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(\(\sqrt[3]{64a^3})^{-1}\) \(\frac{1}{(\sqrt[3]{64a^3})}\) = \(\frac{1}{4a}\)

\(\frac{2 \sqrt{3} - \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\)= m + n√6 \(\frac{2 \sqrt{3} - \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\) x \(\frac{\sqrt{3} - 2 \sqrt{2}}{\sqrt{3} - \sqrt{2}}\) \(\frac{2 \sqrt{3} (\sqrt{3} - 2 \sqrt{2}) - \sqrt{2}(\sqrt{3} - 2 \sqrt{2})}{\sqrt{3}(\sqrt{3} - 2 \sqrt{2}) + 2 \sqrt{2}(\sqrt{3} - 2 \sqrt{2})}\) \(\frac{2 \times 3 - 4\sqrt{6} - 6 + 2 \times 2}{3 - 2 \sqrt{6} + 2 \sqrt{6} - 4 \times 2}\) = \(\frac{6 - 4 \sqrt{6} - \sqrt{6} + 4}{3 - 8}\) = \(\frac{0 - 4 \sqrt{6} - 6}{5}\) = \(\frac{10 - 5 \sqrt{6}}{5}\) = − 2 + √6 ∴ m + n\(\sqrt{6}\) = − 2 + √6 m = − 2, n = 1

1/\alpha + 1/\beta = \beta+\alpha/\alpha\beta 3x2 + 5x - 2 = 0 x2 + 5x/3 - 2/3 = 0 \alpha\beta = -2/3 \beta+\alpha = -5/3 Thus; \beta+\alpha/\alpha\beta = -2/3 -2/3 = -5/2

0.4, −0.4, 0.3, 0.47, −0.53, 0.2, −0.2 Range is the difference between the highest and lowest value i.e Highest − Lowest − 0.53, −0.4, −0.2, 0.2, 0.3, 0.4, 0.47 0.47 is the highest − 0.53 is the lowest ∴ = 0.47 − (− 0.53) ∴0.47 + 0.53 = 1.0

1 − (\(\frac{1}{5}\) x 1\(\frac{2}{3}\)) + (5 + 1\(\frac{2}{3}\)) 1 − (\(\frac{1}{5}\) x \(\frac{5}{3}\)) + (5 + \(\frac{5}{3}\)) 1 − \(\frac{1}{3}\) + \(\frac{20}{3}\) = \(\frac{22}{3}\)

(2x2 - x + 1) × (3 - 2x); 3(2x2 - x + 1) - 2x (2x2 - x + 1) 6x2 - 3x + 3 - 4x3 + 2x2 - 2x -4x3 + 8x2 -5x + 3

Look through the equations of motion

Electromagnetic waves do not require any material medium, but sound waves is a mechanical wave that require a material medium

Electric Field Intensity (E) = \(\frac{v}{d}\) = \(\frac{\text{potential difference}}{\text{distance}}\) = \(\frac{6.5v}{35cm}\) = \(\frac{6.5}{35 \times 10^{- 2}}\) = 18.57NC\(^{-1}\)

Soldiers march disorderly while crossing a bridge to prevent resonance because when soldiers march in orderly manner across a bridge, they generate a rhythmic oscillation of wave on the bridge and at a certain point, would start oscillation to the same rhythm as that of the marching steps. This oscillation would reach a maximum peak when the bridge can no longer sustain its own strength and hence collapses. So for this reasons, soldier are ordered to march disorderly while crossing a bridge.

Upthrust = change in weight density = \(\frac{\text{mass}}{\text{volume}}\) Mass of liquid displaced = Density of Liquid \times Volume; = \(10^{3} \times 50 \times 10^{−6}m^{3}\) = 0.05kg (Note that the volume of \(50cm^{3}\) was converted to \(m^{3}\) by multiplying by \(10^{6}\)) i.e \(50cm^{3} = 50 \times 10^{−6}m^{3}\) Since mass displaced = 0.05kg Upthrust = mg = 0.05 \times 10 = 0.5N

Cathode rays are produced in a discharge tube under Low pressure and high voltage

For work to be done, distance must be involved because workdone = force × distance

Quantity of heat = mass(m) \times specific latent heat (L) Q = mL L = \(\frac{Q}{m}\) = \(\frac{1.13 \times 10^{6}}{15}\) = \(7.53 \times 10^{4}\)Jkg\(^{-1}\)

Internal resistance determine the maximum current that can be supplied Efficiency = \(\frac{R}{R+ r}\) \times 100 = \(\frac{6}{8}\) \times 100 = 75%