A body of mass 2 kg is placed on a rough surface with coefficient of friction $\sqrt{3}$ as given in diagram.
Then $θ$ is :

\frac{π}{4}

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\frac{π}{2}

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\frac{π}{3}

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\frac{π}{6}

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Solution

The correct option is D $\frac{π}{6}$
Given,

$m = 2 k g$

$μ = \sqrt{3}$

From the free body diagram,

$f_{r} = μ N = m g s i n ϕ$ . . .(1)

$N = m g c o s ϕ$ . . . .(2)

equation (1) becomes

$μ m g c o s ϕ = m g s i n ϕ$

$t a n ϕ = μ = \sqrt{3}$

$ϕ = t a n^{- 1} (\sqrt{3})$

$ϕ = 60^{0}$

In triangle, the sum of all angle is $180^{0}$

$ϕ + θ + 90^{0} = 180^{0}$

$θ = 180^{0} - 90^{0} - 60^{0} = 30^{0}$

In radian, $θ = \frac{π}{6}$

The correct option is D.

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Similar questions

Q. The general solution of

(\sqrt{3} - 1) sin θ + (\sqrt{3} + 1) cos θ = 2

is
(where

n \in Z

)

Let $θ \in (0, \frac{π}{2}) a n d x = X c o s θ + y s i n θ, y = X s i n θ - Y c o s θ$ such that $x^{2} + 2 x y + y^{2} = a X^{2} + b Y^{2}$ Where a and b are constants then

List-I	List-II
$1) {tan}^{2} θ = 1$	a $) n π \pm π / 6$
$2) {cos}^{2} θ = \frac{1}{4}$	b $) n π \pm π / 4$
$3) {sin}^{2} θ = \frac{1}{4}$	c $) n π \pm π / 3$
4) cosec $θ = 1$	d) $n π \pm π / 2$ $e) n π \pm π / 8$

Q. If

x = e^{θ} (sin θ + cos θ)

and

y = e^{θ} (sin θ - cos θ),

where

θ

is a real parameter, then

\frac{d^{2} y}{d x^{2}}

θ = \frac{π}{6}

Q. If 2 sin

(θ + \frac{π}{3}) = cos (θ - \frac{π}{6}),

prove that

tan θ + \sqrt{3} = 0

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A body of mass 2 kg is placed on a rough surface with coefficient of friction √3 as given in diagram.Then θ is :

A body of mass 2 kg is placed on a rough surface with coefficient of friction $\sqrt{3}$ as given in diagram.
Then $θ$ is :