A bead starts sliding from a point P on a frictionless wire with initial velocity of 5 m s $^{- 1}$ .
Find the velocity of bead at point R (take g $=$ 10 m s $^{- 2}$ )

3 \sqrt{2} m s^{- 1}

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6 \sqrt{2} m s^{- 1}

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12 \sqrt{2} m s^{- 1}

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18 \sqrt{2} m s^{- 1}

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Solution

The correct option is B $6 \sqrt{2} m s^{- 1}$
At point P, u $=$ 0.5 m s $^{- 1}$
Total energy at point
P $=$ Total energy at point R
$m g h_{p} + \frac{1}{2} m u^{2} = \frac{1}{2} m v^{2} + m g h_{R}$
$10 \times 4 + \frac{1}{2} 5^{2} = \frac{1}{2} \times v^{2} + (10 \times 1.65)$
$\frac{1}{2} v^{2} = (40 + 12.5) (16.5)$
$v^{2} = 36 \times 2$
$v = \sqrt{36 \times 2} m s^{- 1}$
$v = 6 \sqrt{2} m s^{- 1}$

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A bead starts sliding from a point P on a frictionless wire with initial velocity of 5 m s−1.Find the velocity of bead at point R (take g = 10 m s−2)

The correct option is B 6√2ms−1At point P, u = 0.5 m s−1Total energy at pointP = Total energy at point Rmghp+12mu2=12mv2+mghR10×4+1252=12×v2+(10×1.65)12v2=(40+12.5)(16.5)v2=36×2v=√36×2ms−1v=6√2ms−1

A bead starts sliding from a point P on a frictionless wire with initial velocity of 5 m s $^{- 1}$ .
Find the velocity of bead at point R (take g $=$ 10 m s $^{- 2}$ )