A) 4 F
B) 6 F
C) 9 F
D) F
Correct Answer: C
Solution :
As shown in the figure, the wires will have the same Young's modulus (same material) and the length of the wire of area of cross-section 3A will be \[\ell /3\] (same volume as wire 1). For wire 1, \[Y=\frac{F/A}{\Delta x/\ell }\] ?(i) For wire 2, \[Y=\frac{F'/3A}{\Delta x/(\ell /3)}\] ?(ii) From (i) and (ii), \[\frac{F}{A}\times \frac{\ell }{\Delta x}=\frac{F'}{3A}\times \frac{\ell }{3\Delta x}\Rightarrow F'=9F\]You need to login to perform this action.
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