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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 length \(L\) is drawn such that its diameter is reduced to half of its original diameter. If the resistance of the wire were \(10 \Omega\), its new resistance would be. (A) \(40 \Omega\) (B) \(60 \Omega\) (C) \(120 \Omega\) (D) \(160 \Omega\)

There are n resistors having equal value of resistance \(\mathrm{r}\). First they are connected in such a way that the possible minimum value of resistance is obtained. Then they are connected in such a way that possible maximum value of resistance is obtained the ratio of minimum and maximum values of resistances obtained in these way is.... (A) \((1 / n)\) (B) \(\mathrm{n}\) (C) \(\mathrm{n}^{2}\) (D) \(\left(1 / \mathrm{n}^{2}\right)\)

A potentiometer wire of length \(1 \mathrm{~m}\) and resistance \(10 \Omega\) is connected in series with a cell of e.m.f \(2 \mathrm{~V}\) with internal resistance \(1 \Omega\) and a resistance box of a resistance \(R\) if potential difference between ends of the wire is \(1 \mathrm{~V}\) the value of \(R\) is. (A) \(4.5 \Omega\) (B) \(9 \Omega\) (C) \(15 \Omega\) (D) \(20 \Omega\)

The reading of ammeter shown in figure is.... (A) \(2.18 \mathrm{~A}\) (B) \(3.28 \mathrm{~A}\) (C) \(6.56 \mathrm{~A}\) (D) \(1.09 \mathrm{~A}\)

The network is made of uniform wire. The resistance of portion EL is $2 \Omega\(. Find the resistance of star between points \)F \& C .$ (A) \(0.985 \Omega\) (B) \(1.25 \Omega\) (C) \(1.946 \Omega\) (D) \(1.485 \Omega\)
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