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W = \(\frac{1}{2}Fe\) W = \(\frac{1}{2}\times 20 \times 0.05\) W = 0.5J

Mp = ep x Vp Mp = 7.2 x 102 x 4.2 = 3024 VA = \(\frac{M}{e_A}\) VA = \(\frac{3024}{8.4 \times 10^2}\) VA = 3.6

W = \(\frac{W_1 \times V_1}{V_2}\) W = \(\frac{10 \times 0.6}{1.2}\) W = 6.0 x 10 4J

\(V \propto \sqrt{T}\) ⇒ \(\frac{V_{1}}{\sqrt{T_{1}}} = \frac{V_{2}}{\sqrt{T_{2}}}\) Using the above formula, we get \(V_{2} \approxeq 397ms^{-1}\)

F = qE F = 1.5 x 10-19 x 105 F = 1.5 x 10-14N

No expanded working has been attached for this question yet. The verified answer is Option C.

\(E_3 = \frac{E}{n^2}\) \(E_3 = \frac{-24.8}{3^2}\) \(E_3 = 2.75eV\)

E = \(\frac{hc}{\lambda}\) E = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^18}{10^{-10}}\) E = 2 x 10-15J

a = g sin \(\theta\) = 10 sin 60 = 8.7 ms\(^{-2}\)

Power = Force x Velocity. Therefore, the area under the force- velocity graph gives Power.

To find the speed of projection, we can use the formula for the time of flight of an object projected vertically upwards: \(t = \frac{2u}{g}\) where: - \( t \) is the total time of flight (1.2 s), - \( u \) is the initial speed of projection, - \( g \) is the acceleration due to gravity (10 m/s²). Rearranging the formula to solve for \( u \): \(u = \frac{gt}{2}\) Substituting the values: \(u = \frac{10 \, \text{m/s}^2 \times 1.2 \, \text{s}}{2} = \frac{12}{2} = 6 \, \text{m/s}\) Thus, the speed of projection is \( 6 \, \text{m/s} \).

P.E at max = K.E on projection = \(\frac{1}{2}\)mv2 = \(\frac{1}{2} \times 0.02 \times 6^2 = 0.36J\)

Capillarity, or capillary action, occurs when a liquid rises in a narrow space due to the adhesive forces between the liquid and the surface of the tube and the cohesive forces within the liquid itself.

\(Power = \frac{\text{Work done}}{time}\) = \(\frac{Force \times distance}{time}\) = \(\frac{mass \times acceleration \times distance}{time}\) = \(\frac{kg \times m \times m}{s^{2} \times s}\) = \(kgm^{2} s^{-3}\)

At different depths, the pressure of a liquid differs.