A) 1
B) 4
C) 2
D) 3
Correct Answer: B
Solution :
[b] 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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