A) 1
B) 4
C) 2
D) 3
Correct Answer: B
Solution :
In 1st case Using the formula, \[P=\frac{{{V}^{2}}}{R}\] ... (i) where, R is resistance of wire, V is voltage across wire and P is power dissipation in wire and \[R=\frac{\rho l}{A}\] ... (ii) From Eqs. (i) and-(ii), we get \[{{P}_{1}}=\frac{{{V}^{2}}}{\rho l/A}=\frac{{{V}^{2}}}{\rho l}.\,A\] \[{{P}_{1}}=\frac{{{V}^{2}}}{\rho l}.\,A\] ... (iii) In 2nd case Let \[{{R}_{2}}\] be net resistance. \[{{R}_{2}}=\frac{R\times R}{R+R}=\frac{R}{2}\] where, R is the resistance of half wire. \[\therefore \] \[{{R}_{2}}=\frac{\rho .\left( \frac{l}{2} \right)}{A.\,2}=\frac{\rho l}{4A}\] (as \[l=\frac{l}{2}\Rightarrow A=2A\]) \[\therefore \] \[{{P}_{2}}=\frac{{{V}^{2}}}{\rho l}.4\,A\] ... (iv) Hence, from Eqs. (iii) and (iv), we get \[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{1}{4}\Rightarrow \frac{{{P}_{2}}}{{{P}_{1}}}=\frac{4}{1}\]
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