A metallic conductor at 100C connected in the left gap of meter bridge gives balancing length 40 cm. When the conductor is at 600C, the balancing point shifts by ---cm, (temperature coefficient of resistance of the material of the wire is (1/220)/0C)

R1R=ℓ100−ℓ=4060=23...(1)From αR=R2−R1R1t2−R2t1⇒1220=R2−R160R1−10R2⇒60R1−10R2=220R2−220R1⇒280R1=230R2⇒R1R2=2328...(2)From R1R0=23⇒R=32R1Substituting in (1) R232R1=ℓ1100−ℓ1 ⇒23×2823=ℓ1100−ℓ1⇒23×2823=ℓ1100−ℓ1⇒5600−556ℓ1=69ℓ1;125ℓ1=5600ℓ1=5600125=44.8cm;11−l=44.8−40=4.8cm

A metallic conductor at 100C connected in the left gap of meter bridge gives balancing length 40 cm. When the conductor is at 600C, the balancing point shifts by ---cm, (temperature coefficient of resistance of the material of the wire is (1/220)/0C)

R1R=ℓ100−ℓ=4060=23...(1)From αR=R2−R1R1t2−R2t1⇒1220=R2−R160R1−10R2⇒60R1−10R2=220R2−220R1⇒280R1=230R2⇒R1R2=2328...(2)From R1R0=23⇒R=32R1Substituting in (1) R232R1=ℓ1100−ℓ1 ⇒23×2823=ℓ1100−ℓ1⇒23×2823=ℓ1100−ℓ1⇒5600−556ℓ1=69ℓ1;125ℓ1=5600ℓ1=5600125=44.8cm;11−l=44.8−40=4.8cm

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A metallic conductor at 10⁰C connected in the left gap of meter bridge gives balancing length 40 cm. When the conductor is at 60⁰C, the balancing point shifts by ---cm, (temperature coefficient of resistance of the material of the wire is (1/220)/⁰C)

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Detailed Solution

$\frac{R_{1}}{R} = \frac{ℓ}{100 - ℓ} = \frac{40}{60} = \frac{2}{3} . . . (1)$
From $α_{R} = \frac{R_{2} - R_{1}}{R_{1} t_{2} - R_{2} t_{1}} \Rightarrow \frac{1}{220} = \frac{R_{2} - R_{1}}{60 R_{1} - 10 R_{2}}$
$\begin{array}{l} \Rightarrow 60 R_{1} - 10 R_{2} = 220 R_{2} - 220 R_{1} \\ \Rightarrow 280 R_{1} = 230 R_{2} \Rightarrow \frac{R_{1}}{R_{2}} = \frac{23}{28} . . . (2) \end{array}$
From $\frac{R_{1}}{R 0} = \frac{2}{3} \Rightarrow R = \frac{3}{2} R_{1}$
Substituting in (1) $\frac{R_{2}}{\frac{3}{2} R_{1}} = \frac{ℓ^{1}}{100 - ℓ^{1}}$
$\begin{array}{l} \Rightarrow \frac{2}{3} \times \frac{28}{23} = \frac{ℓ^{1}}{100 - ℓ^{1}} \Rightarrow \frac{2}{3} \times \frac{28}{23} = \frac{ℓ^{1}}{100 - ℓ^{1}} \\ \Rightarrow 5600 - 556 ℓ^{1} = 69 ℓ^{1}; 125 ℓ^{1} = 5600 \\ ℓ^{1} = \frac{5600}{125} = 44 .8 cm; 1^{1} - l = 44 .8 - 40 = 4 .8 cm \end{array}$