Mechanical Prop...

To determine the Young's modulus of a wire, the formula is $$Y = \dfrac {F}{A}, \dfrac {L}{\triangle l}$$; where $$L = length, A =$$ area of cross-section of the wire, $$\triangle L =$$ Change in length of the wire when stretched with a force $$F$$. The conversion factor to change it from $$CGS$$ to $$MKS$$ system is

The area of cross section of a steel wire ($$y=2\times { 10 }^{ 11 }N/{ m }^{ 2 }$$) is $$0.1{cm}^{2}$$. The force required to double its length will be:

Identical springs of steel and copper $$\left( { Y }_{ steel }>{ Y }_{ copper } \right) $$ are equally stretched then:

Figure shows the strain-stress curve for a given material. The Young's modulus of the material is 

Let $${Y}_{S}$$ and $${Y}_{A}$$ represent Young's modulus for steel and aluminium respectively It is said that steel is more elastic than aluminium. Therefore, it follows that

If the work done in stretching a wire by $$1mm$$ is $$2J$$, the work necessary for stretching another wire of same material but with double radius of cross-section and half the length by $$1mm$$ is:

Stress is a ______ quantity.

A steel rod of length $$1m$$ and radius $$10mm$$ is stretched by a force $$100kN$$ along its length. The percentage strain in the rod is then
 $$\left( { Y }_{ steel }=2\times { 10 }^{ 11 }N\quad { m }^{ -2 } \right) $$

Match the Column I with Column II

Which of the following statements is correct regarding Poisson's ratio?