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PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369797 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 modulii of copper and steel are respectively
$1.0 \times 10^{11} N m^{- 2} and 2.0 \times 10^{11} N m^{- 2}$ , the total extension of the composite wire is:

1

1.75 m m

2

2.0 m m

3

1.50 m m

4

1.25 m m

Explanation:

Stress is same in both wires.
$Y_{c} \times (Δ L_{c} / L_{c}) = Y_{s} \times (Δ L_{s} / L_{s})$
$\Rightarrow 1 \times 10^{11} \times (\frac{1 \times 10^{- 3}}{1}) = 2 \times 10^{11} \times (\frac{Δ L_{s}}{0.5})$
$∴ Δ L_{s} = \frac{0.5 \times 10^{- 3}}{2} = 0.25 m m$
Total extension of composite wire
$= Δ L_{c} + Δ L_{s} = 1.25 m m$

JEE - 2013

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369798 A stone of $0.5 k g$ mass is attached to one end of a metre long aluminium wire of $1 m m$ diameter and is suspended vertically. The stone is now rotated in a horizontal plane at a rate, such that the wire makes an angle of $75^{\circ}$ with the vertical. Find the increase in length of the wire. (Take Young's modulus of aluminium $= 7 \times 10^{10} N m^{- 2}, \sin 75^{\circ}$ $= 0.97, \cos 75^{\circ} = 0.26,$ $π = 3.14, g = 10 m / s^{2}$

1

0.35 m

2

0.73 m

3

0.92 m

4

0.15 m

Explanation:

When the stone is rotated so that it makes an angle $75^{\circ}$ with the vertical as shown in the figure below, it moves along a circle of radius $O A = r$ (say).
supporting img
$T \cos 75^{\circ}$ is along the vertical and balances the weight of the stone i.e.,
$T \cos 75^{\circ} = M g$
$T = \frac{M g}{\cos 75^{\circ}} = \frac{0.5 \times 10}{0.26} = 19.23 N$
Now, $Y = \frac{F / a}{l / L} = \frac{4 T L}{π D^{2} l}$
$∴ l = \frac{4 T L}{π D^{2} Y}$
$= \frac{4 \times 19.23 \times 1}{π \times {(10^{- 3})}^{2} \times 7 \times 10^{10}} = \frac{76.92}{21.98} \times 10^{- 4}$
$\approx 3.50 \times 10^{- 4} m$
$(R o u n d i n g o f f t o 2 d e c i m a l p l a c e s)$
$∴ l = 0.35 m m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369799 A rod of length $1 m$ is connected to a ceiling and the lower end is connected to a load $20 N$ . The extension produced in the rod is $10^{- 4} m$ . If the cross-sectional area of the wire is $10^{- 6} m^{2}$ , calculate the Young's modulus of the material of the wire.

1

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

2

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

3

2 \times 10^{- 13} N / m^{2}

4

3 \times 10^{- 12} N / m^{2}

Explanation:

Given that
$Δ L = 10^{- 4} m, F = 20 N, A = 10^{- 6} m^{2}, L = 1 m$
$∴ Y = \frac{F L}{A Δ L} = \frac{20 \times 1}{10^{- 6} \times 10^{- 4}}$
$= 20 \times 10^{10} = 2 \times 10^{11} N / m^{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369800 A fixed volume of iron is drawn into a wire of length $l$ . The extension produced in this wire by a constant force $F$ is proportional to -

1

\frac{1}{ℓ^{2}}

2

\frac{1}{ℓ}

3

ℓ^{2}

4

ℓ

Explanation:

Volume $=$ constant
$A \times l = constant$
$Δ l = \frac{F ℓ}{A Y} or Δ l \propto \frac{l}{A} or Δ l \propto l^{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369797 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 modulii of copper and steel are respectively
$1.0 \times 10^{11} N m^{- 2} and 2.0 \times 10^{11} N m^{- 2}$ , the total extension of the composite wire is:

1

1.75 m m

2

2.0 m m

3

1.50 m m

4

1.25 m m

Explanation:

Stress is same in both wires.
$Y_{c} \times (Δ L_{c} / L_{c}) = Y_{s} \times (Δ L_{s} / L_{s})$
$\Rightarrow 1 \times 10^{11} \times (\frac{1 \times 10^{- 3}}{1}) = 2 \times 10^{11} \times (\frac{Δ L_{s}}{0.5})$
$∴ Δ L_{s} = \frac{0.5 \times 10^{- 3}}{2} = 0.25 m m$
Total extension of composite wire
$= Δ L_{c} + Δ L_{s} = 1.25 m m$

JEE - 2013

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369798 A stone of $0.5 k g$ mass is attached to one end of a metre long aluminium wire of $1 m m$ diameter and is suspended vertically. The stone is now rotated in a horizontal plane at a rate, such that the wire makes an angle of $75^{\circ}$ with the vertical. Find the increase in length of the wire. (Take Young's modulus of aluminium $= 7 \times 10^{10} N m^{- 2}, \sin 75^{\circ}$ $= 0.97, \cos 75^{\circ} = 0.26,$ $π = 3.14, g = 10 m / s^{2}$

1

0.35 m

2

0.73 m

3

0.92 m

4

0.15 m

Explanation:

When the stone is rotated so that it makes an angle $75^{\circ}$ with the vertical as shown in the figure below, it moves along a circle of radius $O A = r$ (say).
supporting img
$T \cos 75^{\circ}$ is along the vertical and balances the weight of the stone i.e.,
$T \cos 75^{\circ} = M g$
$T = \frac{M g}{\cos 75^{\circ}} = \frac{0.5 \times 10}{0.26} = 19.23 N$
Now, $Y = \frac{F / a}{l / L} = \frac{4 T L}{π D^{2} l}$
$∴ l = \frac{4 T L}{π D^{2} Y}$
$= \frac{4 \times 19.23 \times 1}{π \times {(10^{- 3})}^{2} \times 7 \times 10^{10}} = \frac{76.92}{21.98} \times 10^{- 4}$
$\approx 3.50 \times 10^{- 4} m$
$(R o u n d i n g o f f t o 2 d e c i m a l p l a c e s)$
$∴ l = 0.35 m m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369799 A rod of length $1 m$ is connected to a ceiling and the lower end is connected to a load $20 N$ . The extension produced in the rod is $10^{- 4} m$ . If the cross-sectional area of the wire is $10^{- 6} m^{2}$ , calculate the Young's modulus of the material of the wire.

1

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

2

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

3

2 \times 10^{- 13} N / m^{2}

4

3 \times 10^{- 12} N / m^{2}

Explanation:

Given that
$Δ L = 10^{- 4} m, F = 20 N, A = 10^{- 6} m^{2}, L = 1 m$
$∴ Y = \frac{F L}{A Δ L} = \frac{20 \times 1}{10^{- 6} \times 10^{- 4}}$
$= 20 \times 10^{10} = 2 \times 10^{11} N / m^{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369800 A fixed volume of iron is drawn into a wire of length $l$ . The extension produced in this wire by a constant force $F$ is proportional to -

1

\frac{1}{ℓ^{2}}

2

\frac{1}{ℓ}

3

ℓ^{2}

4

ℓ

Explanation:

Volume $=$ constant
$A \times l = constant$
$Δ l = \frac{F ℓ}{A Y} or Δ l \propto \frac{l}{A} or Δ l \propto l^{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369797 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 modulii of copper and steel are respectively
$1.0 \times 10^{11} N m^{- 2} and 2.0 \times 10^{11} N m^{- 2}$ , the total extension of the composite wire is:

1

1.75 m m

2

2.0 m m

3

1.50 m m

4

1.25 m m

Explanation:

Stress is same in both wires.
$Y_{c} \times (Δ L_{c} / L_{c}) = Y_{s} \times (Δ L_{s} / L_{s})$
$\Rightarrow 1 \times 10^{11} \times (\frac{1 \times 10^{- 3}}{1}) = 2 \times 10^{11} \times (\frac{Δ L_{s}}{0.5})$
$∴ Δ L_{s} = \frac{0.5 \times 10^{- 3}}{2} = 0.25 m m$
Total extension of composite wire
$= Δ L_{c} + Δ L_{s} = 1.25 m m$

JEE - 2013

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369798 A stone of $0.5 k g$ mass is attached to one end of a metre long aluminium wire of $1 m m$ diameter and is suspended vertically. The stone is now rotated in a horizontal plane at a rate, such that the wire makes an angle of $75^{\circ}$ with the vertical. Find the increase in length of the wire. (Take Young's modulus of aluminium $= 7 \times 10^{10} N m^{- 2}, \sin 75^{\circ}$ $= 0.97, \cos 75^{\circ} = 0.26,$ $π = 3.14, g = 10 m / s^{2}$

1

0.35 m

2

0.73 m

3

0.92 m

4

0.15 m

Explanation:

When the stone is rotated so that it makes an angle $75^{\circ}$ with the vertical as shown in the figure below, it moves along a circle of radius $O A = r$ (say).
supporting img
$T \cos 75^{\circ}$ is along the vertical and balances the weight of the stone i.e.,
$T \cos 75^{\circ} = M g$
$T = \frac{M g}{\cos 75^{\circ}} = \frac{0.5 \times 10}{0.26} = 19.23 N$
Now, $Y = \frac{F / a}{l / L} = \frac{4 T L}{π D^{2} l}$
$∴ l = \frac{4 T L}{π D^{2} Y}$
$= \frac{4 \times 19.23 \times 1}{π \times {(10^{- 3})}^{2} \times 7 \times 10^{10}} = \frac{76.92}{21.98} \times 10^{- 4}$
$\approx 3.50 \times 10^{- 4} m$
$(R o u n d i n g o f f t o 2 d e c i m a l p l a c e s)$
$∴ l = 0.35 m m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369799 A rod of length $1 m$ is connected to a ceiling and the lower end is connected to a load $20 N$ . The extension produced in the rod is $10^{- 4} m$ . If the cross-sectional area of the wire is $10^{- 6} m^{2}$ , calculate the Young's modulus of the material of the wire.

1

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

2

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

3

2 \times 10^{- 13} N / m^{2}

4

3 \times 10^{- 12} N / m^{2}

Explanation:

Given that
$Δ L = 10^{- 4} m, F = 20 N, A = 10^{- 6} m^{2}, L = 1 m$
$∴ Y = \frac{F L}{A Δ L} = \frac{20 \times 1}{10^{- 6} \times 10^{- 4}}$
$= 20 \times 10^{10} = 2 \times 10^{11} N / m^{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369800 A fixed volume of iron is drawn into a wire of length $l$ . The extension produced in this wire by a constant force $F$ is proportional to -

1

\frac{1}{ℓ^{2}}

2

\frac{1}{ℓ}

3

ℓ^{2}

4

ℓ

Explanation:

Volume $=$ constant
$A \times l = constant$
$Δ l = \frac{F ℓ}{A Y} or Δ l \propto \frac{l}{A} or Δ l \propto l^{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369797 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 modulii of copper and steel are respectively
$1.0 \times 10^{11} N m^{- 2} and 2.0 \times 10^{11} N m^{- 2}$ , the total extension of the composite wire is:

1

1.75 m m

2

2.0 m m

3

1.50 m m

4

1.25 m m

Explanation:

Stress is same in both wires.
$Y_{c} \times (Δ L_{c} / L_{c}) = Y_{s} \times (Δ L_{s} / L_{s})$
$\Rightarrow 1 \times 10^{11} \times (\frac{1 \times 10^{- 3}}{1}) = 2 \times 10^{11} \times (\frac{Δ L_{s}}{0.5})$
$∴ Δ L_{s} = \frac{0.5 \times 10^{- 3}}{2} = 0.25 m m$
Total extension of composite wire
$= Δ L_{c} + Δ L_{s} = 1.25 m m$

JEE - 2013

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369798 A stone of $0.5 k g$ mass is attached to one end of a metre long aluminium wire of $1 m m$ diameter and is suspended vertically. The stone is now rotated in a horizontal plane at a rate, such that the wire makes an angle of $75^{\circ}$ with the vertical. Find the increase in length of the wire. (Take Young's modulus of aluminium $= 7 \times 10^{10} N m^{- 2}, \sin 75^{\circ}$ $= 0.97, \cos 75^{\circ} = 0.26,$ $π = 3.14, g = 10 m / s^{2}$

1

0.35 m

2

0.73 m

3

0.92 m

4

0.15 m

Explanation:

When the stone is rotated so that it makes an angle $75^{\circ}$ with the vertical as shown in the figure below, it moves along a circle of radius $O A = r$ (say).
supporting img
$T \cos 75^{\circ}$ is along the vertical and balances the weight of the stone i.e.,
$T \cos 75^{\circ} = M g$
$T = \frac{M g}{\cos 75^{\circ}} = \frac{0.5 \times 10}{0.26} = 19.23 N$
Now, $Y = \frac{F / a}{l / L} = \frac{4 T L}{π D^{2} l}$
$∴ l = \frac{4 T L}{π D^{2} Y}$
$= \frac{4 \times 19.23 \times 1}{π \times {(10^{- 3})}^{2} \times 7 \times 10^{10}} = \frac{76.92}{21.98} \times 10^{- 4}$
$\approx 3.50 \times 10^{- 4} m$
$(R o u n d i n g o f f t o 2 d e c i m a l p l a c e s)$
$∴ l = 0.35 m m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369799 A rod of length $1 m$ is connected to a ceiling and the lower end is connected to a load $20 N$ . The extension produced in the rod is $10^{- 4} m$ . If the cross-sectional area of the wire is $10^{- 6} m^{2}$ , calculate the Young's modulus of the material of the wire.

1

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

2

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

3

2 \times 10^{- 13} N / m^{2}

4

3 \times 10^{- 12} N / m^{2}

Explanation:

Given that
$Δ L = 10^{- 4} m, F = 20 N, A = 10^{- 6} m^{2}, L = 1 m$
$∴ Y = \frac{F L}{A Δ L} = \frac{20 \times 1}{10^{- 6} \times 10^{- 4}}$
$= 20 \times 10^{10} = 2 \times 10^{11} N / m^{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369800 A fixed volume of iron is drawn into a wire of length $l$ . The extension produced in this wire by a constant force $F$ is proportional to -

1

\frac{1}{ℓ^{2}}

2

\frac{1}{ℓ}

3

ℓ^{2}

4

ℓ

Explanation:

Volume $=$ constant
$A \times l = constant$
$Δ l = \frac{F ℓ}{A Y} or Δ l \propto \frac{l}{A} or Δ l \propto l^{2}$