An 8m long copper wire and a 4m long steel wire, each of cross-section 0.5 cm² are fastened end to end and stretched by a 500 N force. The elastic potential energy of the system is $(Y_{C o p p e r} = 1 \times 10^{11} N m^{- 2}; Y_{s t e e l} = 2 \times 10^{11} N m^{- 2})$

$\frac{1}{4}$ J

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$\frac{1}{3}$ J

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$\frac{1}{2}$ J

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4 J

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Solution

The correct option is A
$\frac{1}{4}$ J

The potential energy of an elongated rod is given by $U = \frac{F^{2} L}{2 A Y}$ . In this problem the total

potential energy is the sum of the individual potential energies. Therefore,

$U = \frac{F^{2}}{2 A} (\frac{L_{1}}{Y_{1}} + \frac{L_{2}}{Y_{2}})$

On substitution,

$U = 0.25 J = \frac{1}{4} J$

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Similar questions

An 8m long copper wire and a 4m long steel wire, each of cross-section 0.5 cm² are fastened end to end and stretched by a 500 N force. The elastic potential energy of the system is $(Y_{C o p p e r} = 1 \times 10^{11} N m^{- 2}; Y_{s t e e l} = 2 \times 10^{11} N m^{- 2})$

Q. A copper wire of length

2.4 m

and a steel wire of length

1.6 m

, both the diameter

3 m m

, are connected end to end. The ratio fo elongation of steel to the copper wires is then

(Y_{c o p p e r} = 1.2 \times 10^{11} N m^{- 2}, Y_{s t e e l} = 2 \times 10^{11} N m^{- 2})

Q. A copper wire of length

2.4 m

and a steel wire of length

1.6 m

, both the diameter

3 m m

, are connected end to end. When stretched by a load, the net elongation is found to be

0.7 m m

. The load applied is

(Y_{c o p p e r} = 1.2 \times 10^{11} N m^{- 2}, Y_{s t e e l} = 2 \times 10^{11} N m^{- 2})

A copper wire of length 1.0 m and a steel wire of length 0.5 m having equal cross-sectional areas are joined end to end. The composite wire is stretched by a certain load which stretches the copper wire by 1 mm. If the Young's moduli of copper and steel are $1.0 \times 10^{11} N m^{- 2}$ and $2.0 \times 10^{11} N m^{- 2}$ , the total extension of the composite wire is

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An 8m long copper wire and a 4m long steel wire, each of cross-section 0.5 cm2 are fastened end to end and stretched by a 500 N force. The elastic potential energy of the system is (YCopper=1×1011Nm−2;Ysteel=2×1011Nm−2)

The correct option is A 14J The potential energy of an elongated rod is given by U=F2L2AY. In this problem the total potential energy is the sum of the individual potential energies. Therefore, U=F22A(L1Y1+L2Y2) On substitution, U=0.25J=14J

An 8m long copper wire and a 4m long steel wire, each of cross-section 0.5 cm² are fastened end to end and stretched by a 500 N force. The elastic potential energy of the system is $(Y_{C o p p e r} = 1 \times 10^{11} N m^{- 2}; Y_{s t e e l} = 2 \times 10^{11} N m^{- 2})$