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In an electrolytic cell, electric current enters and leaves the cells through

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6. In an electrolytic cell, electric current enters and leaves the cells through
In an electrolytic cell, electric current is carried by electrons in the external circuit and by ions within the electrolyte. The points of entry and exit for the current into and out of the electrolyte are the electrodes.

  • The correct option is B.

7. A weightlifter lifts a weight of 2000 N2000 \text{ N}. If he holds the weight above his head, what form of energy does the weight have?
When an object is lifted to a certain height and held there, it possesses potential energy due to its position in the Earth's gravitational field.

  • The correct option is D.

8. All of the following constants are dimensional except
A dimensional constant has units, while a dimensionless constant does not.

  • Planck's constant (hh) has units of J⋅s\text{J} \cdot \text{s}.
  • Gravitational constant (GG) has units of N⋅m2/kg2\text{N} \cdot \text{m}^2/\text{kg}^2.
  • Speed of light in a vacuum (cc) has units of m/s\text{m/s}.
  • Refractive index is the ratio of the speed of light in a vacuum to the speed of light in a medium, so it is a ratio of two speeds and thus has no units (it is dimensionless).
  • The correct option is D.

9. SONAR is a device used by submarines to detect underwater objects and water depth. The primary characteristic utilized by this device is
SONAR (Sound Navigation And Ranging) operates by emitting sound waves and detecting the echoes that reflect off objects or the seabed.

  • The correct option is B.

10. The lost volt of a cell is the potential difference across
The "lost volt" refers to the voltage drop that occurs within the cell itself due to its internal resistance when current flows. It is the potential difference across the internal resistance of the cell.

  • The correct option is C.

11. A girl observes an elevated image of a piece of stone at the bottom of a swimming pool 6 m6 \text{ m} deep, when she looks vertically downward. Determine the apparent displacement of the stone from the bottom of the pool. [refractive index of water = 1.33]

  1. Identify the given values.

    Real depth, dreal=6 md_{real} = 6 \text{ m}
    Refractive index of water, n=1.33n = 1.33

  2. Calculate the apparent depth using the formula n=drealdapparentn = \frac{d_{real}}{d_{apparent}}.

    dapparent=drealnd_{apparent} = \frac{d_{real}}{n} dapparent=6 m1.33d_{apparent} = \frac{6 \text{ m}}{1.33} dapparent≈4.51 md_{apparent} \approx 4.51 \text{ m}

  3. Calculate the apparent displacement.

    Apparent displacement = Real depth - Apparent depth

    Apparent displacement=6 m−4.51 m=1.49 m\text{Apparent displacement} = 6 \text{ m} - 4.51 \text{ m} = 1.49 \text{ m}

    Rounding to one decimal place, the displacement is 1.5 m1.5 \text{ m}.

    • The correct option is A.

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

  4. Recall the lens formula.

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

  5. Combine the terms on the right side.

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

  6. Invert both sides to solve for ff.

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

  7. Analyze the given graph.

    The y-axis is uvuv and the x-axis is u+vu+v. The slope of this graph is Δ(uv)Δ(u+v)\frac{\Delta(uv)}{\Delta(u+v)}.

  8. Compare the slope to the focal length formula.

    The slope of the graph, uvu+v\frac{uv}{u+v}, is equal to the focal length ff.

    • The correct option is C.

    13. If the temperature of a piece of metal of mass 100 g100 \text{ g} and specific heat capacity of 460 J kg−1 K−1460 \text{ J kg}^{-1} \text{ K}^{-1} falls from 70 ∘C70 \text{ }^\circ\text{C} to 20 ∘C20 \text{ }^\circ\text{C}, calculate the quantity of heat released.

  9. Identify the given values and convert units if necessary.

    Mass, m=100 g=0.1 kgm = 100 \text{ g} = 0.1 \text{ kg}
    Specific heat capacity, c=460 J kg−1 K−1c = 460 \text{ J kg}^{-1} \text{ K}^{-1}
    Initial temperature, T1=70 ∘CT_1 = 70 \text{ }^\circ\text{C}
    Final temperature, T2=20 ∘CT_2 = 20 \text{ }^\circ\text{C}

  10. 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{ }^\circ\text{C} - 20 \text{ }^\circ\text{C} = 50 \text{ }^\circ\text{C}

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

  11. Calculate the quantity of heat released using the formula Q=mcΔTQ = mc\Delta T.

    Q=(0.1 kg)×(460 J kg−1 K−1)×(50 K)Q = (0.1 \text{ kg}) \times (460 \text{ J kg}^{-1} \text{ K}^{-1}) \times (50 \text{ K}) 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} Q=2.3×103 JQ = 2.3 \times 10^3 \text{ J}
    • The correct option is A.
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