A uniform meter scale of mass $1 k g$ is placed on table such that a part of the scale is beyond the edge. If a body of mass $0.25 k g$ is hung at the end of the scale then the minimum length of scale that should lie on the table so that it does not tilt is :

90 c m

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80 c m

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70 c m

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60 c m

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Solution

The correct option is B $60 c m$
the normal will shift to the edge of table to prevent toppling.
Torque about edge,
$m g (1 - d - 0.5 m) = (0 \cdot 25) g d$
$4 (0.5 - d) = d$
$2 - 4 d = d$
$\frac{2}{5} = d = 0 \cdot 4$ m
part lying on table: $= 1 - \cdot 4 = 0.6 m$

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A uniform metre scale is kept in equilibrium. It is supported at the 60 cm mark and a mass M is suspended from the 90 cm mark as shown in the figure. What is the relation between the weight of the scale and the weight of mass M?

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A uniform meter scale of mass 1kg is placed on table such that a part of the scale is beyond the edge. If a body of mass 0.25kg is hung at the end of the scale then the minimum length of scale that should lie on the table so that it does not tilt is :

The correct option is B 60cmthe normal will shift to the edge of table to prevent toppling.Torque about edge,mg(1−d−0.5m)=(0⋅25)g d4(0.5−d)=d2−4d=d25=d=0⋅4mpart lying on table: =1−⋅4=0.6m

A uniform meter scale of mass $1 k g$ is placed on table such that a part of the scale is beyond the edge. If a body of mass $0.25 k g$ is hung at the end of the scale then the minimum length of scale that should lie on the table so that it does not tilt is :

The correct option is B $60 c m$
the normal will shift to the edge of table to prevent toppling.
Torque about edge,
$m g (1 - d - 0.5 m) = (0 \cdot 25) g d$
$4 (0.5 - d) = d$
$2 - 4 d = d$
$\frac{2}{5} = d = 0 \cdot 4$ m
part lying on table: $= 1 - \cdot 4 = 0.6 m$