Question Library

Answer: A. air (Air is a poor conductor ~0.025 W/mK; water ~0.6, glass ~1, ash varies but is higher.) Air is a poor conductor of heat compared to solids and liquids like ash, glass, and water, making it a good insulator.

Answer: D Food stores energy in chemical bonds, released during metabolism.

Answer: A Energy in an inductor: E = \(\frac{1}{2}\)LI\(^2\) ⇒ I ≈ 1.29 A.

\(\text{Efficiency} = \frac{\text{Output Work}}{\text{Input Work}}\) Given: \(\text{Efficiency} = 80\% = 0.8\), \(\text{Input Work} = 200 \text{ J}\) Then, the output work is calculated as follows: \(\text{Output Work} = 0.8 \times 200 = 160 \, \text{J}\) Thus, the output work is: \(\text{Output Work} = 160 \, \text{J}\).

Answer: A Perforated materials absorb sound, reducing reverberation.

Answer: A An object sinks when its density is greater than that of water.

Brownian motion is the random motion of particles in fluids.

Answer: D To calculate the mass of silver deposited during electrolysis, we use the formula: \(m = \frac{Q \cdot M}{n \cdot F}\) Where: \( I = 0.8 \, \text{A} \), \( t = 25 \, \text{minutes} = 25 \times 60 = 1500 \, \text{s} \), \( Q = I \cdot t = 0.8 \, \text{A} \cdot 1500 \, \text{s} = 1200 \, \text{C} \) \( M = 107.87 \, \text{g/mol} \), \( n = 1 \), \( F \approx 96485 \, \text{C/mol} \) Substituting the values: \(m = \frac{1200 \, \text{C} \cdot 107.87 \, \text{g/mol}}{1 \cdot 96485 \, \text{C/mol}} = \frac{129444}{96500} \approx 1.34 \, \text{g}\)

The thermal capacity (or heat capacity) of a body is the amount of heat required to raise its temperature by 1^\circC (or 1 K). It is calculated as: C = m.c where: ( m ) is the mass of the body, ( c ) is the specific heat capacity, which depends on the nature (material or substance) of the body. Thus, thermal capacity depends on both the mass and the nature of the material.

The ideal gas law is \( PV = nRT \). Making \( P \) the subject: \[ P = \frac{nRT}{V} \] Substitute the values: - \( n = 8.3 \, \mathrm{mol} \), \( R = 8.31 \, \mathrm{J \, K^{-1} \, mol^{-1}} \), \( T = 17^\circ \mathrm{C} = 290 \, \mathrm{K} \) - \( V = 4.5 \, \mathrm{m}^3 \) \( P = \frac{8.3 \times 8.31 \times 290}{4.5} = 4444.93 \, \mathrm{Pa} \)

Specific gravity (SG) is the ratio of the density of the liquid (\(\rho_l\)) to the density of water (\(\rho_w\)): \( \mathrm{SG} = \frac{\rho_l}{\rho_w} \) Rearranging for density: \((\rho_l = \mathrm{SG} \times \rho_w = 0.76 \times 1000 = 760 \, \mathrm{kg \, m^{-3}}\)

\(C = \frac{Q}{\Delta T} = \frac{48000 \, \text{J}}{53 \, \text{ºC}} \approx 905.66 \, \text{J/ºC}\)

\(F = m \cdot g = 1.5 \, \text{kg} \times 10 \, \text{m/s}^2 = 15 \, \text{N}\) \(A = \pi r^2 = \pi (0.003 \, \text{m})^2 = \pi (9 \times 10^{-6}) \approx 2.827 \times 10^{-5} \, \text{m}^2\) \(\text{Stress} = \frac{F}{A} = \frac{15 \, \text{N}}{2.827 \times 10^{-5} \, \text{m}^2} \approx 5.299 \times 10^5 \, \text{Pa}\)

The apparent depth of an object in water (as viewed from air) is less than the real depth due to refraction. The relationship is given by: \(\mu = \frac{\text{Real depth}}{\text{Apparent depth}}\) where \(\mu\) is the refractive index of water (\(\mu = 1.33\)). Rearranging for real depth: \(\text{Real depth} = \mu \times \text{Apparent depth} = 1.33 \times 10 \, \text{cm} = 13.3 \, \text{cm}\)

\(B = \frac{F}{q \cdot v \cdot \sin \theta}\) \(\theta = 45^\circ \implies \sin 45^\circ = \frac{\sqrt{2}}{2}\) Substituting the values: \(B = \frac{6}{(5 \times 10^{-6})(3 \times 10^4) \left(\frac{\sqrt{2}}{2}\right)}\) Simplifying: \(B = \frac{6}{(15 \times 10^{-2}) \left(\frac{\sqrt{2}}{2}\right)} \approx \frac{6}{0.1061} \approx 56.5 \, \text{T}\)