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Read the main scale reading.

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Answer

C

Step-by-step solution

7 steps

For question 28:

  1. Read the main scale reading.

    The main scale shows markings in millimetres. The last visible mark on the main scale before the thimble is 5 mm5 \text{ mm}. There is also a half-millimetre mark visible after the 5 mm5 \text{ mm} mark.
    Main scale reading = 5.5 mm5.5 \text{ mm}

  2. Read the thimble scale reading.

    The thimble scale shows divisions from 0 to 50. The line on the thimble scale that aligns with the main scale line is 32.
    Thimble scale reading = 32×0.01 mm=0.32 mm32 \times 0.01 \text{ mm} = 0.32 \text{ mm}

  3. Calculate the total reading.

    Total reading = Main scale reading + Thimble scale reading
    Total reading = 5.5 mm+0.32 mm=5.82 mm5.5 \text{ mm} + 0.32 \text{ mm} = 5.82 \text{ mm}

    The correct option is C.

    C 5.82 mm\boxed{\text{C } 5.82 \text{ mm}}

    For question 29:

    The unit of force is the newton (N).

    • A watt (W) is the unit of power.
    • A joule (J) is the unit of energy or work.
    • An ampere (A) is the unit of electric current.

    The correct option is D.

    D newton\boxed{\text{D newton}}

    For question 30:

  4. Identify the given values.

    Load (weight of the object), W=900 NW = 900 \text{ N}
    Height raised, h=1 mh = 1 \text{ m}
    Effort applied, E=300 NE = 300 \text{ N}
    Length of the inclined plane, L=4 mL = 4 \text{ m}

  5. Calculate the work output (useful work).

    Work output = Load ×\times Vertical height

    Wout=W×h=900 N×1 m=900 JW_{out} = W \times h = 900 \text{ N} \times 1 \text{ m} = 900 \text{ J}
  6. Calculate the work input (work done by effort).

    Work input = Effort ×\times Distance moved along the plane

    Win=E×L=300 N×4 m=1200 JW_{in} = E \times L = 300 \text{ N} \times 4 \text{ m} = 1200 \text{ J}
  7. Calculate the efficiency of the inclined plane.

    Efficiency (η\eta) is the ratio of work output to work input, expressed as a percentage.

    η=WoutWin×100%\eta = \frac{W_{out}}{W_{in}} \times 100\% η=900 J1200 J×100%\eta = \frac{900 \text{ J}}{1200 \text{ J}} \times 100\% η=0.75×100%\eta = 0.75 \times 100\% η=75%\eta = 75\%

    The correct option is D.

    D 75%\boxed{\text{D } 75\%}
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