Question

A car is crossing a road-turn with a speed of 10 metre/
second. If the coefficient of friction be 0.5, then the
radius of the turn of the car is (g = 10 m/s):
(a) 20 metre
(b) 10 metre
(c) 5 metre
(d) 2 metre​

Answer 1

$\underline{\underline{\sf{\color{orange}{GIVEN:-}}}}$

A car is crossing a road-turn with a speed, v = 10 m/s

The coefficient of friction, μ = 0.5

Acceleration due to gravity, g = 10 m/s²

$\underline{\underline{\sf{\color{orange}{TO\ DETERMINE:-}}}}$

The radius of the turn(R) of the car.

$\underline{\underline{\sf{\color{orange}{SOLUTION:-}}}}$

We know,

$\boxed{\bf{v=\sqrt{\mu Rg} }}$

Solving further by substituting values :-

$\sf{\longmapsto 10=\sqrt{\mu Rg} }$

$\sf{\longmapsto 10=\sqrt{0.5 \times R \times 10} }$

$\sf{\longmapsto 10=\sqrt{5R} }$

$\sf{\longmapsto (10)^2=(\sqrt{5R})^2}$

$\sf{\longmapsto 100=5R}$

$\sf{\longmapsto R=\dfrac{100}{5}}$

$\sf{\longmapsto R=20\ m}$

∴ The radius of the turn of the car is 20 m.

The answer is (a) 20 meter.

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Answer 2

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Answer:

a) 20 metre

Step-by-step explanation:

Given, v=10m/sμ=0.5

We know that Vmax = √μgR

On squaring both sides, we get

R= v^2/μg

= 200/10

=20m

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