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What are the solutions for questions 17 through 21, including the equivalent resistance calculation?

Physics

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Step-by-step solution

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Here are the solutions for questions 17 through 21, and the equivalent resistance calculation.

  1. Given: Focal length, f=20 cmf = 20 \text{ cm}
  1. Convert the focal length from centimeters to meters.

    $$
    f = 20 \text{ cm} = 0.20 \text{ m}
    $$
    
  2. Use the formula for the power of a lens (PP), which is the reciprocal of the focal length in meters.

    For a converging lens, the focal length is positive. P=1f P = \frac{1}{f}

  3. Substitute the value of ff and calculate the power.

    $$
    P = \frac{1}{0.20 \text{ m}}
    $$
    $$
    P = +5.0 \text{ D}
    $$
    The correct option is **B**.  
    The final answer is $\boxed{+5.0 \text{ D}}$.
    

    18. Given: Radius of curvature, R=2 mR = 2 \text{ m}

  4. Recall the relationship between the focal length (ff) and the radius of curvature (RR) for a mirror.

    $$
    f = \frac{R}{2}
    $$
    
  5. For a convex mirror, the focal length is conventionally negative.

    $$
    f = -\frac{R}{2}
    $$
    
  6. Substitute the value of RR and calculate the focal length.

    $$
    f = -\frac{2 \text{ m}}{2}
    $$
    $$
    f = -1 \text{ m}
    $$
    The correct option is **C**.  
    The final answer is $\boxed{-1 \text{ m}}$.
    

    19. Ohm's law states that the voltage (VV) across a conductor is directly proportional to the current (II) flowing through it, provided all physical conditions and temperature remain constant. The constant of proportionality is the resistance (RR). The correct expression for Ohm's law is V=IRV = IR. The correct option is A. The final answer is V=IR\boxed{V = IR}.

    1. The resistance (RR) of a wire is given by the formula R=ρLAR = \rho \frac{L}{A}, where ρ\rho is the resistivity of the material, LL is the length of the wire, and AA is its cross-sectional area. From this formula, resistance is directly proportional to length. Therefore, an increase in length will increase the resistance of the wire. The correct option is D. The final answer is increase in length\boxed{\text{increase in length}}.

    2. The SI unit of electric current is the ampere (A). The correct option is B. The final answer is ampere\boxed{\text{ampere}}.

    For the equivalent resistance calculation shown in the image:
    The circuit shows three 2Ω2 \Omega resistors connected in parallel.

  7. Use the formula for equivalent resistance (ReqR_{eq}) of resistors in parallel.

    1Req=1R1+1R2+1R3\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
  8. Substitute the given resistance values.

    1Req=12Ω+12Ω+12Ω\frac{1}{R_{eq}} = \frac{1}{2 \Omega} + \frac{1}{2 \Omega} + \frac{1}{2 \Omega}
  9. Add the fractions.

    1Req=1+1+12Ω\frac{1}{R_{eq}} = \frac{1+1+1}{2 \Omega} 1Req=32Ω\frac{1}{R_{eq}} = \frac{3}{2 \Omega}
  10. Invert the result to find ReqR_{eq}.

    Req=23ΩR_{eq} = \frac{2}{3} \Omega
  11. Convert the fraction to a decimal.

    Req≈0.666...ΩR_{eq} \approx 0.666... \Omega Req≈0.67ΩR_{eq} \approx 0.67 \Omega

    The calculation shown in the image for ReqR_{eq} is incorrect in its initial setup (1−12+12+12Ω1 - \frac{1}{2} + \frac{1}{2} + \frac{1}{2} \Omega), but the final numerical value of 0.67Ω0.67 \Omega is correct for three 2Ω2 \Omega resistors in parallel.

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