A car turns on a horizontal curved road of radius 40 m. If coefficient of friction is 0.4, then maximum speed with which car can take turn is : (take g= $10 {m / s}^{2}$ )

4 m / s

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2 \sqrt{5} m / s

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4 \sqrt{10} m / s

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10 \sqrt{4} m / s

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Solution

The correct option is C $4 \sqrt{10} m / s$
Given,

$r = 40 m$

$μ = 0.4$

$g = 10 m / s^{2}$

The maximum speed with which car can take turns,

$v = \sqrt{μ r g}$

$v = \sqrt{0.4 \times 40 \times 10}$

$v = 4 \sqrt{10} m / s$

The correct option is C.

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A car turns on a horizontal curved road of radius 40 m. If coefficient of friction is 0.4, then maximum speed with which car can take turn is : (take g=10m/s2)

The correct option is C 4√10m/sGiven,r=40mμ=0.4g=10m/s2The maximum speed with which car can take turns,v=√μrgv=√0.4×40×10v=4√10m/sThe correct option is C.

A car turns on a horizontal curved road of radius 40 m. If coefficient of friction is 0.4, then maximum speed with which car can take turn is : (take g= $10 {m / s}^{2}$ )

The correct option is C $4 \sqrt{10} m / s$
Given,

$r = 40 m$

$μ = 0.4$

$g = 10 m / s^{2}$

The maximum speed with which car can take turns,

$v = \sqrt{μ r g}$

$v = \sqrt{0.4 \times 40 \times 10}$

$v = 4 \sqrt{10} m / s$

The correct option is C.