A truck of mass 2000 ,kg is moving along a ci | vedclass

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(D) Given:Mass of the truck, $M = 2000 \,kg$Radius of curvature, $R = 10 \,m$Banking angle, $	heta = 39^{\circ}$Acceleration due to gravity, $g = 10

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) Given:Mass of the truck, $M = 2000 \,kg$Radius of curvature, $R = 10 \,m$Banking angle, $	heta = 39^{\circ}$Acceleration due to gravity, $g = 10 \,ms^{-2}$Value of $	an 39^{\circ} = 0.81$The formula for the maximum permissible speed $v$ on a banked road (without friction) is given by:$v = \sqrt{Rg 	an 	heta}$Substituting the given values into the formula:$v = \sqrt{10 	imes 10 	imes 0.81}$$v = \sqrt{100 	imes 0.81}$$v = \sqrt{81}$$v = 9 \,ms^{-1}$Thus, the maximum permissible speed of the truck is $9 \,ms^{-1}$.

$A$ truck of mass $2000 \,kg$ is moving along a circular path having a radius of curvature $10 \,m$. If the banking angle is $39^{\circ}$, then the maximum permissible speed of the truck is (Acceleration due to gravity $= 10 \,ms^{-2}$, take $\tan 39^{\circ} = 0.81$). (in $\,ms^{-1}$)

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