HomeLearn Fast select Course
Show Unit/ChapterChange UnitChange SubjectLast Menu
NEET-XII-Physics

33: Thermal and Chemical Effects of Electric Current

with Solutions - page 3
Qstn# iv-4 Prvs-QstnNext-Qstn
  • #4
    A heater coil is to be constructed with a nichrome wire (ρ = 1.0 × 10-6 Ωm) that can operate at 500 W when connected to a 250 V supply. (a) What would be the resistance of the coil? (b) If the cross-sectional area of the wire is 0.5 mm2, what length of the wire will be needed? (c) If the radius of each turn is 4.0 mm, how many turns will be there in the coil? (a) What would be the resistance of the coil? (b) If the cross-sectional area of the wire is 0.5 mm2, what length of the wire will be needed? (c) If the radius of each turn is 4.0 mm, how many turns will be there in the coil?
    Ans : (a) Let R be the resistance of the coil.
    The power P consumed by a coil of resistance R when connected across a supply V is given by
    `` P=\frac{{V}^{2}}{R}``
    `` \Rightarrow R=\frac{{V}^{2}}{P}``
    `` \Rightarrow R=\frac{{\left(250\right)}^{2}}{500}=125\,\mathrm{\,\Omega \,}`` (b) We know:
    `` R=\,\mathrm{\,\rho \,}\frac{l}{A}``
    `` \Rightarrow l=\frac{RA}{\,\mathrm{\,\rho \,}}``
    `` \Rightarrow l=\frac{125\times 0.5\times {10}^{-6}}{{10}^{-6}}=62.5\,\mathrm{\,m\,}`` (c) Let n be the number of turns in the coil. Then,
    `` l=2\,\mathrm{\,\pi \,}rn``
    `` \mathit{\Rightarrow }n\mathit{=}\frac{\mathit{l}}{\mathit{2}\mathit{\pi }\mathit{r}}``
    `` \mathit{\Rightarrow }n\mathit{=}\frac{\mathit{62}\mathit{.}\mathit{5}}{\mathit{2}\mathit{\times }\mathit{3}\mathit{.}\mathit{14}\mathit{\times }\mathit{4}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{3}}}\mathit{\approx }2500``
    Page No 219: (a) Let R be the resistance of the coil.
    The power P consumed by a coil of resistance R when connected across a supply V is given by
    `` P=\frac{{V}^{2}}{R}``
    `` \Rightarrow R=\frac{{V}^{2}}{P}``
    `` \Rightarrow R=\frac{{\left(250\right)}^{2}}{500}=125\,\mathrm{\,\Omega \,}`` (b) We know:
    `` R=\,\mathrm{\,\rho \,}\frac{l}{A}``
    `` \Rightarrow l=\frac{RA}{\,\mathrm{\,\rho \,}}``
    `` \Rightarrow l=\frac{125\times 0.5\times {10}^{-6}}{{10}^{-6}}=62.5\,\mathrm{\,m\,}`` (c) Let n be the number of turns in the coil. Then,
    `` l=2\,\mathrm{\,\pi \,}rn``
    `` \mathit{\Rightarrow }n\mathit{=}\frac{\mathit{l}}{\mathit{2}\mathit{\pi }\mathit{r}}``
    `` \mathit{\Rightarrow }n\mathit{=}\frac{\mathit{62}\mathit{.}\mathit{5}}{\mathit{2}\mathit{\times }\mathit{3}\mathit{.}\mathit{14}\mathit{\times }\mathit{4}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{3}}}\mathit{\approx }2500``
    Page No 219:
  • #4-a
    What would be the resistance of the coil?
    Ans : Let R be the resistance of the coil.
    The power P consumed by a coil of resistance R when connected across a supply V is given by
    `` P=\frac{{V}^{2}}{R}``
    `` \Rightarrow R=\frac{{V}^{2}}{P}``
    `` \Rightarrow R=\frac{{\left(250\right)}^{2}}{500}=125\,\mathrm{\,\Omega \,}``
  • #4-b
    If the cross-sectional area of the wire is 0.5 mm2, what length of the wire will be needed?
    Ans : We know:
    `` R=\,\mathrm{\,\rho \,}\frac{l}{A}``
    `` \Rightarrow l=\frac{RA}{\,\mathrm{\,\rho \,}}``
    `` \Rightarrow l=\frac{125\times 0.5\times {10}^{-6}}{{10}^{-6}}=62.5\,\mathrm{\,m\,}``
  • #4-c
    If the radius of each turn is 4.0 mm, how many turns will be there in the coil?
    Ans : Let n be the number of turns in the coil. Then,
    `` l=2\,\mathrm{\,\pi \,}rn``
    `` \mathit{\Rightarrow }n\mathit{=}\frac{\mathit{l}}{\mathit{2}\mathit{\pi }\mathit{r}}``
    `` \mathit{\Rightarrow }n\mathit{=}\frac{\mathit{62}\mathit{.}\mathit{5}}{\mathit{2}\mathit{\times }\mathit{3}\mathit{.}\mathit{14}\mathit{\times }\mathit{4}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{3}}}\mathit{\approx }2500``
    Page No 219: