Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
One end of a wire of $8 \mathrm{~mm}$ radius and $100 \mathrm{~cm}$ length is fixed and the other end is twisted through an angle of $45^{\circ}$. The angle of shear is
PhysicsMechanical Properties of SolidsAP EAMCETAP EAMCET 2021 (23 Aug Shift 1)
Options:
  • A $0.36^{\circ}$
  • B $0.12^{\circ}$
  • C $3.6^{\circ}$
  • D $1.2^{\circ}$
Solution:
2330 Upvotes Verified Answer
The correct answer is: $0.36^{\circ}$
Given, radius of wire
$$
\begin{aligned}
r & =8 \mathrm{~mm} \\
& =8 \times 10^{-3} \mathrm{~m}
\end{aligned}
$$
Length of wire, $l=100 \mathrm{~cm}=1 \mathrm{~m}$
Twist, $\phi=45^{\circ}$
Let shear angle be $\theta$.
and
$$
\begin{aligned}
\tan \theta & =\frac{r \phi}{l} \\
\theta & =\frac{r \phi}{l} \quad(\because \tan \theta \simeq \theta) \\
& =\frac{8 \times 10^{-3} \times 45}{1}=0.36^{\circ}
\end{aligned}
$$
$$
\therefore \quad \theta=\frac{r \phi}{l}
$$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.