Question #4dd5f

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Question #4dd5f

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Answer 1 · 2016-11-06T06:33:21

the ans. will be $1 o h m$ $1 ohm$ .

Explanation:

we know R=s(l/a).where s=specific resistance of the element, l=length of the element,a=circular base area of the element.
now for the first case if resistance was $R_{1}$ $R_1$ and for the second case resistance will be $R_{2}$ $R_2$ .
so $R_{1}$ $R_1$ = $s (\frac{l_{1}}{a_{1}})$ $s(l_1/a_1)$ and $R_{2} = s (\frac{l_{2}}{a_{2}})$ $R_2=s(l_2/a_2)$
now the diameter of the element is doubled ,so the radius of the circular base is also doubled. $sin c e, a r e a = π r^{2}$ $since, area=pi r^2$ ,the area will be 4times.so, $a_{2} = 4 a_{1}$ $a_2=4a_1$ .
here the volume of the element will remain constant,so the product of the area & length remains the same i.e. $a_{1} l_{1} = a_{2} l_{2}$ $a_1l_1=a_2l_2$
again $a_{2} = 4 a_{1}$ $a_2=4a_1$ hence $l_{2} = \frac{l_{1}}{4}$ $l_2=l_1/4$ .
since the element remains same s will be same for both the cases.

now $R_{2} = s (\frac{l_{2}}{a_{2}})$ $R_2=s(l_2/a_2)$ = $s ⎛ ⎝ \frac{\frac{l_{1}}{4}}{4 a_{1}} ⎞ ⎠$ $s((l_1/4)/(4a_1))$ = $s (\frac{l_{1}}{16 a_{1}})$ $s(l_1/(16a_1))$ = $\frac{R_{1}}{16}$ $R_1/16$ = $\frac{16}{16}$ $16/16$ = $1 o h m$ $1 ohm$ .

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Question #4dd5f

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