SCIENCE 1

SIXTH FORM ENTRANCE TEST - SCIENCE

GENERAL INFORMATION

  • Entrance test for Lower Sixth (Year 12) starting September 2022.

  • Duration: 1 Hour.

  • Instructions for candidates:

    • Spend 5 minutes reading the papers.

    • Complete any two whole subject sections from Physics, Chemistry, and Biology.

    • Questions can be attempted in any order.

    • Name must be written on every section.

    • Two out of three sections must be completed.

    • All questions in the two chosen sections must be answered.

    • Advisable time for each section: 25 minutes.

    • Each section is worth 20 marks.

    • A calculator may be used throughout.

BIOLOGY SECTION

Candidate Information:

  • Name (in capitals): ___

  • Candidate number:

  • Current school:

  • Current science syllabus:
    (e.g., triple or double science; type of qualification; any science not currently studied)
    Total Marks: 20

Question 1: Enzyme Activity

  1. Trypsin: an enzyme that catalyzes the breakdown of proteins into amino acids.

    • Experiment: 9 cm³ of liquid egg white (protein) mixed with 1 cm³ of 1% trypsin in a small beaker.

    • pH Measurement: Initial pH of the solution was 7.5.

      • a) Graph Sketching: Sketch a graph to show the change of pH over 30 minutes. [3 Marks]

Investigation of Enzyme Concentration

  • Experiment Repetition: Different concentrations of trypsin ranging from 1% to 5%.

    • b) Controlled Variable: State and describe how one variable should be controlled to obtain valid results. [2 Marks]

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Question 2: Photosynthesis

  1. Apparatus Diagram: Measures the rate of photosynthesis in pond weed.

    • a) Gas Produced: Identify the name of the gas produced in this experiment. [1 Mark]

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    • b) Methodology: Describe how to use this apparatus to measure the rate of photosynthesis. [2 Marks]

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Light Intensity Investigation

  • A student observed changes in the rate of photosynthesis:

    • Initially increased with light intensity, then plateaued.

    • c) Plateau Reasoning: Suggest one reason for the plateau in the rate of photosynthesis. [2 Marks]

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Question 3: Root Hair Cell Structure

  1. Root Hair Cell: Specialised for absorption of water and minerals from soil.

    • a) Adaptation: Describe how the shape of this cell is adapted to its function. [1 Mark]

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    • b) Magnification: Given cell is magnified 600 times. Calculate the actual length of the cell in µm (1 mm = 1000 µm). [2 Marks]

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    • c) Organelle Statement: State the name of one plant cell organelle not found in this cell. [1 Mark]

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Watering Experiment

  • d) Basil Plant with 1M NaCl Solution:

    • Explain why the plant wilted after watering with this solution. [3 Marks]

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  • e) Waterlogged Basil Plant:

    • Explain why this plant could not absorb minerals like nitrates and phosphates. [3 Marks]

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END OF BIOLOGY SECTION

CHEMISTRY SECTION

Candidate Information:

  • Name (in capitals): ___

  • Candidate number:

  • Current school:

  • Current science syllabus:
    (e.g., triple or double science; type of qualification; any science not currently studied)
    Total Marks: 20

Question 1: Reaction Rate Law

  1. Rate Law Definition: Mathematical expression relating the rate of a reaction to the concentrations of the reactants.

    • Reaction of iodine, I2, with propanone, (CH3)2CO, in dilute acid, H+:

      • Reaction Equation: I<em>2+(CH</em>3)<em>2COightleftharpoonsCH</em>3COCH2I+I−I<em>2 + (CH</em>3)<em>2CO ightleftharpoons CH</em>3COCH_2I + I^-

    • a) Role of H+: Suggest the role of H+ in the reaction (not included in the overall equation). [1 Mark]

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    • Rate Law for the Reaction:

      • extRate=kimesext(concentrationof(CH<em>3)</em>2CO)imesext(concentrationofH+)ext{Rate} = k imes ext{(concentration of (CH<em>3)</em>2CO)} imes ext{(concentration of H^+)}

Measuring the Rate Constant

  • b) Calculating Rate Constant (k): Given rate of reaction is 0.01 mol dm⁻³ s⁻¹, concentration of (CH3)2CO is 0.02 mol dm⁻³, and concentration of H+ is 0.04 mol dm⁻³. Calculate k using the Rate Law. [2 Marks]

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  • c) Rate Calculation: Use calculated k and Rate Law to find rate of reaction at both concentrations 0.06 mol dm⁻³. [2 Marks]

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  • d) Effect of Concentration Change: Explain how doubling the concentration of I2 affects the rate of reaction using the Rate Law. [2 Marks]

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  • e) Mathematical Relationship: Describe the relationship between the rate of reaction and concentration of (CH3)2CO considering H+ concentration as constant. [1 Mark]

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Question 2: Activation Energy

  • Definition of Activation Energy (Ea): The energy barrier to be overcome for a successful reaction.

  • Calculation Methods:
    i. via simultaneous equations
    ii. via graphical methods

    • Activation Energy Equation:

      • extlnrack<em>2k</em>1=racE<em>a8.314(rac1T</em>1−rac1T2)ext{ln} rac{k<em>2}{k</em>1} = rac{E<em>a}{8.314} \bigg( rac{1}{T</em>1} - rac{1}{T_2} \bigg)

      • - Where:

      • k1: rate constant at temperature T1

      • k2: rate constant at temperature T2

      • Temperature: in Kelvin (K)

      • ln: natural logarithm function

    • Experiment Results:

      • 1. Experiment No. with Temperature /K and k values:

      • 293 K; k = 4.2

      • 313 K; k = 12.3

    • a) Calculate ln(k2/k1): .

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    • b) Ea Calculation: Use (a) and Equation 2 to calculate Ea with three significant figures. [3 Marks]

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Graphical Method

  • Plot a graph of ln k vs (1/T) and calculate the gradient.

  • Data Summary:

    • Complete table calculating missing values of T /K, (1/T) /K⁻¹, k, ln k. [2 Marks]

    • c) Data Points to Plot:

      • 1. T: 293 K; (1/T): 0.0034; k: 3.2; ln k: 1.16

      • 2. T: 313 K; (1/T): 0.0032; k: 12.3; ln k: ?

      • 3. T: 333 K; (1/T): ?; k: 20.7; ln k: ?

      • 4. T: 353 K; (1/T): ?; k: 54; ln k: 3.99

    • d) Graph: Plot data on axes with (1/T) on x-axis and ln k on y-axis. [3 Marks]

    • e) Line of Best Fit: Draw through points. [1 Mark]

    • f) Calculate Gradient: The gradient is equal to -Ea/8.314. Use this to calculate Ea with two significant figures. [1 Mark]

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END OF CHEMISTRY SECTION

PHYSICS SECTION

Candidate Information:

  • Name (in capitals): ___

  • Candidate number:

  • Current school:

  • Current science syllabus:
    (e.g., triple or double science; type of qualification; any science not currently studied)
    Total Marks: 20

Question 1: Measuring Kinetic Energy

  1. Definition: Kinetic energy (KE) is the energy due to movement.

    • Formula: EK=rac12mv2E_K = rac{1}{2} mv^2

    • Describe how to measure the maximum kinetic energy of a person sprinting 100 metres:

      • Identify all equipment used. [6 Marks]

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Question 2: Kinetic Energy and Temperature

  1. a) Kinetic Energy and Temperature Relationship: When warmer, atoms exhibit more kinetic energy and vibrate faster.

    • Equation relating temperature in Kelvin (T) to average kinetic energy (Ek):

      • E<em>K=rac32k</em>BTE<em>K = rac{3}{2} k</em>B T

    • Calculate average kinetic energy at 300 Kelvin given average KE at 200 Kelvin is 4.14imes10−21J4.14 imes 10^{-21} J. [2 Marks]

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    • b) Calculate kB using data from previous part. [2 Marks]

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Question 3: Resultant Force and Motion

  1. a) Force and Acceleration: Resultant force causes acceleration.

    • Formula: FR=maF_R = ma

    • Given an 8 kg object experiences 32 N of air resistance while falling in gravitational field of 10 N/kg: Calculate acceleration of object. [3 Marks]

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    • b) Falling Speed Calculation: If the object is falling at 40 m/s, how fast would it be traveling 5 seconds later with constant 32 N air resistance? [2 Marks]

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Question 4: Air Resistance Scenario

  1. a) Falling Object and Air Resistance: A different object dropped from rest experiences air resistance equal to half its velocity.

    • After 12 seconds of falling, its acceleration is 5 m/s².

    • Show that after 12 seconds, it could assume it was traveling at 90 m/s (clearly state any assumptions made). [2 Marks]

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    • b) Mass Calculation: Estimate mass of object based on information from 4a. Show workings clearly. [3 Marks]

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END OF PHYSICS SECTION