A mass of 10 kg is suspended vertically by a rope of length 5 m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is θ= tan -1(x × 10-1). The value of x is . (Given .g=10 m / s 2)

Question

A mass of 10 kg is suspended vertically by a rope of length 5 m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is θ= tan -1(x × 10-1). The value of x is . (Given .g=10 m / s 2) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f1665b5d540a624525637be5b08e3e18-.png" /> T sin θ=30 T cos θ=100 ⇒ tan θ=0.3

Q. A mass of 10kg is suspended vertically by a rope of length 5m from the roof. A force of 30N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is θ=tan−1(x×10−1). The value of x is _____. (Given g=10m/s2)

Tsinθ=30 Tcosθ=100 ⇒tanθ=0.3

$T sin θ = 30$
$T cos θ = 100$
$\Rightarrow tan θ = 0.3$