A) \[\frac{(x+y)(x-y)\rho L}{2k\theta }\]
B) \[\frac{(x-y)\rho L}{2k\theta }\]
C) \[\frac{(x+y)(x-y)\rho L}{k\theta }\]
D) \[\frac{(x-y)\rho Lk}{2\theta }\]
Correct Answer: A
Solution :
Since \[t=\frac{\rho L}{2k\theta }(x_{2}^{2}-x_{1}^{2})\] \[\therefore \]\[t=\frac{\rho L}{2k\theta }({{x}^{2}}-{{y}^{2}})=\frac{\rho L(x+y)(x-y)}{2K\theta }\]
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