o wire is 10 S2. Its length is 12. ...
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o wire is 10 S2. Its length is 12. The resistance of a 10 m long wire is 10 . It increased by 25% by stretching the wire uniformly. The resistance of wire will change to (approximately) (a) 12.5 12 (b) 14.522 (c)15.612 (d) 16.622

JEE/Engineering Exams
Physics
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Let resstance before steetching be ( R_{1}=10 Omega ) The Founula of remstarce is ( R=k frac{L}{A} ) L = Length of remitance A theien of consection ( v= ) Volume of wrise It in comtant ( V=A ) A ( x L ) ( Rightarrow A=underline{V} ) [ begin{array}{l} therefore R_{1}=kleft(frac{L^{2}}{v}right) text { Aflee cliet thing length becomes } (L+0 cdot 25 L)=1.25 L end{array} ] Now rens [ begin{array}{l} qquad begin{array}{c} R_{2}=kleft(frac{(1 cdot 25 L)^{2}}{V}right) R_{1}=frac{K(1 cdot 25)^{2} k^{2}}{X}=(1.25)^{2} frac{R_{2}}{R_{1}}=1.5625 end{array} end{array} ] ( begin{aligned} R_{2} &=1 cdot 5625 times R_{1} &=1.5625 times 10 &=15.625 Omega end{aligned} )
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