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\(\frac{V_s}{V_p}\) = \(\frac{N_s}{N_p}\) = \(\frac{I_p}{I_s}\) \(\frac{100}{400}\) = \(\frac{V_S}{12}\) V = 3V

Work function = 5 - 2 = 3eV

Capacitors in parallel = 4\(\mu\)f + 2 \(\mu\) f = 6\(\mu\) f Capacitors in series =\(\frac{1}{C_{eff}}\) = \(\frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \frac{1}{C_4}\) \(\frac{1}{C_{eff}}\) = \(\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \frac{1}{4}\) = \(\frac{6 + 4 + 2 + 3}{12}\) \(\frac{1}{C_{eff}}\) = \(\frac{15}{12}\) C\(_{eff}\) = \(\frac{12}{15}\) = 0.8\(\mu\) f

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

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

The total upward vertical force must balance the downward weight of 20 N. There are two vertical components: 2 \times (T \times sin(30^\circ)) = 20 N Sin(30^\circ) = \(\frac{1}{2}\) = 0.5 So: 2 \times T \times 0.5 = 20 T = 20

M\(_2\)g - M\(_2\)a = M\(_1\)a + M\(_1\)g Sin\(\theta\) (30 x 10) - 30a = 20a + 20 x 10 x Sin 30 300 - 30a = 20a + 100 - 30a - 20a = 100 - 300 - 50a = - 200 a = \(\frac{-200}{-50}\) = 4m/s\(^2\)

Force required to move the object forward = Fcos 60 = 300 x 0.5 = 150N Frictional Force = 100N Total force acting on the object = 150 - 100 = 50N Work Done = Total force x distance = 50 x 4 = 200J.

Diagram above shows a Parallel circuit If one of the parallel path is broken, current will continue to flow in other parts. Voltage is the same across each component of the parallel circuit. The brightness of a light bulb is given by its power. P = I\(^2\)R, and so brightness depends on current and resistance. Hence; the brightness of Y will not be affected except for a variation in resistance property.

Chromatography is a technique that separates the components of a mixture based on their different rates of migration through a stationary phase and a mobile phase. Mobile phase: The moving part of the chromatography setup. Stationary phase: The non-moving part of the chromatography setup. The rate at which a component moves through the chromatography setup depends on how well it absorbs to the stationary phase and how soluble it is in the mobile phase.

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

First write a dissociation equation: H\(_2\)SO\(_4\) ⇌ 2H\(^+\) + SO\(_4\)\(^{2-}\) Number of moles = \(\frac{Number of ions}{Avogadro number}\) But we need to get the number of moles first by applying n = \(\frac{m}{M}\) n = \(\frac{4.9}{98}\) (Molar mass of H\(_2\)SO\(_4\) = 98g/mol) n = 0.05 mole Number of moles = \(\frac{Number of ions}{Avogadro number}\) Number of hydrogen ions = ( n x L ) X 2 → We multiplied by 2 because 2H\(^+\) were formed. = 0.05 x 2 x 6.02 x 10\(^{23}\) = 6.02 x 10\(^{22}\) ions

First write a balanced equation for the chemical reaction: 2H\(_2\) + O\(_2\) → 2H\(_2\)O Mole ratio: 2 moles : 1 mole Expected reacted volume: 8 cm\(^3\) : 20 cm\(^3\) Actual reacted volume: 8 cm\(^3\) : 4 cm\(^3\) Residual Volume(Expected - Actual): 0 : 16 cm\(^3\) Therefore, the volume of oxygen remaining is 16 cm\(^3\)

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

Avogadro's Law states that, at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas. In simpler terms, equal volumes of different gases contain the same number of molecules when they are at the same temperature and pressure. Since both Oxygen and Nitrogen contain the same volume of gases at the same temperature and pressure, their number of moles will also be the same. i.e If Nitrogen contains 5.0 moles, then Oxygen will also contain 5.0moles. Thus, number of moles of oxygen, n = 5.0 moles Molar mass of O\(_2\) = 16 x 2 = 32g/mol But n = \(\frac{m}{M}\) m = n x M = 5 X 32 = 160.0g