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Rotational Motion question

2020 · 2 Sep · Shift 1 · Q49

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JEE Main
Physics
Rotational Motion
2020 · 2 Sep · Shift 1 · Q49

Rotational Motion question

2020 · 2 Sep · Shift 1 · Q49

JEE MainPhysicsRotational MotionMCQ+4 / −1

JEE Main 2020 (Online) 2nd September Morning Slot Physics - Rotational Motion Question 148 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 :

A
0.5 mg
B
2 mg
C
0.75 mg
D
1 mg

View written solutionFree

Correct answer: D

Identify all forces on the rod

A uniform rod $AB$ of length $1\text{ m}$ is held horizontally by strings at ends $A$ and $B$ .

Let tensions be:

$T_A$ at end $A$
$T_B$ at end $B$

Downward forces:

Weight of rod $= mg$ , acting at its center, i.e. $0.5\text{ m}$ from $A$
Additional weight $= 2mg$ , acting at $0.75\text{ m}$ from $A$

Apply translational equilibrium

Since the rod is in equilibrium, $T_A + T_B = mg + 2mg = 3mg$

So, $T_A + T_B = 3mg \quad ...(1)$

Apply rotational equilibrium about point $A$

Taking moments about $A$ :

$T_A$ has zero moment about $A$
$T_B$ acts upward at distance $1\text{ m}$
Rod's weight $mg$ acts downward at $0.5\text{ m}$
Extra weight $2mg$ acts downward at $0.75\text{ m}$

For equilibrium, $T_B(1) = mg(0.5) + 2mg(0.75)$

$T_B = 0.5mg + 1.5mg = 2mg$

Thus, $T_B = 2mg$

Find $T_A$ using equation (1)

$T_A + 2mg = 3mg$

$T_A = mg$

Match with options

$T_A = 1\,mg$

So the correct option is D.

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