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Chapter 1: Problem 135

Dimensional formula for capacitance (C) (a) \(\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\) (c) \(\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \mathrm{~A}^{1}\) (d) \(\mathrm{M}^{3} \mathrm{~L}^{1} \mathrm{~T}^{-1} \mathrm{~A}^{-2}\)

Short Answer

Expert verified
The correct dimensional formula for capacitance (C) is \( \mathrm{M}^{-1}\mathrm{L}^{-2}\mathrm{T}^{4}\mathrm{A}^{2} \).

Step by step solution

01

Determine the dimensions of charge

Charge (Q) is the product of electric current (I) and time (t), so we have \(Q = It\). Since the dimension for the electric current (I) is A (for amperes) and the dimension for time (t) is T (for seconds), the dimensions of charge (Q) are $$[Q] = [I][t] = \mathrm{A}^{1}\mathrm{T}^{1}$$
02

Determine the dimensions of voltage

Voltage (V) can be expressed in terms of work (W) and charge (Q), i.e., \(V = \frac{W}{Q}\). Work (W) can also be expressed as the product of force (F) and distance (d), i.e., \(W = Fd\). Force (F) has the dimensions of mass (M), length (L), and time (T), given as \(\mathrm{M}^{1}\mathrm{L}^{1}\mathrm{T}^{-2}\). Thus, the dimensions of work (W) are $$[W] = [F][d] = \mathrm{M}^{1}\mathrm{L}^{2}\mathrm{T}^{-2}$$ Now, we can find the dimensions of voltage (V) using the dimensions of work (W) and charge (Q), so we have $$[V] = \frac{[W]}{[Q]} = \frac{\mathrm{M}^{1}\mathrm{L}^{2}\mathrm{T}^{-2}}{\mathrm{A}^{1}\mathrm{T}^{1}} = \mathrm{M}^{1}\mathrm{L}^{2}\mathrm{T}^{-3}\mathrm{A}^{-1}$$
03

Determine the dimensions of capacitance

Now, substitute the dimensions of charge (Q) and voltage (V) into the defining equation for capacitance $$[C] = \frac{[Q]}{[V]} = \frac{\mathrm{A}^{1}\mathrm{T}^{1}}{\mathrm{M}^{1}\mathrm{L}^{2}\mathrm{T}^{-3}\mathrm{A}^{-1}}$$ After canceling out the numerator and the denominator dimensions, we have $$[C] = \mathrm{M}^{-1}\mathrm{L}^{-2}\mathrm{T}^{4}\mathrm{A}^{2}$$
04

Identify the correct option

Looking at the given options, we find that the correct dimensional formula for capacitance (C) is option (a): $$\mathrm{M}^{-1}\mathrm{~L}^{-2}\mathrm{~T}^{4}\mathrm{~A}^{2}$$

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Most popular questions from this chapter

In the relation $P=(\alpha / \beta) \mathrm{e}[\\{-\alpha z\\} /\\{(\mathrm{k}) \beta \theta\\}], \mathrm{P}\( is pressure, \)\mathrm{z}$ is distance, \(\mathrm{k}\) is Boltzmann constant and \(\theta\) is the temperature. The dimensional formula of \(B\) will be (a) \(\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{1}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{0}\) Copyright \(\odot\) StemEZ.com. All rights reserved.

The dimensional formula of plank's constant is ........... (a) \(\mathrm{M}^{3} \mathrm{~L}^{2} \mathrm{~T}^{-1}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\) (c) \(\mathrm{M}^{2} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3}\)

Pressure \(P=\left(a t^{2} / b x\right)\) where \(x=\) distance, \(t=\) time find the dimensional formula for \(\mathrm{a} / \mathrm{b}\) (a) \(\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{-4}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{-2}\) (d) \(\mathrm{M}^{+1} \mathrm{~L}^{0} \mathrm{~T}^{-2}\)

The radius of circle is \(1.26 \mathrm{~cm}\). According to the concept of significant figures area of it can be represented as - (a) \(4.9850 \mathrm{~cm}^{2}\) (b) \(4.985 \mathrm{~cm}^{2}\) (c) \(4.98 \mathrm{~cm}^{2}\) (d) \(9.98 \mathrm{~cm}^{2}\)

950 dyne \(=\ldots \ldots \ldots \ldots \ldots\) newton (a) \(9.5 \times 10^{-3}\) (b) \(95 \times 10^{-5}\) (c) \(950 \times 10^{-7}\) (d) \(9.5 \times 10^{-4}\)
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