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Q. A mass of 0.1 kg is hung at the 20 cm mark from a 1 m rod weighing 0.25 kg pivoted at its centre. The rod will not topple if

COMEDKCOMEDK 2012System of Particles and Rotational Motion

Solution:

Given situation is shown in the figure.
Here, $m_1 = 0.1 \, kg,$
x = 50 - 20 = 30 cm = 0.3 m
Apply law of moments about centre of the rod,
$m_1 gx = m_2 gx'$
$m_2x' = 0.1 \times 0.3 = 0.03 \, kg\, m$
Hence according to options
$m_2 = 0.15\, kg, x' = 0.2\, m = 20 \, cm$
Hence rod will not topple if second mass of
0.15 kg is hung at 70 cm mark.

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