Question

the centripetal force on a particle moving on a circular path depends on its mass, speed and radius of path. use dimensional method to find a relation for force in terms of mass, speed and radius ( PLEASE SOLVE WHOLE WORKSHEET FOR ME IT IS VERY IMPORTANT ) IHAVE SENT A PHOTO WITH THIS QUESTION

Answer 1

The centripetal force, F acting on a particle moving uniformly in a circle depend upon the mass (m), velocity (v) and radius (r) of the circle. The formula for F using the method of dimensions.

Let F = km a v b r c … (i)

Where, k is the dimensionless constant of proportionality, and a, b, c are the powers of m,v, r respectively.

On writing the dimensions of various quantities in (i), we get

[M 1 L 1 T −2 ] = M a [LT −1 ] b L c 

= M a L b T −b L c 

M 1 L 1 T −2 = M a L b + c T −b 

On applying the principle of homogeneity of dimensions, we get

a = 1,

b= 2,

b + c = 1 …(ii)

From (ii), c = 1 − b = 1 − 2 = −1

On putting these values in (i), we get

F = km 1 v 2 r −1

OR



This is the required relation for centripetal force.

Answer 2

22) Unit of energy is joule = kg*m²/s²

Joule is a MKS unit and erg is a CGS unit

MKS = CGS

1 m = 100 cm

1 s = 1 s

1 kg = 1000 gm

From above :

1 joule = 1 kg*m²/s² = 1000 * 10000 / 1 cm*m²/s²= 10^7 gm*cm²/s² = 10^7 erg

So, Answer : 1 joule = 10^7 erg

11)Let F = k m^a v^b r^c…..... (i)

On writing the dimensions we get

[M^1L^1T^-2] = M^a [LT^-1]^bL^c

=> [M^1L^1T^-2] = M^aL^bL^cT^-b

=> [M^1L^1T^-2] = M^aL^b+c T^-b

on comparing both sides we get

a = 1

b+c = 1

-b = -2 => b = 2

Then c = -1

Onputting these values in (i),

we get

F = k m^1 v^2 r^-1

Or

F=kmv^2/r

20) P + a/V^2) (V - b) = RT
P is pressure, V is volume and T is temperature.
[P] = [F/A] = [MLT^ −2/L^2] = [ML^−1T ^−2]
[V] = [L^3]

We cannot add or subtract quantities of different dimensions. Therefore
[P] = [a/V^2] ⇒ [a] = [PV^2] = [ML^−1T ^−2 L^6] = [M L^5 T^ −2]
∴ [a] = [M L^5 T ^−2]

Similarly,
[b] = [V] = [L^3]
a/b= (ML^5T^-2)/(L^3)=(M^1L^2T^-2) is answer


12) a) Q=IT
(Q)=M^0L^0A^1T^1
Where A is amphere

b) W=VIT
(V)=(M^1L^2T^-2)/(AT^1)=(M^1L^2T^-3A^-1) is answer

c) Q=CV
C=Q/V=IT/W/IT=I²T²/W
C=A²T²/ML^2T^-2=(M^-1L^-2T^4A^2) is answer

d) V= IR
R= V/I =W/IT/I= W/I^2T
R =(M^1L^2T^-2)/(A^2T^1)
R=( M^1L^2T^-3A^-2) is answer


13) G=force×(distance²)/m×m
G=(MLT^-2)(L^2)/(M)(M)
G=(M^-1L^3T^-2) is answer

14) h=energy/frequency
h=(ML^2T^-2)/(T^-1)
h=(ML^2T^-1) is answer


19) y = A sin (w t - k x)
wt=radian⇒w=1/t=T^-1

k x = radians => k = 1/x = L^-1
w/k=(L^-1T-1) is answer


23) f=6×π×eta ×r×v
eta=f/(r×v)
eta=(MLT^-2)/(L^2)(LT^-1/L)
eta=(ML^-1T^-1) is answer


16) N = no. of particles per unit area per unit time = [M^0L^−2T^−1]

n1 and n2 are no of particles per unit volume = [M^0L^−3T^0]

x1 and x2 are distance from some reference point = [M^0L^1T^0]

N = −D(n2 − n1/x2 − x1)
[M^0L^-2T^-1] = -D[M^0L^-3T^0][M^0L^1T^0]
Dimension of D = [M^0L^2T^-1] is answer


17) Let  v = k * λ^αρ^β g^gamma
    v = [L] [T]⁻¹   λ = [ L]       ρ = density = [M] [L]⁻³     g = [ L] [T]⁻²
Asumming α=a ,β=b and gamma=c
Hence   [L][T]⁻¹ = [M]^b [L]^{a-3b+c} [T]^{-2c}
Comparing the dimensions:
             b = 0.    c = 1/2       => a = 1/2
So   v = k √(gλ)
So α=gamma≠β

19) answer is (M^0L^2T^0)

25) p = p0exp(–αt^2)
As, αt^2 is dimensionless
Therefore, α = 1/t^2
= 1/[T^2]
= [T^–2] is answer

26) a) n= 1.54+1.53+1.44+1.54+1.56+1.45/6=1.51
Δn1=1.51-1.54=-0.03
Δn2=1.51-1.53=-0.02
Δn3=1.51-1.44=0.07
Δn4=-0.03, Δn5=-0.05, Δn6=0.06
To calculate mean absolute error
Δn=|Δn1|+|Δn2|+-----/6=0.26/6=0.043=0.04
Refractive index of glass with absolute error n=1.51plus minu0.04
b)Relative error=Δn/n=0.04/1.51=0.02649=0.03
c) % error= 0.03×100=3%

27) Question is similar to 26 bus value change ha.

18) p=(M^0L^1T^1A^1) is answer

Ques 15 and 24 samajh nahi aaya