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Chapter 1: Problem 148

From $\left[\mathrm{p}+\left(\mathrm{a} / \mathrm{v}^{2}\right)\right](\mathrm{v}-\mathrm{b})=$ constant equation is dimensionally correct find the dimensional formula for \(\mathrm{b}\) ? where \(\mathrm{P}=\) pressure \(\mathrm{V}=\) volume (a) \(\mathrm{M}^{0} \mathrm{~L}^{3} \mathrm{~T}^{0}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{3} \mathrm{~T}^{0}\) (c) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{3}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\)

Short Answer

Expert verified
The dimensional formula for \(b\) is \(\mathrm{M}^{0} \mathrm{~L}^{3} \mathrm{~T}^{0}\).

Step by step solution

01

Identify the dimensions of known terms

Pressure (P) has dimensions of force per unit area, which can be represented as: \(\mathrm{M}^{1}\mathrm{L}^{-1}\mathrm{T}^{-2}\). Volume (V) has dimensions of length cubed: \(\mathrm{M}^{0}\mathrm{L}^{3}\mathrm{T}^{0}\). 2.
02

Determine dimensions of a

In order to have a dimensionally correct equation, the dimensions of P and a / V^2 must be the same. Since P has dimensions of \(\mathrm{M}^{1}\mathrm{L}^{-1}\mathrm{T}^{-2}\), a must have dimensions of: \[\mathrm{M}^{1}\mathrm{L}^{-1}\mathrm{T}^{-2} \times \mathrm{V}^{2} = \mathrm{M}^{1}\mathrm{L}^{-1}\mathrm{T}^{-2} \times (\mathrm{M}^{0}\mathrm{L}^{3}\mathrm{T}^{0})^{2}\] 3.
03

Calculate dimensions of a

Multiply the dimensions of P and V^2: \[\mathrm{M}^{1}\mathrm{L}^{-1}\mathrm{T}^{-2} \times (\mathrm{M}^{2}\mathrm{L}^{6}\mathrm{T}^{0})\] 4.
04

Simplify the dimensions of a

The final dimensions of a are: \(\mathrm{M}^{1}\mathrm{L}^{5}\mathrm{T}^{-2}\). 5.
05

Make the whole equation dimensionally correct

Since the equation [P + (a / V^2)](V - b) = constant is said to be dimensionally correct, the dimensions of (V - b) must be the same as the dimensions of P + (a / V^2). To find the dimensions of b, we need to find the dimensions of (V - b): We know the dimensions of V are: \(\mathrm{M}^{0}\mathrm{L}^{3}\mathrm{T}^{0}\) We need to find the dimensions of b such that: \((\mathrm{M}^{0}\mathrm{L}^{3}\mathrm{T}^{0} - \mathrm{M}^{x}\mathrm{L}^{y}\mathrm{T}^{z}) = \mathrm{M}^{1}\mathrm{L}^{-1}\mathrm{T}^{-2}\) 6.
06

Compare dimensions and find b

For the equation to be dimensionally correct, the dimensions of V and b must be the same, i.e., \(x = 0, y = 3, z = 0\). So, the dimensions of b are: \(\mathrm{M}^{0}\mathrm{L}^{3}\mathrm{T}^{0}\). The correct answer is option (a): \(\mathrm{M}^{0} \mathrm{~L}^{3} \mathrm{~T}^{0}\).

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Most popular questions from this chapter

One planet is observed from two diametrically opposite point \(\mathrm{A}\) and \(\mathrm{B}\) on the earth the angle subtended at the planet by the two directions of observations is \(1.8^{\circ}\). Given the diameter of the earth to be about \(1.276 \times 10^{7} \mathrm{~m}\). What will be distance of the planet from the earth? (a) \(40.06 \times 10^{8} \mathrm{~m}\) (b) \(4.06 \times 10^{8} \mathrm{~m}\) (c) \(400.6 \times 10^{13} \mathrm{~m}\) (d) \(11 \times 10^{8} \mathrm{~m}\)

What is the ratio of 10 micron to 1 nanometer? (a) \(10^{4}\) (b) \(10^{3}\) (c) \(10^{16}\) (d) \(10^{15}\)

Which physical quantity is represented by $\mathrm{M}^{1} \mathrm{~L}^{3} \mathrm{~T}^{-3} \mathrm{~A}^{-2}$ ? (a) Resistivity (b) Resistance (c) conductance (d) conductivity

Dimensional formula of \(\mathrm{CV}\) ? where C - capacitance and \(\mathrm{V}\) - potential difference (a) \(\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{1}\) (c) \(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1} \mathrm{~A}^{-1}\) (d) \(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1} \mathrm{~A}^{1}\)

In an experiment to determine the density of a cube the percentage error in the measurement of mass is \(0.25 \%\) and the percentage error in the measurement of length is \(0.50 \%\) what will be the percentage error in the determination of its density? (a) \(2.75 \%\) (b) \(1.75 \%\) (c) \(0.75 \%\) (d) \(1.25 \%\)
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