JEE Challenger
More from Units and Measurements
PhyMedAdv 2020 • Paper 1

Dimensions of Quantities in a Single Base Quantity System

Assume a hypothetical unit system where all physical quantities are defined using a single base quantity XX. In this system, the dimensional formulas are: [position]=[Xα][\text{position}] = [X^\alpha], [speed]=[Xβ][\text{speed}] = [X^\beta], [acceleration]=[Xp][\text{acceleration}] = [X^p], [linear momentum]=[Xq][\text{linear momentum}] = [X^q], and [force]=[Xr][\text{force}] = [X^r]. Which of the following relations are correct?

Options

A

α+p=2β\alpha + p = 2\beta

Correct
B

p+q−r=βp + q - r = \beta

Correct
C

p−q+r=αp - q + r = \alpha

D

p+q+r=βp + q + r = \beta

Topics & Concepts

Units and MeasurementsMeasuring Instruments

Step-by-Step Solution

To determine the correct relationships between the powers of the base quantity XX, we use standard physical relations between the quantities.

Given the dimensions in terms of XX:

  • [position]=[Xα][\text{position}] = [X^\alpha]
  • [speed]=[Xβ][\text{speed}] = [X^\beta]
  • [acceleration]=[Xp][\text{acceleration}] = [X^p]
  • [linear momentum]=[Xq][\text{linear momentum}] = [X^q]
  • [force]=[Xr][\text{force}] = [X^r]

Step 1: Relation between Position, Speed, and Acceleration

The dimensional relationship between acceleration, speed, and position is: [acceleration]=[speed]2[position][\text{acceleration}] = \frac{[\text{speed}]^2}{[\text{position}]}

Substituting the given powers of XX: [Xp]=[Xβ]2[Xα]=[X2β−α][X^p] = \frac{[X^\beta]^2}{[X^\alpha]} = [X^{2\beta - \alpha}]

Comparing the exponents of XX on both sides: p=2β−α  ⟹  α+p=2βp = 2\beta - \alpha \implies \alpha + p = 2\beta

Thus, Option (A) is correct.

Step 2: Relation between Force, Linear Momentum, Acceleration, and Speed

The relationships for force and linear momentum in terms of mass (mm) are:

  • force=mass×acceleration  ⟹  [force]=[mass][acceleration]\text{force} = \text{mass} \times \text{acceleration} \implies [\text{force}] = [\text{mass}][\text{acceleration}]
  • linear momentum=mass×speed  ⟹  [linear momentum]=[mass][speed]\text{linear momentum} = \text{mass} \times \text{speed} \implies [\text{linear momentum}] = [\text{mass}][\text{speed}]

Dividing the dimensional formula of force by linear momentum eliminates mass: [force][linear momentum]=[acceleration][speed]\frac{[\text{force}]}{[\text{linear momentum}]} = \frac{[\text{acceleration}]}{[\text{speed}]}

Substituting the given powers of XX: [Xr][Xq]=[Xp][Xβ]\frac{[X^r]}{[X^q]} = \frac{[X^p]}{[X^\beta]} [Xr−q]=[Xp−β][X^{r - q}] = [X^{p - \beta}]

Comparing the exponents of XX on both sides: r−q=p−β  ⟹  p+q−r=βr - q = p - \beta \implies p + q - r = \beta

Thus, Option (B) is correct.

Step 3: Verifying Options (C) and (D)

  • Option (C): p−q+r=αp - q + r = \alpha From p+q−r=βp + q - r = \beta, we have r−q=p−βr - q = p - \beta. Substituting this into p−q+rp - q + r gives 2p−β2p - \beta, which is not equal to α=2β−p\alpha = 2\beta - p in general. Therefore, Option (C) is incorrect.

  • Option (D): p+q+r=βp + q + r = \beta Comparing this with p+q−r=βp + q - r = \beta implies r=0r = 0, which is not generally true for force. Therefore, Option (D) is incorrect.

Conclusion

The correct options are (A) and (B).

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