Length and cross sectional area of a iron rod is 1 m and `1 cm^2` respectively. `Y of iron = 10^11 N//m^2`. Calculate the amount of force at two end to increase its length by 1 mm. [Y = Young's Modulus].

A wire (Y =2xx10^11N//m^2) has length 1m and area 1mm^2 . Calculate the work required to increase its length by 2mm.

Two rods of equal length and cross- sectional area have their Young's moduli Y_1 and Y_2 respectively. IF the rods are joined end to end, then the equivalent Young's modulus of the combined rod system is

A load of 20 kg is hung from one end of a wire of length 6 m and of cross sectional area 1mm^2 . IF the load is withdrawn, the length of the wire becomes 5.995 m. Calculate the (i) longitudinal strain, (ii) longitudinal stress and (iii) Young's modulus for the material of the wire.

The area of cross section of a steel wire (Y=2.0xx10^11 N//m^2) is 0.1 cm^2 . The force required to double its length will be

An 8kg mass is suspended from one end of an iron wire of length 2 m and diameter 1mm. IF the young's modulus of iron is 2 times 10^12 dyn. cm^-2 , then what is the increase in length of the wire? [ g=980 cm.s^-2 ]

For a uniform wire of length 3 m and cross sectional area 1mm^2 , 0.021 J of work is necessary to stretch it through 1mm. Calculate the Young's modulus for its material.

A uniform rod weights W, has length L and cross sectional area a. The rod is suspended from one of its ends. The young's modulus of its material is Y. Increase in length of the rod will be