Question

3. What is the dimensional
formula of torque and energy
O (a) [ML-3T-2) and (MLT-2]
(b) [ML2T-2) and (MLT-2]
O (c) [ML2T-2) and (ML2T2]
O (d) (MLT2) and (ML2T2)​

Answer 1

Explanation:

$\red{\boxed{\rm Solution}}$

We know that,

$\rm \tau \: = rFsin\theta$

Here $\tau$ is the magnitude of torque, r is the magnitude of the position vector which tells the distance between the body and the pivot and θ is the angle between the force vector and the lever arm

We know that the SI unit of radius is the same as the SI unit of length which is metre (m)

Therefore,

The dimensional formula of radius = [M⁰L¹T⁰]

We know that the SI unit of force is N

We know that Force = Mass x acceleration

We know that the SI unit of Mass is kg

We know that the SI unit of acceleration is m/s²

Therefore,

The SI unit of force = kg m/s²

Therefore,

We can conclude that the unit Newton (N) is same as the unit kg m/s².

Therefore,

The dimensional formula of Force = [M¹L¹$\rm {T}^{-2}$]

We know that the dimensional formula of angle is [M⁰L⁰T⁰]

Therefore,

Dimensional formula of torque = [M⁰L¹T⁰] x [M¹L¹$\rm {T}^{-2}$] x [M⁰L⁰T⁰]

From the laws of exponents,

Dimensional formula of torque = [M¹L²$\rm {T}^{-2}$]

We know that,

The SI unit of energy is same as the SI unit of work, i.e Joule (J)

We know that,

Work = Force x Displacement x cosθ

We know that the SI unit of force is N (or) kg m/s²

We know that the SI unit of displacement is the same as the SI unit of Distance which is metre (m).

Therefore,

The SI unit of Work is kg m/s² x m = kg m²/s²

Therefore,

We can conclude that the unit Joule is the same as the unit kg m²/s²

We know that dimensional formula of Displacement = [M⁰L¹T⁰]

We know that dimensional formula of Force = [M¹L¹$\rm {T}^{-2}$]

Therefore,

Dimensional formula of Work = [M⁰L¹T⁰] x [M¹L¹$\rm {T}^{-2}$]

From the laws of exponents,

Dimensional formula of Work = [M¹L²$\rm {T}^{-2}$].