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Answer:

Mass of the object hanging at one end of the metre scale = m1 =60gms

Distance of the above object from the Pivot support = x1 = 40 Cms

 Therefore the other end of the metre scale is at a distance of 100-40 = 60cms from the pivot

Mass of the insect crawling on the metre scale = m2 = 100gms

Let the scale be in equilibrium when the insect travels a distance of x2 cms from the other end of the  metre scale towards the pivot. Therefore distance of the insect from the pivot = (60-x) cms

When the scale is in equilibrium,the moments(of the forces) taken about the pivot will be equal and opposite.

Moment of the mass of the object = m1*g*x1 =  60g*40 = 2400g , where g is acceleration due to gravity

Moment of the mass of the insect = m2*g*x2 =  100*g*(60-x2)

Equating the moments we get  100*g*(60-x2) = 2400g

Or 100*(60-x2) = 2400 ,that is 6000 - 100x2 =2400 . or 100*x2 = 6000-2400 =3600  

Therefore x2 =  3600/100 =  36 Cms.

The insect travels this distance in 6 seconds.Therefore speed of the insect =v = 36/6 = 6 Cms/s

Or v =  0.06*3600/1000  = 0.216Km/hr

Speed of the insect = 0.216Km/hr 

 

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