Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityMicroscopicEasy2 minQB

PhysicsEasysingle choice

A wire of length $l$ has a resistance $R$ . If half of the length is stretched to make the radius half of its original value, then the final resistance of the wire is

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Answer:

Solution:

$R = \frac{ρ l}{A} = \frac{ρ l}{π r^{2}}$

$π r^{2} \frac{l}{2} = π {(\frac{r}{2})}^{2} l^{'}$

$l^{'} = 2 l$

$R' = \frac{ρ l'}{A'} = \frac{ρ (2 l)}{π {(\frac{r}{2})}^{2}} = 8 R$

$R'' = \frac{ρ (\frac{l}{2})}{π r^{2}} = \frac{R}{2}$

$R_{eq} = R' + R'' = \frac{17 R}{2}$

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Microscopic

⏱ 2mℹ️ Source: QB

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