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Byju's Answer

Standard XII

Physics

Young's Modulus of Elasticity

A copper wire...

Question

A copper wire having $Y = 1 \times 10^{11} N / m^{2}$ with length $6 m$ and a steel wire having $Y = 2 \times 10^{11} N / m^{2}$ with length $4 m$ each of cross section $10^{- 5} m^{2}$ are fastened end to end stretched by a tension of $100 N$ . The elongation produced in the copper wire is :

0.2 m m

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0.4 m m

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0.6 m m

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0.8 m m

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Solution

The correct option is D $0.8 m m$
Elongation. $Δ l = \frac{F L}{Y A}$

$Δ l = \frac{100 \times 6}{10^{11} \times 10^{- 5}} + \frac{100 \times 4}{2 \times 10^{11} \times 10^{- 5}} = 0.8 m m$

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Similar questions

Q. A copper wire (

Y = 10^{11} N / m^{2}

) of length

8

m and a steel wire

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

of length

4

m, each of

0.5 c m^{2}

cross-section are fastened end to end and streatched with a tension of

500

Q. A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of the stresses developed in steel and copper wire

Y

of steel is

2 \times 10^{11} {N/m}^{2}

and

Y

of copper is

1.3 \times 10^{11} {N/m}^{2}

c o p p e r s t e e l

A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension.

If the ratio of the strains developed is

\frac{x}{20}

. Find

x

. (Y of steel

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

and Y of copper =

1.3 \times 10^{11} N / m^{2})

Q. A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm are connected end to end. When stretched by a force, the elonation in length 0.50 mm is produced in the copper wire. The stretching force is

(Y_{c u} = 1.1 \times 10^{11} N / m^{2}, Y_{s t e e l} = 2.0 \times 10^{11} N / m^{2})

A block of weight $100$ N is suspended by copper and steel wires of same cross sectional area $0.5 c m^{2}$ and length $\sqrt{3} m$ and $1 m$ respectively. Their other ends are fixed on a ceiling as shown in figure. The angles subtended by copper and steel wires with ceiling are $30^{o}$ and $60^{o}$ respectively. If elongation in copper wire is $Δ l_{C}$ and elongation in steel wire is $Δ l_{S}$ , then the ratio $\frac{Δ l_{C}}{Δ l_{S}}$ is

(Young's modulus for copper and steel are $1 \times 10^{11} N / m^{2}$ and $2 \times 10^{11} N / m^{2}$ respectively)

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A copper wire having Y=1×1011N/m2 with length 6m and a steel wire having Y=2×1011N/m2 with length 4m each of cross section 10−5m2 are fastened end to end stretched by a tension of 100N. The elongation produced in the copper wire is :

The correct option is D 0.8mmElongation.Δl=FLYAΔl=100×61011×10−5+100×42×1011×10−5=0.8mm

A copper wire having $Y = 1 \times 10^{11} N / m^{2}$ with length $6 m$ and a steel wire having $Y = 2 \times 10^{11} N / m^{2}$ with length $4 m$ each of cross section $10^{- 5} m^{2}$ are fastened end to end stretched by a tension of $100 N$ . The elongation produced in the copper wire is :

The correct option is D $0.8 m m$
Elongation. $Δ l = \frac{F L}{Y A}$

$Δ l = \frac{100 \times 6}{10^{11} \times 10^{- 5}} + \frac{100 \times 4}{2 \times 10^{11} \times 10^{- 5}} = 0.8 m m$