Dimensions and Equations in Physics

LO1
a) Dimension of p = kg/m/sec = ML-1T
Dimension of rho = ML-3
Dimension of volume = L3
Dimension of velocity = LT-1
L T−1 =¿
Equating x, y and z
x + y=0=¿ y=1
−x−3 y +3 z=1=¿1−3+3 z=1=¿ z=1
x=−1
b) Coefficient of viscosity = ML-1T-1
Dimension of velocity = LT-1
Dimension of Radius = L
M 1 L1 T−2=¿
Equating x, y and z,
x=1
−x + y + z=1=¿ z=1
−x− y =−2=¿ y =1
F=η V R
(MIT, 2019)
c) 8
2 ( 2 ×3+7 × d ) =2 × 5
2 ( 2 ×3+ 4 × d )
4 ( 6+7 d )=5 ( 6 +4 d )
24+ 28 d=30+20 d
d= 3
4
Sum of the geometric series,
S= a (rn−1)
(r −1) =8 ¿ ¿
d) d= h
tan20 − h
tan 60
d= h
tan20 − h
tan 60

h
tan 20 − h
tan 60 /( 1
60 )=600
h
tan 20 − h
tan 60 =10
h=4.61 miles
e) 2) Radioactive decay equation
dN
dt =−λ N
dN
N =−λ dt
N=N 0 e−λt
Half-life is defined as,
t0.5= ln 2
λ
And half-life is given to be 1 week, so λ
λ=0.693/week
3) N=N 0 e−λt
N=20 e−0.693× 3
N=20× 0.125=2.5
After three weeks it will be 2.5 counts per second.
1)

f) 1) Since it is an exponential in nature
N=N 0 eλt
From the data we can estimate λ=0.255
2) In 2025, t =25 years
N2025=225 e0.255× 25
N2025=132071.7
3)
N2010=225 e0.255× 10
N2010=2881.59
Average rate of change from 2006 to 2010 is,
(N ¿¿ 2010−N2006)/4 ¿
Rate of change=461.15
4) Instantaneous change is = 0.255 * 225 = 57.375
LO2
a) Mean = 49.2
Standard deviation = 16.12328
b) If x is a random variable with distribution B(n, p), then for sufficiently large n, the following
random variable has a standard normal distribution:
z= x−μ
σ N ( 0,1 ) , μ=np , σ 2=np(1− p)
Proof: