Distance And Displacement

The angular momentum

%5Cvec%7BL%7D

of a particle about the origin is given by the cross product of the position vector

%5Cvec%7Br%7D

and the linear momentum

%5Cvec%7Bp%7D

%5Cvec%7BL%7D%20%3D%20%5Cvec%7Br%7D%20%5Ctimes%20%5Cvec%7Bp%7D

The linear momentum

%5Cvec%7Bp%7D

is the product of mass

m

and velocity

%5Cvec%7Bv%7D

%5Cvec%7Bp%7D%20%3D%20m%5Cvec%7Bv%7D

Given

%5Cvec%7Bv%7D%20%3D%205%20%5Chat%7Bi%7D%20%2B%204%20%5Chat%7Bj%7D%20%2B%206%20%5Chat%7Bk%7D

and

%5Cvec%7Br%7D%20%3D%20-2%20%5Chat%7Bi%7D%20%2B%204%20%5Chat%7Bj%7D%20%2B%206%20%5Chat%7Bk%7D

%5Cvec%7Bp%7D%20%3D%20m(5%20%5Chat%7Bi%7D%20%2B%204%20%5Chat%7Bj%7D%20%2B%206%20%5Chat%7Bk%7D)

The cross product

%5Cvec%7Br%7D%20%5Ctimes%20%5Cvec%7Bv%7D

is calculated as follows:

%5Cvec%7Br%7D%20%5Ctimes%20%5Cvec%7Bv%7D%20%3D%20%5Cleft%7C%5Cbegin%7Barray%7D%7Bccc%7D%5Chat%7Bi%7D%20%26%20%5Chat%7Bj%7D%20%26%20%5Chat%7Bk%7D%20%5C%5C-2%20%26%204%20%26%206%20%5C%5C5%20%26%204%20%26%206%5Cend%7Barray%7D%5Cright%7C

Expanding the determinant, we get:

%5Cvec%7Br%7D%20%5Ctimes%20%5Cvec%7Bv%7D%20%3D%20%5Chat%7Bi%7D(4%20%5Ccdot%206%20-%206%20%5Ccdot%204)%20-%20%5Chat%7Bj%7D(-2%20%5Ccdot%206%20-%206%20%5Ccdot%205)%20%2B%20%5Chat%7Bk%7D(-2%20%5Ccdot%204%20-%204%20%5Ccdot%205)

Simplifying further:

%5Cvec%7Br%7D%20%5Ctimes%20%5Cvec%7Bv%7D%20%3D%20%5Chat%7Bi%7D(24%20-%2024)%20-%20%5Chat%7Bj%7D(-12%20-%2030)%20%2B%20%5Chat%7Bk%7D(-8%20-%2020)

%5Cvec%7Br%7D%20%5Ctimes%20%5Cvec%7Bv%7D%20%3D%20%5Chat%7Bi%7D(0)%20%2B%20%5Chat%7Bj%7D(42)%20%2B%20%5Chat%7Bk%7D(-28)

%5Cvec%7Br%7D%20%5Ctimes%20%5Cvec%7Bv%7D%20%3D%2042%20%5Chat%7Bj%7D%20-%2028%20%5Chat%7Bk%7D

Multiplying by

m

, the angular momentum is:

%5Cvec%7BL%7D%20%3D%20m(42%20%5Chat%7Bj%7D%20-%2028%20%5Chat%7Bk%7D)

Therefore, the correct answer is:

m(42%20%5Chat%7Bi%7D%20-%2028%20%5Chat%7Bk%7D)

To determine the Vernier constant, we need to find the difference between one main scale division and one Vernier scale division.

Given:

1 main scale division (MSD) = 1 mm
100 Vernier scale divisions (VSD) = 99 main scale divisions

We need to find the length of 1 Vernier scale division (1 VSD):

$1%20%5Ctext%7B%20VSD%7D%20%3D%20%5Cfrac%7B%5Ctext%7Bvalue%20of%2099%20MSDs%7D%7D%7B100%7D%20%3D%20%5Cfrac%7B99%20%5Ctext%7B%20mm%7D%7D%7B100%7D%20%3D%200.99%20%5Ctext%7B%20mm%7D$

Therefore, the difference between one main scale division and one Vernier scale division (the Vernier constant) is:

$%5Ctext%7BVernier%20Constant%7D%20%3D%201%20%5Ctext%7B%20MSD%7D%20-%201%20%5Ctext%7B%20VSD%7D$

$%3D%201%20%5Ctext%7B%20mm%7D%20-%200.99%20%5Ctext%7B%20mm%7D%20%3D%200.01%20%5Ctext%7B%20mm%7D$

Since 0.01 mm is equivalent to 10 micrometers (µm), the Vernier constant is:

10 µm

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Distance And Displacement

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs