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Byju's Answer

Standard XII

Physics

Relation between Linear and Angular Speed

A car of mass...

Question

A car of mass $1000 k g$ negotiates a banked curve of radius $90 m$ on a frictionless road. If the banking angle is $45^{\circ}$ , the speed of the car is:-

5 m / s

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10 m / s

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20 m / s

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30 m / s

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Solution

The correct option is B. $30 m / s$
Given,
$r = 90 m$
$m = 1000 k g$
$θ = 45^{0}$
$g = 10 m / s^{2}$
The speed of car,
The relation between permissible speed and banking angle is,
$t a n θ = \frac{v^{2}}{r g}$
$\Rightarrow v = \sqrt{r g t a n θ}$
$v = \sqrt{90 \times 10 \times 1} = 30 m / s$
Thus the correct option is B.

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Similar questions

Q. A car of mass

1000 kg

negotiates a banked curve of radius

90 m

on a frictionless road. If the banking angle is

45^{\circ}

, the speed of the car is:
Take

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

Q. A car of mass

1000 kg

negotiates a banked curve of radius

90 m

on a frictionless road. If the banking angle is

45^{\circ}

, the speed of car is (Take

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

)

Q. A car of mass

1000 kg

negotiates a banked curve of radius

90 m

on a frictionless road. If the banking angle is

45^{\circ}

, the speed of car is (Take

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

)

Q. A frictionless track has a curve of radius

90 \sqrt{3} m

. If the speed limit on this road is

30 m/s

, what should be the banking angle for this road to ensure safe driving ? Take

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

A car is negotiating a curve of radius R = 20 m on a banked road with banking angle $θ$ and coefficient of friction $μ$ . Take speed of car as v.

$\begin{matrix} Column 1 & Column 2 & Column 3 (I) & v = 10 \sqrt{2} m / s & (i) & μ = 0.2 & (p) & θ = 30^{\circ} (I I) & v = 20 m / s & (i i) & μ = 0.4 & (Q) & θ = 45^{\circ} (I I I) & v = 50 m / s & (i i i) & μ = 0.6 & (R) & θ = 37^{\circ} (I V) & v = 5 m / s & (i v) & μ = 0.8 & (S) & θ = 60^{\circ} \end{matrix}$

In which of the following cases, sliding occurs?

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A car of mass 1000 kg negotiates a banked curve of radius 90 m on a frictionless road. If the banking angle is 45∘, the speed of the car is:-

The correct option is B. 30 m/sGiven,r=90 mm=1000 kgθ=450g=10 m/s2The speed of car,The relation between permissible speed and banking angle is,tanθ=v2rg⇒v=√rgtanθv=√90×10×1=30 m/sThus the correct option is B.

A car of mass $1000 k g$ negotiates a banked curve of radius $90 m$ on a frictionless road. If the banking angle is $45^{\circ}$ , the speed of the car is:-