Question

The force acting on an object is given by \(F = A + \frac{B}{t}\) where t is the time. Then the dimensions of A and B respectively are:
1. [M¹ L¹ T^-2], [M¹L¹ T^-1]
2. [M1 L0 T-2], [M1L3. [M1 L0 T-2], [M1L4. [M1 L2 T-2], [M1L<span style="position: rela

Answer 1

Correct Answer - Option 1 : [M¹ L¹ T^-2], [M¹L¹ T^-1]

The correct answer is option 1) i.e. [M1 L1 T-2], [M1L1 T-1]

CONCEPT:

• Force is defined as the push or pull acting on an object.
  • It's SI unit is the newton (N)

  • The force acting on an object of mass (m) with an acceleration (a) is given as:

Force = m × a​

• The dimensional formula is used to express any physical quantity in terms of fundamental quantities.
  • Fundamental quantity: A quantity that can be measured is called a physical quantity.
  • The seven fundamental quantities along with their dimensions are as follows:

Fundamental quantity	Dimension
Time	T
Mass	M
Electric current	A
Thermodynamic temperature	K
Amount of substance	mol
Luminous intensity	cd
Length	L

EXPLANATION:

Dimensional unit of mass = [M1]

Dimensional unit of acceleration (unit: m/s2) = [L1T-2]

We know, Force = m × a. Therefore, the dimensional formula of force = [M1L1T-2]

Given that: \(F = A \: + \frac{B}{t}\), the terms A and \(\frac{B}{t}\) are added. So, all of them will have the dimensional formula of force.

• A is independent of any other variables and thus its dimensional formula is [M1L1T-2].
• \(\frac{B}{t}\) is added to A and hence it will have the same dimensional formula as A. The dimension of time t is [T].

⇒ \(\frac{B}{[T]} = B[T^{-1}]= [M^1L^1T^{-2}]\)

⇒ B = [M1L1 T-1] 

Thus, the dimensions of A and B respectively are [M1 L1 T-2], [M1L1 T-1].