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 ) [2023]

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Solution

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 $m / s^{2}$ ) [2023]

1 $(4 \sqrt{3} - 1) N$

2 $4 (\sqrt{3} - 1) N$

3 $(4 \sqrt{3} + 1) N$

4 $4 (\sqrt{3} + 1) N$

Ans.
(4)

$4 g \sin 60 ° - T = 4 a$ ...(i)

$T - g \sin 30 ° = a$ ...(ii)

Solving (i) and (ii), we get

$20 \sqrt{3} - T = 4 T - 20$

$\Rightarrow T = 4 (\sqrt{3} + 1) N$

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