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Chapter 1: Problem 125

Dimensional formula for Resistance (R) is ............. (a) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{0} \mathrm{~A}^{-1}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\)

Short Answer

Expert verified
The short answer is: The dimensional formula for Resistance (R) is (c) \(\mathrm{M}^{1}\mathrm{~L}^{2}\mathrm{~T}^{-3}\mathrm{~A}^{-2}\).

Step by step solution

01

Write Ohm's law

The equation that relates voltage (V), current (I), and resistance (R) is known as Ohm's law. This equation is: \(V = IR\) Step 2: Identify the dimensions of voltage and current
02

Identify the dimensions of voltage and current

Before finding the dimensions of resistance, we need to know the dimensions of voltage and current: - The dimensions of voltage (V) are \(\mathrm{M}^{1}\mathrm{~L}^{2}\mathrm{~T}^{-3}\mathrm{~A}^{-1}\) - The dimensions of current (I) are \(\mathrm{A}^{1}\) Step 3: Find the dimensions of resistance using the dimensional formula
03

Find the dimensions of resistance using the dimensional formula

To find the dimensions of resistance (R), we need to rearrange Ohm's law to solve for R and replace V and I with their dimensions: \(R = \dfrac{V}{I}\) Replace V and I with their dimensions: \(R = \dfrac{\mathrm{M}^{1}\mathrm{~L}^{2}\mathrm{~T}^{-3}\mathrm{~A}^{-1}}{\mathrm{A}^{1}}\) Now, divide the dimensions by cancelling any common terms: \(R = \mathrm{M}^{1}\mathrm{~L}^{2}\mathrm{~T}^{-3}\mathrm{~A}^{-2}\) Hence, the dimensional formula of resistance (R) is \(\mathrm{M}^{1}\mathrm{~L}^{2}\mathrm{~T}^{-3}\mathrm{~A}^{-2}\). The correct choice is: (c) \(\mathrm{M}^{1}\mathrm{~L}^{2}\mathrm{~T}^{-3}\mathrm{~A}^{-2}\)

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Most popular questions from this chapter

Write the dimensional formula of the ratio of linear momentum to angular momentum. (a) \(\mathrm{M}^{0} \mathrm{~L}^{-1} \mathrm{~T}^{0}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{0}\) (c) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}\) (d) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{1}\)

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) \(11 \times 10^{8} \mathrm{~m}\)

Kinetic energy \(\mathrm{K}\) and linear momentum \(\mathrm{P}\) are related as \(\mathrm{K}=\left(\mathrm{P}^{2} / 2 \mathrm{~m}\right) .\) What is the equation of the relative error \(\Delta \mathrm{k} / \mathrm{k}\) in measurement of the \(\mathrm{K} ?\) (mass in constant) (a) \((\mathrm{P} / \Delta \mathrm{P})\) (b) \(2(\Delta \mathrm{P} / \mathrm{P})\) (c) \((\mathrm{P} / 2 \Delta \mathrm{P})\) (d) \(4(\Delta \mathrm{P} / \mathrm{P})\)

For a sphere having volume is given by \(\mathrm{V}=(4 / 3) \pi \mathrm{r}^{3}\) What is the equation of the relative error \((\Delta \mathrm{V} / \mathrm{V})\) in measurement of the volume \(\mathrm{V}\) ? (a) \(3(\Delta \mathrm{r} / \mathrm{r})\) (b) \(4(\Delta \mathrm{r} / \mathrm{r})\) (c) \((4 / 3)(\Delta \mathrm{r} / \mathrm{r})\) (d) \((1 / 3)(\Delta \mathrm{r} / \mathrm{r})\)

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) \(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1} \mathrm{~A}^{1}\)
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