A road is 10,, m wide. Its radius of curvature is 50,,m . outer edge is above lower edge by a

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4-1.Newton's Laws of Motion

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A road is $10\,\, m$ wide. Its radius of curvature is $50\,\,m$ . The outer edge is above the lower edge by a distance of $1.5\,\,m$ . this road is most suited for the velocity ......... $m/\sec$

A

$2.5$

B

$4.5$

C

$6.5$

D

$8.5$

Solution

$\tan \theta=\frac{1.5}{10}=0.15$

$\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}} \Rightarrow \mathrm{v}=\sqrt{\mathrm{rg} \tan \theta}$

$=\sqrt{50 \times 9 \cdot 8 \times 0 \cdot 15}=8 . 57 \mathrm{m} / \mathrm{s}$

Standard 11

Physics

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