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Physics-

General

Easy

Question

A wheel of radius ‘r’ and mass ‘m’ stands in front of a step of height 'h’ The least horizontal force which should be applied to the axle of the wheel to allow it to raise onto the step is

$\frac{m g h (2 r - h)}{r - h}$
$m g h (r - h)$
$\frac{m g \sqrt{h (2 r - h)}}{r - h}$
$\frac{m g h}{r}$

The correct answer is: $fraction numerator m g square root of h left parenthesis 2 r minus h right parenthesis end root over denominator r minus h end fraction$

Applying the condition of rotational equilibrium,
$F left parenthesis r minus h right parenthesis equals m g x$
But $r to the power of 2 end exponent equals x to the power of 2 end exponent plus left parenthesis r minus h right parenthesis to the power of 2 end exponent rightwards double arrow x equals square root of h left parenthesis 2 r minus h right parenthesis end root$
$therefore F equals fraction numerator m g square root of h left parenthesis 2 r minus h right parenthesis end root over denominator r minus h end fraction$

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