If a wire of resistance R is stretched to twice its length (volume conserved), its new resistance becomes:
A4R
BR/2
C2R
DR/4
Answer & Solution
Correct answer: A. 4R
l → 2l, A → A/2 ⇒ R' = ρ(2l)/(A/2) = 4R.
Related questions
For a battery of emf ε and internal resistance r driving current I through external resistA galvanometer of resistance G is converted into a voltmeter reading full-scale voltage V A galvanometer of resistance G shows full-scale deflection at current I_g. To convert it iThe relation between current density j, drift velocity v_d, number density n and charge e A wire has a resistance R. It is stretched uniformly so that its length becomes 2L. The neKirchhoff current law at a junction is a statement ofTwo cells of emf ε and internal resistance r each are connected in parallel. The equivalenA potentiometer of length L is used to compare emfs. If the balance lengths for cells of e