Question

In order to increases the resistance of a given wire of uniform cross-section to four times its value, a fraction of its length is stretched uniformly till the full length of the wire becomes $\dfrac{3}{2}$ times the original length what is the value of this fraction?

Answer 1

Hint: For finding a small amount of an entire number, we increase the numerator of the division by the given number and afterward partition the item by the denominator of the portion. Understood models for finding a small amount of an entire number. Obstruction is a proportion of the resistance to flow stream in an electrical circuit. By the same logic apply basic formula of resistance.

Obstruction is estimated in ohms, represented by the Greek letter omega $\Omega$.The estimation of a portion. Since a part speaks to a division, we ought to do that division to recognize what the estimation of a portion is.

Formula used:
$R_{al}$ = $\dfrac{{\rho al}}{A} = a{R_1}l$
$al$ = The length of wire which is stretched
$R$ = The length of wire
$\rho $ = Numerator of the wire
$A$ = Denominator of the wire

Complete step by step answer:
Let the length of the wire = $\dfrac{{\rho l}}{a}{\dfrac{{\left( {\dfrac{1}{2} + a} \right)}}{a}^2}$ $l$ unit
Fraction required = $a$
Then, the length of wire which is stretched = $al$
Eliminate all $ l $ value then
Volume of wire $ al $ is same
$l {\rm A} $ = $\left( {al} \right)A$
$l' = l\left( {\dfrac{1}{2} + a} \right)$
Now, resistance of stretched parts,
According to the question,
${R_{l - al}} + R $ = $4{R_1}$
$ \Rightarrow$ ${R_{l - al}} + {R_1}$ $\dfrac{{{R_l}{{\left( {\dfrac{1}{2} + a} \right)}^{^2}}}}{a}$ = $4{R_1}$
On putting the values and we get,
$ \Rightarrow $$a - {a^2} + \dfrac{1}{4} + {a^2} + a = 4a$
Cancelling the terms and adding we get
$ \Rightarrow$ $\dfrac{1}{4} + 2a$ = $4a$
Let us subtract \[2a\] on both side we get
$ \Rightarrow \dfrac{1}{4} + 2a - 2a = 4a - 2a$
On cancel the term and subtract the LHS, we get
$ \Rightarrow$ $\dfrac{1}{4}$ = $2a$
Taking cross multiplication we get, $ \Rightarrow$ $a$ = $\dfrac{1}{8}$.

Note: The higher the obstruction, the lower the current stream. If unusually high, one potential reason (among many) could be harmed conductors because of consuming or consumption. All conductors emit some level of warmth, so overheating is an issue regularly connected with resistance. Make sure that resistivity of a material is constant.