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PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369771 Two parallel and opposite forces, each of magnitude $4000 N$ , are applied tangentially to the upper and lower faces of a cubical metal block $25 c m$ on a side. Find the displacement of the upper surface relative to the lower surface . The shear modulus for the metal is $80 G p a$ .

1

2.0 \times 10^{- 5} c m

2

5.0 \times 10^{- 5} c m

3

9.0 \times 10^{- 5} c m

4

8.0 \times 10^{- 5} c m

Explanation:

We use the approximate from $S = F / (A ϕ)$ .
Putting $S = 8 \times 10^{10} N / m^{2}, F = 4000 N$ and $A = (0.25 m)^{2}$
$= 6.25 \times 10^{- 2} m^{2}$ in the above equation and solving for $ϕ$ , we get
$ϕ = \frac{(4000 N)}{(6.25 \times 10^{- 2} m^{2}) (8 \times 10^{10} N / m^{2})}$
$= 8.0 \times 10^{- 7} r a d$
The displacement of the upper surface is given by $d = L ϕ$ , where $L$ is an edge of the cube.
$∴ d = (8.0 \times 10^{- 7}) (25 c m) = 2.0 \times 10^{- 5} c m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369772 Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass $M$ is attached to each of the free ends at the centre of the rods.

1 Both the rods will elongate but there shall be no perceptible change in shape

2 The steel rod will elongate and change shape but the rubber rod will only elongate

3 The steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse.

4 The steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.

NCERT Exemplar

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369773 The upper end of a wire of radius $4 m m$ and length $100 c m$ is clamped and its other end is twisted through an angle of $30^{\circ}$ . Then angle of shear is

1

{0.12}^{\circ}

2

12^{\circ}

3

{0.012}^{\circ}

4

{1.2}^{\circ}

Explanation:

Angle of shear
$ϕ = \frac{r θ}{L} = \frac{4 \times 10^{- 1}}{100} \times 30^{\circ} = {0.12}^{\circ}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369774 A $5.0 c m$ cube of substance has its upper face displaced by $0.65 c m$ by a tangential force of $0.25 N$ Calculate the modulus of rigidity of the substance.

1

769.2 N m^{- 2}

2

754.3 N m^{- 2}

3

735.2 N m^{- 2}

4

714.3 N m^{- 2}

Explanation:

$η = \frac{F L}{A ℓ}, A = L^{2}, η = \frac{F L}{L^{2} ℓ} = \frac{F}{L ℓ}$
Here, $L = 5.0 \times 10^{- 2} m$
$ℓ = 0.65 \times 10^{- 2} m, F = 0.25 N$
$η = \frac{0.25}{5.0 \times 10^{- 2} \times 0.65 \times 10^{- 2}} = \frac{0.25 \times 10^{4}}{3.25}$
$= 769.2 N m^{- 2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369771 Two parallel and opposite forces, each of magnitude $4000 N$ , are applied tangentially to the upper and lower faces of a cubical metal block $25 c m$ on a side. Find the displacement of the upper surface relative to the lower surface . The shear modulus for the metal is $80 G p a$ .

1

2.0 \times 10^{- 5} c m

2

5.0 \times 10^{- 5} c m

3

9.0 \times 10^{- 5} c m

4

8.0 \times 10^{- 5} c m

Explanation:

We use the approximate from $S = F / (A ϕ)$ .
Putting $S = 8 \times 10^{10} N / m^{2}, F = 4000 N$ and $A = (0.25 m)^{2}$
$= 6.25 \times 10^{- 2} m^{2}$ in the above equation and solving for $ϕ$ , we get
$ϕ = \frac{(4000 N)}{(6.25 \times 10^{- 2} m^{2}) (8 \times 10^{10} N / m^{2})}$
$= 8.0 \times 10^{- 7} r a d$
The displacement of the upper surface is given by $d = L ϕ$ , where $L$ is an edge of the cube.
$∴ d = (8.0 \times 10^{- 7}) (25 c m) = 2.0 \times 10^{- 5} c m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369772 Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass $M$ is attached to each of the free ends at the centre of the rods.

1 Both the rods will elongate but there shall be no perceptible change in shape

2 The steel rod will elongate and change shape but the rubber rod will only elongate

3 The steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse.

4 The steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.

NCERT Exemplar

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369773 The upper end of a wire of radius $4 m m$ and length $100 c m$ is clamped and its other end is twisted through an angle of $30^{\circ}$ . Then angle of shear is

1

{0.12}^{\circ}

2

12^{\circ}

3

{0.012}^{\circ}

4

{1.2}^{\circ}

Explanation:

Angle of shear
$ϕ = \frac{r θ}{L} = \frac{4 \times 10^{- 1}}{100} \times 30^{\circ} = {0.12}^{\circ}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369774 A $5.0 c m$ cube of substance has its upper face displaced by $0.65 c m$ by a tangential force of $0.25 N$ Calculate the modulus of rigidity of the substance.

1

769.2 N m^{- 2}

2

754.3 N m^{- 2}

3

735.2 N m^{- 2}

4

714.3 N m^{- 2}

Explanation:

$η = \frac{F L}{A ℓ}, A = L^{2}, η = \frac{F L}{L^{2} ℓ} = \frac{F}{L ℓ}$
Here, $L = 5.0 \times 10^{- 2} m$
$ℓ = 0.65 \times 10^{- 2} m, F = 0.25 N$
$η = \frac{0.25}{5.0 \times 10^{- 2} \times 0.65 \times 10^{- 2}} = \frac{0.25 \times 10^{4}}{3.25}$
$= 769.2 N m^{- 2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369771 Two parallel and opposite forces, each of magnitude $4000 N$ , are applied tangentially to the upper and lower faces of a cubical metal block $25 c m$ on a side. Find the displacement of the upper surface relative to the lower surface . The shear modulus for the metal is $80 G p a$ .

1

2.0 \times 10^{- 5} c m

2

5.0 \times 10^{- 5} c m

3

9.0 \times 10^{- 5} c m

4

8.0 \times 10^{- 5} c m

Explanation:

We use the approximate from $S = F / (A ϕ)$ .
Putting $S = 8 \times 10^{10} N / m^{2}, F = 4000 N$ and $A = (0.25 m)^{2}$
$= 6.25 \times 10^{- 2} m^{2}$ in the above equation and solving for $ϕ$ , we get
$ϕ = \frac{(4000 N)}{(6.25 \times 10^{- 2} m^{2}) (8 \times 10^{10} N / m^{2})}$
$= 8.0 \times 10^{- 7} r a d$
The displacement of the upper surface is given by $d = L ϕ$ , where $L$ is an edge of the cube.
$∴ d = (8.0 \times 10^{- 7}) (25 c m) = 2.0 \times 10^{- 5} c m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369772 Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass $M$ is attached to each of the free ends at the centre of the rods.

1 Both the rods will elongate but there shall be no perceptible change in shape

2 The steel rod will elongate and change shape but the rubber rod will only elongate

3 The steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse.

4 The steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.

NCERT Exemplar

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369773 The upper end of a wire of radius $4 m m$ and length $100 c m$ is clamped and its other end is twisted through an angle of $30^{\circ}$ . Then angle of shear is

1

{0.12}^{\circ}

2

12^{\circ}

3

{0.012}^{\circ}

4

{1.2}^{\circ}

Explanation:

Angle of shear
$ϕ = \frac{r θ}{L} = \frac{4 \times 10^{- 1}}{100} \times 30^{\circ} = {0.12}^{\circ}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369774 A $5.0 c m$ cube of substance has its upper face displaced by $0.65 c m$ by a tangential force of $0.25 N$ Calculate the modulus of rigidity of the substance.

1

769.2 N m^{- 2}

2

754.3 N m^{- 2}

3

735.2 N m^{- 2}

4

714.3 N m^{- 2}

Explanation:

$η = \frac{F L}{A ℓ}, A = L^{2}, η = \frac{F L}{L^{2} ℓ} = \frac{F}{L ℓ}$
Here, $L = 5.0 \times 10^{- 2} m$
$ℓ = 0.65 \times 10^{- 2} m, F = 0.25 N$
$η = \frac{0.25}{5.0 \times 10^{- 2} \times 0.65 \times 10^{- 2}} = \frac{0.25 \times 10^{4}}{3.25}$
$= 769.2 N m^{- 2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369771 Two parallel and opposite forces, each of magnitude $4000 N$ , are applied tangentially to the upper and lower faces of a cubical metal block $25 c m$ on a side. Find the displacement of the upper surface relative to the lower surface . The shear modulus for the metal is $80 G p a$ .

1

2.0 \times 10^{- 5} c m

2

5.0 \times 10^{- 5} c m

3

9.0 \times 10^{- 5} c m

4

8.0 \times 10^{- 5} c m

Explanation:

We use the approximate from $S = F / (A ϕ)$ .
Putting $S = 8 \times 10^{10} N / m^{2}, F = 4000 N$ and $A = (0.25 m)^{2}$
$= 6.25 \times 10^{- 2} m^{2}$ in the above equation and solving for $ϕ$ , we get
$ϕ = \frac{(4000 N)}{(6.25 \times 10^{- 2} m^{2}) (8 \times 10^{10} N / m^{2})}$
$= 8.0 \times 10^{- 7} r a d$
The displacement of the upper surface is given by $d = L ϕ$ , where $L$ is an edge of the cube.
$∴ d = (8.0 \times 10^{- 7}) (25 c m) = 2.0 \times 10^{- 5} c m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369772 Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass $M$ is attached to each of the free ends at the centre of the rods.

1 Both the rods will elongate but there shall be no perceptible change in shape

2 The steel rod will elongate and change shape but the rubber rod will only elongate

3 The steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse.

4 The steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.

NCERT Exemplar

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369773 The upper end of a wire of radius $4 m m$ and length $100 c m$ is clamped and its other end is twisted through an angle of $30^{\circ}$ . Then angle of shear is

1

{0.12}^{\circ}

2

12^{\circ}

3

{0.012}^{\circ}

4

{1.2}^{\circ}

Explanation:

Angle of shear
$ϕ = \frac{r θ}{L} = \frac{4 \times 10^{- 1}}{100} \times 30^{\circ} = {0.12}^{\circ}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369774 A $5.0 c m$ cube of substance has its upper face displaced by $0.65 c m$ by a tangential force of $0.25 N$ Calculate the modulus of rigidity of the substance.

1

769.2 N m^{- 2}

2

754.3 N m^{- 2}

3

735.2 N m^{- 2}

4

714.3 N m^{- 2}

Explanation:

$η = \frac{F L}{A ℓ}, A = L^{2}, η = \frac{F L}{L^{2} ℓ} = \frac{F}{L ℓ}$
Here, $L = 5.0 \times 10^{- 2} m$
$ℓ = 0.65 \times 10^{- 2} m, F = 0.25 N$
$η = \frac{0.25}{5.0 \times 10^{- 2} \times 0.65 \times 10^{- 2}} = \frac{0.25 \times 10^{4}}{3.25}$
$= 769.2 N m^{- 2}$

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