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PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369780 A wire of length $L$ is hanging from a fixed support. The length changes from $L_{1}$ to $L_{2}$ when masses $M_{1}$ and $M_{2}$ are suspended respectively from its free end. Then $L$ is equal to

1

\sqrt{L_{1} L_{2}}

2

\frac{L_{1} + L_{2}}{2}

3

\frac{L_{1} M_{2} + L_{2} M_{1}}{M_{1} + M_{2}}

4

\frac{L_{1} M_{2} - L_{2} M_{1}}{M_{2} - M_{1}}

Explanation:

L be original length of the wire. are tensions in the wire when masses are hung. When a mass is suspended from the wire, change in length of wire is $Δ L_{1} = L_{1} - L$
When a mass is suspended from it, change in length of wire is $Δ L_{2} = L_{2} - L$
From figure (2), $T_{1} = M_{1} g (1)$
From figure (3), $T_{2} = M_{2} g (2)$
As Young’s modulus, $Y = \frac{T_{1} L}{A Δ L_{1}} = \frac{T_{2} L}{A Δ L_{2}}$
$\frac{M_{1} g}{L_{1} - L} = \frac{M_{2} g}{L_{2} - L} [Using (1) and (2)]$ $L = \frac{L_{1} M_{2} - L_{2} M_{1}}{M_{2} - M_{1}} .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369781 A rigid bar of mass $M$ is supported symmetrically by three wires each of length $l$ . Those at each end are of copper and the middle one is of iron. The ratio of their diameter, if each is to have the same tension, is equal to

1

γ_{copper} / γ_{iron}

2

\sqrt{\frac{γ_{iron}}{γ_{copper}}}

3

\frac{γ_{iron}^{2}}{γ_{copper}^{2}}

4

\frac{γ_{iron}}{γ_{copper}}

Explanation:

As,
$\begin{aligned} Y = \frac{m g L}{π r^{2} Δ l} (Y = Young's modulus) \\ \Rightarrow Y \propto \frac{1}{d^{2}} \\ \Rightarrow d \propto \sqrt{\frac{1}{Y}} \end{aligned}$
$\Rightarrow \frac{d_{copper}}{d_{iron}} = \sqrt{\frac{Y_{iron}}{Y_{copper}}}$

NCERT Exemplar

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369782 Two wires are made of the same material and have the same volume. The first wire has crosssectional area A and the second wire has crosssectional area $3 A$ . If the length of the first wire is increased by $l$ on applying a force $F$ , how much force is needed to stretch the second wire by the same amount ?

1

4 F

2

F

3

6 F

4

9 F

Explanation:

Volume $V = A \times L_{1} = 3 A \times L_{2}$
$L_{2} = \frac{L_{1}}{3}$
$Stress = Y (strain)$
$F = (\frac{A Y}{L}) Δ L$
$F^{'} = [\frac{3 A Y}{(\frac{L}{3})}] \cdot Δ L = 9 F$

NEET - 2018

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369783 A force is applied to a steel wire ' $A$ ', rigidly clamped at one end. As a result elongation in the wire is $0.2 m m$ . If same force is applied to another steel wire ' $B$ ' of double the length and a diameter 2.4 times that of the wire ' $A$ ', the elongation in the wire ' $B$ ' will be (wires having uniform circular cross sections)

1

2.77 \times 10^{- 2} m m

2

6.9 \times 10^{- 2} m m

3

6.06 \times 10^{- 2} m m

4

3.0 \times 10^{- 2} m m

Explanation:

As, elongation in wire is $= 0.2 m m$ .
So, $Y = \frac{F l}{π r_{a}^{2} (0.2)} (1)$
[where $r_{a}$ is the radius of wire $a$ ]
Let ' $x$ ' be the elongation produced in wire $b$ .
$Y = \frac{F (2 l)}{π {(2.4 r_{a})}^{2} x} (2)$
$[as, d_{b} = 2.4 d_{a} or r_{b} = 2.4 r_{a}]$
From eq. (1) and (2), we have
$\frac{F l}{π r_{a}^{2} (0.2)} = \frac{F (2 l)}{π {(2.4 r_{a})}^{2} x}$
$x = \frac{0.2 \times 2}{5.76}$ or $x = 6.9 \times 10^{- 2} m m$

JEE - 2023

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369780 A wire of length $L$ is hanging from a fixed support. The length changes from $L_{1}$ to $L_{2}$ when masses $M_{1}$ and $M_{2}$ are suspended respectively from its free end. Then $L$ is equal to

1

\sqrt{L_{1} L_{2}}

2

\frac{L_{1} + L_{2}}{2}

3

\frac{L_{1} M_{2} + L_{2} M_{1}}{M_{1} + M_{2}}

4

\frac{L_{1} M_{2} - L_{2} M_{1}}{M_{2} - M_{1}}

Explanation:

L be original length of the wire. are tensions in the wire when masses are hung. When a mass is suspended from the wire, change in length of wire is $Δ L_{1} = L_{1} - L$
When a mass is suspended from it, change in length of wire is $Δ L_{2} = L_{2} - L$
From figure (2), $T_{1} = M_{1} g (1)$
From figure (3), $T_{2} = M_{2} g (2)$
As Young’s modulus, $Y = \frac{T_{1} L}{A Δ L_{1}} = \frac{T_{2} L}{A Δ L_{2}}$
$\frac{M_{1} g}{L_{1} - L} = \frac{M_{2} g}{L_{2} - L} [Using (1) and (2)]$ $L = \frac{L_{1} M_{2} - L_{2} M_{1}}{M_{2} - M_{1}} .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369781 A rigid bar of mass $M$ is supported symmetrically by three wires each of length $l$ . Those at each end are of copper and the middle one is of iron. The ratio of their diameter, if each is to have the same tension, is equal to

1

γ_{copper} / γ_{iron}

2

\sqrt{\frac{γ_{iron}}{γ_{copper}}}

3

\frac{γ_{iron}^{2}}{γ_{copper}^{2}}

4

\frac{γ_{iron}}{γ_{copper}}

Explanation:

As,
$\begin{aligned} Y = \frac{m g L}{π r^{2} Δ l} (Y = Young's modulus) \\ \Rightarrow Y \propto \frac{1}{d^{2}} \\ \Rightarrow d \propto \sqrt{\frac{1}{Y}} \end{aligned}$
$\Rightarrow \frac{d_{copper}}{d_{iron}} = \sqrt{\frac{Y_{iron}}{Y_{copper}}}$

NCERT Exemplar

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369782 Two wires are made of the same material and have the same volume. The first wire has crosssectional area A and the second wire has crosssectional area $3 A$ . If the length of the first wire is increased by $l$ on applying a force $F$ , how much force is needed to stretch the second wire by the same amount ?

1

4 F

2

F

3

6 F

4

9 F

Explanation:

Volume $V = A \times L_{1} = 3 A \times L_{2}$
$L_{2} = \frac{L_{1}}{3}$
$Stress = Y (strain)$
$F = (\frac{A Y}{L}) Δ L$
$F^{'} = [\frac{3 A Y}{(\frac{L}{3})}] \cdot Δ L = 9 F$

NEET - 2018

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369783 A force is applied to a steel wire ' $A$ ', rigidly clamped at one end. As a result elongation in the wire is $0.2 m m$ . If same force is applied to another steel wire ' $B$ ' of double the length and a diameter 2.4 times that of the wire ' $A$ ', the elongation in the wire ' $B$ ' will be (wires having uniform circular cross sections)

1

2.77 \times 10^{- 2} m m

2

6.9 \times 10^{- 2} m m

3

6.06 \times 10^{- 2} m m

4

3.0 \times 10^{- 2} m m

Explanation:

As, elongation in wire is $= 0.2 m m$ .
So, $Y = \frac{F l}{π r_{a}^{2} (0.2)} (1)$
[where $r_{a}$ is the radius of wire $a$ ]
Let ' $x$ ' be the elongation produced in wire $b$ .
$Y = \frac{F (2 l)}{π {(2.4 r_{a})}^{2} x} (2)$
$[as, d_{b} = 2.4 d_{a} or r_{b} = 2.4 r_{a}]$
From eq. (1) and (2), we have
$\frac{F l}{π r_{a}^{2} (0.2)} = \frac{F (2 l)}{π {(2.4 r_{a})}^{2} x}$
$x = \frac{0.2 \times 2}{5.76}$ or $x = 6.9 \times 10^{- 2} m m$

JEE - 2023

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369780 A wire of length $L$ is hanging from a fixed support. The length changes from $L_{1}$ to $L_{2}$ when masses $M_{1}$ and $M_{2}$ are suspended respectively from its free end. Then $L$ is equal to

1

\sqrt{L_{1} L_{2}}

2

\frac{L_{1} + L_{2}}{2}

3

\frac{L_{1} M_{2} + L_{2} M_{1}}{M_{1} + M_{2}}

4

\frac{L_{1} M_{2} - L_{2} M_{1}}{M_{2} - M_{1}}

Explanation:

L be original length of the wire. are tensions in the wire when masses are hung. When a mass is suspended from the wire, change in length of wire is $Δ L_{1} = L_{1} - L$
When a mass is suspended from it, change in length of wire is $Δ L_{2} = L_{2} - L$
From figure (2), $T_{1} = M_{1} g (1)$
From figure (3), $T_{2} = M_{2} g (2)$
As Young’s modulus, $Y = \frac{T_{1} L}{A Δ L_{1}} = \frac{T_{2} L}{A Δ L_{2}}$
$\frac{M_{1} g}{L_{1} - L} = \frac{M_{2} g}{L_{2} - L} [Using (1) and (2)]$ $L = \frac{L_{1} M_{2} - L_{2} M_{1}}{M_{2} - M_{1}} .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369781 A rigid bar of mass $M$ is supported symmetrically by three wires each of length $l$ . Those at each end are of copper and the middle one is of iron. The ratio of their diameter, if each is to have the same tension, is equal to

1

γ_{copper} / γ_{iron}

2

\sqrt{\frac{γ_{iron}}{γ_{copper}}}

3

\frac{γ_{iron}^{2}}{γ_{copper}^{2}}

4

\frac{γ_{iron}}{γ_{copper}}

Explanation:

As,
$\begin{aligned} Y = \frac{m g L}{π r^{2} Δ l} (Y = Young's modulus) \\ \Rightarrow Y \propto \frac{1}{d^{2}} \\ \Rightarrow d \propto \sqrt{\frac{1}{Y}} \end{aligned}$
$\Rightarrow \frac{d_{copper}}{d_{iron}} = \sqrt{\frac{Y_{iron}}{Y_{copper}}}$

NCERT Exemplar

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369782 Two wires are made of the same material and have the same volume. The first wire has crosssectional area A and the second wire has crosssectional area $3 A$ . If the length of the first wire is increased by $l$ on applying a force $F$ , how much force is needed to stretch the second wire by the same amount ?

1

4 F

2

F

3

6 F

4

9 F

Explanation:

Volume $V = A \times L_{1} = 3 A \times L_{2}$
$L_{2} = \frac{L_{1}}{3}$
$Stress = Y (strain)$
$F = (\frac{A Y}{L}) Δ L$
$F^{'} = [\frac{3 A Y}{(\frac{L}{3})}] \cdot Δ L = 9 F$

NEET - 2018

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369783 A force is applied to a steel wire ' $A$ ', rigidly clamped at one end. As a result elongation in the wire is $0.2 m m$ . If same force is applied to another steel wire ' $B$ ' of double the length and a diameter 2.4 times that of the wire ' $A$ ', the elongation in the wire ' $B$ ' will be (wires having uniform circular cross sections)

1

2.77 \times 10^{- 2} m m

2

6.9 \times 10^{- 2} m m

3

6.06 \times 10^{- 2} m m

4

3.0 \times 10^{- 2} m m

Explanation:

As, elongation in wire is $= 0.2 m m$ .
So, $Y = \frac{F l}{π r_{a}^{2} (0.2)} (1)$
[where $r_{a}$ is the radius of wire $a$ ]
Let ' $x$ ' be the elongation produced in wire $b$ .
$Y = \frac{F (2 l)}{π {(2.4 r_{a})}^{2} x} (2)$
$[as, d_{b} = 2.4 d_{a} or r_{b} = 2.4 r_{a}]$
From eq. (1) and (2), we have
$\frac{F l}{π r_{a}^{2} (0.2)} = \frac{F (2 l)}{π {(2.4 r_{a})}^{2} x}$
$x = \frac{0.2 \times 2}{5.76}$ or $x = 6.9 \times 10^{- 2} m m$

JEE - 2023

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369780 A wire of length $L$ is hanging from a fixed support. The length changes from $L_{1}$ to $L_{2}$ when masses $M_{1}$ and $M_{2}$ are suspended respectively from its free end. Then $L$ is equal to

1

\sqrt{L_{1} L_{2}}

2

\frac{L_{1} + L_{2}}{2}

3

\frac{L_{1} M_{2} + L_{2} M_{1}}{M_{1} + M_{2}}

4

\frac{L_{1} M_{2} - L_{2} M_{1}}{M_{2} - M_{1}}

Explanation:

L be original length of the wire. are tensions in the wire when masses are hung. When a mass is suspended from the wire, change in length of wire is $Δ L_{1} = L_{1} - L$
When a mass is suspended from it, change in length of wire is $Δ L_{2} = L_{2} - L$
From figure (2), $T_{1} = M_{1} g (1)$
From figure (3), $T_{2} = M_{2} g (2)$
As Young’s modulus, $Y = \frac{T_{1} L}{A Δ L_{1}} = \frac{T_{2} L}{A Δ L_{2}}$
$\frac{M_{1} g}{L_{1} - L} = \frac{M_{2} g}{L_{2} - L} [Using (1) and (2)]$ $L = \frac{L_{1} M_{2} - L_{2} M_{1}}{M_{2} - M_{1}} .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369781 A rigid bar of mass $M$ is supported symmetrically by three wires each of length $l$ . Those at each end are of copper and the middle one is of iron. The ratio of their diameter, if each is to have the same tension, is equal to

1

γ_{copper} / γ_{iron}

2

\sqrt{\frac{γ_{iron}}{γ_{copper}}}

3

\frac{γ_{iron}^{2}}{γ_{copper}^{2}}

4

\frac{γ_{iron}}{γ_{copper}}

Explanation:

As,
$\begin{aligned} Y = \frac{m g L}{π r^{2} Δ l} (Y = Young's modulus) \\ \Rightarrow Y \propto \frac{1}{d^{2}} \\ \Rightarrow d \propto \sqrt{\frac{1}{Y}} \end{aligned}$
$\Rightarrow \frac{d_{copper}}{d_{iron}} = \sqrt{\frac{Y_{iron}}{Y_{copper}}}$

NCERT Exemplar

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369782 Two wires are made of the same material and have the same volume. The first wire has crosssectional area A and the second wire has crosssectional area $3 A$ . If the length of the first wire is increased by $l$ on applying a force $F$ , how much force is needed to stretch the second wire by the same amount ?

1

4 F

2

F

3

6 F

4

9 F

Explanation:

Volume $V = A \times L_{1} = 3 A \times L_{2}$
$L_{2} = \frac{L_{1}}{3}$
$Stress = Y (strain)$
$F = (\frac{A Y}{L}) Δ L$
$F^{'} = [\frac{3 A Y}{(\frac{L}{3})}] \cdot Δ L = 9 F$

NEET - 2018

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369783 A force is applied to a steel wire ' $A$ ', rigidly clamped at one end. As a result elongation in the wire is $0.2 m m$ . If same force is applied to another steel wire ' $B$ ' of double the length and a diameter 2.4 times that of the wire ' $A$ ', the elongation in the wire ' $B$ ' will be (wires having uniform circular cross sections)

1

2.77 \times 10^{- 2} m m

2

6.9 \times 10^{- 2} m m

3

6.06 \times 10^{- 2} m m

4

3.0 \times 10^{- 2} m m

Explanation:

As, elongation in wire is $= 0.2 m m$ .
So, $Y = \frac{F l}{π r_{a}^{2} (0.2)} (1)$
[where $r_{a}$ is the radius of wire $a$ ]
Let ' $x$ ' be the elongation produced in wire $b$ .
$Y = \frac{F (2 l)}{π {(2.4 r_{a})}^{2} x} (2)$
$[as, d_{b} = 2.4 d_{a} or r_{b} = 2.4 r_{a}]$
From eq. (1) and (2), we have
$\frac{F l}{π r_{a}^{2} (0.2)} = \frac{F (2 l)}{π {(2.4 r_{a})}^{2} x}$
$x = \frac{0.2 \times 2}{5.76}$ or $x = 6.9 \times 10^{- 2} m m$

JEE - 2023

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