Find the ratio of breaking force required to break the wire of radius r to that of 2r.

Radius of the wire is given by r and 2r. 
The breaking force of wire is given by, 

                         F = σA

where,
σ is the breaking stress (constant of material) and A is the area of cross section.

Let F1 and F2 be the breaking force for wires respectively.

    F1=σπr2and  F2=σπ(2r)2 

Therefore, 
                      F2=4 F1

Hence the breaking force required to break the thick wire is four times the thin wire. 

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Two wires of same material but different diameters d1 and d2 (d1 > d2) are loaded. Which can support more weight?

Breaking stress is defined as breaking load per unit area. Breaking stress is constant of a material.

Therefore, 

Breaking load = Breaking stress X Area

                        = Breaking stress x πd24

Breaking load is directly proportional to the diameter. Therefore, thick wire can support more weight than the thin wire. 
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A cable is cut into two equal lengths. How the breaking force will change, if two parts are connected in parallel?

Breaking force is given by breaking stress x Area of cross section

If F is the force required to break the wire of area of cross section a, then breaking force becomes 2F. 
 
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The figure shows the load extension graph of two wires of material A and B having same dimensions. Which of the two materials has greater value of Young’s modulus?



Young's modulus is given by, 

           Y = F×LA×L 

Therefore, 
            F = YALL
Thus the slope of load v/s extension graph is  YAL.

For wires of same dimensions, the slope of graph is directly proportional to Young's modulus. Since slope of graph for A is greater than that of B, therefore the value of Young's modulus of material of A is greater than B. 
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Plot the stress versus strain curve for rubber and what does the area of the loop of the curve represent? 


When the rubber undergoes the cycle of loading and unloading some of the mechanical


When the rubber undergoes the cycle of loading and unloading some of

energy is lost in the cycle. The area under the graph is equal to the work done to complete the cycle of loading and unloading.
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