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Pressure in liquid = e x h x g Pressure in solid = \(\frac{force}{area}\) \(\frac{force}{area}\) = e x h x g force = 0.5 x 10 x 200 x 1.03 = 1030N

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

\(\frac{P_1}{T_1}\) = \(\frac{P_2}{T_2}\) at constant volume \(\frac{150}{27 + 273}\) = \(\frac{P_1}{57 + 273}\) P2 = 165cmHg

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

Using, V2 = U2 + 2gs V2 = 02 + (2 x 10 x 10) V = 10√2ms-1

Density Bottle (Pycnometer): This is a specialized laboratory flask designed to have a very precise volume. It is the standard tool for accurately determining the relative density of liquids and fine solids like powders or granules

m = Itz density = \(\frac{\text{mass}}{\text{area x thickness}}\) Density = \(\frac{Itz}{A \times h}\) \(10^4 = \frac{0.5 \times 100 \times 60 \times 0.0003}{100 \times h}\) h = 9 x 10-7cm = 9 x 10-6mm

F =\(\frac{m(v - u)}{T}\) = \(\frac{0.01(5000 - 0)}{1}\) = 50N

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

Heat energy = Electrical energy mL = power x time 0.005 x L = 50 x 50 L = 5 x 105J/kg

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

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.

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 B.