A triangular set square of angles 30°, 60°, 90° and of negligible mass is suspended freely from the right angled corner and weights are hung at the two corners. If the hypotenuse of the set square sets horizontally, then the ratio of the weights W1 / W2 is

Question

A triangular set square of angles 30°, 60°, 90° and of negligible mass is suspended freely from the right angled corner and weights are hung at the two corners. If the hypotenuse of the set square sets horizontally, then the ratio of the weights W1 / W2 is (A) 1: 1 (B) 1: 3 (C) √3: 1 (D) 1: √3

Tardigrade Admin · Accepted Answer

A B=l, A D=A cos 30°=l cos 30° cos 30° B D=B C cos 60°=l cos 60° cos 60° <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/24f94894b963c34c3487ce65d8cd13b1-.png" /> W1(A D)=W2(B D) ⇒ (W1/W2)=(1/3)

Q. A triangular set square of angles $3 0^{\circ}, 6 0^{\circ}, 9 0^{\circ}$ and of negligible mass is suspended freely from the right angled corner and weights are hung at the two corners. If the hypotenuse of the set square sets horizontally, then the ratio of the weights $W_{1} / W_{2}$ is

1 : 1

1 : 3

$3 : 1$

$1 : 3$

$A B = l, A D = A cos 3 0^{\circ} = l cos 3 0^{\circ} cos 3 0^{\circ}$
$B D = BC cos 6 0^{\circ} = l cos 6 0^{\circ} cos 6 0^{\circ}$

$W_{1} (A D) = W_{2} (B D)$
$\Rightarrow \frac{W _{1}}{W _{2}} = \frac{1}{3}$

Q. A triangular set square of angles 30∘,60∘,90∘ and of negligible mass is suspended freely from the right angled corner and weights are hung at the two corners. If the hypotenuse of the set square sets horizontally, then the ratio of the weights W1​/W2​ is

1 : 1

1 : 3

3​:1

1:3​

AB=l,AD=Acos30∘=lcos30∘cos30∘ BD=BCcos60∘=lcos60∘cos60∘ W1​(AD)=W2​(BD) ⇒W2​W1​​=31​

Q. A triangular set square of angles $3 0^{\circ}, 6 0^{\circ}, 9 0^{\circ}$ and of negligible mass is suspended freely from the right angled corner and weights are hung at the two corners. If the hypotenuse of the set square sets horizontally, then the ratio of the weights $W_{1} / W_{2}$ is

$3 : 1$

$1 : 3$

$A B = l, A D = A cos 3 0^{\circ} = l cos 3 0^{\circ} cos 3 0^{\circ}$
$B D = BC cos 6 0^{\circ} = l cos 6 0^{\circ} cos 6 0^{\circ}$

$W_{1} (A D) = W_{2} (B D)$
$\Rightarrow \frac{W _{1}}{W _{2}} = \frac{1}{3}$