If the ratio of lengths, radii and Young’s modulii of steel and brass wires in the figures are a, b and c respectively. Then the corresponding ratio of increase in their lengths would be :

Young’s modulus of the material of a wire of length ‘L’ and radius ‘r’ in ‘Y’ \[N/m^{2}\]. If the length in reduced to L/2 and radius to r/2, the Young modulus will be :

The Young’s Modulus of a wire in numerically equal to the stress which will :

The interatomic distance for a metal is

\[3.6\times 10^{-10}m\].

If the interatomic force constant is

\[3.6\times 10^{-9}\]

N/Å then the young’s modulus in

\[N/m^{2}\]

will be :

The load versus elongation graph for four wires of the same material is shown in the figure. The thickest wire is represented by the line :

The stress versus strain graphs for wires of two materials A and B are as shown in the figure. If \[Y_{A}andY_{B}\] are the Young’s modulus of the materials then :

An elevator cable’s maximum stress is \[11\times 10^{7}N/m^{2}\] Its maximum upward acceleration is \[1.2m/s^{2}\]. If the cable has to support the total weight of 2000kg of a loaded elevator, the area of cross-section of the cable should be [Take g = \[9.8m/s^{2}\]]

Young modulus of elasticity of brass is 10^11 N/m2. The increase in its energy on pressing a rod of length 0.1 m and cross–sectional area 1 cm2 made of brass with a force of 10 kg along its length, will be …………

The diameter of a brass rod is 4 mm and Young’s modulus of brass is \[9\times 10^{10}N/m^{2}\]. The force required to stretch by 0.1% of its length is :

The approximate depth of an ocean is 2700 m. The compressibility of water is \[45.4\times10^{-11}Pa^{-1}\] and density of water is \[10^{3}kg/m^{3}\]. What fractional compression of water will be obtained at the bottom of the ocean ?