You visited us 1 times! Enjoying our articles? Unlock Full Access!

Deformation of Tapered Bars

A mild steel ...

Question

A mild steel rod tapers uniformly from $30 m m$ diameter to $12 m m$ diameter in a length of $300 m m$ . The rod is subjected to an axial load of $12 k N$ . $E = 2 \times 10^{5} N / m m^{2}$ . What is the extension of the rod in mm?

\frac{4 π}{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{4}{5 π}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{5 π}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{1}{5 π}$
Total extension in the rod will be
$Δ L = \frac{4 P L}{π E D_{1} D_{2}}$
$= \frac{4 \times 12 \times 10^{3} \times 300}{π \times 2 \times 10^{5} \times 12 \times 30} = \frac{1}{5 π}$

Suggest Corrections

Similar questions

24 m m

12 m m

400 m m

12 k N

E = 2 \times 10^{5} N / m m^{2} .

m m

40 m m

20 m m

800 m m

20 k N

E = 2 \times 10^{5} N / m m^{2}

450 m m

36 m m

18 m m

12 k N

E = 2 \times 10^{5} N / m m^{2}

E = 2 \times 10^{5} N / m m^{2}

30 m m

15 m m

15 m m

E = 2 \times 10^{5} N / m m^{2}

100 k N

m m

Join BYJU'S Learning Program