A body of mass m = 10 kg is attached to one
end of a wire of length 0.3 m. The maximum
angular speed (in rad s–1) with which it can be
rotated about its other end in space station is :
(Breaking stress of wire = 4.8 × 107 Nm–2 and
area of cross-section of the wire = 10–2 cm2) is:
Answer (integer)
4
Solution
T = m${\omega ^2}l$
<br><br>Breaking stress = $\sigma$ = ${{m{\omega ^2}l} \over A}$
<br><br>$\Rightarrow$ ${\omega ^2}$ = $${{4.8 \times {{10}^7} \times \left( {{{10}^{ - 2}} \times {{10}^{ - 4}}} \right)} \over {10 \times 0.3}}$$ = 16
<br><br>$\Rightarrow$ $\omega$ = 4
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
This question is part of PrepWiser's free JEE Main question bank. 158 more solved questions on Rotational Motion are available — start with the harder ones if your accuracy is >70%.