Ten identical cells connected in series are needed to heat a wire of length 1 m and radius r by 10°C in time t. How many cells will be required to heat the wire of length 2 m of same radius by the same temperature in time t??

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Ten identical cells connected in series are needed to heat a wire of length 1 m and radius r by 10°C in time t. How many cells will be required to heat the wire of length 2 m of same radius by the same temperature in time t??

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Answer 1 · 2018-01-31T15:10:59

$20$ $20$

Explanation:

Suppose, $10$ $10$ identical cells each had voltage of $V$ $V$ ,so in series,total voltage will be $10 V$ $10V$

Now, resistance( $R$ $R$ ) is directly proportional to the length( $l$ $l$ ),so as length is doubled,resistance will also become doubled.(ref. $R \propto \frac{l}{A}$ $R prop l/A$ )

So,we can say,Heat generated $= (\frac{V^{2}}{R}) \cdot t$ $=( V^2/R) *t$ ,which will be taken up by the wire, So, we can write, $(\frac{V^{2}}{R}) \cdot t = m s \partial θ$ $(V^2/R)*t = ms del theta$ (where, $m$ $m$ is the mass of the wire, $s$ $s$ is specific heat and $\partial θ$ $del theta$ is the increase in temperature)

Again,mass is directly proportional to length.(ref. mass =volumedensity= lengtharea*density)

So, for the initial case,we can write,

$\frac{{(10 V)}^{2}}{R} \cdot t = m s \partial θ$ $(10V)^2/R *t = ms del theta$ ...1

and,finally after the changes made,if we need $n$ $n$ cells,then,

$\frac{{(n V)}^{2}}{2 R} \cdot t = (2 m) s \partial θ$ $(nV)^2/(2R)*t = (2m)s del theta$ ....2

putting the value of $m s \partial θ$ $ms del theta$ in equation 2 from 1,we get, $n = 20$ $n=20$

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Ten identical cells connected in series are needed to heat a wire of length 1 m and radius r by 10°C in time t. How many cells will be required to heat the wire of length 2 m of same radius by the same temperature in time t??

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Explanation:

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