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Physics-

General

Easy

Question

If the shear modulus of a wire material is 5.9 $blank cross times 10 to the power of 11 end exponent d y n e blank c m to the power of negative 2 end exponent$ then the potential energy of a wire of $4 cross times 10 to the power of 3 end exponent c m$ in diameter and 5 cm long twisted through an angle of 10’ , is

$1.253 \times 10^{- 12} J$
$2.00 \times 10^{- 12} J$
$1.00 \times 10^{- 12} J$
$0.8 \times 10^{- 12} J$

The correct answer is: $1.253 blank cross times 10 to the power of negative 12 end exponent blank J$

To twist the wire through the angle $d theta comma blank$ it is necessary to do the work
$d W equals blank tau d theta$
And $theta equals 10 to the power of ´ end exponent equals fraction numerator 10 over denominator 60 end fraction cross times fraction numerator pi over denominator 180 end fraction equals fraction numerator pi over denominator 1080 end fraction r a d$
$W equals blank not stretchy integral from 0 to theta of tau blank d theta equals blank not stretchy integral from 0 to theta of fraction numerator eta pi r to the power of 4 end exponent theta d theta over denominator 2 l end fraction equals blank fraction numerator eta pi r to the power of 4 end exponent theta over denominator 4 l end fraction$
$W equals blank fraction numerator 5.9 blank cross times 10 to the power of 11 end exponent cross times 10 to the power of negative 5 end exponent cross times blank pi open parentheses 2 cross times 10 to the power of negative 5 end exponent close parentheses to the power of 4 end exponent pi to the power of 2 end exponent over denominator 10 to the power of negative 4 end exponent cross times 4 cross times 5 cross times 10 to the power of negative 2 end exponent cross times open parentheses 1080 close parentheses to the power of 2 end exponent end fraction$
$W equals 1.253 blank cross times 10 to the power of negative 12 end exponent blank J$

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