NTSE Stage 2 SAT 2017 Question Paper

Question : 36 of 100

Marks: +1, -0

The diameter of a wire is reduced to one fifth of its original value by stretching it. If its initial resistance is R, what would be its resistance after reduction of the diameter ?

$\R/625$
$\R/25$
25 R
625 R

Validate

Solution:

\R={\ρ \L}/\A

\R'={\ρ' \L'}/\A'= {\ρ×25\L}/{\π×\d^2/{4×5^2}}

[If A reduces by 25 times, length increases by 25 times]

={625×\ρ\L}/\A

=625×\R

Go to Question:

Prev Question

Next Question