Question

Write the dimensions of a and b in the relation P=b-x2/at where P is power, x is distance and t is time.

Answer 1

Given:

P = b-x²/at

P is power

x is distance

t is time.

To Find :

Find the dimensions of a and b.

Solution:

1.Power (P) = Work × time-1 = Joule × second-1 . . . . . (1)

Since, Work (J) = Force (M x a) × displacement = M¹ L¹ T⁻² × [L]

Therefore, the dimensional formula of work = M¹ L² T⁻³ . . . . (2)

On substituting equation (2) in equation (1) we get,

Power (P) = Work × time⁻¹

Or, P = [M¹ L² T⁻²] × [T⁻¹] = M¹ L² T⁻³.

Therefore, power is dimensionally represented as [ M¹ L² T⁻³].

2. x is distance. So its dimension of x² is [L²].

3. t is time therefore its dimension is [T].

By the Principal of Homogenity, which states that every term of the physical relation must have the same dimensions i. e. Power , b and

x²/at will have the same dimension [ M¹ L² T⁻³].

So dimensions of b is [ M¹ L² T⁻³].

Dimensions of x²/at=   [ M¹ L² T⁻³]

                        [L²]/a[T]=  [ M¹ L² T⁻³]

                        [L²]/[T]. [ M¹ L² T⁻³]= a

∴                        a=[M⁻¹T²]

Hence the dimension of b is [ M¹ L² T⁻³] and

            the dimension of a is [M⁻¹T²]

Answer 2

Answer:

The dimensions of a and b are  $[a]=M^{-1} L^{0} T^{2}$and $[b] =M^{0} L^{2} T^{0}$

Explanation:

Given:

• P=POWER
• $P=\frac{b-x^{2} }{at}$
• x= distance
• Dimension of x=L
• Dimensions of Power: $\mathrm{P}=\mathrm{ML}^{2} \mathrm{~T}^{-3}$

Since the given expression is dimensionally correct, each term of the expression must have same dimensions as that of power according to PRINCIPLE OF HOMOGENITY.

Therefore,

$\frac{\left[\mathrm{x}^{2}\right]}{[\mathrm{a}][\mathrm{t}]}=\frac{\mathrm{L}^{2}}{[\mathrm{a}] \mathrm{T}}=\mathrm{ML}^{2} \mathrm{~T}^{-3} \Rightarrow[\mathrm{a}]=\mathrm{M}^{-1} \mathrm{~L}^{0} \mathrm{~T}^{2} \frac{[\mathrm{b}]}{[\mathrm{a}][\mathrm{t}]}=\frac{[\mathrm{b}]}{\left(\mathrm{M}^{-1} \mathrm{~L}^{0} \mathrm{~T}^{2}\right)(\mathrm{T})}=\mathrm{ML}^{2} \mathrm{~T}^{-3} \Rightarrow[\mathrm{b}]=\mathrm{L}^{2}$

• $[a]=M^{-1} L^{0} T^{2}$

• $[b] =M^{0} L^{2} T^{0}$