The length of a germanium rod is 0.58 cm and its area of cross-section is `1mm^(2)`. If for germanium `n_(i)=2.5xx10^(19)m^(-3),mu_(h)=0.19 m^(2)//V-s,mu_(e)=0.39 m^(2)//V-s`, then the resistance of the rod will be-

The following data are for intrinsic germanium at 300 K. n_(i) = 2.4 xx 10^(19)//m^(3), mu_(e) = 0.39 m^(2)//Vs, mu_(h) = 0.19 m^(2)//Vs . Calculate the coductivity of intrinsic germanium.

Find the resistance of an ebonite rod of length 1m and area of cross section 2mm^(2) .The resistance of ebonite = 10^(16)Ohm-m .

Resistance of the conductor of length 5m and area of cross -section 4 mm^(2) s 0.02 Omega . Its resistivity is

Resistance of the conductor of length 5m and area of cross -section 4 mm^(2) s 0.02 Omega . Its resistivity is

Find the load on wire of cross section 16 mm to increase its length by 0.3%(Y=5 xx 10^(9)N//m^(2),g=10m//s^(2))

The length of a rod is 20 cm and area of cross-section 2 cm^(2) . The Young's modulus of the material of wire is 1.4 xx 10^(11) N//m^(2) . If the rod is compressed by 5 kg-wt along its length, then increase in the energy of the rod in joules will be