A wore of certain material is stretched slowly by 10 %. Its new resistance and specific resistance becomes respectively

Question

A wore of certain material is stretched slowly by 10 %. Its new resistance and specific resistance becomes respectively (A) 1.21 times, same (B) both remains the same (C) 1.1 times, 1.1 times (D) 1.2 times, 1.1 times

Tardigrade Admin · Accepted Answer

Let 11=100,12=110 R propto 12 ( R 2/ R 1)=((12/11))2=((110/100))2=1.21 R 2=1.21 R 1 Specific resistance remains same

Q. A wore of certain material is stretched slowly by $10%$ . Its new resistance and specific resistance becomes respectively

$1.21$ times, same

both remains the same

$1.1$ times, $1.1$ times

$1.2$ times, $1.1$ times

Let $1_{1} = 100, 1_{2} = 110$
$R \propto 1^{2}$
$\frac{R _{2}}{R _{1}} = (\frac{1 _{2}}{1 _{1}})^{2} = (\frac{110}{100})^{2} = 1.21$
$R_{2} = 1.21 R_{1}$
Specific resistance remains same

Q. A wore of certain material is stretched slowly by 10%. Its new resistance and specific resistance becomes respectively

1.21 times, same

both remains the same

1.1 times, 1.1 times

1.2 times, 1.1 times

Let 11​=100,12​=110 R∝12 R1​R2​​=(11​12​​)2=(100110​)2=1.21 R2​=1.21R1​ Specific resistance remains same

Q. A wore of certain material is stretched slowly by $10%$ . Its new resistance and specific resistance becomes respectively

$1.21$ times, same

$1.1$ times, $1.1$ times

$1.2$ times, $1.1$ times

Let $1_{1} = 100, 1_{2} = 110$
$R \propto 1^{2}$
$\frac{R _{2}}{R _{1}} = (\frac{1 _{2}}{1 _{1}})^{2} = (\frac{110}{100})^{2} = 1.21$
$R_{2} = 1.21 R_{1}$
Specific resistance remains same