JEE MAIN - Physics (2020 - 2nd September Morning Slot - No. 12)

JEE Main 2020 (Online) 2nd September Morning Slot Physics - Rotational Motion Question 134 English
Shown in the figure is rigid and uniform one meter long rod AB held in horizontal position by two strings tied to its ends and attached to the ceiling. The rod is of mass ‘m’ and has another weight of mass 2 m hung at a distance of 75 cm from A. The tension in the string at A is :
0.5 mg
2 mg
0.75 mg
1 mg

Explanation



$$\tau $$B = 0 (torque about point B is zero)

(T A) × 100 – (mg) × 50 – (2mg) × 25 = 0

$$ \Rightarrow $$ 100T A = 100 mg

$$ \Rightarrow $$ TA = 1 mg

Comments (0)

Advertisement