Question

Question 26 of 31 > Attempt 1 Two steel wires are connected together, end to end, and attached to a wall. The two wires have the same length and elastic modulus, but the ratio of the radius of the first wire n to the radius of the second wire r is 9: 3. As the wires are initially the same length, the midpoint of the combination coincides with the connection point. An applied force then stretches the combination by 3.150 mm while the two wires stay connected together. After the wires are stretched, what is 6, the magnitude of the distance from the midpoint to the connection point? 8 = 0.525 mm Incorrect about us privacy policy terms of use RE 4 1:45 PM 6/18/2020

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

0.315171772=IIܠ
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$\times$10¹¹ 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 $F = mg+\rho Av^{2}=mg+\frac{m}{AL}Av^{2}=0.7\times 9.81+\frac{0.7}{0.5}\times 9.9045^{2}$

Now,

: ܐܠ FL AY FL 1R27 144.2057x0.5 x0,00552x1.5x1011

  $\Delta l=5.058\times 10^{-3}mm$