First slide

Elastic Modulie

Question

Two wires of diameter 0.25cm. One made of steel and the other made of brass are loaded as shown in figure. The length of steel wire 1.5m and that of brass is 1.0m. The elongation in steel wire is (10^-4m) ( Y_steel = 2.0 × 10¹¹Pa, Y_brass = 0.91 × 10¹¹Pa)

Moderate

A

1.5
B

3.0
C

2.5
D

5.0

Solution

$\large 2r = 0.25 = 25 \times {10^{ - 2}} \times {10^{ - 2}}m$
$\large r = \frac{{25}}{2} \times {10^{ - 4}} = 12.5 \times {10^{ - 4}} = 125 \times {10^{ - 5}}m$
$\large {l_{\rm{s}}}{\rm{ = 1}}{\rm{.5m; }}{l_{\rm{b}}}{\rm{ = 1}}{\rm{.0 cm}};\;e = \frac{{({m_1} + {m_2})g{l_s}}}{{\pi {r^2}{y_s}}}$
$\large e = \frac{{(6 + 4) \times 9.8 \times 1.5}}{{3.14 \times 125 \times 125 \times {{10}^{ - 10}} \times 2 \times {{10}^{11}}}}m$
$\large e = 1.498 \times {10^{ - 4}} \approx 1.5 \times {10^{ - 4}}m$

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