Home 1st Year Physics Notes FSC 1st Year Physics Chapter 5 Exercise & Numericals Notes PDF

# FSC 1st Year Physics Chapter 5 Exercise & Numericals Notes PDF

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In this post, I am sharing FSC 1st Year Physics Chapter 5 Solved Exercise & Numericals Notes PDF for the students of Intermediate Part 1. This is the 5th chapter in the 11th physics textbook and its name is Circular Motion. Students can download these 11th class chapter 5 Circular Motion Solved Exercise and Numericals Notes PDF on their laptop or mobile. These Physics Notes for Class 11 are for all the boards working under Punjab Board like Gujranwala Board, Lahore Board, Faisalabad Board, Multan Board, Rawalpindi Board, Sargodha Board, DG Kahn Board, and Sahiwal Board. On ilmihub, I have already shared the complete FSC part 1 Physics Notes PDF of all chapters.

## FSC Part 1 Physics Chapter 5 Circular Motion Solved Exercise & Numericals Notes PDF Download

### A tiny laser beam is directed from the Earth to the moon. If the beam is to have a diameter of 2.50 m at the moon, how small must the divergence angle be for the beam? The distance of the moon from the Earth is 3.8 x 108 m

Solution.
data:                      diameter of beam = length of arc = S = 2.50 m
distance of moon from the earth = radius of circular arc = r = 3.8 x 108
to find                   divergence angle = q = ?
Calculation:

Using the formula
S = rq
q = S/r
putting the value we get
q = 2.50/ 3.8 x 108 = 6.6 x 10 -9 rad
q = 6.6 x 10 -9 rad

### A body of moment of inertia l = 0.80 kg m2 about a fixed axis rotates with a constant angular velocity of 100 rad s-1. Calculate its angular momentum L and the torque to sustain the motion.

Solution:
Data:                     moment of inertia = l = 0.80Kgm2
angular velocity = w = 100 rads-1
to find                   angular momentum = L = ?
Torque =?
Calculation:
L = lw
putting the value we have
L = 0.80 x 100
= 80kgm2s-1
or                            L = 80 kgm2s-1 X x/x
= 80 kgm2s-2s
L = 80Jstorque can be found by using the relation
i = la

= 1×0 = 0
i = 0

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