Which object? More information is needed to answer this question
Because,
In left image pin is not touch to the wire.
In right image pin is touch to the wire.
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Answer:
Explanation:
The two charges are q and Q - q. Let the distance between them is r
Use the formula for coulomb's law for the force between the two charges

So, the force between the charges q and Q - q is given by

For maxima and minima, differentiate the force with respect to q.

For maxima and minima, the value of dF/dq = 0
So, we get
q = Q /2
Now 
the double derivate is negative, so the force is maxima when q = Q / 2 .
Answer:
d)no unit
Explanation:
refractive index is a unit less quantity.
Object true weight is given as

now we know that g = 9.8 m/s^2


now when it is complete submerged in water its apparent weight is given as 82 N
apparent weight = weight - buoyancy force
apparent weight = 82 N
weight = 123 N
now we have
82 = 123 - buoyancy force
buoyancy force = 123 - 82 = 41 N
now we also know that buoyancy force is given as




now as we know that mass of the object is 12.55 kg
its volume is 4.18 * 10^-3 m^3
now we know that density will be given as mass per unit volume



so here density of object is 3002.4 kg/m^3