Answer: 
Explanation:
Given
Magnitude of charge is 
Force experienced is 
Electric field intensity is the electrostatic force per unit charge
Thus, the electric field intensity is
Answer:
The potential difference between the ends of a wire is 60 volts.
Explanation:
It is given that,
Resistance, R = 5 ohms
Charge, q = 720 C
Time, t = 1 min = 60 s
We know that the charge flowing per unit charge is called current in the circuit. It is given by :
I = 12 A
Let V is the potential difference between the ends of a wire. It can be calculated using Ohm's law as :
V = IR
V = 60 Volts
So, the potential difference between the ends of a wire is 60 volts. Hence, this is the required solution.
Answer:
a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
Explanation:
Let n₁ and n₂ be no of lines per unit length of grating A and B respectively.
λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,
Distance of first maxima for green light
= λ₁ D/ d₁
Distance of first maxima for red light
= λ₂ D/ d₂
Given that
λ₁ D/ d₁ = λ₂ D/ d₂
λ₁ / d₁ = λ₂ / d₂
λ₁ / λ₂ = d₁ / d₂
But
λ₁ < λ₂
d₁ < d₂
Therefore no of lines per unit length of grating A will be more because
no of lines per unit length ∝ 1 / d
If grating B is illuminated with green light first maxima will be at distance
λ₁ D/ d₂
As λ₁ < λ₂
λ₁ D/ d₂ < λ₂ D/ d₂
λ₁ D/ d₂ < 1 m
In this case position of first maxima will be less than 1 meter.
Option a is correct .
