Chapter 11 · Electricity

NCERT Class 10 Science Chapter 11 Intext Questions (Page 190)

All 3 questions of NCERT Class 10 Science Chapter 11 Intext Questions (Page 190) — Electricity — in the book's order, each with its answer and a step-by-step solution.

1. Q1
   Why ​does the cord of an electric heater ​not glow while ​the ​heating ​element ​does?
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   Both carry the same current, but the element's far higher resistance makes it produce much more heat, so only it glows.

   The cord and the heating element are joined in series, so exactly the same current $I$ flows through both for the same time $t$. By Joule's law the heat produced is $H=I^{2}Rt$, so with $I$ and $t$ common to both, the heat produced in each part is proportional to its resistance. The element is a long, thin coil of nichrome, an alloy of very high resistivity, so its resistance is large (a few tens of ohms). The cord is a short, thick copper wire of very low resistivity, so its resistance is only a small fraction of an ohm. Almost all the heat is therefore produced in the element; its temperature rises until it is red hot and it glows, and nichrome withstands this because it has a high melting point and does not oxidise readily. The heat produced in the cord is negligible, its temperature hardly rises, and so it does not glow.

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2. Q2
   Compute the heat generated while transferring $96000$ coulomb ​of ​charge ​in ​one ​hour ​through a potential difference of $50$ V.
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   The heat generated is $4.8\times 10^6$ J (4.8 MJ), since each coulomb gives up 50 J.

   Each coulomb falls through $50$ V and so gives up $50$ J, and the whole of that energy appears as heat. Therefore $H=VQ=50\text{V}\times 96000\text{C}=4.8\times 10^6$ J $=4.8$ MJ (equal to $1.33$ kW h). The time of one hour is not needed for the heat; it only fixes the current, $I=Q/t=96000/3600=26.67$ A.

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3. Q3
   An ​electric ​iron of resistance $20\Omega$ takes ​a current of $5$ A. Calculate the heat developed ​in $30$ ​s.
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   $H=1.5\times 10^4$ J: the iron develops 15 kJ of heat in 30 s.

   $H=I^{2}Rt=(5\text{A}{)}^2\times 20\Omega \times 30\text{s}=25\times 20\times 30=15000$ J $=1.5\times 10^4$ J. The iron produces heat at the rate $P=I^{2}R=500$ W, so in $30$ s it develops $15$ kJ.

   Step-by-step solution →