A) 1
B) 4
C) 2
D) 3
Correct Answer: B
Solution :
In 1st case: Using the formula \[P=\frac{{{V}^{2}}}{R}\] where\[R\]is resistance of wire, V is voltage across wire and P is power dissipation in wire and \[R=\frac{\rho l}{A}\] From eqs. (1) and (2) \[=\frac{{{V}^{2}}}{\frac{\rho l}{A}}=\frac{{{V}^{2}}}{\rho l}.A\] \[{{P}_{1}}=\frac{{{V}^{2}}}{\rho l}.A\] ...(1) In 2nd case: Let \[{{R}_{2}}\]is net resistance \[{{R}_{2}}=\frac{R.R}{R+R}=\frac{R}{2}\] \[{{R}_{2}}=\frac{\rho .\left( \frac{l}{2} \right)}{A.2}\] \[=\frac{\rho l}{4A}\] \[\therefore \] \[{{P}_{2}}=\frac{{{V}^{2}}}{\rho l}.4A\] ...(2) Hence, from eqs. (3) and (4) \[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{1}{4}\] \[\Rightarrow \] \[\frac{{{P}_{2}}}{{{P}_{1}}}=\frac{4}{1}\]You need to login to perform this action.
You will be redirected in
3 sec