Question

Q6. If Force (F), Velocity (v) and Acceleration(a) are taken as fundamental have units, then the linear momentum is represented as:
A) Fv2a-2
B)Fv3a-2
C)Fva-1
D)F2v3a-1
​

Answer 1

Given :-

▪ Force(F), Velocity(V) and acceleration(a) are fundamental units.

To Find :-

▪ Dimensional formula of linear momentum.

Solution :-

Let the dimensional formula of linear momentum in terms of Force, velocity, and acceleration be [ Fˣ Vʸ aᶻ ]

Where, x, y and z are real numbers.

We know, Dimensional formula of,

• Force = [ MLT⁻² ]
• Velocity = [ LT⁻¹ ]
• Acceleration = [ LT⁻² ]
• Linear Momentum = [ MLT⁻¹ ]

so,

⇒ [MLT⁻¹] = [F]ˣ [V]ʸ [a]ᶻ

⇒ [MLT⁻¹] = [MLT⁻²]ˣ [LT⁻¹]ʸ [LT⁻²]ᶻ

⇒ [MLT⁻¹] = [M]ˣ [L]ˣ [T]⁻²ˣ [L]ʸ [T]⁻ʸ [L]ᶻ [T]⁻²ᶻ

⇒ [MLT⁻¹] = [M]ˣ [L]⁽ˣ⁺ʸ⁺ᶻ⁾ [T]⁽⁻²ˣ⁻ʸ⁻²ᶻ⁾

Comparing the powers of the exponents, we have,

⇒ [M]¹ = [M]ˣ

⇒ x = 1 ...(1)

Also,

⇒ [L]¹ = [L]⁽ˣ⁺ʸ⁺ᶻ⁾

⇒ 1 + y + z = 1 [ from (1), x = 1 ]

⇒ y + z = 0 ...(2)

Further,

⇒ [T]⁻¹ = [T]⁽⁻²ˣ⁻ʸ⁻²ᶻ⁾

⇒ -1 = -2(1) - y - 2z [ from (1), x = 1 ]

⇒ 2 - 1 = -y - 2z

⇒ -y - 2z = 1 ...(3)

Adding (2), and (3)

⇒ y + z - y - 2z = 1 + 0

⇒ z - 2z = 1

⇒ -z = 1

⇒ z = -1

Substituting z = 1 in (2) , we get

⇒ y = 1

So,

⇒ [p] = [F]ˣ [V]ʸ [a]ᶻ

⇒ [p] = [F]¹ [V]¹ [a]⁻¹

⇒ [p] = [ FVa⁻¹ ]

Hence, Option (C) is correct.