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Weight of glycerine displaced = weight of water dispalced Mgh glycerine = Mgh water Pvgh glycerine = Pvgh water 0.4V P glycerine = (0.5v)/v.4v P water = 1.25 (Polycerine)/p water = But density of substance = relative density x density of water = 1. 25 x 1000kgm3 = 1250kgm-3

Neutral Equilibrium: If the body neither returns to its original position nor increases its displacement further, it will simply adopt its new position. Examples of neutral equilibrium Ball moving on the ground. Motion of sphere. A roller. A pencil lying horizontally. An egg lying horizontally on a flat surface.

At maximum height, the final velocity, v = 0 thus from v = u - gt 0 = u - gt ∴ U = at = 10 x 4 = 40 mls

The sum of momentum before collision M\(_1\) U\(_1\) + M\(_2\)U\(_2\) 12 x 4.2 + 18 x 0 12 x 4.2 = 50.4 The sum of momentum after collision (m\(_1\) + m\(_2\))v = (12 + 18)v = 30v V = 50.4/30 = 1.7ms-1

Simple harmonic motion is the motion of a body whose acceleration is directed towards a fixed point and is directly proportional to its displacement from point. Statements ii, iii and v are clear examples

At vt = terminal velocity = zero acceleration = uniform velocity

When the adhesion of liquid molecules to the glass is more than the cohesion of the liquid, the liquid will wet glass.

Anomalous expansion of water is at temperature ranges 0^\circc to 4^\circc u. water contract when heated from 0^\circc to 4^\circc. thus the volume of water is least at 4^\circc

Work done on a stretched cord = Average force x extension = \(\frac{1 \times 10 \times 0.012}{2}\) = 6.0 x 10\(^{-2}\)J

F = Ke (Hooke's law) K = F/e = 80/4 = 20 Nm-1

\(R^2 = 100^2 + 60^2\) 10000+3600 = 13600 R = \(\sqrt{13600}\) = 117km 60km & 100km = tan\theta = \(\frac{60}{100}\) \(tan^{-1}\) 0.6 = 31º west of the North 117km N31ºw

pressure in fluid is given by P = hpg. ∴ fluid pressure depends on height of liquid column, h, density of liquid p and acceleration of free fall g. it is therefore independent of type of container and constituents of the liquid.

Liquid boils when its saturated vapor pressure is equal to the (external) atmospheric pressure. I.e. Boiling point at 100^\circc

The energy supplied by the heater = heater gained by the water pt = Mc\(\Delta\)\theta 1000 x 300 = m x 4200(100 - 30) m = \(\frac{(1000 \times 300)}{4200 \times 70}\) = 1.02 kg

F ∝ \(\frac{M_1 M_2}{r^2}\) F = \(\frac{GM_1M_2}{r^2}\) where F is the force of attraction M\(_1\) and M\(_2\) are the masses of the two bodies, r is the distance between them, and G is the universal gravitational constant Therefore, G = \(\frac{Fr^2}{M_1M_2}\) Dimension for F = \(\frac{\text{ML}}{T.T}\) = MLT\(^{-2}\) The dimension for G = \(\frac{MLT^{-2}L^2}{M.M}\) = \(\frac{L^3T^{-2}}{M}\) = M\(^{-1}\)L\(^3\)T\(^{-2}\) In comparison with M\(^x\)L\(^y\)T\(^z\), the values for x, y and z = -1, 3, -2