A cable is replaced by another cable of the same length and material but of double the diameter.
(i) Under a given load which cable will show greater extension?
(ii) How many times the second cable can support the maximum load without exceeding the elastic limit?

A cable is replaced by another cable of the same length and material but of duouble the diameter. (i) Under a given load which cable wil show greater extension? (ii) How many times the second cable can support the maxium load without exceeding the elastic limit?

A wire is roplaced by another wire of same length and material but of twice diameter. (i) What will be the effect on the increases in its length under a given load? (ii) What will be the effect on the maximum load which it can bear?

The stress for elastic limit of a material is 3.5times10^(8)N/m^(2) .The minimum diameter of a rod made of this material which can support 500N load without exceeding elastic limit will be

A cable stretched by an amount d when it supports a crate of mass M. The cable is replaced by another cable of the same material having the same length and twice the diameter. If the same crate is supported by the thicker cable, by how much will the cable stretch ?

A cable that can support a load of 800 N is cut into two equal parts. The maximum load that can be supported by either part is

A cable that can support a load W is cut into two equal parts .T he maximum load that can be supported by either part is

When a tensile or compressive load 'P' is applied to rod or cable, its length changes. The change length x which, for an elastic material is proportional to the force Hook' law). P alpha x or P = kx The above equation is similar to the equation of spring. For a rod of length L , area A and young modulue Y . the extension x can be expressed as. x = (PL)/(AY) or P = (AY)/(x) , hence K = (AY)/(L) Thus rod or cables attached to lift can be treated as springs. The energy stored in rod is called stra energy & equal to (1)/(2)Px . The loads placed or dropped on the floor of lift cause stresses in the cable and can be evaluted by spring analogy. If the cable of lift cause stresses and load is placed dropped, then maximum extension in cable can be calculated by enerfy conservation. In the above problem, the maximum stress developed in the rod is equal to - (N//m^(2))

There are two wires of same material and same length while the diameter of second wire is 2 times the diameter of first wire, then ratio of extension produced in the wires by applying same load will be

There are two wires of same material and same length while the diameter of second wire is 2 times the diameter of first wire, then ratio of extension produced in the wires by applying same load will be

A steel wire of length l and cross sectional area A is stretched by 1 cm under a given load. When the same load is applied to another steel wire of double its length and half of its cross section area, the amount of stretching (extension) is