Question Library

1\muf and 5\muf are parallel; Ceff = 1 + 5 = 6\muf 6\muf, 4\muf and 4\muf are in series. total capacitance of the circuit in series( \(\frac{1}{C_T}\)) = \(\frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}\) = \(\frac{1}{4} + \frac{1}{4} + \frac{1}{6}\) = \(\frac{ 6 + 6 + 4}{ 24}\) = \(\frac{ 16}{ 24}\) \(\frac{1}{C_T}\) = \(\frac{ 16}{ 24}\) \(C_T\) = \(\frac{24}{ 16}\) = 1.5\muf.

The 2\(\Omega\) resistors are in parallel R\(_{para}\) = \(\frac{R_1 \times R_2}{R_1 + R_2}\) = \(\frac{2 \times 2}{2 + 2}\) = \(\frac{4}{4}\) = 1 \(\Omega\) This 1 \(\Omega\) resistor is in series with both the 4 \(\Omega\) and 3 \(\Omega\) resistors Therefore, R\(_{total}\) = 1 + 4 + 3 = 8 \(\Omega\)

2\sqrt13N in a direction N 56◦ E

The downward forces acting on the system are the weight of the man (800 N) and the weight of the beam (200 N). Thus, the total downward force is: \(800 \, \text{N} + 200 \, \text{N} = 1000 \, \text{N}\) The total upward forces are represented by reactions \( P \) and \( Q \): \(P + Q = 1000 \, \text{N}\) Since the beam is uniform, its weight acts at the center, which is located 2.5 m from either end. To satisfy the condition of equilibrium, the total clockwise moments must equal the total anticlockwise moments. Taking moments about point \( Q \): \((200 \, \text{N} \times 2.5 \, \text{m}) + (800 \, \text{N} \times 2 \, \text{m}) = P \times 5 \, \text{m}\) Calculating the moments: \(500 \, \text{N}\cdot\text{m} + 1600 \, \text{N}\cdot\text{m} = 5P\) This simplifies to: \(2100 \, \text{N}\cdot\text{m} = 5P\) Solving for \( P \): \(P = \frac{2100}{5} = 420 \, \text{N}\) Now, substituting \( P \) back into the upward force equation: \(420 \, \text{N} + Q = 1000 \, \text{N}\) Solving for \( Q \): \(Q = 1000 - 420 = 580 \, \text{N}\) Thus, the reactions at \( P \) and \( Q \) are: 420N and 580N respectively.

The value of the resistor R = \(\frac{V_{fs}}{I_s}\) - R\(_g\) R = \(\frac{15}{0.015}\) - 5 R = 1000 - 5 = 995\(\Omega\)

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

Given: - \( m_1 = 10 \, \text{kg} \) (on incline) - \( m_2 = 15 \, \text{kg} \) (hanging) - \( \theta = 30^\circ \) Inclined mass: \( F_{\text{incline}} = m_1 g \sin(\theta) = 10g \cdot 0.5 = 5g\) Hanging mass: \(F_{\text{weight}} = m_2 g = 15g\) Net Force: \(F_{\text{net}} = F_{\text{weight}} - F_{\text{incline}} = 15g - 5g = 10g\) Total Mass \(m_{\text{total}} = m_1 + m_2 = 25 \, \text{kg}\) Using \( F_{\text{net}} = m_{\text{total}} \cdot a \): \(10g = 25a \quad \Rightarrow \quad a = \frac{10g}{25} = \frac{2g}{5}\) The acceleration is: \(a = \frac{2g}{5} \, \text{m/s}^2\)

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

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

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 C.

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

During the electrolysis of NaCl using carbon electrodes, hydrogen is usually deposited at the cathode because hydrogen is less reactive and lower than sodium in the electrochemical series(sodium and hydrogen are the two cations in the solution)

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 E.