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PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369788 A uniform rod of length $L$ and density $ρ$ is rotated in a horizontal plane about a vertical axis through one of its ends. The angular speed of rotation is $ω$ . If increase in the length is found to be $\frac{ρ ω^{2} L^{n}}{n Y}$ where, $Y$ is Young's modulus of rod, determine value of $n$ .

1 2

2 5

3 1

4 3

Explanation:

supporting img
$Y = \frac{F L}{A l}$
Here, $F =$ centrifugal force due to rotation.
$∴ l = \frac{F L}{A Y}$
From the figure, considering small elongation $d x$ of length segment $x$
$d l = \frac{d F \times x}{A Y} (1)$
where, $d F =$ centrifugal force of differential mass $d M$
but $d F = d M ω^{2} x = (ρ A d x) ω^{2} x$
$= ρ A ω^{2} x d x$
Substituting in equation (1),
$∴ d l = \frac{(ρ A ω^{2} x d x) \times x}{A Y}$
Integrating over limits,
$∴ \int_{0}^{1} d l = \frac{ρ ω^{2}}{Y} \int_{0}^{L} x^{2} d x$
$∴ l = \frac{ρ ω^{2}}{Y} {[\frac{x^{3}}{3}]}_{0}^{L} = \frac{ρ ω^{2}}{3 Y} [L^{3} - 0]$
$∴ l = \frac{ρ ω^{2} L^{3}}{3 Y}$
$\Rightarrow n = 3$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369789 If the interatomic spacing in a steel wire is $3.0 \overset{o}{A}$ and $Y_{s t e e l} = 20 \times 10^{10} N / m^{2}$ then force constant is

1

225 K N

2

6 \times 10^{- 2} N / \overset{o}{A}

3

6 \times 10^{- 5} N / \overset{o}{A}

4

4 \times 10^{- 5} N / \overset{o}{A}

Explanation:

$K = Y r_{0} = 20 \times 10^{10} \times 3 \times 10^{- 10} = 60 N / m$
$= 6 \times 10^{- 9} \frac{N}{\overset{o}{A}}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369790 The graph shows the extension of a wire of length $1 m$ suspended from the top of a roof at one end and with a load $W$ connected to the other end. If the cross sectional area of the wire is $1 {m m}^{2}$ , then the Young's modulus of the material of the wire is
supporting img

1

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

2

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

3

\frac{1}{2} \times 10^{11} {N m}^{- 2}

4

4 \times 10^{10} N m^{- 2}

Explanation:

$Y = \frac{F / A}{Δ l / l} = \frac{W l}{A Δ l} \Rightarrow \frac{W}{Δ l} = \frac{Y A}{l} =$ slope
$\Rightarrow Y = \frac{l}{A} \times ($ slope $) = \frac{1}{10^{- 6}} (\frac{40 - 20}{(2 - 1) \times 10^{- 3}})$
$= 2 \times 10^{10} {N m}^{- 2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369791 The following four wires of length $L$ and radius $r$ are made of the same material. Which of these will have the largest extension, when the same tension is applied?

1

L = 100 c m, r = 0.1 m m

2

L = 100 c m, r = 0.4 m m

3

L = 100 c m, r = 0.6 m m

4

L = 200 c m, r = 0.8 m m

Explanation:

$Δ L = \frac{L T}{A Y} = \frac{L T}{π r^{2} Y} = (\frac{T}{π Y}) \times \frac{L}{r^{2}}$
$\frac{L}{r^{2}}$ $\to$ max for $L = 100 c m & r = 0.1 m m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369792 A uniformly tapering conical wire is made from a material of Young's modulus Y and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values $R$ and $3 R$ , respectively, The upper end of the wire is fixed to a rigid support and a mass $M$ is suspended from its lower end. The equilibrium extended length, of this wire, would equal

1

L (1 + \frac{2}{9} \frac{M g}{π Y R^{2}})

2

L (1 + \frac{1}{9} \frac{M g}{π Y R^{2}})

3

L (1 + \frac{1}{3} \frac{M g}{π Y R^{2}})

4

L (1 + \frac{2}{3} \frac{M g}{π Y R^{2}})

Explanation:

Consider a thin disc of radius $r$ and width $dx$ . The small extension produced in the disc is dL.
supporting img

$\begin{matrix} \tan θ = \frac{r - R}{x} = \frac{3 R - R}{L} \Rightarrow r = R (1 + \frac{2 x}{L}); \\ Y = \frac{M g}{π r^{2} \frac{d L}{d x}} \\ Δ L = \frac{M g}{Y π R^{2}} \int_{0}^{L} \frac{d x}{{(1 + \frac{2 x}{L})}^{2}} = \frac{M g L}{3 π R^{2} Y} \\ L^{'} = L + Δ L = L (1 + \frac{1}{3} \frac{M g}{π R^{2} Y}) \end{matrix}$

JEE - 2016

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369788 A uniform rod of length $L$ and density $ρ$ is rotated in a horizontal plane about a vertical axis through one of its ends. The angular speed of rotation is $ω$ . If increase in the length is found to be $\frac{ρ ω^{2} L^{n}}{n Y}$ where, $Y$ is Young's modulus of rod, determine value of $n$ .

1 2

2 5

3 1

4 3

Explanation:

supporting img
$Y = \frac{F L}{A l}$
Here, $F =$ centrifugal force due to rotation.
$∴ l = \frac{F L}{A Y}$
From the figure, considering small elongation $d x$ of length segment $x$
$d l = \frac{d F \times x}{A Y} (1)$
where, $d F =$ centrifugal force of differential mass $d M$
but $d F = d M ω^{2} x = (ρ A d x) ω^{2} x$
$= ρ A ω^{2} x d x$
Substituting in equation (1),
$∴ d l = \frac{(ρ A ω^{2} x d x) \times x}{A Y}$
Integrating over limits,
$∴ \int_{0}^{1} d l = \frac{ρ ω^{2}}{Y} \int_{0}^{L} x^{2} d x$
$∴ l = \frac{ρ ω^{2}}{Y} {[\frac{x^{3}}{3}]}_{0}^{L} = \frac{ρ ω^{2}}{3 Y} [L^{3} - 0]$
$∴ l = \frac{ρ ω^{2} L^{3}}{3 Y}$
$\Rightarrow n = 3$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369789 If the interatomic spacing in a steel wire is $3.0 \overset{o}{A}$ and $Y_{s t e e l} = 20 \times 10^{10} N / m^{2}$ then force constant is

1

225 K N

2

6 \times 10^{- 2} N / \overset{o}{A}

3

6 \times 10^{- 5} N / \overset{o}{A}

4

4 \times 10^{- 5} N / \overset{o}{A}

Explanation:

$K = Y r_{0} = 20 \times 10^{10} \times 3 \times 10^{- 10} = 60 N / m$
$= 6 \times 10^{- 9} \frac{N}{\overset{o}{A}}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369790 The graph shows the extension of a wire of length $1 m$ suspended from the top of a roof at one end and with a load $W$ connected to the other end. If the cross sectional area of the wire is $1 {m m}^{2}$ , then the Young's modulus of the material of the wire is
supporting img

1

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

2

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

3

\frac{1}{2} \times 10^{11} {N m}^{- 2}

4

4 \times 10^{10} N m^{- 2}

Explanation:

$Y = \frac{F / A}{Δ l / l} = \frac{W l}{A Δ l} \Rightarrow \frac{W}{Δ l} = \frac{Y A}{l} =$ slope
$\Rightarrow Y = \frac{l}{A} \times ($ slope $) = \frac{1}{10^{- 6}} (\frac{40 - 20}{(2 - 1) \times 10^{- 3}})$
$= 2 \times 10^{10} {N m}^{- 2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369791 The following four wires of length $L$ and radius $r$ are made of the same material. Which of these will have the largest extension, when the same tension is applied?

1

L = 100 c m, r = 0.1 m m

2

L = 100 c m, r = 0.4 m m

3

L = 100 c m, r = 0.6 m m

4

L = 200 c m, r = 0.8 m m

Explanation:

$Δ L = \frac{L T}{A Y} = \frac{L T}{π r^{2} Y} = (\frac{T}{π Y}) \times \frac{L}{r^{2}}$
$\frac{L}{r^{2}}$ $\to$ max for $L = 100 c m & r = 0.1 m m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369792 A uniformly tapering conical wire is made from a material of Young's modulus Y and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values $R$ and $3 R$ , respectively, The upper end of the wire is fixed to a rigid support and a mass $M$ is suspended from its lower end. The equilibrium extended length, of this wire, would equal

1

L (1 + \frac{2}{9} \frac{M g}{π Y R^{2}})

2

L (1 + \frac{1}{9} \frac{M g}{π Y R^{2}})

3

L (1 + \frac{1}{3} \frac{M g}{π Y R^{2}})

4

L (1 + \frac{2}{3} \frac{M g}{π Y R^{2}})

Explanation:

Consider a thin disc of radius $r$ and width $dx$ . The small extension produced in the disc is dL.
supporting img

$\begin{matrix} \tan θ = \frac{r - R}{x} = \frac{3 R - R}{L} \Rightarrow r = R (1 + \frac{2 x}{L}); \\ Y = \frac{M g}{π r^{2} \frac{d L}{d x}} \\ Δ L = \frac{M g}{Y π R^{2}} \int_{0}^{L} \frac{d x}{{(1 + \frac{2 x}{L})}^{2}} = \frac{M g L}{3 π R^{2} Y} \\ L^{'} = L + Δ L = L (1 + \frac{1}{3} \frac{M g}{π R^{2} Y}) \end{matrix}$

JEE - 2016

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369788 A uniform rod of length $L$ and density $ρ$ is rotated in a horizontal plane about a vertical axis through one of its ends. The angular speed of rotation is $ω$ . If increase in the length is found to be $\frac{ρ ω^{2} L^{n}}{n Y}$ where, $Y$ is Young's modulus of rod, determine value of $n$ .

1 2

2 5

3 1

4 3

Explanation:

supporting img
$Y = \frac{F L}{A l}$
Here, $F =$ centrifugal force due to rotation.
$∴ l = \frac{F L}{A Y}$
From the figure, considering small elongation $d x$ of length segment $x$
$d l = \frac{d F \times x}{A Y} (1)$
where, $d F =$ centrifugal force of differential mass $d M$
but $d F = d M ω^{2} x = (ρ A d x) ω^{2} x$
$= ρ A ω^{2} x d x$
Substituting in equation (1),
$∴ d l = \frac{(ρ A ω^{2} x d x) \times x}{A Y}$
Integrating over limits,
$∴ \int_{0}^{1} d l = \frac{ρ ω^{2}}{Y} \int_{0}^{L} x^{2} d x$
$∴ l = \frac{ρ ω^{2}}{Y} {[\frac{x^{3}}{3}]}_{0}^{L} = \frac{ρ ω^{2}}{3 Y} [L^{3} - 0]$
$∴ l = \frac{ρ ω^{2} L^{3}}{3 Y}$
$\Rightarrow n = 3$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369789 If the interatomic spacing in a steel wire is $3.0 \overset{o}{A}$ and $Y_{s t e e l} = 20 \times 10^{10} N / m^{2}$ then force constant is

1

225 K N

2

6 \times 10^{- 2} N / \overset{o}{A}

3

6 \times 10^{- 5} N / \overset{o}{A}

4

4 \times 10^{- 5} N / \overset{o}{A}

Explanation:

$K = Y r_{0} = 20 \times 10^{10} \times 3 \times 10^{- 10} = 60 N / m$
$= 6 \times 10^{- 9} \frac{N}{\overset{o}{A}}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369790 The graph shows the extension of a wire of length $1 m$ suspended from the top of a roof at one end and with a load $W$ connected to the other end. If the cross sectional area of the wire is $1 {m m}^{2}$ , then the Young's modulus of the material of the wire is
supporting img

1

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

2

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

3

\frac{1}{2} \times 10^{11} {N m}^{- 2}

4

4 \times 10^{10} N m^{- 2}

Explanation:

$Y = \frac{F / A}{Δ l / l} = \frac{W l}{A Δ l} \Rightarrow \frac{W}{Δ l} = \frac{Y A}{l} =$ slope
$\Rightarrow Y = \frac{l}{A} \times ($ slope $) = \frac{1}{10^{- 6}} (\frac{40 - 20}{(2 - 1) \times 10^{- 3}})$
$= 2 \times 10^{10} {N m}^{- 2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369791 The following four wires of length $L$ and radius $r$ are made of the same material. Which of these will have the largest extension, when the same tension is applied?

1

L = 100 c m, r = 0.1 m m

2

L = 100 c m, r = 0.4 m m

3

L = 100 c m, r = 0.6 m m

4

L = 200 c m, r = 0.8 m m

Explanation:

$Δ L = \frac{L T}{A Y} = \frac{L T}{π r^{2} Y} = (\frac{T}{π Y}) \times \frac{L}{r^{2}}$
$\frac{L}{r^{2}}$ $\to$ max for $L = 100 c m & r = 0.1 m m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369792 A uniformly tapering conical wire is made from a material of Young's modulus Y and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values $R$ and $3 R$ , respectively, The upper end of the wire is fixed to a rigid support and a mass $M$ is suspended from its lower end. The equilibrium extended length, of this wire, would equal

1

L (1 + \frac{2}{9} \frac{M g}{π Y R^{2}})

2

L (1 + \frac{1}{9} \frac{M g}{π Y R^{2}})

3

L (1 + \frac{1}{3} \frac{M g}{π Y R^{2}})

4

L (1 + \frac{2}{3} \frac{M g}{π Y R^{2}})

Explanation:

Consider a thin disc of radius $r$ and width $dx$ . The small extension produced in the disc is dL.
supporting img

$\begin{matrix} \tan θ = \frac{r - R}{x} = \frac{3 R - R}{L} \Rightarrow r = R (1 + \frac{2 x}{L}); \\ Y = \frac{M g}{π r^{2} \frac{d L}{d x}} \\ Δ L = \frac{M g}{Y π R^{2}} \int_{0}^{L} \frac{d x}{{(1 + \frac{2 x}{L})}^{2}} = \frac{M g L}{3 π R^{2} Y} \\ L^{'} = L + Δ L = L (1 + \frac{1}{3} \frac{M g}{π R^{2} Y}) \end{matrix}$

JEE - 2016

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369788 A uniform rod of length $L$ and density $ρ$ is rotated in a horizontal plane about a vertical axis through one of its ends. The angular speed of rotation is $ω$ . If increase in the length is found to be $\frac{ρ ω^{2} L^{n}}{n Y}$ where, $Y$ is Young's modulus of rod, determine value of $n$ .

1 2

2 5

3 1

4 3

Explanation:

supporting img
$Y = \frac{F L}{A l}$
Here, $F =$ centrifugal force due to rotation.
$∴ l = \frac{F L}{A Y}$
From the figure, considering small elongation $d x$ of length segment $x$
$d l = \frac{d F \times x}{A Y} (1)$
where, $d F =$ centrifugal force of differential mass $d M$
but $d F = d M ω^{2} x = (ρ A d x) ω^{2} x$
$= ρ A ω^{2} x d x$
Substituting in equation (1),
$∴ d l = \frac{(ρ A ω^{2} x d x) \times x}{A Y}$
Integrating over limits,
$∴ \int_{0}^{1} d l = \frac{ρ ω^{2}}{Y} \int_{0}^{L} x^{2} d x$
$∴ l = \frac{ρ ω^{2}}{Y} {[\frac{x^{3}}{3}]}_{0}^{L} = \frac{ρ ω^{2}}{3 Y} [L^{3} - 0]$
$∴ l = \frac{ρ ω^{2} L^{3}}{3 Y}$
$\Rightarrow n = 3$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369789 If the interatomic spacing in a steel wire is $3.0 \overset{o}{A}$ and $Y_{s t e e l} = 20 \times 10^{10} N / m^{2}$ then force constant is

1

225 K N

2

6 \times 10^{- 2} N / \overset{o}{A}

3

6 \times 10^{- 5} N / \overset{o}{A}

4

4 \times 10^{- 5} N / \overset{o}{A}

Explanation:

$K = Y r_{0} = 20 \times 10^{10} \times 3 \times 10^{- 10} = 60 N / m$
$= 6 \times 10^{- 9} \frac{N}{\overset{o}{A}}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369790 The graph shows the extension of a wire of length $1 m$ suspended from the top of a roof at one end and with a load $W$ connected to the other end. If the cross sectional area of the wire is $1 {m m}^{2}$ , then the Young's modulus of the material of the wire is
supporting img

1

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

2

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

3

\frac{1}{2} \times 10^{11} {N m}^{- 2}

4

4 \times 10^{10} N m^{- 2}

Explanation:

$Y = \frac{F / A}{Δ l / l} = \frac{W l}{A Δ l} \Rightarrow \frac{W}{Δ l} = \frac{Y A}{l} =$ slope
$\Rightarrow Y = \frac{l}{A} \times ($ slope $) = \frac{1}{10^{- 6}} (\frac{40 - 20}{(2 - 1) \times 10^{- 3}})$
$= 2 \times 10^{10} {N m}^{- 2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369791 The following four wires of length $L$ and radius $r$ are made of the same material. Which of these will have the largest extension, when the same tension is applied?

1

L = 100 c m, r = 0.1 m m

2

L = 100 c m, r = 0.4 m m

3

L = 100 c m, r = 0.6 m m

4

L = 200 c m, r = 0.8 m m

Explanation:

$Δ L = \frac{L T}{A Y} = \frac{L T}{π r^{2} Y} = (\frac{T}{π Y}) \times \frac{L}{r^{2}}$
$\frac{L}{r^{2}}$ $\to$ max for $L = 100 c m & r = 0.1 m m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369792 A uniformly tapering conical wire is made from a material of Young's modulus Y and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values $R$ and $3 R$ , respectively, The upper end of the wire is fixed to a rigid support and a mass $M$ is suspended from its lower end. The equilibrium extended length, of this wire, would equal

1

L (1 + \frac{2}{9} \frac{M g}{π Y R^{2}})

2

L (1 + \frac{1}{9} \frac{M g}{π Y R^{2}})

3

L (1 + \frac{1}{3} \frac{M g}{π Y R^{2}})

4

L (1 + \frac{2}{3} \frac{M g}{π Y R^{2}})

Explanation:

Consider a thin disc of radius $r$ and width $dx$ . The small extension produced in the disc is dL.
supporting img

$\begin{matrix} \tan θ = \frac{r - R}{x} = \frac{3 R - R}{L} \Rightarrow r = R (1 + \frac{2 x}{L}); \\ Y = \frac{M g}{π r^{2} \frac{d L}{d x}} \\ Δ L = \frac{M g}{Y π R^{2}} \int_{0}^{L} \frac{d x}{{(1 + \frac{2 x}{L})}^{2}} = \frac{M g L}{3 π R^{2} Y} \\ L^{'} = L + Δ L = L (1 + \frac{1}{3} \frac{M g}{π R^{2} Y}) \end{matrix}$

JEE - 2016

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369788 A uniform rod of length $L$ and density $ρ$ is rotated in a horizontal plane about a vertical axis through one of its ends. The angular speed of rotation is $ω$ . If increase in the length is found to be $\frac{ρ ω^{2} L^{n}}{n Y}$ where, $Y$ is Young's modulus of rod, determine value of $n$ .

1 2

2 5

3 1

4 3

Explanation:

supporting img
$Y = \frac{F L}{A l}$
Here, $F =$ centrifugal force due to rotation.
$∴ l = \frac{F L}{A Y}$
From the figure, considering small elongation $d x$ of length segment $x$
$d l = \frac{d F \times x}{A Y} (1)$
where, $d F =$ centrifugal force of differential mass $d M$
but $d F = d M ω^{2} x = (ρ A d x) ω^{2} x$
$= ρ A ω^{2} x d x$
Substituting in equation (1),
$∴ d l = \frac{(ρ A ω^{2} x d x) \times x}{A Y}$
Integrating over limits,
$∴ \int_{0}^{1} d l = \frac{ρ ω^{2}}{Y} \int_{0}^{L} x^{2} d x$
$∴ l = \frac{ρ ω^{2}}{Y} {[\frac{x^{3}}{3}]}_{0}^{L} = \frac{ρ ω^{2}}{3 Y} [L^{3} - 0]$
$∴ l = \frac{ρ ω^{2} L^{3}}{3 Y}$
$\Rightarrow n = 3$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369789 If the interatomic spacing in a steel wire is $3.0 \overset{o}{A}$ and $Y_{s t e e l} = 20 \times 10^{10} N / m^{2}$ then force constant is

1

225 K N

2

6 \times 10^{- 2} N / \overset{o}{A}

3

6 \times 10^{- 5} N / \overset{o}{A}

4

4 \times 10^{- 5} N / \overset{o}{A}

Explanation:

$K = Y r_{0} = 20 \times 10^{10} \times 3 \times 10^{- 10} = 60 N / m$
$= 6 \times 10^{- 9} \frac{N}{\overset{o}{A}}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369790 The graph shows the extension of a wire of length $1 m$ suspended from the top of a roof at one end and with a load $W$ connected to the other end. If the cross sectional area of the wire is $1 {m m}^{2}$ , then the Young's modulus of the material of the wire is
supporting img

1

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

2

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

3

\frac{1}{2} \times 10^{11} {N m}^{- 2}

4

4 \times 10^{10} N m^{- 2}

Explanation:

$Y = \frac{F / A}{Δ l / l} = \frac{W l}{A Δ l} \Rightarrow \frac{W}{Δ l} = \frac{Y A}{l} =$ slope
$\Rightarrow Y = \frac{l}{A} \times ($ slope $) = \frac{1}{10^{- 6}} (\frac{40 - 20}{(2 - 1) \times 10^{- 3}})$
$= 2 \times 10^{10} {N m}^{- 2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369791 The following four wires of length $L$ and radius $r$ are made of the same material. Which of these will have the largest extension, when the same tension is applied?

1

L = 100 c m, r = 0.1 m m

2

L = 100 c m, r = 0.4 m m

3

L = 100 c m, r = 0.6 m m

4

L = 200 c m, r = 0.8 m m

Explanation:

$Δ L = \frac{L T}{A Y} = \frac{L T}{π r^{2} Y} = (\frac{T}{π Y}) \times \frac{L}{r^{2}}$
$\frac{L}{r^{2}}$ $\to$ max for $L = 100 c m & r = 0.1 m m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369792 A uniformly tapering conical wire is made from a material of Young's modulus Y and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values $R$ and $3 R$ , respectively, The upper end of the wire is fixed to a rigid support and a mass $M$ is suspended from its lower end. The equilibrium extended length, of this wire, would equal

1

L (1 + \frac{2}{9} \frac{M g}{π Y R^{2}})

2

L (1 + \frac{1}{9} \frac{M g}{π Y R^{2}})

3

L (1 + \frac{1}{3} \frac{M g}{π Y R^{2}})

4

L (1 + \frac{2}{3} \frac{M g}{π Y R^{2}})

Explanation:

Consider a thin disc of radius $r$ and width $dx$ . The small extension produced in the disc is dL.
supporting img

$\begin{matrix} \tan θ = \frac{r - R}{x} = \frac{3 R - R}{L} \Rightarrow r = R (1 + \frac{2 x}{L}); \\ Y = \frac{M g}{π r^{2} \frac{d L}{d x}} \\ Δ L = \frac{M g}{Y π R^{2}} \int_{0}^{L} \frac{d x}{{(1 + \frac{2 x}{L})}^{2}} = \frac{M g L}{3 π R^{2} Y} \\ L^{'} = L + Δ L = L (1 + \frac{1}{3} \frac{M g}{π R^{2} Y}) \end{matrix}$

JEE - 2016