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A metal wire of circular cross-section has a resistance R. The wire is now stretched without breaking so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now becomes R_2 then R_2 : R_1 is |
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Answer» `1:1` here` RHO ` and Vare CONSTANT`thereforeR propl^2therefore(R_1)/( R_2) =(l_(1)^(2))/( l_(2)^(2))` given`l_2 = 2l_1`. Hence` (R_1)/( R_2)=(l_1^(2) )/( 4l_(1)^(2)) = 1/4implies R_2: R_1= 4 :1` |
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