As per given figure, a weightless pulley P is attached on a double inclined frictionless surfaces. The tension in the string (massless) will be (if g =10 / s 2 )

Question

As per given figure, a weightless pulley P is attached on a double inclined frictionless surfaces. The tension in the string (massless) will be (if g =10 / s 2 ) (A) (4 √3-1) N (B) 4(√3-1) N (C) (4 √3+1) N (D) 4(√3+1) N

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f4e67949193b366737dbf195cfc891d5-.png" /> Solving (1) and (2) we get. 20 √3-T=4 T -20 T =4(√3+1) N

Q. As per given figure, a weightless pulley $P$ is attached on a double inclined frictionless surfaces. The tension in the string (massless) will be (if $g = 10 / s^{2}$ )

$(43 - 1) N$

$4 (3 - 1) N$

$(43 + 1) N$

$4 (3 + 1) N$

Solving (1) and (2) we get.
$203 - T = 4 T - 20$
$T = 4 (3 + 1) N$

Q. As per given figure, a weightless pulley P is attached on a double inclined frictionless surfaces. The tension in the string (massless) will be (if g=10/s2 )

(43​−1)N

4(3​−1)N

(43​+1)N

4(3​+1)N