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Chapter 12: Problem 1761

Masses of three conductors of same material are in the proportion of \(1: 2: 3\) their lengths are in the proportion of \(3: 2: 1\) then their resistance will be in the proportion of... (A) \(1: 1: 1\) (B) \(1: 2: 3\) (C) \(9: 4: 1\) (D) 27:6:1

Short Answer

Expert verified
The proportion of their resistances will be \(9 : 4 : 1\).

Step by step solution

01

Write the given information as ratios

The given information states that the masses are in the proportion of \(1:2:3\), and the lengths are in the proportion of \(3:2:1\). Let the masses of the conductors be \(M_1\), \(M_2\), and \(M_3\), and their lengths be \(L_1\), \(L_2\), and \(L_3\). We can represent this information as: \[M_1 : M_2 : M_3 = 1 : 2 : 3\] \[L_1 : L_2 : L_3 = 3 : 2 : 1\]
02

Find the ratio of cross-sectional areas

Since the conductors are of the same material, their volume ratios are equal to their mass ratios: \[V_1 : V_2 : V_3 = 1 : 2 : 3\] Volume is given by the product of cross-sectional area and length: \[V = A \times L\] So, the area ratios are: \[A_1 : A_2 : A_3 = \frac{V_1}{L_1} : \frac{V_2}{L_2} : \frac{V_3}{L_3}\] Substituting the given mass and length ratios: \[A_1 : A_2 : A_3 = \frac{1}{3} : \frac{2}{2} : \frac{3}{1} = 1 : 1 : 3\]
03

Find the ratio of resistances

Using the resistance formula, \[R = \rho \frac{L}{A}\], we can find the ratio of resistances: \[R_1 : R_2 : R_3 = \frac{\rho \frac{L_1}{A_1}}{\rho \frac{L_2}{A_2}} : \frac{\rho \frac{L_2}{A_2}}{\rho \frac{L_3}{A_3}} : \frac{\rho \frac{L_3}{A_3}}{\rho \frac{L_1}{A_1}}\] Since the resistivity \(\rho\) is constant for all conductors: \[R_1 : R_2 : R_3 = \frac{L_1}{A_1} : \frac{L_2}{A_2} : \frac{L_3}{A_3}\] Using the given length ratio and the area ratio we found in step 2: \[R_1 : R_2 : R_3 = \frac{3}{1} : \frac{2}{1} : \frac{1}{3} = 9 : 4 : 1\] So, the ratio of resistances is \(9 : 4 : 1\), which corresponds to the answer (C).

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Most popular questions from this chapter

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) \((4 \pi / \mathrm{R} \alpha)(2 \pi-\alpha)\)

How would you arrange 48 cells each of e.m.f \(2 \mathrm{~V}\) and internal resistance \(1.5 \Omega\) so as to pass maximum current through the external resistance of \(2 \Omega\) ? (A) 2 cells in 24 groups (B) 4 cells in 12 groups (C) 8 cells in 6 groups (D) 3 cells in 16 groups

The temperature co-efficient of resistance of a wire is $0.00125^{\circ} \mathrm{k}^{-1}\(. Its resistance is \)1 \Omega\( at \)300 \mathrm{~K}$. Its resistance will be \(2 \Omega\) at. (A) \(1400 \mathrm{~K}\) (B) \(1200 \mathrm{~K}\) (C) \(1000 \mathrm{~K}\) (D) \(800 \mathrm{~K}\)

For a cell of e.m.f \(2 \mathrm{~V}\), a balance is obtained for $50 \mathrm{~cm}\( of the potentiometer wire If the cell is shunted by a \)2 \Omega$ resistor and the balance is obtained across \(40 \mathrm{~cm}\) of the wire, then the internal resistance of the cell is. (A) \(1 \Omega\) (B) \(0.5 \Omega\) (C) \(1.2 \Omega\) (D) \(2.5 \Omega\)

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) \(\mathrm{M}^{1} \mathrm{~L}^{-3} \mathrm{~T}^{3} \mathrm{~A}^{2}\)
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