A) \[\frac{1}{4KA}\]
B) \[\frac{7l}{4KA}\]
C) \[\frac{7l}{12KA}\]
D) \[\frac{l}{12KA}\]
Correct Answer: C
Solution :
[c] \[{{R}_{1}}=\frac{l}{KA},{{R}_{2}}=\frac{l}{2KA},{{R}_{3}}=\frac{l}{4KA}\] \[{{R}_{eq}}=\left( \frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}+{{R}_{3}} \right)=\frac{7l}{12KA}\] [as rods 1 and 2 are in parallel and equivalent is in series with 3].
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