QBManager

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369668 A stress of $6 \times 10^{6} N m^{- 2}$ required for breaking a material. The density $ρ$ of the material is $3 \times 10^{3} k g m^{- 3} .$ If the wire is to break under its own weight, the length of the wire made of that material should be (take, $g = 10 m s^{- 2}$ )

1

20 m

2

200 m

3

100 m

4

2000 m

Explanation:

The minimum stress after which the wire
breaks, i.e.
$S t r e s s = \frac{F o r c e}{A r e a} = \frac{m g}{A}$
$= \frac{V ρ g}{A} = \frac{L A ρ g}{A} = L ρ g$
Given, stress $= 6 \times 10^{6} N m^{- 2},$
$ρ = 3 \times 10^{3} k g m^{- 3}$
and $g = 10 m s^{- 2}$
$L = \frac{S t r e s s}{ρ g} = \frac{6 \times 10^{6}}{3 \times 10^{3} \times 10}$
$= 2 \times 10^{2} = 200 m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369669 A ring of radius $R$ made of lead wire breaking strength $σ$ and density $ρ$ , roatated about stationary vertical axis passing through its centre and perpendicular to the plane of the ring. Calculate the number of rotations per second at which the ring ruptures

1

n = \frac{1}{π R} \sqrt{\frac{σ}{ρ}}

2

n = \frac{1}{2 π R} \sqrt{\frac{σ}{ρ}}

3

n = \frac{1}{R} \sqrt{\frac{σ}{ρ}}

4

n = \frac{1}{2 R} \sqrt{\frac{σ}{ρ}}

Explanation:

Let us consider small part of ring, which subtend an angle $θ$ at the centre. Due to rotation, each part of ring experience an outward force.

supporting img

Mass of the element $(d m) = ρ d V = ρ (A R θ)$
$2 T \sin θ / 2 = (d m) ω^{2} R$
$\Rightarrow T = ρ A ω^{2} R^{2}$
The stress at any section of the ring
$\Rightarrow f = \frac{T}{A} = \frac{ρ A ω^{2} R^{2}}{A} = ρ ω^{2} R^{2}$
Rupture takes place when $f = σ$
$\Rightarrow σ = ρ ω^{2} R^{2} \Rightarrow ω = \sqrt{\frac{σ}{ρ R^{2}}}$
and $n = \frac{1}{2 π} \sqrt{\frac{ω}{ρ R^{2}}} = \frac{1}{2 π R} \sqrt{\frac{σ}{ρ}}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369670 The elastic limit of brass is $3.5 \times 10^{10} N / m^{2}$ . Find the maximum load that can be applied to a brass wire of $0.75 m m$ diameter without exceeding the elastic limit.

1

4.12 \times 10^{4} N

2

5.15 \times 10^{4} N

3

0.55 \times 10^{4} N

4

1.55 \times 10^{4} N

Explanation:

Stress $= \frac{F}{A}$ ; For elastic limit, $s t r e s s = 3.5 \times 10^{10} N / m^{2} (g i v e n) .$
$T h u s, F = (π r^{2}) \times s t r e s s$
$= 3.14 \times {(\frac{0.75}{2} \times 10^{- 3})}^{2} \times 3.5 \times 10^{10}$
$= 3.14 \times \frac{{(0.75)}^{2}}{4} \times 3.5 \times 10^{4}$
$= 1.55 \times 10^{4} N$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369671 A steel wire of cross-sectional area $3 \times 10^{- 6} m^{2}$ can withstand a maximum strain of $10^{- 3}$ . Young's modulus of steel is $2 \times 10^{11} N / m^{2} .$ The maximum mass the wire can hold is (Take $g = 10 m / s^{2}$ )

1

60 k g

2

40 kg

3

100 k g

4

80 k g

Explanation:

Breaking force $\propto$ Cross-sectional area.
$max strain = \frac{max stress}{Y} = \frac{m g / A}{Y}$
$A = 3 \times 10^{- 6} m^{2}$
$∴ m = \frac{Y strain A}{g} = \frac{2 \times 10^{11} \times 10^{- 3} \times 3 \times 10^{- 6}}{10}$
$= 60 k g .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369668 A stress of $6 \times 10^{6} N m^{- 2}$ required for breaking a material. The density $ρ$ of the material is $3 \times 10^{3} k g m^{- 3} .$ If the wire is to break under its own weight, the length of the wire made of that material should be (take, $g = 10 m s^{- 2}$ )

1

20 m

2

200 m

3

100 m

4

2000 m

Explanation:

The minimum stress after which the wire
breaks, i.e.
$S t r e s s = \frac{F o r c e}{A r e a} = \frac{m g}{A}$
$= \frac{V ρ g}{A} = \frac{L A ρ g}{A} = L ρ g$
Given, stress $= 6 \times 10^{6} N m^{- 2},$
$ρ = 3 \times 10^{3} k g m^{- 3}$
and $g = 10 m s^{- 2}$
$L = \frac{S t r e s s}{ρ g} = \frac{6 \times 10^{6}}{3 \times 10^{3} \times 10}$
$= 2 \times 10^{2} = 200 m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369669 A ring of radius $R$ made of lead wire breaking strength $σ$ and density $ρ$ , roatated about stationary vertical axis passing through its centre and perpendicular to the plane of the ring. Calculate the number of rotations per second at which the ring ruptures

1

n = \frac{1}{π R} \sqrt{\frac{σ}{ρ}}

2

n = \frac{1}{2 π R} \sqrt{\frac{σ}{ρ}}

3

n = \frac{1}{R} \sqrt{\frac{σ}{ρ}}

4

n = \frac{1}{2 R} \sqrt{\frac{σ}{ρ}}

Explanation:

Let us consider small part of ring, which subtend an angle $θ$ at the centre. Due to rotation, each part of ring experience an outward force.

supporting img

Mass of the element $(d m) = ρ d V = ρ (A R θ)$
$2 T \sin θ / 2 = (d m) ω^{2} R$
$\Rightarrow T = ρ A ω^{2} R^{2}$
The stress at any section of the ring
$\Rightarrow f = \frac{T}{A} = \frac{ρ A ω^{2} R^{2}}{A} = ρ ω^{2} R^{2}$
Rupture takes place when $f = σ$
$\Rightarrow σ = ρ ω^{2} R^{2} \Rightarrow ω = \sqrt{\frac{σ}{ρ R^{2}}}$
and $n = \frac{1}{2 π} \sqrt{\frac{ω}{ρ R^{2}}} = \frac{1}{2 π R} \sqrt{\frac{σ}{ρ}}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369670 The elastic limit of brass is $3.5 \times 10^{10} N / m^{2}$ . Find the maximum load that can be applied to a brass wire of $0.75 m m$ diameter without exceeding the elastic limit.

1

4.12 \times 10^{4} N

2

5.15 \times 10^{4} N

3

0.55 \times 10^{4} N

4

1.55 \times 10^{4} N

Explanation:

Stress $= \frac{F}{A}$ ; For elastic limit, $s t r e s s = 3.5 \times 10^{10} N / m^{2} (g i v e n) .$
$T h u s, F = (π r^{2}) \times s t r e s s$
$= 3.14 \times {(\frac{0.75}{2} \times 10^{- 3})}^{2} \times 3.5 \times 10^{10}$
$= 3.14 \times \frac{{(0.75)}^{2}}{4} \times 3.5 \times 10^{4}$
$= 1.55 \times 10^{4} N$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369671 A steel wire of cross-sectional area $3 \times 10^{- 6} m^{2}$ can withstand a maximum strain of $10^{- 3}$ . Young's modulus of steel is $2 \times 10^{11} N / m^{2} .$ The maximum mass the wire can hold is (Take $g = 10 m / s^{2}$ )

1

60 k g

2

40 kg

3

100 k g

4

80 k g

Explanation:

Breaking force $\propto$ Cross-sectional area.
$max strain = \frac{max stress}{Y} = \frac{m g / A}{Y}$
$A = 3 \times 10^{- 6} m^{2}$
$∴ m = \frac{Y strain A}{g} = \frac{2 \times 10^{11} \times 10^{- 3} \times 3 \times 10^{- 6}}{10}$
$= 60 k g .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369668 A stress of $6 \times 10^{6} N m^{- 2}$ required for breaking a material. The density $ρ$ of the material is $3 \times 10^{3} k g m^{- 3} .$ If the wire is to break under its own weight, the length of the wire made of that material should be (take, $g = 10 m s^{- 2}$ )

1

20 m

2

200 m

3

100 m

4

2000 m

Explanation:

The minimum stress after which the wire
breaks, i.e.
$S t r e s s = \frac{F o r c e}{A r e a} = \frac{m g}{A}$
$= \frac{V ρ g}{A} = \frac{L A ρ g}{A} = L ρ g$
Given, stress $= 6 \times 10^{6} N m^{- 2},$
$ρ = 3 \times 10^{3} k g m^{- 3}$
and $g = 10 m s^{- 2}$
$L = \frac{S t r e s s}{ρ g} = \frac{6 \times 10^{6}}{3 \times 10^{3} \times 10}$
$= 2 \times 10^{2} = 200 m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369669 A ring of radius $R$ made of lead wire breaking strength $σ$ and density $ρ$ , roatated about stationary vertical axis passing through its centre and perpendicular to the plane of the ring. Calculate the number of rotations per second at which the ring ruptures

1

n = \frac{1}{π R} \sqrt{\frac{σ}{ρ}}

2

n = \frac{1}{2 π R} \sqrt{\frac{σ}{ρ}}

3

n = \frac{1}{R} \sqrt{\frac{σ}{ρ}}

4

n = \frac{1}{2 R} \sqrt{\frac{σ}{ρ}}

Explanation:

Let us consider small part of ring, which subtend an angle $θ$ at the centre. Due to rotation, each part of ring experience an outward force.

supporting img

Mass of the element $(d m) = ρ d V = ρ (A R θ)$
$2 T \sin θ / 2 = (d m) ω^{2} R$
$\Rightarrow T = ρ A ω^{2} R^{2}$
The stress at any section of the ring
$\Rightarrow f = \frac{T}{A} = \frac{ρ A ω^{2} R^{2}}{A} = ρ ω^{2} R^{2}$
Rupture takes place when $f = σ$
$\Rightarrow σ = ρ ω^{2} R^{2} \Rightarrow ω = \sqrt{\frac{σ}{ρ R^{2}}}$
and $n = \frac{1}{2 π} \sqrt{\frac{ω}{ρ R^{2}}} = \frac{1}{2 π R} \sqrt{\frac{σ}{ρ}}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369670 The elastic limit of brass is $3.5 \times 10^{10} N / m^{2}$ . Find the maximum load that can be applied to a brass wire of $0.75 m m$ diameter without exceeding the elastic limit.

1

4.12 \times 10^{4} N

2

5.15 \times 10^{4} N

3

0.55 \times 10^{4} N

4

1.55 \times 10^{4} N

Explanation:

Stress $= \frac{F}{A}$ ; For elastic limit, $s t r e s s = 3.5 \times 10^{10} N / m^{2} (g i v e n) .$
$T h u s, F = (π r^{2}) \times s t r e s s$
$= 3.14 \times {(\frac{0.75}{2} \times 10^{- 3})}^{2} \times 3.5 \times 10^{10}$
$= 3.14 \times \frac{{(0.75)}^{2}}{4} \times 3.5 \times 10^{4}$
$= 1.55 \times 10^{4} N$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369671 A steel wire of cross-sectional area $3 \times 10^{- 6} m^{2}$ can withstand a maximum strain of $10^{- 3}$ . Young's modulus of steel is $2 \times 10^{11} N / m^{2} .$ The maximum mass the wire can hold is (Take $g = 10 m / s^{2}$ )

1

60 k g

2

40 kg

3

100 k g

4

80 k g

Explanation:

Breaking force $\propto$ Cross-sectional area.
$max strain = \frac{max stress}{Y} = \frac{m g / A}{Y}$
$A = 3 \times 10^{- 6} m^{2}$
$∴ m = \frac{Y strain A}{g} = \frac{2 \times 10^{11} \times 10^{- 3} \times 3 \times 10^{- 6}}{10}$
$= 60 k g .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369668 A stress of $6 \times 10^{6} N m^{- 2}$ required for breaking a material. The density $ρ$ of the material is $3 \times 10^{3} k g m^{- 3} .$ If the wire is to break under its own weight, the length of the wire made of that material should be (take, $g = 10 m s^{- 2}$ )

1

20 m

2

200 m

3

100 m

4

2000 m

Explanation:

The minimum stress after which the wire
breaks, i.e.
$S t r e s s = \frac{F o r c e}{A r e a} = \frac{m g}{A}$
$= \frac{V ρ g}{A} = \frac{L A ρ g}{A} = L ρ g$
Given, stress $= 6 \times 10^{6} N m^{- 2},$
$ρ = 3 \times 10^{3} k g m^{- 3}$
and $g = 10 m s^{- 2}$
$L = \frac{S t r e s s}{ρ g} = \frac{6 \times 10^{6}}{3 \times 10^{3} \times 10}$
$= 2 \times 10^{2} = 200 m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369669 A ring of radius $R$ made of lead wire breaking strength $σ$ and density $ρ$ , roatated about stationary vertical axis passing through its centre and perpendicular to the plane of the ring. Calculate the number of rotations per second at which the ring ruptures

1

n = \frac{1}{π R} \sqrt{\frac{σ}{ρ}}

2

n = \frac{1}{2 π R} \sqrt{\frac{σ}{ρ}}

3

n = \frac{1}{R} \sqrt{\frac{σ}{ρ}}

4

n = \frac{1}{2 R} \sqrt{\frac{σ}{ρ}}

Explanation:

Let us consider small part of ring, which subtend an angle $θ$ at the centre. Due to rotation, each part of ring experience an outward force.

supporting img

Mass of the element $(d m) = ρ d V = ρ (A R θ)$
$2 T \sin θ / 2 = (d m) ω^{2} R$
$\Rightarrow T = ρ A ω^{2} R^{2}$
The stress at any section of the ring
$\Rightarrow f = \frac{T}{A} = \frac{ρ A ω^{2} R^{2}}{A} = ρ ω^{2} R^{2}$
Rupture takes place when $f = σ$
$\Rightarrow σ = ρ ω^{2} R^{2} \Rightarrow ω = \sqrt{\frac{σ}{ρ R^{2}}}$
and $n = \frac{1}{2 π} \sqrt{\frac{ω}{ρ R^{2}}} = \frac{1}{2 π R} \sqrt{\frac{σ}{ρ}}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369670 The elastic limit of brass is $3.5 \times 10^{10} N / m^{2}$ . Find the maximum load that can be applied to a brass wire of $0.75 m m$ diameter without exceeding the elastic limit.

1

4.12 \times 10^{4} N

2

5.15 \times 10^{4} N

3

0.55 \times 10^{4} N

4

1.55 \times 10^{4} N

Explanation:

Stress $= \frac{F}{A}$ ; For elastic limit, $s t r e s s = 3.5 \times 10^{10} N / m^{2} (g i v e n) .$
$T h u s, F = (π r^{2}) \times s t r e s s$
$= 3.14 \times {(\frac{0.75}{2} \times 10^{- 3})}^{2} \times 3.5 \times 10^{10}$
$= 3.14 \times \frac{{(0.75)}^{2}}{4} \times 3.5 \times 10^{4}$
$= 1.55 \times 10^{4} N$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369671 A steel wire of cross-sectional area $3 \times 10^{- 6} m^{2}$ can withstand a maximum strain of $10^{- 3}$ . Young's modulus of steel is $2 \times 10^{11} N / m^{2} .$ The maximum mass the wire can hold is (Take $g = 10 m / s^{2}$ )

1

60 k g

2

40 kg

3

100 k g

4

80 k g

Explanation:

Breaking force $\propto$ Cross-sectional area.
$max strain = \frac{max stress}{Y} = \frac{m g / A}{Y}$
$A = 3 \times 10^{- 6} m^{2}$
$∴ m = \frac{Y strain A}{g} = \frac{2 \times 10^{11} \times 10^{- 3} \times 3 \times 10^{- 6}}{10}$
$= 60 k g .$

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: