Two particles of masses in the ratio $1 : 2$ are moving in circles of radii in the ratio $2 : 3$ with time periods in the ratio $3 : 4$ . The ratio of their centripetal forces is:

9 : 4

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1 : 4

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9 : 16

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16 : 27

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Solution

The correct option is D $16 : 27$
Given, $\frac{m_{1}}{m_{2}} = \frac{1}{2}, \frac{R_{1}}{R_{2}} = \frac{2}{3}, \frac{T_{1}}{T_{2}} = \frac{3}{4}$
$\Rightarrow \frac{ω_{1}}{ω_{2}} = \frac{T_{2}}{T_{1}} = \frac{4}{3}$
So, $\frac{F_{1}}{F_{2}} = \frac{m_{1} ω_{1}^{2} R_{1}}{m_{2} ω_{2}^{2} R_{2}}$
$= \frac{1}{2} \times \frac{2}{3} \times \frac{4^{2}}{3^{2}} = \frac{16}{27}$

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Similar questions

The circumferences of two circles are in the ratio 3 : 4. The ratio of their area is

(a) 3 : 4 (b) 4 : 3 (c) 9 : 16 (d) 16 : 9

The radii of two circles are in the ratio $2 : 3$ . The ratio of their areas is
$(a)$ $2 : 3$

$(b)$ $4 : 9$

$(c)$ $3 : 2$

$(d)$ $9 : 4$

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