JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 3)
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:
(Breaking stress of wire = 4.8 × 107 Nm–2 and
area of cross-section of the wire = 10–2 cm2) is:
Answer
4
Explanation
T = m$${\omega ^2}l$$
Breaking stress = $$\sigma $$ = $${{m{\omega ^2}l} \over A}$$
$$ \Rightarrow $$ $${\omega ^2}$$ = $${{4.8 \times {{10}^7} \times \left( {{{10}^{ - 2}} \times {{10}^{ - 4}}} \right)} \over {10 \times 0.3}}$$ = 16
$$ \Rightarrow $$ $$\omega $$ = 4
Breaking stress = $$\sigma $$ = $${{m{\omega ^2}l} \over A}$$
$$ \Rightarrow $$ $${\omega ^2}$$ = $${{4.8 \times {{10}^7} \times \left( {{{10}^{ - 2}} \times {{10}^{ - 4}}} \right)} \over {10 \times 0.3}}$$ = 16
$$ \Rightarrow $$ $$\omega $$ = 4
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