Question

Find the dimensions of the following derived physical quantities :- Acceleration,pressure,Momentum,work and Density​

Answer 1

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:

✧ To find dimensions:

☞We have know its S.I units

✪ Acceleration

S.I unit of acceleration is m/s²

Therefore, its dimension is:

${m}^{0} {l}^{1} {t}^{ - 2}$

✪ Pressure

S.I unit of pressure is N/m²

Therefore, its dimension is:

${m}^{1} {l}^{ - 1} {t}^{ - 2}$

✪ Momentum

S.I unit of momentum is kgm/sec

Therefore, its dimension is:

${m}^{1} {l}^{ 1} {t}^{ - 1}$

✪ Work

S.I unit of work is joule

Therefore, its dimension is:

${m}^{1} {l}^{ 2} {t}^{ - 2}$

✪ Density

S.I unit of density is kg/m³

Therefore, its dimension is:

${m}^{1} {l}^{ - 3} {t}^{0}$

Answer 2

Answer:

Acceleration = [M° L¹ T-²] ( M zero L one T minus two)

Pressure (P) = M¹ L-¹ T -² (M one L minus one T minus two)

Momentum = [M¹ L¹ T-¹] (M one L one T minus one)

Work = = [M¹ L² T-²] (M one L two T Minus two)

Density = [M¹ L-³ T°] (M one L minus three T zero)

Explanation:

Acceleration = Velocity × Time-1. Or, a = [M0 L1 T-1] × [T] = [M0 L1 T-2] Therefore, acceleration is dimensionally represented as [M0 L1 T-2].

Pressure (P) = Force × Area-1. Or, P = [M1 L1 T-2] × [L2]-1 = M1 L-1 T -2. Therefore, the pressure is dimensionally represented as M1 L-1 T -2.

Therefore, momentum is dimensionally represented as [M1 L1 T-1].

Work = Force × Displacement. Or, W = [M1 L1 T-2] × [M0 L1 T0] = [M1 L2 T-2]. Therefore, work is dimensionally represented as [M1 L2 T-2].

Density = Mass × Volume-1. Or, (ρ) = [M1 L0 T0] × [M0 L3 T0]-1 = [M1 L-3 T0] Therefore, density is dimensionally represented as [M1 L-3 T0].