In the arrangement as shown, tension T2 is (g=10 m / s 2)

Question

In the arrangement as shown, tension T2 is (g=10 m / s 2) (A) 50 N (B) 100 N (C) 50 √3 N (D) 100 √3 N

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a075fe0e54ff54c3579f8629f38e13d2-.png" /> For pulley F=2 T for block T=10 g =100 N F=200 N For horizontal equilibrium T1 cos 60°=T2 cos 30° T1=2 T2 (√3/2)=√3 T2 ...(i) For vertical equilibrium T1= sin 60°+T2 sin 30°=200 ...(ii) Using (i) and (ii) solve for T 2 2 T2=200 T2=100 N

Q. In the arrangement as shown, tension $T_{2}$ is $(g = 10 m / s^{2})$

$50 N$

$100 N$

$503 N$

$1003 N$

Q. In the arrangement as shown, tension T2​ is (g=10m/s2)

50N

100N

503​N

1003​N

For pulley F=2T for block T=10g=100N F=200N For horizontal equilibrium T1​cos60∘=T2​cos30∘ T1​=2T2​23​​=3​T2​ ...(i) For vertical equilibrium T1​=sin60∘+T2​sin30∘=200 ...(ii) Using (i) and (ii) solve for T2​ 2T2​=200 T2​=100N

Q. In the arrangement as shown, tension $T_{2}$ is $(g = 10 m / s^{2})$

$50 N$

$100 N$

$503 N$

$1003 N$