If the radius of curvature of the path of two particles of same masses are in the ratio 1 : 2, then in order to have constant centripetal force, their velocity, should be in the ratio of

$1 : 4$

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$4 : 1$

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$\sqrt{2} : 1$

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$1 : \sqrt{2}$

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Solution

The correct option is D
$1 : \sqrt{2}$

The centripetal force, $F = \frac{m v^{2}}{r} \Rightarrow r = \frac{m v^{2}}{F}$
$∴ r \propto v^{2} o r v \propto \sqrt{r}$ (If m and F are constant),
$\Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{r_{1}}{r_{2}}} = \sqrt{\frac{1}{2}}$

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If the radius of curvature of the path of two particles of same masses are in the ratio 1 : 2, then in order to have constant centripetal force, their velocity, should be in the ratio of

The correct option is D 1:√2 The centripetal force, F=mv2r⇒r=mv2F ∴r∝v2 or v∝√r (If m and F are constant), ⇒v1v2=√r1r2=√12

The correct option is D
$1 : \sqrt{2}$

The centripetal force, $F = \frac{m v^{2}}{r} \Rightarrow r = \frac{m v^{2}}{F}$
$∴ r \propto v^{2} o r v \propto \sqrt{r}$ (If m and F are constant),
$\Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{r_{1}}{r_{2}}} = \sqrt{\frac{1}{2}}$