The force required to stretch a steel wire of $1 c m^{2}$ cross - section to 1.1 times its length would be $(Y = 2 \times 10^{11} N m^{- 2})$

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Solution

The correct option is A
$L_{2} = 1.1 L_{1}$
$∴$ Strain = $\frac{l}{L_{1}} = \frac{L_{2} - L_{1}}{L_{1}} = \frac{1.1 L_{1} - L_{1}}{L_{1}} = 0.1.$
$F = Y A \frac{l}{L} = 2 \times 10^{11} \times 1 \times 10^{- 4} \times 0.1 = 2 \times 10^{6} N .$

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Similar questions

Q. The force required to stretch a steel wire of

1 c m^{2}

cross - section to 1.1 times its length would be

(Y = 2 \times 10^{11} N m^{- 2})

Q. The force required to stretch a steel wire of

1

sq. cm cross-section to

1.1

times, its length would be

(Y = 2 \times 10^{11} N m^{- 2}

)

Q. What force is required to stretch a steel wire

1 c m^{2}

in cross section to increase its length by

1 %

given

[Y = 2 \times 10^{11} N / m^{2}]

Q. What force is required to stretch a copper wire

1 c m^{2}

in cross-section to double its length?

Y

for copper is

1.26 \times 10^{12} d y n e c m^{- 2}

What force is required to stretch a steel wire $1 c m^{2}$ in cross section to double its length? (Youngs modulus for steel is $2 \times 10^{11} P a$ )

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The force required to stretch a steel wire of 1cm2 cross - section to 1.1 times its length would be (Y=2×1011Nm−2)

The correct option is A L2=1.1L1∴ Strain = lL1=L2−L1L1=1.1L1−L1L1=0.1. F=YAlL=2×1011×1×10−4×0.1=2×106N.

The force required to stretch a steel wire of $1 c m^{2}$ cross - section to 1.1 times its length would be $(Y = 2 \times 10^{11} N m^{- 2})$

The correct option is A
$L_{2} = 1.1 L_{1}$
$∴$ Strain = $\frac{l}{L_{1}} = \frac{L_{2} - L_{1}}{L_{1}} = \frac{1.1 L_{1} - L_{1}}{L_{1}} = 0.1.$
$F = Y A \frac{l}{L} = 2 \times 10^{11} \times 1 \times 10^{- 4} \times 0.1 = 2 \times 10^{6} N .$