M(g) \(W_1\)(g) \(W_2\)(g) \(W_3\)(g) U=(\(W_1\) - \(W_2\))g V=(\(W_1\)-\(W_3\))g 5.0 5.00 4.20 4.70 0.80 0.80 10.0 10.00 7.90 8.60 2.10 1.40 15.0 15.00 12.75 13.50 2.25 1.50 20.0 20.00 17.10 16.40 2.90 3.60 25.0 25.00 22.10 21.40 2.90 3.60 Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{2.35-1.00}{3.25-1.65} = \frac{1.35}{1.60}\) = 0.84 Precautions: I avoided parallax error in reading spring balance. I noted zero error of spring balance. l avoided splashing of liquid. I avoided object touching the bottom or side beaker. I cleaned the mass before dipping it into the liquid. (bi) Weight of kerosine displaced = volume of object x density of kerosine x g \(\frac{\text {Mass of object}}{\text {density of of object}}\) x density of kerosine x g = \(\frac{20 \times 10^{-3}kg \times 8.0 \times 10^{2} kgm^{-3} \times 10ms^{-2}}{ 8.0 \times 10^{-3} kgm^{-3}}\) = 0.02N = upthrust of the weight of object in air = mass x g = 20 x 10\(^{-3}\) X 10 = 0.2N Tension in the spring = weight of an object in the air - upthrust = 0.2N - 0.02N=0.18N The reading on the spring balance= 0.18 ii. Archimede's principle states that when a body is wholly or partially immersed in a liquid, it experiences an upthrust equal to the weight of the liquid displaced. On the other hand, the law of floatation states that a floating body displaces its own weight of the liquid in which it floats.