Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 23

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Question 1
1 / -0

A wire elongates by l mm when a load W is hanged from it. If the wire goes over a pulley and two weights W each are hung at the two ends,the elongation of the wire will be (in mm)

A
zero

B
l / 2

C
l

D
2l

Solution

Here, when load was hanging both ends of the wire was haring a tension w.
Now also the both ends of the wire have tension w.
So, tension is same So, elongation will be same i.e. 1mm.
Question 2
1 / -0

A wire extends by $$1 mm$$ when a force is applied. Double the force is applied to another wire of same material and length but of half the radius of cross section.The elongation of wire in mm will be

A
$$8$$

B
$$4$$

C
$$2$$

D
$$1$$

Solution

$$Y=\dfrac{Fl}{A\Delta l}$$

$$\Delta l=\dfrac{Fl}{\pi r^{2}Y}$$

$$\dfrac{\Delta l_{1}}{\Delta l_{2}}=\dfrac{Fl}{\pi r^{2}Y}\times \dfrac{Y\pi \left ( \dfrac{r}{2} \right )^{2}}{2Fl}$$

$$\dfrac{\Delta l_{1}}{\Delta l_{2}}=\dfrac{1}{8}$$

$$8\Delta l_{1}=\Delta l_{2}$$

So, $$\Delta l_{2}=8(1)$$
$$=8mm$$
Question 3
1 / -0

A copper solid cube of $$60mm$$ side is subjected to a compressible pressure of 2.5 x 10$$^{7}$$ Pa. If the bulk modulus of copper is 1.25 x 10$$^{11}$$ pascals, the change in the volume of cube is

A
$$-43.2 mm$$$$^{3}$$

B
$$-43.2 m$$$$^{3}$$

C
$$-43.2 cm^{3}$$

D
$$-432 m m^{3}$$

Solution

$$B=\dfrac{P}{\Delta v/v}$$

$$\Delta v=\dfrac{pv}{B}$$

$$=\dfrac{2.5\times 10^{7}\times (60)^{3}}{1.25\times 10^{11}}$$

$$=43.2mm^{3}$$

So, $$\Delta v=43.2mm^{3}$$ and it is negative since volume decreases.
Question 4
1 / -0

A sphere contracts in volume by $$0.01$$% when taken to the bottom of lake $$1km$$ deep. If the density of water is $$1gm/cc$$, the bulk modulus of water is

A
9.8 x 10$$^{5}$$N/m$$^{2}$$

B
9.8 x 10$$^{8}$$N/m$$^{2}$$

C
9.8 x 10$$^{10}$$N/m$$^{2}$$

D
9.8 x 10$$^{6}$$N/m$$^{2}$$

Solution

$$B=\dfrac{\Delta p}{\Delta v/v}=\dfrac{\rho g\Delta h}{\Delta v/v}$$

$$=\dfrac{1000\times 9.8\times 1000}{0.00\times 10^{-2}}$$

Bulk Modulus $$=9.8\times 10^{10}N/m^{2}$$
Question 5
1 / -0

The Y of a material having a cross sectional area of $$1 cm$$ $$^{2}$$ is 2 x 10$$^{12}$$ dynes/cm$$^{2}$$. The force required to double the length of the wire is:

A
1 x 10$$^{12}$$dynes

B
2 x 10$$^{12}$$dynes

C
0.5 x 10$$^{12}$$dynes

D
4 x 10$$^{12}$$dynes

Solution

$$Y=\dfrac{Fl}{A\Delta l}$$
$$F=\dfrac{YA\Delta l}{l}$$
$$=2\times 10^{12}\times 1$$ $$\left ( \therefore \dfrac{\Delta l}{l}=1 \right )$$
$$=2\times 10^{12}$$ dynes.
Question 6
1 / -0

An elongation of $$0.1$$% in a wire of cross-section $$10^{-6}m^{2}$$ causes a tension of $$100N$$. $$Y$$ for the wire is

A
$$10^{12}N/m^{2}$$

B
$$10^{11}N/m^{2}$$

C
$$10^{10}N/m^{2}$$

D
$$100\ N/m^{2}$$

Solution

$$ Y = \dfrac{F l}{A\triangle l}$$
$$ \triangle l = \dfrac{Fl}{AY}$$
$$So, \triangle l \alpha \dfrac{1}{\triangle}$$
$$ \therefore \dfrac{\triangle l_{1}}{\triangle L_{2}} = \dfrac{A_{2}}{A_{1}} = 3$$
$$\dfrac{\triangle l_{1}}{3} = \triangle l_{2}$$
$$\triangle l_{2} = \dfrac{0.1}{3} = 0.033mm $$
Question 7
1 / -0

One end of a wire $$2m$$ long and $$0.2cm$$$$^{2}$$ in cross section is fixed to a ceiling and a load of $$4.8kg$$ is attached to its free end. The elongation in the wire in mm is (if y $$=$$ 2.0 x 10$$^{11}$$ N.m$$^{-2}$$, g $$=$$10ms$$^{-2}$$)

A
$$2.4$$ x $$10$$$$^{-5}$$

B
$$2.4$$

C
$$0.024$$

D
$$0.0024$$

Solution

$$Y=\dfrac{Fl}{A\Delta l}$$

$$\Delta l=\dfrac{Fl}{\Delta Y}$$

$$=\dfrac{45\times 2}{0.2\times 10^{-4}\times 2\times 10^{11}}$$

$$=24\times 10^{-6}m$$

$$=24\times 10^{-3}mm$$

$$=0.024mm$$
Question 8
1 / -0

The increase in pressure required to decrease the $$200$$ litres volume of a liquid by $$0.004$$% in kPa is : (bulk modulus of the liquid $$=$$ $$2100 MPa$$)

A
$$8.4$$

B
$$84$$

C
$$92.4$$

D
$$168$$

Solution

$$B=P\Delta v/v$$

$$P=B\times \dfrac{\Delta v}{v}$$

$$=2100\times 10^{3}\times \dfrac{0.004}{100}$$

$$=84k$$ pa.
Question 9
1 / -0

An iron wire of length $$4m$$ and diameter $$2 mm$$ is loaded with a weight of $$8kg$$. If the young's modulus '$$Y$$' for iron is $$2$$ x $$10$$$$^{11}$$ N/m$$^{2}$$, then the increase in the length of the wire is

A
$$0.2 mm$$

B
$$0.5mm$$

C
$$2mm$$

D
$$1 mm$$

Solution

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

\Delta l } $$

$$ \Delta l= \dfrac {

F l }{ A Y} $$

$$\quad = \dfrac { 80 \times

l }{ r { (\dfrac { 2 }{

2 } ) }^{ 2 }\times { 10 }^{ -6 }\times 2\times { 10 }^{ 11 } } $$

$$\quad = 5.0 \times {

10 }^{ -4 }m $$
$$\quad = 0.5\ mm$$
Question 10
1 / -0

The extension of a wire by the application of a load is $$0.3 cm$$. The extension in the wire of the same material but of double the length and half the radius of cross section in cm is

A
$$3$$

B
$$0.3$$

C
$$1.2$$

D
$$2.4$$

Solution

$$\Delta l=\dfrac{Fl}{Y\pi r^{2}}$$

$$\dfrac{\Delta l_{1}}{\Delta l_{2}}=\dfrac{Fl}{Y\pi r^{2}}\times \dfrac{Y\pi \left ( \dfrac{r}{2} \right )^{2}}{F2l}$$

$$\dfrac{\Delta l_{1}}{\Delta l_{2}}=\dfrac{1}{8}$$

$$8\Delta l_{1}=\Delta l_{2}$$

So, $$\Delta l_{2}=8(0.3)$$

$$=2.4cm$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 23

Result

Mechanical Properties of Solids Test - 23

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SHARING IS CARING

Question 1

zero

l / 2

l

2l

Solution

Question 2

$$8$$

$$4$$

$$2$$

$$1$$

Solution

Question 3

$$-43.2 mm$$$$^{3}$$

$$-43.2 m$$$$^{3}$$

$$-43.2 cm^{3}$$

$$-432 m m^{3}$$

Solution

Question 4

9.8 x 10$$^{5}$$N/m$$^{2}$$

9.8 x 10$$^{8}$$N/m$$^{2}$$

9.8 x 10$$^{10}$$N/m$$^{2}$$

9.8 x 10$$^{6}$$N/m$$^{2}$$

Solution

Question 5

1 x 10$$^{12}$$dynes

2 x 10$$^{12}$$dynes

0.5 x 10$$^{12}$$dynes

4 x 10$$^{12}$$dynes

Solution

Question 6

$$10^{12}N/m^{2}$$

$$10^{11}N/m^{2}$$

$$10^{10}N/m^{2}$$

$$100\ N/m^{2}$$

Solution

Question 7

$$2.4$$ x $$10$$$$^{-5}$$

$$2.4$$

$$0.024$$

$$0.0024$$

Solution

Question 8

$$8.4$$

$$84$$

$$92.4$$

$$168$$

Solution

Question 9

$$0.2 mm$$

$$0.5mm$$

$$2mm$$

$$1 mm$$

Solution

Question 10

$$3$$

$$0.3$$

$$1.2$$

$$2.4$$

Solution

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