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Q1. State two functions each of the following structures during pregnancy in humans: (i) Placenta (ii) Amniotic fluid. Explain each of the following terms: (i) Nursery bed (ii) Cultural practices (iii

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Q1a. State two functions each of the following structures during pregnancy in humans:

(i) Placenta

  • Facilitates the exchange of nutrients, oxygen, and waste products between the mother and the fetus.
  • Produces hormones such as progesterone and estrogen, which are essential for maintaining pregnancy.

(ii) Amniotic fluid

  • Protects the fetus from physical shock and injury by cushioning it.
  • Helps maintain a constant temperature around the fetus.

Q1b. Explain each of the following terms:

(i) Nursery bed
A nursery bed is a specially prepared plot of land or a container where young plants (seedlings) are grown and cared for during their early stages before being transplanted to their permanent growing location.

(ii) Cultural practices
Cultural practices refer to the traditional customs, beliefs, and behaviors that are shared by a particular group of people or society, often passed down through generations, influencing various aspects of their lives.

(iii) Transplanting
Transplanting is the process of moving a plant from one location to another, typically from a nursery bed or a smaller pot to a larger pot or its final growing site in a field or garden.

Q1c. Balance the following nuclear equations:

(i)

  1. Balance the mass numbers (superscripts).

    226=222+4226 = 222 + 4
  2. Balance the atomic numbers (subscripts).

    P=86+2P = 86 + 2 P=88P = 88

    The balanced equation is:

    88226Ra→86222Rn+24He\boxed{_{88}^{226}Ra \to _{86}^{222}Rn + _2^4He}

    (ii)

  3. Balance the mass numbers (superscripts).

    247=A+0247 = A + 0 A=247A = 247
  4. Balance the atomic numbers (subscripts).

    Z=97+(−1)Z = 97 + (-1) Z=96Z = 96

    The balanced equation is:

    96247Cm→97247Th+−10e\boxed{_{96}^{247}Cm \to _{97}^{247}Th + _{-1}^0e}

    (Note: The element with atomic number 97 is Berkelium (Bk), not Thorium (Th). The symbol 'Th' in the question is inconsistent with the atomic number 97.)

    (iii)

  5. Balance the mass numbers (superscripts).

    59=59+059 = 59 + 0
  6. Balance the atomic numbers (subscripts).

    27=28+(−1)27 = 28 + (-1)

    The balanced equation is:

    2759Co→2859NL+−10e\boxed{_{27}^{59}Co \to _{28}^{59}NL + _{-1}^0e}

    (Note: The element with atomic number 28 is Nickel (Ni). 'NL' is not a standard element symbol.)

    Q1d. State the type of radiation emitted in each reaction in question c:

    (i) The emitted particle is 24He_2^4He, which is an alpha particle. Therefore, the type of radiation is alpha decay.

    (ii) The emitted particle is −10e_{-1}^0e, which is a beta particle. Therefore, the type of radiation is beta decay.

    (iii) The emitted particle is −10e_{-1}^0e, which is a beta particle. Therefore, the type of radiation is beta decay.

    Q1e.

    (i) Define light energy.
    Light energy is a form of electromagnetic radiation that is visible to the human eye. It travels in waves and consists of photons, carrying energy that can be perceived as brightness and color.

    (ii) An electrical appliance is rated 1500W, 240V and left running for 2 hours. Calculate the:

    (a) Current drawn

  7. Identify the given values.

    Power (PP) = 1500 W1500 \text{ W}
    Voltage (VV) = 240 V240 \text{ V}

  8. Use the formula for power, P=V×IP = V \times I, to find the current (II).

    I=PVI = \frac{P}{V} I=1500 W240 VI = \frac{1500 \text{ W}}{240 \text{ V}} I=6.25 AI = 6.25 \text{ A}

    The current drawn is 6.25 A\boxed{6.25 \text{ A}}.

    (b) Electrical energy consumed

  9. Identify the given values.

    Power (PP) = 1500 W1500 \text{ W}
    Time (tt) = 2 hours2 \text{ hours}

  10. Calculate energy consumed using the formula E=P×tE = P \times t.

    To express energy in Joules (J), convert time to seconds:

    t=2 hours×3600 s/hour=7200 st = 2 \text{ hours} \times 3600 \text{ s/hour} = 7200 \text{ s} E=1500 W×7200 sE = 1500 \text{ W} \times 7200 \text{ s} E=10,800,000 JE = 10,800,000 \text{ J} E=1.08×107 JE = 1.08 \times 10^7 \text{ J}

    Alternatively, to express energy in kilowatt-hours (kWh), convert power to kilowatts:

    P=1500 W=1.5 kWP = 1500 \text{ W} = 1.5 \text{ kW} E=1.5 kW×2 hoursE = 1.5 \text{ kW} \times 2 \text{ hours} E=3.0 kWhE = 3.0 \text{ kWh}

    The electrical energy consumed is 1.08×107 J\boxed{1.08 \times 10^7 \text{ J}} or 3.0 kWh\boxed{3.0 \text{ kWh}}.

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