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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 D.

E - V1 + V2 24V = 14V + VQP VQP = 24 + 14V VQP = 10V

F = \(\frac{GM_1M_2}{12^2}\) F = 6.7 x 10-11x F = \(\frac{M_1M_2}{R^2}\) F = \(\frac{6.7 \times 10^{-11} \times 6.0 \times 10^{24} \times 7.0 \times 10^{22}}{(4 \times 10^8)^2}\) = 1.8 x 1020N

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 C.

The diagram given simple illustrates that a pentagon contains three \(\Delta\)s. Two angles which are in the ratio 2:3 will have actual values 2xo, 3xo respectively. Thus 2xo + 3xo + 3 x 60 = 3x sum of angles of \(\Delta\)a i.e. 5xo + 180o = 3 x 180o 5xo + 180o = 540o 5xo = 540o - 180o 5xo = 360o xo = \(\frac{360^o}{5}\) = 72o Hence, the smaller of the two angles is 2 x 72o = 144o

(a) Projectile is an object thrown into the air, and left to move under the influence of gravity. (b) the acceleration of free fall due to gravity does not act in the horizontal direction.

Given: L = 2m r = 1mm = 1 x \(10^{-3}\)m. e = 25mm = 25 x \(10^{-3}\)m F = 1000N Young modulus, E = \(\frac{F \times l}{A \times e}\) Area, A = \pir^2 = \(3.142 x (10^{-3})^2\) = \(3.142x10^{-6}\) E = \(\frac{1000 \times 2}{3.142 \times 10^{-6} \times 25 \times 10^{-3}}\) E = \(2.55 x 10^{10} NM^{-2}\)

Ways of increasing the cooling rate of a cup of tea are: (i) Stirring the tea continuously. (ii) Increasing the rate of evaporation. (iii) Reducing the ambient temperature.

(a) Doping in a semi conductor is the addition of impurity to a pure semi condutor to increase its electrical conductivity through increase in the number of holes and electrons or charge carriers.

a) \(\lambda = \frac{h}{p} or \alpha = \frac{h}{mV}\) b) The significance of the equation is tha, a particle of mass (m) moving with a velocity (v) has an associated wavelength (\(\lambda\)). The wavelength decreases with increase in momentum or velocity.

(i) Uniform acceleration is a constant time rate of increase in velocity. (ii) Linear velocity = angular velocity x radius of path V = \(\omega\)r (b) Scale: 1 cm to 5 N 30N = 6cm and 40N = 8cm (c) A floating ship is balanced by its weight and the upthrust on it. Hence when off-loaded the upthrust becomes greater than the weight of the ship, therefore when refilled with sand and water, equilibrium is restored for stability. d)(i) \(\mu = tan \theta = tan 30^o = 0.5774\) (ii) F = \(\mu g cos \theta\) = 0.5774 x 20 x 10 x Cos 30\(^o\) = 100N

a) Specific latent heat of vaporization in the quantity of heat required to change a unit mass of a liquid at its boiling point to gas. (b)(i) Renewable energy sources are sources of energy that cannot be exhausted. (ii) Renewable energy sources are sun or solar, wave, wind, hydroelectric, tidal, biomass or vegetable oils, geothermal, etc. (c) Clay pot is porous, therefore molecules of water in the moist jute bag will extract latent heat of vaporization from the tomatoes which will escape through the pores of the pot, thereby causing cooling of the tomatoes. (i) Mass = density x volume = 8.00 x 10\(^3\) x \(\frac{70}{80}\) x 0.28 = 1568 kg. (ii) Energy required Q = MC (\(\theta_2 - \theta_1\)) + mL = \(\frac{10}{100}\) x 1568 x 460(1500 - 25) + \(\frac{10}{100}\) x 1568 x 1.38 x 10\(^5\) = 106, 388, 800 + 21,638, 400 = 128,027,200J or 1.28 x 10\(^8\)J