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Typist - /ˈtaɪ.pɪst/ refuse (noun) - /ˈref.juːs/ superb - /suːˈpɜːb/ propose - /prəˈpəʊz/ rebel - /ri.'bel/

they are certain rules and exceptions to stressing a word. you need to first of all be able to differentiate between nouns, verb, adjectives, compound words etc. with the knowledge of this, you can easily place the stress on the right syllable. perfect is stressed as PERfect. include is stressed as inCLUDE interest is stressed as inTErest employ is stressed at the second syllable as emPLOY Therefore, perfect has the same word stress as cement on the first syllable. NOTE; generally speaking words beginning with ‘un’ or ‘in’ have the stress on the second syllable.

In general, the energy released during nuclear fission is given by Einstein's mass-energy equation, given by \( E = \Delta MC^2 \) Where \( \Delta m\) is the mass defect, and C is the speed of light. Therefore \( \Delta m = 0.01% \text{of} 1.0g = \frac{0.01}{100}\times 1.0g \) = 1.0 \(\times 10^{-4} \) = 1.0 \(\times 10^{-7}kg\) \(\text{Energy Released } = \Delta MC^2 \) = 1.0 \(\times 10^{-7} \times (3.0 \times 10^8)^2\) = 1.0 \(\times 10^{-7} \times 9.0 \times 10^{16} \) = 9.0 \(\times 10^9 J \)

K = 500Nm-1; M = 5 x 10-3kg; e = 0.02m; g = 10ms-2. The elastic potential energy stored in the compressed spring is given by = 1/2Ke2, and the energy is transformed into gravitational potential energy = mgh, thus from the principle of energy conversion, we have that mgh = 1/2Ke2 5 x 10-3 x 10 x h = 1/2 x 500 x (0.02)2 ∴ h = 1 x 500 x 0.02 x 0.022 x 5 x 10-3 x 10 = 5 x 2 x 0.02___10-1 = 10 x 0.02 x 10 = 2 ∴ h = 2m

To find the depth to which a test tube sinks when placed upright in water, we apply the principle of buoyancy. Given: - Radius of the test tube, \( r = 1.0 \, \text{cm} = 0.01 \, \text{m} \) - Mass of the test tube, \( m = 8.8 \, \text{g} = 0.0088 \, \text{kg} \) - Density of water, \( \rho = 1000 \, \text{kg/m}^3 \) - Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) Calculate the weight of the test tube The weight \( W \) of the test tube is given by: \(W = mg = 0.0088 \, \text{kg} \times 10 \, \text{m/s}^2 = 0.088 \, \text{N}\) Calculate the buoyant force The buoyant force \( F_b \) is given by: \(F_b = \rho g V_{\text{displaced}} = \rho g \pi r^2 h\) But, the buoyant force is equal to the weight: \(\rho g \pi r^2 h = W\) Substituting the known values: \(1000 \times 10 \times \pi \times (0.01)^2 \times h = 0.088\) Solve for \( h \) Rearranging gives: \(h = \frac{0.088}{1000 \times 10 \times \pi \times (0.01)^2}\) Calculating: \(h = \frac{0.088}{1000 \times 10 \times \pi \times 0.0001} = \frac{0.088}{3.142} \approx 0.028 \, \text{m}\) ≈ 2.8cm

The length of the glass block = The main scale reading of the vernier calliper + the vernier scale reading = 1.6 + 0.04 = 1.64cm.

Vol = cross-sectional area x length ∴ vol. of water disharged per sec = 60ms-1 x 0.5m = 30m3 5-1 => Vol. of water discharged in the said 20s = 30m3 x 20 = 600m3 ∴ vol. of the container = 600m3

A body sliding down a plane inclined at 30o to the horizontal, has coefficient of friction = tan 30o = 1/√3

Surface tension - The tendency of a liquid surface to behave as an elastic skin

Pressure P = \(\frac{F}{A}\); Pressure P = \(\frac{F}{\pi r^2}\) = \(\frac{7.7 \times 10^{-2}}{\pi \times 10^{-10}}\) = 2.45 x 10\(10^8\)Nm\(^{-2}\)

AS the stone is whirled round and te string suddenly cuts, say at A, B, C or D. The stone will fly off in the direction indicated VA, VB, VC or VD respectively which is tangential to the circular path

Since the lever is a uniform one, its center of gravity will be acting half way (i.e 45cm) ∴from the principle of moments, F x 45 = 30 x 30 ∴ F = (30 x 30) / 45 = 20N

A satellite is said to be in a parking orbit if its period is exactly equal to the period of the earth as it turns its own orbit;. which is about 24hrs

R.H = \(\frac{Partial Pressure of Water Vapour Present}{S.V.P at temperature of atmosphere}\) x 100 = \(\frac{10}{17.50}\) x 100 = 57.14% ≈ 57%

In vibration in a closed pipe/tube, the first resonance occurs at a position L = V/4f, When reading from the top of the tube, L\(_1\) = \(\frac{340}{4 \times 340}\) = 0.25m = 0.25m L\(_2\) = \(\frac{3V}{4f}\) L\(_2\) = \(\frac{3 \times 340}{4 \times 340}\) = 0.75m L\(_3\) = \(\frac{5V}{4f}\) L\(_3\) = \(\frac{5\times 340}{4\times 340}\) = 1.25m However, the effective length of the tube is 1.2m. Thus reading from below, the first position of resonance will be 1.20 - 0.25m = 0.95m