Two identical objects each of radii R and masses $m_{1}$ and $m_{2}$ are suspended using two strings of equal length L as shown in the figure (R<<L). The angle $θ$ which mass $m_{2}$ makes with the vertical is approximately

\frac{m_{1} R}{(m_{1} + m_{2}) L}

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\frac{2 m_{1} R}{(m_{1} + m_{2}) L}

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\frac{2 m_{2} R}{(m_{1} + m_{2}) L}

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\frac{m_{2} R}{(m_{1} + m_{2}) L}

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Solution

The correct option is B $\frac{2 m_{1} R}{(m_{1} + m_{2}) L}$

Distance of center of mass from $m_{2}$
$d = \frac{m_{1}}{(m_{1} + m_{2})} \times 2 R θ = \frac{d i s t a n c e (d)}{L} θ = \frac{m_{1}}{(m_{1} + m_{2})} \times \frac{2 R}{L}$

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Similar questions

Q. Two identical objects each of radii

R

and masses

m_{1}

and

m_{2}

are suspended using two strings of equal length

L

as shown in the figure

(R << L)

. The angle

θ

which mass

m_{2}

makes with the vertical is approximately.

Two positively charged spheres of masses $m_{1}$ and $m_{2}$ , are suspended from a common point at the ceiling by identical insulating massless strings of length l. Charges on the two spheres are $q_{1}$ and $q_{2}$ , respectively. At equilibrium both strings make the same angle $θ$ with the vertical. Then