The point in which it originates.
A) 4.7 cm
The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

where
n is the order of the maximum
is the wavelength
is the distance between the slits
In this problem,
n = 5


So we find

And given the distance of the screen from the slits,

The distance of the 5th bright fringe from the central bright fringe will be given by

B) 8.1 cm
The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

And the distance of the 8th dark fringe from the central bright fringe will be given by

Given:
1st run: 20 meters North
2nd run: 15 meters East
time: 15 seconds
Average speed = total distance covered / total time taken
Ave. Speed = (20m + 15m) / 15s
Ave. Speed = 35m / 15s
Ave. Speed = 2 1/3 meters per second
Answer: 0.01 m
Explanation: The formulae for capillarity rise or fall is given below as
h = (2T×cosθ)/rpg
Where θ = angle mercury made with glass = 50°
T = surface tension = 0.51 N/m
g = acceleration due gravity = 9.8 m/s²
r = radius of tube = 0.5mm = 0.0005m
p = density of mercury.
h = height of rise or fall
From the question, specific gravity of density = 13.3
Where specific gravity = density of mercury/ density of water, where density of water = 1000 kg/m³
Hence density of mercury = 13.3×1000 = 13,300 kg/m³.
By substituting parameters, we have that
h = 2×0.51×cos 50/0.0005×9.8×13,300
h = 0.6556/65.17
h = 0.01 m
Answer:
The right answer is:
(a) 63.83 kg
(b) 0.725 m/s
Explanation:
The given query seems to be incomplete. Below is the attachment of the full question is attached.
The given values are:
T = 3 sec
k = 280 N/m
(a)
The mass of the string will be:
⇒ 
or,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
(b)
The speed of the string will be:
⇒ 
then,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
⇒ 