If a uniform cylindrical copper rod is pulled from its ends to elongate its
length by 20%,keeping its volume constant, find the ratio of the resistance of
the original rod and that of the elongated rod.
Answers
Given:
Elongation in the length of the rod, Δl = 20%
To Find:
The ratio of the resistance of the original rod and that of the elongated rod.
Calculation:
- Let the initial resistance be R1 and final resistance be R2.
- Let the resistivity of copper be ρ and the volume of the rod be V.
- Let original length and area be L1 and A1 respectively and the final length and area be L2 and A2 respectively.
- ATQ: L2 = L1 + 20% of L1
⇒ L2 = 1.2 L1
- The volume of copper rod remains constant.
⇒ A1L1 = A2L2
⇒ A2 = A1L1/1.2L1
⇒ A2 = 0.83 A1
- The ratio of resistances:
R1:R2 = (ρL1/A1) / (ρL2/A2)
⇒ R1/R2 = (L1/A1) / (1.2 L1/0.83 A1)
⇒ R1/R2 = 0.83/1.2
⇒ R1/R2 = 1 : 1.446
- So, the ratio of the resistance of the original rod and that of the elongated rod is approximately 1 : 1.446
Answer:
The Answer for this question would be, R₁:R₂ = 25:36
Explanation:
Let the original length of the rod, l₁, be equal to l. ...(i)
It is given that the length is elongated by 20%.
So, the new length of the rod, l₂, would be equal to,
l + 20l/100
= (100l + 20l)/100 [on LCM]
= 120l/100
= 12l/10
= 1.2l
⇒ the new length of the rod, l₂ = 1.2l. ...(ii)
It is given that the volume of the rod is constant.
So, volume of the rod before elongating (V₁) = Volume after elongating (V₂)
⇒ V₁ = V₂
⇒ A₁ x l₁ = A₂ x l₂ [where, A₁ and A₂ are the areas of cross-section of the two rods and l₁ and l₂ are the lengths of the two rods respectively; and since area x length = volume, for figures with constant cross-sections]
⇒ A₁/A₂ = l₁/l₂ [on cross-multiplication]
⇒ A₁/A₂ = 1.2l/l [from (i) and (ii)]
⇒ A₁/A₂ = 1.2 [since l gets divided]
⇒ A₁ = 1.2A₂. ...(iii)
We know that,
R₁ = p.l₁/A₁ and R₂ = p.l₂/A₂ [where R₁ and R₂ are the resistances of the two rods respectively, and p is the proportionality constant called specific resistance or resistivity]
Therefore, R₁/R₂ = p.l₁/A₁ ÷ p.l₂/A₂ = p.l₁/A₁ x A₂/p.l₂
⇒ R₁/R₂ = l₁/A₁ x A₂/l₂ [since p gets divided]
⇒ R₁/R₂ = l/1.2A₂ x A₂/1.2l [from (i), (ii) and (iii)]
⇒ R₁/R₂ = 1/1.44 [since l and A₂ gets divided]
⇒ R₁/R₂ = 100/144 = 25/36
⇒ R₁:R₂ = 25:36.
Thus, the ratio of resistance of the original and the elongated rod is,
25:36.
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