CHAPTER 01
Question 1.1
Speed
of light, v = 3.0 x 108 ms-1
Travel
time, t = 1 year
= 365 x 24 x 60 x 60 sec
= 31536000 sec
Since
S = v . t
Therefore
S = 3.0 x 108 x 31536000
S = 9.5 x 1015 m Answer
Question 1.2
Question 1.3
Length,
L = 15.3 cm
Width,
W = 12.80 cm
Volume,
V = ?
Since
Volume = Length x Width
V = L x W
V = 15.3 x 12.80
V = 196 cm2 Answer
Question 1.4
Question 1.5
Here
Length
of simple pendulum, l = 100 cm
Time for
20 vibrations = 40.2 s
Meter
scale accuracy = 1 mm
Stop watch
accuracy = 0.1 s
Now
l = 100
cm
= 100/100
m
= 1 m
T =
40.2/20
= 2.01 s
FOR
LENGTH
Absolute uncertainty = 1 mm = 0.1 cm
%age uncertainty = (0.1/100) x (100/100)
= 0.1 %
FOR
TIME
Absolute uncertainty = 0.1 sec
Average uncertainty = 0.1/20 = 0.005 sce
%age uncertainty = (0.005/2.01) x (100/100)
= 0.25 %
Total
uncertainty = 2 x 0.025 + 0.1 = 0.6 %
Total
uncertainty for g = 9.76 x 0.6 /100 = 0.06
Thus
g = 9.76 ± 0.06 ms-2
Question 1.6
Question 1.7
vf = vi + a t
[LT-1] = [LT-1] + [LT-2 T]
[LT-1] = [LT-1] + [LT-1]
[LT-1] = [LT-1]
LHS = RHS Hence Proved
Question 1.8
According
to question
v
; ρa Eb
v
= constant . ρa Eb -
Eq (1)
For ρ
ρ = m/v = [ML-3] - Eq (2)
For E
E = stress/strain
E = (F/A)/(∆l/l) = [MLT-2/L2] / [L/L] = [ML-1T-2]
- Eq (3)
For v
v
= [LT-1] -
Eq (4)
Putting
the value of ρ, E and v from Eq (2), (3)
and (4) in Eq (1), we get
v
= constant . ρa Eb
[LT-1] = constant [ML-3]a
[ML-1T-2]b
[LT-1] = constant Ma+b
L-3a-b T-2b
Equating
powers of corresponding quantities on both sides
a + b =0; - 3a - b = 1; -2b = -1
By
solving, we get,
a = -½, b = ½
Therefore
from Eq (1)
v
= constant . ρa Eb
v
= constant . ρ-½ E½
Question 1.9
E = m C2
mgh = m v2
[M LT-2 L] = [M (L T-1)2]
[M L2 T-2] = [M
L2 T-2]
LHS = RHS Hence Proved
Question 1.10
a
; rn vm
a
= constt. rn vm
[L T-2] = constt. [L]n
[L T-1]m
L T-2 = constt. Ln+m
T-m
Equating
powers of corresponding quantities on both sides
n + m = 1; -m = -2
Þ m = 2; n = -1 Answer
