Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 21

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Question 1
1 / -0

Two exactly similar wires of steel (y$$=$$20 x 10$$^{11}$$dyne/cm$$^{2}$$) and copper (y $$=$$ 12 x 10$$^{11}$$ dyne/cm$$^{2}$$)are stretched by equal forces. If the total elongation is 1cm, elongation of copper wire is

A
$$3/5 cm$$

B
$$5/3 cm$$

C
$$3/8 cm$$

D
$$5/8 cm$$

Solution

$$ \triangle l = \dfrac{F l}{A Y} $$
$$ \triangle l_{s} = \dfrac{F l}{A Y_{s}} - 1$$ (different cases have different subscripts)
$$ \triangle l_{c} = \dfrac{F l}{A Y_{c}} - 2$$
Dividing 1 and 2
$$\dfrac{\triangle l_{s}}{\triangle l_{c}} = \dfrac{Y_{c}}{Y_{s}} = \dfrac{12 \times 10^{11}}{20 \times 10^{11}} = \dfrac{3}{5}$$
$$ 1 cm = \triangle l_{s} + \triangle l_{c}$$
$$ 1 cm = \dfrac{3}{5} \triangle l_{c} + \triangle l_{c}$$
$$ 1 cm = \dfrac{8}{5} \triangle l_{c}$$
$$ So, \triangle l_{c} = \dfrac{5}{8}cm$$
Question 2
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The radii and Young's modulus of two uniform wires $$A$$ & $$B$$ are in the ratio $$2:\ 1$$ and $$1:\ 2$$ respectively. Both the wires are subjected to the same longitudinal force. If increase in the length of wire $$A$$ is $$1\%$$ . Then the percentage increase in length of wire $$B$$ is :

A
$$1$$

B
$$1.5$$

C
$$2$$

D
$$3$$

Solution

$$ Y = \dfrac{F/A}{\triangle l/l}$$
$$ \triangle l = \dfrac{F l}{\pi r^{2}Y}$$
$$ \triangle l_{A} = \dfrac{F l}{\pi r_{A}^{2}Y_{A}} .............(1)$$

$$ \triangle l_{B} = \dfrac{F l}{\pi r_{B}^{2}Y_{B}} .............(2)$$

Dividing 1 by 2

$$ \dfrac{\triangle l_{A}}{\triangle l_{B}} =\dfrac{r_{B}^{2}Y_{B}}{r_{A}^{2}Y_{A}} = (\dfrac{r_{B}}{r_{A}})^{2} (\dfrac{Y_{B}}{Y_{A}}) = (\dfrac{1}{2})^{2} (\dfrac{2}{1}) = \dfrac{1}{2}$$
$$ \triangle l_{B} = 2\triangle l_{A} = 2 \times 1$$ %
$$ = 2$$ %
Question 3
1 / -0

A load of $$4.0\ kg$$ is suspended from a ceiling through a steel wire of length $$20\ m$$ and radius $$2.0\ mm$$. It is found that the length of the wire increases by $$0.031\ mm$$ as equilibrium is achieved. If $$g=3.1\ \pi\ ms^{-2}$$, the value of young's modulus in $$Nm^{-2}$$ is

A
$$2.0 \times 10^{12}$$

B
$$4.0 \times 10^{11}$$

C
$$2.0 \times 10^{11}$$

D
$$0.02 \times 10^{9}$$

Solution

For equilibrium
Weight = Tension
$$mg = T$$
$$ \therefore T = 4 \times 3.1 \pi $$$$ = 12.4\pi N$$ (as can be inferred from the question)
$$ Y = \dfrac{T/A}{\triangle l/l}$$
$$ = \dfrac{12.4 \pi / \pi (\dfrac{2}{1000})^{2}}{\dfrac{0.031}{1000}/20}$$
$$ = \dfrac{12.4 \times 20 \times 1000 \times (1000)^{2}}{4 \times 0.031}$$
$$=2 \times 10^{12} N/m^{2}$$
Question 4
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A uniform wire of length $$4m$$ and area of cross section $$2mm$$$$^{2}$$ is subjected to longitudinal force produced an elongation of $$1mm$$.If Y$$=$$0.2 x 10$$^{11}$$ NM$$^{-2}$$, elastic potential energy stored in the body is

A
$$0.5J$$

B
$$0.05J$$

C
$$0.005J$$

D
$$5.0J$$

Solution

$$\text{ Work done} = \int _{0}^{1mm} \dfrac{AY}{L} xdx$$ (using standard result $$Y=\dfrac{(\dfrac{F}{A})}{(\dfrac{\text{change in length}}{\text{actual length}})}$$)

$$= \dfrac{AY}{2L} (\dfrac{1}{1000})^{2}$$

$$ = \dfrac{2 \times 10^{-6}\times 0.2\times 10^{11}}{2 \times 4} \times 10^{-6}$$

$$ = 0.005 J$$
Question 5
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Two wires of equal cross section, but one made up of steel and the other copper, are joined end to end. When the combination is kept under tension, the elongations in the two wires are found to be equal. If $$Y_{steel} =$$ 2.0 x $$10^{11} Nm^{-2}$$ and $$Y_{copper} =$$ 1.1 x 10$$^{11} Nm^{-2}$$, the ratio of the lengths of the two wires is :

A
$$20 : 11$$

B
$$11:20$$

C
$$5 : 4$$

D
$$4 : 5$$

Solution

$$ Y = \dfrac{F/A}{\triangle l/l}$$ (standard result)
$$ \triangle l = \dfrac{F l}{A Y}$$
$$\triangle l_{S} = \dfrac{F L_{s}} {A Y_{s}}$$ (for different cases different subscripts are used)
$$ \triangle l_{c} = \dfrac{F l_{c}}{A Y_{c}}$$
$$ By\ problem \triangle l_{S} = \triangle l_{c}$$
$$ \dfrac{F l_{s}}{A Y_{s}} = \dfrac{F l_{c}}{A Y_{c}}$$
$$ \dfrac {l_{S}} {l_{c}} = \dfrac {Y_{s}} {Y_{c}} = \dfrac {2 \times 10^{11}} {1.1 \times 10^{11}} = \dfrac{20}{11}$$
Question 6
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A load of 1kg weight is attached to one end of a steel wire of cross sectional area 3mm$$^{2}$$ and Youngs modulus 10$$^{11}$$ N/m$$^{2}$$ . The other end is suspended vertically from a hook on a wall, then the load is pulled horizontally and released.When the load passes through its lowest position the fractional change in length is: (g$$=$$10m/ s$$^{2}$$ )

A
$$0.3 \times 10^{-4}$$

B
$$0.3 \times 10^{-3}$$

C
$$10^{3}$$

D
$$10^{4}$$

Solution

$$ Y = \dfrac{F/A}{\triangle l/l}$$

$$ \dfrac{\triangle l}{l} = \dfrac{F}{YA}$$

$$ = \dfrac{1 \times 10}{10^{11} \times 3 \times 10^{-6}}$$

$$ = \ 0.3 \times 10^{-4}$$
Question 7
1 / -0

Two wires A and B of the same dimensions are under loads of $$4$$ and $$5.5 kg$$ respectively. The ratio of Young's modulii of the materials of the wires for the same elongation is:

A
$$64 : 121$$

B
$$\sqrt{11}:\sqrt{8}$$

C
$$11:8$$

D
$$8 : 11$$

Solution

$$ Y = \dfrac{F/A}{\triangle l/l}$$
$$ Y_{A} = \dfrac{F_A/A}{\triangle l/l} ........................... 1$$
$$ Y_{B} = \dfrac{F_B/A}{\triangle l/l} ........................... 2$$
Dividing 1 and 2
$$ \dfrac{Y_{A}}{Y_{B}} = \dfrac{F_{A}}{F_{B}} = \dfrac{4}{5.5} = \dfrac{8}{11}$$
Question 8
1 / -0

A wire is subjected to a longitudinal strain of $$0.05.$$ If its material has a Poisson's ratio $$0.25$$, the lateral strain experienced by it is

A
0.00625

B
0.125

C
0.0125

D
0.0625

Solution

$$\epsilon x=0.05$$ (given)
$$\sigma =0.25$$
$$\dfrac{\epsilon y}{0.05}=-0.25$$ (standard result)
$$=-0.0125$$
Question 9
1 / -0

A metallic rod undergoes a strain of $$0.05$$%. The energy stored per unit volume is ($$Y$$ = $$2 \times 10$$$$^{11}$$Nm$$^{-2}$$):

A
$$0.5 \times 10$$$$^{4}$$ Jm$$^{-3}$$

B
$$0.5 \times 10$$$$^{5}$$ Jm$$^{-3}$$

C
$$2.5 \times 10$$$$^{5}$$ Jm$$^{-3}$$

D
$$2.5 \times 10$$$$^{4}$$ Jm$$^{-3}$$

Solution

$$ Stress = Y \times strain $$

$$ Potential\ energy = \dfrac{1}{2} \times stress \times strain \times volume$$

$$\dfrac{Potential\ energy}{volume} = \dfrac{1}{2} \times stress \times strain$$

$$ = \dfrac{1}{2} \times Y strain \times strain$$

$$ = \dfrac{1}{2} \times 2\times 10^{11}\times (\dfrac{0.05}{100})^{2}$$

$$ = 2.5 \times 10^{4} J/m^{3}$$
Question 10
1 / -0

A steel wire of length $$4m$$ is stretched by a force of $$100N$$. The work done to increase the length of the wire by $$2mm$$ is

A
$$0.4J$$

B
$$0.2J$$

C
$$0.1J$$

D
$$1J$$

Solution

$$\text{ Work done} = \dfrac{AY}{2L} l^{2}$$ (using standard result $$Y=\dfrac{(\dfrac{F}{A})}{(\dfrac{\text{change in length}}{\text{actual length}})}$$)

$$ = \dfrac{F l}{2} (\because F = \dfrac{AYl}{L})$$

$$ = \dfrac{100 \times 2}{2 \times 1000}$$

$$= 0.1 J$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 21

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Mechanical Properties of Solids Test - 21

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Question 1

$$3/5 cm$$

$$5/3 cm$$

$$3/8 cm$$

$$5/8 cm$$

Solution

Question 2

$$1$$

$$1.5$$

$$2$$

$$3$$

Solution

Question 3

$$2.0 \times 10^{12}$$

$$4.0 \times 10^{11}$$

$$2.0 \times 10^{11}$$

$$0.02 \times 10^{9}$$

Solution

Question 4

$$0.5J$$

$$0.05J$$

$$0.005J$$

$$5.0J$$

Solution

Question 5

$$20 : 11$$

$$11:20$$

$$5 : 4$$

$$4 : 5$$

Solution

Question 6

$$0.3 \times 10^{-4}$$

$$0.3 \times 10^{-3}$$

$$10^{3}$$

$$10^{4}$$

Solution

Question 7

$$64 : 121$$

$$\sqrt{11}:\sqrt{8}$$

$$11:8$$

$$8 : 11$$

Solution

Question 8

0.00625

0.125

0.0125

0.0625

Solution

Question 9

$$0.5 \times 10$$$$^{4}$$ Jm$$^{-3}$$

$$0.5 \times 10$$$$^{5}$$ Jm$$^{-3}$$

$$2.5 \times 10$$$$^{5}$$ Jm$$^{-3}$$

$$2.5 \times 10$$$$^{4}$$ Jm$$^{-3}$$

Solution

Question 10

$$0.4J$$

$$0.2J$$

$$0.1J$$

$$1J$$

Solution

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