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Let the ends of the steel wire be called A and B. length of the wire is 2l = 1 m. The cross-sectional area of the wire is  Let the depression at the midpoint due to the suspended 100 g be y. Change in the length of the wire is  The strain is  The vertical components of the tension in the arms balance the weight of the suspended mass, we have The stress in the wire will be  The Young's...
Cross-Sectional Area of wire A is AA = 1 mm2 Cross-Sectional Area of wire B is AB = 2 mm2 Let the Mass m be suspended at y distance From the wire A Let the Tension in the two wires A and B be FA and FB respectively Since the Strain in the wires is equal Equating moments of the Tension in the wires about the point where mass m is suspended we have The Load should be suspended at a point...
Cross-Sectional Area of wire A is AA = 1 mm2 Cross-Sectional Area of wire B is AB = 2 mm2 Let the Mass m be suspended at x distance From the wire A Let the Tension in the two wires A and B be FA and FB respectively Since the Stress in the wires is equal Equating moments of the Tension in the wires about the point where mass m is suspended we have The Load should be suspended at a point 70 cm...
Change in volume is  Bulk modulus of water is  A pressure of  is to be applied so that a litre of water compresses by 0.1%. Note: The answer is independent of the volume of water taken into consideration. It only depends upon the percentage change.
Bulk modulus of copper is  Edge of copper cube is s = 10 cm = 0.1 m Volume Of copper cube is V = s3                                           V = (0.1)3                                           V = 0.001 m3 Hydraulic Pressure applies is  From the definition of bulk modulus The volumetric strain is  Volume contraction will be The volume contraction has such a small value even under high...
Bulk's Modulus of Glass is  Pressure is P = 10 atm. The fractional change in Volume would be given as The fractional change in Volume is 
Water at the surface is under 1 atm pressure. At the depth, the pressure is 80 atm. Change in pressure is  Bulk Modulus of water is  The negative sign signifies that for the same given mass the Volume has decreased The density of water at the surface  Let the density at the given depth be  Let a certain mass occupy V volume at the surface Dividing the numerator and denominator of RHS by V we...
Pressure Increase, P = 100.0 atm Initial Volume = 100.0 l Final volume = 100.5 l Change in Volume = 0.5 l Let the Bulk Modulus of water be B The bulk modulus of air is  The Ratio of the Bulk Modulus of water to that of air is This ratio is large as for the same pressure difference the strain will be much larger in the air than that in water. 
Mass of the body = 14.5 kg Angular velocity,  = 2 rev/s  The radius of the circle, r = 1.0 m Tension in the wire when the body is at the lowest point is T Cross-Sectional Area of wire, A = 0.065 cm2 Young's Modulus of steel, 
Each wire must support the same load and are of the same length and therefore should undergo the same extension. This, in turn, means they should undergo the same strain.                                         As F, l and  are equal for all wires
Let the maximum Load the Cable Can support be T Maximum Stress Allowed, P = 108 N m-2 Radius of Cable, r = 1.5 cm
Length of the copper piece, l = 19.1 mm The breadth of the copper piece, b = 15.2 mm Force acting, F = 44500 N Modulus of Elasticity of copper,  
Inner radii of each column, r1 = 30 cm = 0.3 m Outer radii of each colum, r2 = 60 cm = 0.6 m Mass of the structure, m = 50000 kg Stress on each column is P Youngs Modulus of steel is 
Edge of the aluminium cube, l = 10 cm = 0.1 m Area of a face of the Aluminium cube, A = l2 = 0.01 m2 Tangential Force is F  Tangential Stress is F/A Shear modulus of aluminium  Let the Vertical deflection be 
Tension in the steel wire is F1 Length of steel wire l1 = 1.5 m The diameter of the steel wire, d = 0.25 cm Area od the steel wire,  Let the elongation in the steel wire be  Young's Modulus of steel, Y1 =  Tension in the Brass wire is F2 Length of Brass wire l2 = 1.5 m Area od the brass wire,  Let the elongation in the steel wire be  Young's Modulus of steel, Y2 = 
True: As the force acts Normal to the parallel planes in which helical parts of the wire lie, the actual length of the wire would not change but it's shape would. Therefore the amount of elongation of the coil taking place for corresponding stress depends upon the Shear Modulus of elasticity.
False: Young's Modulus is defined as the ratio of the stress applied on a material and the corresponding strain that occurs. As for the same amount of pressure applied on a piece of rubber and steel, the elongation will be much lesser in case of steel than that in the case of rubber and therefore the Young's Modulus of rubber is lesser than that of steel.
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