Answer:
Yes, there is such a way.
Explanation:
If currents flow in the same direction in two or more long parallel wires, there will be an attractive force between the wires. If the current flows in different directions, there will be a repulsive force between the wires. In this case, these three parallel wires, can be be made to carry current in the same direction, creating an attractive force between all three wires.
Note that it is not possible to have at the least one of them carry current in the opposite direction and still have an attractive current between them.
Answer with Explanation:
We are given that
Angle of incidence,
Angle of refraction,
a.Refractive index of air,
We know that


b.Wavelength of red light in vacuum,

Wavelength in the solution,

c.Frequency does not change .It remains same in vacuum and solution.
Frequency,
Where 
Frequency,
d.Speed in the solution,

The answer is 22.1925 I hope you get right let me know I got my answer because I multiply it all together.
The equivalent resistance of several devices connected in parallel is given by

where

are the resistances of the various devices. We can see that every time we add a new device in parallel, the term

increases, therefore the equivalent resistance of the circuit

decreases.
But Ohm's law:

tells us that if the equivalent resistance decreases, the total current in the circuit increases. The power dissipated through the circuit (and so, the heat produced) depends on the square of the current:

therefore if there are too many devices connected in parallel, this can be a problem because there could be too much power dissipated (and too much heat) through the circuit.