The force required to punch a hole of diameter 2 mm, will be (If a shearing stress `4xx10^(8)N//m^(2)`.)

The shearing strain produced in a block of metal subjected to a shearing stress of 10^(8)N//m^(2) is ( Modulus of rigidity eta=8 xx 10^(10)N//m^(2) )

Calculate the force required to increase the length of a steel wire of cross- sectional area 10^(-6)m^(2) by 0.5% given: Y_(("for steel"))=2xx10^(11)N-m^(2)

A square steel plate has area 1 m^(2) and thickness 5 cm. The lower surface is fixed. A tangential force applied to top surface displace it through 0.005 cm. Find the modulus of rigidity of steel, if shearing stress of modulus of rigidity of steel. If shearing stress of 4.2 xx 10^(3) N//m^(2) is being applied.

The stress in a wire of diameter 2 mm, if a load of 100 g is applied to a wire, is

A metal wire is observed to stretch by one part in a million when subjected to a stress of 8xx10^(4)N//m^(2) . The Young's modulus of the metal is

The metal cube of side 10 cm is subjected to a shearing stress of 10^(4) N m^(-2) . The modulus of rigidity if the top of the cube is displaced by 0.05 cm with respect to its bottom is

The U -tube with limbs of diameters 6 mm and 3 mm contain water of surface tension 7xx10^(-2) N m^(-1) . The angle of contact is zero and density 10^3 kg m^(-3) . If g is 10 m s^(-2) , then the difference in levels in the two limbes is

Calculate the force F needed to punch at 1.46 cm diameter hole in a steel plate 1.27 cm thick fig. The ultimates shear stress of steel is 3.45 xx 10^(8)Nm^(-2) .

Steel ruptures when a shear of 3.5xx10^(8)Nm^(-2) is applied. The force needed to punch a 1 cm diameter hole in a steel sheet 0.3 cm thick is nearly:

A square brass plate of side 1.0 m and thickness 0.0035m is subjected to a force F on its smaller opposite edges, causing a displacement of 0.02 cm . If the shear modulus o brass is 0.4 xx 10^(11) N//m^(2) , the value on force F is