Question Library

R.d = \(\frac{\text{weight of object in air}}{\text{upthrust in water}}\) = \(\frac{10.5}{10.5 - 6.8} = \frac{10.5}{3.7} = 2.84\)

M\(_{(g)}\) \(W_1\) \(W_2\) \(W_3\) U=(\(W_1\)-\(W_2\)) V=(\(W_1\)-\(W_3\)) 50 0.50 0.43 0.45 0.07 0.05 100 1.00 0.90 0.88 0.10 0.17 150 1.50 1.30 1.33 0.20 0.17 200 2.00 1.73 1.75 0.27 0.25 250 2.50 2.15 2.20 0.35 0.30 SEE THE GRAPH ABOVE Slope, s = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{0.275-0.10}{0.32-0.125}\) s = \(\frac{0.175}{0.195}\) = 0.9 Precautions: i - ensured that I avoided errors due to parallax while taking the readings on the spring balance. - made sure the mass is totally immersed in water and liquid - ensured the retort stand and clamp are well tightened. (b)i. Archimedes' principle states that when a body is totally immersed in a fluid (liquid or gas), it experiences an upthrust that is equal to the weight of the fluid displaced. Mass of brass = 20g = 2 x 10\(^{-2}\)kg Density of kerosine 8.0 x 10\(^{2}\)kgm\(^{-3}\) g = 10ms\(^{-2}\)kg Density of brass = 8.0 x 10\(^{3}\)kgm\(^{-3}\)kg Volume of brass = \(\frac{m}{D} = \frac{2 \times 10}{8 \times 10^{3}}\) = \(\frac{2}{8}\) x 10\(^{-2-3}\) = 0.25 x 10\(^{-5}\)m\(^{3}\) Volume of brass = 2.5 x 10\(^{-6}\)m\(^{3}\) Therefore, the mass of brass in kerosene m = Density x volume of brass = 8.0 x 10\(^{2}\) x 2.5 x 10\(^{-6}\)m\(^{3}\) = 20 x 10\(^{-4}\) Reading on the spring balance 20 x 10\(^{-4}\) = 20 x 10\(^{-3}\) N

S/N M\(_{(g)}\) V(cm\(^{3}\)) V\(^{-1}\)(cm\(^{-3}\)) 1 500 9.30 0.11 2 1000 7.40 0.14 3 1500 6.40 0.16 4 2000 5.40 0.19 5 2500 4.60 0.22 Slope, s = \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{2425 - 725}{0.12 - 0.125}\) \(\frac{1700}{0.086}\) = 20,000 Evaluation S = 20,000 K = s\(^{-1}\) = \(\frac{1}{20,000}\) = 5 x 10\(^{-5}\) Precautions; i - ensured that the retort stand and clamp are well tightened. - avoided error due to parallel when taking my readings. - made sure that the syringe is not on zero error. (b)i. The quantities of the gas above the plunger increase ii. While the quantities of the gas below the plunger decrease. iii. Weight

The resultant force acting on the man is given by w = mg - R Mg - mg = 0 The man feels weightless because the reaction, R is equal to the weight mg

V\(^2\) = U\(^2\) - 2gS 0 = 100\(^2\) - 2 x 10 x S = 10000 - 20S S = \(\frac{10000}{20}\) = 500m Which is independent of mass. The two bodies were acted upon by the same retarding force of gravity, which is independent of the mass of the body; therefore, both will be 500 m from the point of throw.

Centripetal force, F = \( \frac{mv^2}{r} \) = \( \frac{2 \times 10^2}{4} \) = \( \frac{200}{4} \)= 50N

K e just before hitting the gound = P e at the maximum height (conservation of mechanical energy) K e = P e = mgh = 1.5 x 10 x 20 = 300 J

Efficiency = \( \frac{\text{work output}}{\text{work input}} \times \frac{100}{1} \) = 40 = \( \frac{20 \times 10 }{x} \times \frac{100}{1} \) = 500J

Pressure in liquid is independent of cross sectional area of the container/ liquid ∴ the correct options are I & II

Dull, black surface are good emitters or absorbers of energy

V = X/t 330 = ( 4.95 x 10^3)/t t = (4.95 x 10^3/330 = = 15 secs

Regelation means refreezing a phenomenon whereby melted ice free from increased pressure refreezed when they are released from the pressure.

The specific latent heat of vaporization of steam is 2.1 x 106JKg-1 while the energy required to raise the temperature of a unit mass of water from 0^\circc to 100^\circc = C x Dt = 4200 x 100 = 420000 = 4.2 x 105 JKg-1 ∴ steam possesses greater heat energy per unit mass

Energy consumed = power x time = lvt = 2 x 250 x 6hr = 300 wh = 3 kwh but 1 kwh = N4 3 kwh = 3 x N4 = N 12

Under high potential difference and low pressure gases are good conductor of electricity