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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityElectrical instrumentHard2 minQB

PhysicsHardsingle choice

On interchanging the resistances, the balance point of a meter bridge shifts to the left by $10 c m$ . The resistance of their series combination is $1 k Ω$ . How much was the resistance on the left slot before interchanging the resistances?

Options:

Answer:

Solution:

$\frac{a}{b} = \frac{l}{100 - l}$

$a + b = 1000$

$\frac{b}{a} = \frac{l - 10}{110 - l}$

$\frac{100 - l}{l} = \frac{l - 10}{110 - l}$

$l^{2} - 10 l = 11000 + l^{2}$

$- 100 l = 110 l$

$200 l = 11000$

$l = \frac{110}{2} = 55$

$\frac{55}{45} = \frac{a}{b}$

$\Rightarrow a = 550Ω b = 450Ω$

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Electrical instrument

⏱ 2mℹ️ Source: QB

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