Question

A steel wire is 1m long and 2mm² in area cross section.A force of 5 × 10³ N streches the wire by 1mm. A wire of same material and diameter has a length of 10 m the force required to produce an extension of 2mm is​

Answer 1

$\mathfrak{ \huge{ \red{ \underline{ \underline{ANSWER}}}}}$

• Given :

=> Length of wire A = 1m

=> Force acts on wire A = 5×10^3N

=> Change in length of wire A = 1mm = 0.001m

=> Length of wire B = 10m

=> Area of cross section of both wires are same

=> Both wires are made from same material

• To Find :

=> Force required to produce an extension in wire B of 2mm = 0.002m

• Formula :

=> Here material of both wires is same hence young modulus of both wires is also same.

=> Young modulus of material is given by...

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed {\sf{ \bold{ \pink{Y = \frac{stress}{strain} }}}}$

$\: \: \: \boxed{ \sf{ \blue{stress = \frac{F}{A} }}} \: \: \: \: \: \boxed{ \sf{ \blue{strain = \frac{ \delta \: L}{L} }}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf{ \bold{ \pink{Y = \frac{F \times L}{A \times \delta \: L} }}}}$

• Calculation :

=> Here Young modulus and Area of cross section is same for both...

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \therefore \: \: \boxed{ \sf{ \bold{ \orange{F \propto \: \frac{ \delta \: L}{L} }}}}$

$\implies \sf \: \frac{</strong><strong>F</strong><strong>a}{</strong><strong>F</strong><strong>b} = \frac{ \delta \: </strong><strong>L</strong><strong>a}{</strong><strong>L</strong><strong>a} \times \frac{</strong><strong>L</strong><strong>b}{ \delta \: </strong><strong>L</strong><strong>b}$

$\implies \sf \: \frac{5 \times {10}^{3} }{Fb} = \frac{0.001}{1} \times \frac{10}{0.002} \\ \\ \implies \sf \: Fb = 1 \times {10}^{3} N \\ \\ \\ \huge{\star} \: \boxed{ \boxed{ \sf{ \bold{ \green{ \huge{Fb = {10}^{3}N }}}}}}$