 22g = ? ( 22 x 22.4)/ 44 = 11.2dm3

Given: T\(_1\) = 273K V\(_1\) = X P\(_1\) = Y → Let x and y represent the initial volume and initial pressure respectively. T\(_2\) = ? V\(_2\) = 2X P\(_2\) = 2Y → Since the volume and pressure were doubled. Using the formula, \(\frac{P_1 V_1}{T_1}\) = \(\frac{P_2 V_2}{T_2}\) T\(_2\) = \(\frac{P_2 V_2 T_1}{P_1 V_1}\) = \(\frac{2Y\times 2X \times 273}{Y\times X}\) = 2 X 2 X 273 T\(_2\)= 1092K

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 C.

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

The temperature of the gas is proportional to the average kinetic energy of its molecules. Faster moving particles will collide with the container walls more frequently and with greater force. This causes the force on the walls of the container to increase and so the pressure increases.