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In the diagram below, u is the object distance from a lens and v is the distance of the image produced. The slope of the graph represents the

Physics

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Answer

C

focal length

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12. In the diagram, uu is the object distance and vv is the image distance from a lens. The graph plots UVUV against U+VU+V.

  1. Recall the lens formula.

    The lens formula is given by:

    1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

    where ff is the focal length.

  2. Manipulate the lens formula to match the graph axes.

    Combine the terms on the right side:

    1f=v+uuv\frac{1}{f} = \frac{v+u}{uv}
  3. Rearrange the equation to express uvuv in terms of u+vu+v.

    Invert both sides:

    f=uvu+vf = \frac{uv}{u+v}

    Multiply by (u+v)(u+v):

    uv=f(u+v)uv = f(u+v)

    This equation is in the form y=mxy = mx, where y=uvy = uv, x=u+vx = u+v, and mm is the slope.

  4. Identify what the slope represents.

    Comparing uv=f(u+v)uv = f(u+v) with y=mxy = mx, the slope mm is equal to ff.
    Therefore, the slope of the graph represents the focal length.

    The correct option is C. focal length.

    13. To calculate the quantity of heat released, we use the formula Q=mcΔTQ = mc\Delta T.

  5. Identify the given values and convert units to S.I.

    units where necessary.
    Mass (mm) = 100 g=0.1 kg100 \text{ g} = 0.1 \text{ kg}
    Specific heat capacity (cc) = 460 Jkg−1K−1460 \text{ Jkg}^{-1}\text{K}^{-1}
    Initial temperature (T1T_1) = 70 °C70 \text{ °C}
    Final temperature (T2T_2) = 20 °C20 \text{ °C}

  6. Calculate the change in temperature (ΔT\Delta T).

    ΔT=T1−T2=70 °C−20 °C=50 °C\Delta T = T_1 - T_2 = 70 \text{ °C} - 20 \text{ °C} = 50 \text{ °C}

    Since a change of 1 °C1 \text{ °C} is equivalent to a change of 1 K1 \text{ K}, ΔT=50 K\Delta T = 50 \text{ K}.

  7. Substitute the values into the heat released formula.

    Q=mcΔTQ = mc\Delta T Q=(0.1 kg)×(460 Jkg−1K−1)×(50 K)Q = (0.1 \text{ kg}) \times (460 \text{ Jkg}^{-1}\text{K}^{-1}) \times (50 \text{ K})
  8. Calculate the quantity of heat released.

    Q=0.1×460×50 JQ = 0.1 \times 460 \times 50 \text{ J} Q=46×50 JQ = 46 \times 50 \text{ J} Q=2300 JQ = 2300 \text{ J}
  9. Express the answer in scientific notation and compare with the options.

    Q=2.3×103 JQ = 2.3 \times 10^3 \text{ J}

    The correct option is A. 2.3×103 J2.3 \times 10^3 \text{ J}.

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