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Question 26 of 31 > Attempt 1 Two steel wires are connected together, end to end, and attached to a wall. The two wires have
Question 26 of 31 Attempt 1 Wire 1 L L Wire 2 L + AL L + AL2
Question 27 of 31 > Attempt 1 A person drops a cylindrical steel bar (Y = 1.500 x 10 Pa) from a height of 5.00 m (distance be
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Answer #1

26. Answer

Given ratio of radii =r_{1}:r_{2}=9:3

Total extension, \Delta L_{1}+\Delta L_{2}=3.150 mm --------------(1)

\delta =\frac{(L+\Delta L_{1})+(L+\Delta L_{2})}{2}-(L+\Delta L_{1})=\Delta L_{2}-\Delta L_{1} -----------(2)

as we know \Delta L=\frac{FL}{AY}=\frac{FL}{\pi r^{2}Y} or, \frac{\Delta L_{1}}{\Delta L_{2}}= \frac{r_{2}^{2}}{r_{1}^{2}}

\frac{\Delta L_{1}}{\Delta L_{2}}= (\frac{3}{9})^{2} or, \Delta L_{2}=9\Delta L_{1} -------------------(3)

solving equations (1) and (3),we get

\Delta L_{1}=0.315mm and \Delta L_{2}=(3.150-0.315)mm=2.835 mm

By equation(2) \delta =2.835 mm- 0.315 mm =2.52 mm

27.Given,mass of bar m = 0.7 kg,Y of bar = 1.5\times1011 N/m2 ,Length L= 0.5 m,radius R = 0.0055 m

and hight from ground h = 5 m

velocity of bar just before impact v= v=\sqrt{2gh}=\sqrt{2\times 9.81\times 5}=9.9045 m/s

net force acting on cross sectional area of bar during impact F = mg+\rho Av^{2}=mg+\frac{m}{AL}Av^{2}=0.7\times 9.81+\frac{0.7}{0.5}\times 9.9045^{2}

F=6.867+137.338 =144.2057 N

Now,\Delta l=\frac{FL}{AY}=\frac{FL}{\pi R^{2}Y}=\frac{144.2057\times 0.5}{\pi 0.0055^{2}\times 1.5\times 10^{11}}

  \Delta l=5.058\times 10^{-3}mm

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