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Question

A rope of length 30 cm is on a horizontal table with maximum length hanging from edge A of the table. The coefficient of friction between the rope and table is 0.5. The distance of centre of mass of the rope from A is:


A
5153 cm
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B
5173 cm
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C
5193 cm
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D
7173 cm
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Solution

The correct option is B $$\displaystyle \dfrac{5\sqrt{17}}{3}$$ cm
Total length of a rope, $$L=30cm$$

Let mass density $$=\lambda$$

$$\mu=0.5$$ coefficient of friction.

Frictional force,

$$\mu[\lambda(L-l)]g=\lambda lg$$

$$\mu (L-l)=l$$

$$\dfrac{l}{L}=\dfrac{\mu}{\mu +1}$$

$$l=30\dfrac{0.5}{0.5+1}=10cm$$

$$L-l=20cm$$

The center of mass along x-axis is given by

$$X_{CM}=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}$$

$$X_{CM}=\dfrac{20\lambda\times (-10)+10\lambda \times (0)}{20\lambda +10\lambda}$$

$$X_{CM}=-\dfrac{20}{3}cm$$

The center of mass along y-axis is given by

$$Y_{CM}=\dfrac{m_1y_1+m_2y_2}{m_1+m_2}$$

$$Y_{cm}=\dfrac{5\times 10\lambda+20\lambda \times 0}{20\lambda +10\lambda}$$

$$Y_{CM}=-\dfrac{5}{3}cm$$

We know that,

$$CM=\sqrt{(X_{CM})^2+(Y_{CM})^2}$$

$$CM=\sqrt{(\dfrac{-20}{3})^2+(\dfrac{-5}{3})^2}$$

$$CM=\dfrac{5\sqrt{17}}{3}cm$$

The correct option is B.

1447109_40138_ans_fc0fe2f5ea994778a39470a68b29b1a6.png

Physics

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