Question Library

Table of values/observation Number of oscillations (n) = 20 S/N M(g) \(t_1\)(s) \(t_2\)(s) t = \(\frac{t_{1}+t_{2}}{2(s)}\) T = \(\frac{t}{n}\) T\(^{2}\)(sec\(^{2}\)) T\(^{2}\)(sec\(^{2}\))x10\(^{-3}\) 1 50.0 5.80 5.80 5.80 0.290 0.084 84 2 70.0 6.70 6.90 6.80 0.340 0.116 116 3 90.0 7.70 7.90 7.80 0.395 0.156 156 4 110.0 8.50 8.70 8.60 0.430 0.185 185 5 130.0 9.20 9.20 9.20 0.460 0.212 212 (7)The graph of T\(^{2}\)(sec\(^{2}\)) Vs m(g) axis Scale: T\(^{2}\)(sec\(^{2}\)) axis: 2cm = 20 units m(g) axis: 2cm = 20 units (8) Slope/gradient = \(\frac{\bigtriangleup {T}^{2}(sec^{2})}{\bigtriangleup {m(s)}}\) \(\frac{234-10}{140-6} = \frac{224}{134}\) = 1.672 sec\(^{2}\)/g l on the vertical axis is (0) (9) K = \(\frac{4\pi ^{2}}{s} = \frac{4 \times ({\frac{22}{7})^{2}}}{1.672}\) = \(\frac{4 \times (22)^{2}}{7^{2} \times 1.672} = \frac{4 \times 484}{49 \times 1.672}\) = \(\frac{1936}{81.928}\) = 23.61 g/sec\(^{2}\) Precautions are: I ensured that I took care of zero error when taking the period of oscillation. I ensured that there was no conical oscillation when the mass hanger was loaded with different slotted masses. I ensured that I took repeated readings in order to avoid random errors. (b)i. Young modulus is defined as the ratio of tensile or Compressive stress to tensile or compressive strain. It is denoted by E and measured in N/m\(^{2}\). Thus, Young modulus = \(\frac{\text {tensile/compressive stress}}{\text{tensile/compressive strain}}\) E = \(\frac{F/A}{e/L} =\frac{FL}{Ae}\) Force constant is defined as the amount of force that causes a unit extension of an elastic material or the ratio of F/e of an elastic material. It is denoted by K and measured in N/m or N/m\(^{-1}\). Thus, F = Ke, K = \(\frac{f}{e}\) ii. F = 500N, K = 1.0 x 10\(^{4}\) Nm\(^{-1}\), F = ke 500 = 1 x 10\(^{4}\) x e e = \(\frac{500}{1 \times 10^{4}} = \frac{5 \times 10^{2}}{1 \times 10^{4}}\) = 5 x 10\(^{2-4}\) = 5 x 10\(^{-2}\) (0.05)m W = ½Fe = ½ x 500 x 0.05 = 12.5J

Table of values / observation S/N L^\circ e^\circ \(\theta\)^\circ \(\theta\) = i + e 1 5.0 5.0 130.0 10.0 2 10.0 10.0 140.0 20.0 3 15.0 15.0 150.0 30.0 4 20.0 20.0 160.0 40.0 5 25.0 25.0 170.0 50.0 7. Title: The graph of \(\theta\)^\circ vs \(\phi\) Scale: \(\theta\)^\circ axis: 2cm = 20 units \(\phi\) axis: = 2cm = 20 units 8. Slope/gradient = \(\frac{\bigtriangleup \theta ^\circ}{\bigtriangleup \phi ^\circ} = \frac{170-120}{50-0} = \frac{50}{50}\) = 1.0 Intercept on the vertical axis (I) = 120^\circ 9. Precautions are: I ensured that the objective pins and the image pins were erect and in a straight line to avoid parallax error. I ensured that the optical pins were reasonably separated. I ensured that I used a sharp-pointed pencil to obtain neat traces. I ensured that I look directly at the protractor scales when taking my readings to avoid parallax error. (b)i. lt means that the ratio of the speed/wavelength of light in air to glass is 1.5. i.e, n \(\frac{\text {Speed/wavelength in air}}{\text {speed/wavelength in glass}}\) = \(\frac{3}{2}\) = 1.5 pin ii. \(_{a}\)\(n_g\) = \(\frac{1}{\text {sin c}}\), 1.65 = \(\frac{1}{\text {sin c}}\) = Sin C = \(\frac{1}{1.65}\) = 0.6060 C = sin\(^{-1}\) (0.6060), C = 37.3^\circ

Tables of value/observation S/N L 9cm) V (volts) L\(^{-1}\)(cm\(^{-1}\)) V\(^{-1}\)(volts\(^{-1}\)) L\(^{-1}\)(cm\(^{-1}\))x10\(^{-3}\) 1 15.0 0.60 0.067 1.667 67.0 2 25.0 0.80 0.040 1.250 40.0 3 35.0 1.10 0.029 0.909 29.0 4 45.0 1.30 0.022 0.909 22.0 5 55.5 1.60 0.018 0.625 18.0 6 65.0 1.80 0.015 0.556 15.0 (6) The graph of V\(^{-1}\)(volts\(^{-1}\)) vs \(\bigtriangleup\) L\(^{-1}\) (cm\(^{-1}\) Scale: V\(^{-1}\)(volts\(^{-1}\)) axis: 2cm = 0.15units \(\bigtriangleup\) L\(^{-1}\) (cm\(^{-1}\)) axis: 2cm = 7units (7) Slope/gradient = \(\frac{\bigtriangleup {V}^{-1}(volts ^{-1})}{\bigtriangleup {L}^{-1}(cm^{-1})}\) = \(\frac{1.575-0.405}{63-4.2} = \frac{1.17}{58.8}\) = 0.0199cm/volts (8) K = \(\frac{1}{s} = \frac{1}{0.0199}\) = 50.25 volts/cm precautions are: I would ensure that the key is opened when the circuit is not in use to avoid the cell being used up. I would ensure that the electrical wires are tightly connected I would ensure that I take note of the zero error in the pointer instruments to avoid systematic error I would ensure that I take pointer readings within the vertical eye level to avoid parallax I used short connecting wires to ensure that the required energy to dive the electrical charges was not used up in overcoming the resistance of the wire. (b)i. The factors affecting the resistance of a wire are: - length of wire (R&L) - cross-sectional area of the (R \(\propto\)) - Nature of material - Temperature of the wire ii. R = 100cm x 0.02\(\Omega\)cm = 2\(\Omega\), v =1.5v t = 1min = 60s H =p x t = \(\frac{V^{2}t}{R} = \frac{(1.5)^{2}}{2}\) x 60 = 67.5J OR I = \(\frac{V}{R} = \frac{1.5}{2}\) = 0.75A H = P x t = I\(^{2}\)Rt = (0.7)\(^{2}\) x 60 x 2 = \(\frac{9}{16}\)x120 = 67.5J

Option D is the graph of stopping potential Vo against the frequency of the incident light I. The straight line graph cuts the intercept I = threshold frequency. As the frequency of the incident light increases above this point(threshold frequency), the stopping potential corresponding to the kinetic energy of the photoelectron also increases.

To calculate the amount of energy released when 0.5 kg of uranium is completely burnt, we use Einstein's mass-energy equivalence formula: \(E = mc^2\) Given: Mass of uranium (\( m \)) = 0.5 kg, Speed of light (\( c \)) = \( 3 \times 10^8 \, \text{m/s} \) Substituting the values into the formula: \(E = 0.5 \, \text{kg} \times (3 \times 10^8 \, \text{m/s})^2\) Calculating \( c^2 \): \((3 \times 10^8)^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2\) Now, substituting back: \(E = 0.5 \times 9 \times 10^{16} = 4.5 \times 10^{16} \, \text{J}\) The amount of energy released when 0.5 kg of uranium is burnt completely is \(E = 4.5 \times 10^{16} \, \text{J}\)

To determine the deceleration of the motorcyclist, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance: \(v^2 = u^2 + 2as\) Where: \( v \) = final velocity (0 m/s, since he stops at the traffic light), \( u \) = initial velocity (30 m/s), \( a \) = acceleration (deceleration, which will be negative) \( s \) = distance (50 m) Substituting the values into the equation, we rearrange it to solve for \( a \): \(0 = (30)^2 + 2a(50)\) Calculating \( (30)^2 \): \(0 = 900 + 100a\) Now, rearranging for \( a \): \(100a = -900\) \(a = -\frac{900}{100} = -9 \, \text{m/s}^2\) The deceleration of the motorcyclist is \( \mathbf{9 \, m/s^2} \).

Equimolar solution implies all the aqueous solutions have the same concentration. The pH scale ranges from 0 to 14, with 7 being neutral. Values below 7 indicate acidity, and values above 7 indicate basicity. Among all the aqueous solutions we have in the option, NaOH has the highest value, thus it has the highest pH. NaOH is a strong base, meaning it completely dissociates into ions (Na+ and OH-) in water, resulting in a high concentration of hydroxide ions (OH-) and a high pH.

To get the quantity of electricity for discharge at an electrode for each ions, we multiply the number of electrons by the number of moles, since 1 mole = 1F 4.0 moles of Cl\(^-\) → 1e\(^-\) x 4 = 4F 3.0 moles of Na\(^+\) → 3e\(^-\) x 1 = 3F 2.5 moles of Cu\(^{2+}\) → 2e\(^-\) x 2.5 = 5F 2.0 moles of Al\(^{3+}\) → 3e\(^-\) x 2 = 6F. The Ions that require the largest quantity of electricity for discharge at an electrode is the Al\(^{3+}\).

Phenolphthalein, a common acid-base indicator, remains colourless in acidic solutions but turns pink in basic solutions.

The rusting of iron is a spontaneous reaction, meaning it occurs naturally without external energy input, as iron oxidizes to form rust (iron oxide) when exposed to oxygen and moisture. Although, it is a slow process, but there is an increase in entropy (\DeltaS) as well as the emission of heat which makes it a spontaneous reaction as the resulting change in Gibbs free energy will be negative.i.e \DeltaG = - ve

2H\(_2\) + O\(_2\) → 2H\(_2\)O \DeltaH = -571kJ For Hydrogen, Number of moles = \(\frac{mass}{Molar mass}\) Stoichiometrically, 2 = \(\frac{mass}{2}\) Mass of Hydrogen = 2 x 2 = 4g If 4.0g of Hydrogen burns in O\(_2\) to produce - 571 kJ of energy ∴ 12.0g of Hydrogen will burn in O\(_2\) to produce x kJ of energy x = \(\frac{{12}\times{- 571}}{4}\) = - 1713kJ

When sulphur dioxide (SO₂) reacts with a solution of sodium hydroxide (NaOH), it forms sodium sulphite (Na₂SO₃) and water (H₂O). The balanced chemical equation for this reaction is: SO\(_2\)(g) + 2NaOH(aq) → Na\(_2\)SO\(_3\)(aq) + H\(_2\)O(l). When the gas is passed for a long time, the solution becomes saturated with SO\(_2\)(excess). The initially formed sodium sulfite reacts further with the excess SO2 and water to produce sodium hydrogen sulphite, NaHSO\(_3\)(Sodium bisulphite) Na\(_2\)SO\(_3\) + H\(_2\)O + SO\(_2\) → 2NaHSO\(_3\)

NO\(_{2} ^{-}\) = - 1 N\(O_2\) = - 1 N + (-2 x 2) = - 1 N - 4 = - 1 N = - 1 + 4 = + 3 IUPAC name of NO\(_{2} ^{-}\) is dioxonitrate(III) ion

When an aluminum spoon reacts with a solution of iron(II) trioxonitrate(V) (Fe(NO3)2), aluminum, being more reactive than iron, displaces iron from its compound, resulting in the formation of solid iron and aluminum nitrate solution. The chemical reaction can be represented as: 2Al(s) + 3Fe(NO\(_3\))\(_2\)(aq) → 3Fe(s) + 2Al(NO\(_3\))\(_3\)(aq)

The enzyme that converts sucrose into two simple sugars, glucose and fructose, is called sucrase (also known as invertase). Sucrase is a specific enzyme that catalyzes the hydrolysis (breaking down with water) of sucrose, a disaccharide (a sugar made of two monosaccharides), into its two constituent monosaccharides (simple sugars): glucose and fructose. Invertase is another name for sucrase, and it is important in plants for metabolizing sucrose into simpler sugars for growth and energy.