A boy and a man carry a uniform rod of length L , horizontally in such a way that the boy gets 1 4 t h of the load. If the boy is at one end of the rod, the distance of the man from the other end is

Weight of the rod= w Reaction of boy R B = w 4 Reaction of man R M = 3 w 4 As the rod is in rotational equilibrium ∴ ∑ τ = 0 R B × L 2 - R M × x = 0 ⇒ w 4 × L 2 - 3 w 4 × x = 0 ⇒ x = L 6 ∴ Distance from other end, y = L 2 - x ⇒ y = L 2 - L 6 = 2 L 6 = L 3

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Physics - Rotation Question with Solution | TestHub

PhysicsRotationTorque / angular accelerationEasy2 minPYQ_2020

PhysicsEasysingle choice

A boy and a man carry a uniform rod of length $L$ , horizontally in such a way that the boy gets ${\frac{1}{4}}^{t h}$ of the load. If the boy is at one end of the rod, the distance of the man from the other end is

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Answer:

Solution:

Weight of the rod= $w$

Reaction of boy $R_{B} = \frac{w}{4}$

Reaction of man $R_{M} = \frac{3 w}{4}$

As the rod is in rotational equilibrium

$∴$ $\sum τ = 0$

$R_{B} \times \frac{L}{2} - R_{M} \times x = 0$

$\Rightarrow$ $\frac{w}{4} \times \frac{L}{2} - \frac{3 w}{4} \times x = 0$

$\Rightarrow$ $x = \frac{L}{6}$

$∴$ Distance from other end, $y = \frac{L}{2} - x$

$\Rightarrow$ $y = \frac{L}{2} - \frac{L}{6} = \frac{2 L}{6} = \frac{L}{3}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:RotationSubtopic:Torque / angular acceleration

⏱ 2mℹ️ Source: PYQ_2020

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