Answer:
(a) charge q=5.33 nC
(b) charge density σ=10.62 nC/m²
Explanation:
Given data
radius r=0.20 m
potential V=240 V
coulombs constant k=9×10⁹Nm²/C²
To find
(a) charge q
(b) charge density σ
Solution
For (a) charge q
As

For (b) charge density
As charge density σ is given as:
σ=q/(4πR²)
σ=(5.333×10⁻⁹) / (4π×(0.20)²)
σ=10.62 nC/m²
Explanation:
The given data is as follows.
Resistance (R) = 1200 ohm, Area (A) =
m (as
)
Diameter (d) = 2.3 mm =
m
First, we will calculate the length as follows.
R =
Here,
= resistivity of aluminium = 
Putting the given values above and we will calculate the value of length as follows.
R =
1200 = 
L = 
As the circumference of circular wire = 
or, =
= 
And, number of turns will be calculated as follows.
No. of turns × Circumference = Length of wire
No. of turns × 
= 
Thus, we can conclude that
turns of wire are needed.
Answer:
21.67 rad/s²
208.36538 N
Explanation:
= Final angular velocity = 
= Initial angular velocity = 78 rad/s
= Angular acceleration
= Angle of rotation
t = Time taken
r = Radius = 0.13
I = Moment of inertia = 1.25 kgm²
From equation of rotational motion

The magnitude of the angular deceleration of the cylinder is 21.67 rad/s²
Torque is given by

Frictional force is given by

The magnitude of the force of friction applied by the brake shoe is 208.36538 N