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Chapter 12: Problem 1721

Which is the dimensional formula for conductance from the give below? (A) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{2}\) (B) \(\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \mathrm{~A}^{2}\) (C) \(\mathrm{M}^{1} \mathrm{~L}^{-3} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\) (D) \(\mathrm{M}^{1} \mathrm{~L}^{-3} \mathrm{~T}^{3} \mathrm{~A}^{2}\)

Short Answer

Expert verified
The short answer is: (B) \(M^{-1} L^{-2} T^{3} A^{2}\).

Step by step solution

01

Write down the formula for resistance

The formula for resistance is given by Ohm's Law, which states that: Resistance (R) = Voltage (V) / Current (I)
02

Write down the dimensional formulae for voltage and current

Voltage (V) has the dimension of work done per unit charge. The dimensional formula for work is [M^1 L^2 T^{-2}], and the dimensional formula for charge is [A^1 T^1]. Therefore, the dimensional formula for voltage can be represented as [M^1 L^2 T^{-3} A^{-1}]. Current (I) has the dimension of Ampere (A), so the dimensional formula for current is [A^1].
03

Determine the dimensional formula for resistance

Since Resistance (R) = Voltage (V) / Current (I), we find the dimensional formula for resistance by dividing the dimensional formula of voltage by the dimensional formula of current. Which gives us: [\(R = \frac{[M^1 L^2 T^{-3} A^{-1}]}{[A^1]} = M^1 L^2 T^{-3} A^{-2}\)]
04

Determine the dimensional formula for conductance

Conductance (G) is the reciprocal of resistance (R). So, to find the dimensional formula for conductance, we invert the dimensional formula for resistance: \[G = \frac{1}{M^1 L^2 T^{-3} A^{-2}} = M^{-1} L^{-2} T^{3} A^{2}\] The correct answer is (B) \(M^{-1} L^{-2} T^{3} A^{2}\).

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Most popular questions from this chapter

In the circuit shown in fig the potential difference across \(3 \Omega\) is. (A) \(2 \mathrm{~V}\) (B) \(4 \mathrm{~V}\) (C) \(8 \mathrm{~V}\) (D) \(16 \mathrm{~V}\)

Area of cross-section of a copper wire is equal to area of a square of $2 \mathrm{~mm}\( length It carries a current of \)8 \mathrm{~A}$ Find drift velocity of electrons (Density of free electrons in copper $\left.=8 \times 10^{28} \mathrm{~m}^{-3}\right)$ (A) \(1.56 \times 10^{-2} \mathrm{~ms}^{-1}\) (B) \(1.56 \times 10^{-4} \mathrm{~ms}^{-1}\) (C) \(3.12 \times 10^{-2} \mathrm{~ms}^{-1}\) (D) \(3.12 \times 10^{-3} \mathrm{~ms}^{-1}\)

A wire of resistor \(R\) is bent into a circular ring a circular ring of radius \(\mathrm{r}\) Equivalent resistance between two points \(\mathrm{X}\) and \(\mathrm{Y}\) on its circumference, when angle xoy is \(\alpha\), can be given by (A) $\left\\{(\mathrm{R} \alpha) /\left(4 \pi^{2}\right)\right\\}(2 \pi-\alpha)$ (B) \((\mathrm{R} / 2 \pi)(2 \pi-\alpha)\) (C) \(\mathrm{R}(2 \pi-\alpha)\) (D) \((4 \pi / \mathrm{R} \alpha)(2 \pi-\alpha)\)

In the given circuit the equivalent resistance between the points \(\mathrm{A}\) and \(\mathrm{B}\) in \(\mathrm{ohm}\) is. (A) 9 (B) \(11.6\) (C) \(14.5\) (D) \(21.2\)

Match the physical quantities given in column I with their dimensional formulae given in column II - I stands for the dimension of current. \(\begin{array}{ll}\text { Column I } & \text { Column II }\end{array}\) (a) Electromotive force (emf) (p) \(\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\) (b) Resistance (q) \(\mathrm{ML}^{3} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\) (c) Resistivity (r) \(\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{3} \mathrm{~A}^{2}\) (d) Conductivity (s) \(\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\) (A) \(a-s, b-p, c-q, d-r\) (B) \(a-p, b-s, c-r, d-p\) (C) \(a-p, b-s, c-r, d-q\) (D) \(a-r, b-p, c-q, d-s\)
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