Explanation:
It is given that,
Distance between wires, d = 3.5 mm = 0.0035 m
Power of light bulb, P = 100 W
Potential difference, V = 120 V
(a) We need to find the force per unit length each wire of the cord exert on the other. It is given by :

Power, P = V × I

This gives, 


(b) Since, the two wires carry equal currents in opposite directions. So, teh force is repulsive.
(c) This force is negligible.
Hence, this is the required solution.
Answer:
The inside Pressure of the tank is 
Solution:
As per the question:
Volume of tank, 
The capacity of tank, 
Temperature, T' =
= 299.8 K
Temperature, T =
= 288.2 K
Now, from the eqn:
PV = nRT (1)
Volume of the gas in the container is constant.
V = V'
Similarly,
P'V' = n'RT' (2)
Also,
The amount of gas is double of the first case in the cylinder then:
n' = 2n
![\]frac{n'}{n} = 2](https://tex.z-dn.net/?f=%5C%5Dfrac%7Bn%27%7D%7Bn%7D%20%3D%202)
where
n and n' are the no. of moles
Now, from eqn (1) and (2):


If the distance between two charges is halved, the electrical force between them increases by a factor 4.
In fact, the magnitude of the electric force between two charges is given by:

where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:

the magnitude of the force changes as follows:

so, the force increases by a factor 4.