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A ring of radius `R` is made of a thin wire of material of density `rho` having cross section area `a.` The ring rotates with angular velocity `omega` about an axis passing through its centre and perpendicular to the plane. If we consider a small element of the ring,it rotates in a circle. The required centripetal force is provided by the component of tensions on the element towards the centre. A small element of length `dl` of angular width `d theta` is shown in the figure. If for a given mass of the ring and angular velocity, the radius `R` of the ring is increased to `2R`, the new tension will beA. `T//2`B. `T`C. `2T`D. `4T` |
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Answer» Correct Answer - C As the small element `(dm =a.rho.dl)` in rotating int the circle, centripetal force `F_(C)=d m omega^(2)R=a rho d l . Omega^(2)R` `T= a rho R^(2) omega^(2)=(m)/(2pi) R omega^(2) prop R` Radius is doubled, tension is doubled . `(2T)` `T=a rho R^(2) omega^(2)=(m)/(2pi)R omega^(2) prop R` |
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