A steel rod of cross-sectional area 4cm^2 and length 2m shrinks by 0.1cm as the temperature decreases in night.

(d)1/2 mw2l. Explanation:  Since the rod rotates on the horizontal surface, the horizontal component of the force applied by the clamp is the centripetal force = mω²r = mω²(l/2) = ½mω²l.

Length of wire  L= 3 m,  Load F = 10*10 N =100 N, {taking g = 10 m/s²}  Area of cross-section A = 4 mm²  →A = 4x10⁻⁶ m²  (a) The stress = Load...

(a) Since both types of wires are joined end to end, the same load will be on the cross-sections of them. If the load = F and the area of...

Cross-sectional area A =0.01 cm²   →A = 0.01/10000 =1x10⁻⁶ m²  Stress = 20/1x10⁻⁶ N/m² =2x10⁷ N/m²  Hence the longitudinal strain =Stress/Y =2x10⁷/1.1x10¹¹  =1.82x10⁻⁴  Poisson's ratio = 0.32  Lateral strain = 0.32*1.82x10⁻⁴ =5.82x10⁻⁵  Taking the cross-section as...

The correct answer is (b) (c) (b) may be equal to 20V/cm (c) may be greater than 20V/cm

(b) The rod and the experimenter move with the same speed v in opposite directions.

Magnetic field inside the solenoid B =μ0nI n → number of turns per unit length, I → current flowing, ϕ' = Magnetic field through each turn = BA = μ0nIA Total magnetic flux linked...

23minutes. A+B =2 minutes A returns=1 minute A+C=7 minutes A returns= 1 minute A+D=10 minutes A comes back =1 minute Leaves lamp and crosses again=1 minute Total 23 minutes.

K = 45w/m-°C ℓ = 60cm = 60 × 10–2m A = 0.2cm2 = 0.2 × 10–4m2 Rate of heat flow, = KA( θ1 + θ2 )/l = (45 x 0.2 x 10-4 x...