A) \[=4\times 100\mu s=400\mu s\]
B) \[=\frac{1}{2}\times Y\times {{(\text{Strain})}^{2}}\]
C) \[=\frac{1}{2}Y\times {{\left( \frac{l}{L} \right)}^{2}}\]
D) \[l=L\times 2%\]
Correct Answer: A
Solution :
The given circles are \[{{x}^{2}}+{{y}^{2}}-3x-6y+14=0\] ...(i) \[{{x}^{2}}+{{y}^{2}}-x-4y+8=0\] .. .(ii) and \[{{x}^{2}}+{{y}^{2}}+2x-6y+9=0\] ...(iii) The radical axis of Eqs. (i), (ii), (iii) are respectively \[x+y-3=0\] ...(iv) and \[3x-2y+1=0\] ...(v) On solving Eqs. (iv) and (v), we get \[x=1,y=2\] Thus, the coordinates of the radical centre are (1,2). The length of the tangent from (1, 2) to Eq. (i) is \[r=\sqrt{1+4-3-12+14}=2\] Hence, the required circle is \[{{(x-1)}^{2}}+{{(y-2)}^{2}}={{2}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-2x-4y+1=0\]
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