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Standard VIII

Physics

Normal Force

A mass of 1...

Question

A mass of $10 k g$ is suspended by a rope of length $2.8 m$ from a ceiling. A force of $98 N$ is applied at the midpoint of the rope as shown in the figure. The angle made by rope with the vertical in equlibrium is

30^{0}

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60^{0}

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45^{0}

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90^{0}

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Solution

The correct option is D $45^{0}$

Let the angle made by the rope with the vertical be $θ$
Now resolving the components of forces along vertical and horizontal direction we have
$T c o s θ = 10 g N$
$T s i n θ = 98 N$
Dividing we have,
$t a n θ = \frac{98}{98} = 1$
$θ = 45$ degrees.

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A mass of 10kg is suspended by a rope of length 2.8m from a ceiling. A force of 98N is applied at the midpoint of the rope as shown in the figure. The angle made by rope with the vertical in equlibrium is

The correct option is D 450Let the angle made by the rope with the vertical be θNow resolving the components of forces along vertical and horizontal direction we have Tcosθ=10gNTsinθ=98NDividing we have,tanθ=9898=1θ=45 degrees.

A mass of $10 k g$ is suspended by a rope of length $2.8 m$ from a ceiling. A force of $98 N$ is applied at the midpoint of the rope as shown in the figure. The angle made by rope with the vertical in equlibrium is

The correct option is D $45^{0}$

Let the angle made by the rope with the vertical be $θ$
Now resolving the components of forces along vertical and horizontal direction we have
$T c o s θ = 10 g N$
$T s i n θ = 98 N$
Dividing we have,
$t a n θ = \frac{98}{98} = 1$
$θ = 45$ degrees.