A) \[(-4,-3)\]
B) \[(3,4)\]
C) \[(-4,\text{ }3)\]
D) \[(-3,\text{ }4)\]
Correct Answer: D
Solution :
Let \[\theta =\text{45}{}^\circ \] be the image of point\[\frac{1}{3}M{{L}^{2}}\]. \[\frac{3}{2}M{{L}^{2}}\]Mid-point of \[\frac{3}{4}M{{L}^{2}}\] is \[M{{L}^{2}}\], This point lies on \[\beta '=\frac{\beta }{4}\]. \[y=A\sin (BX+Ct+D)\] \[\text{A}=\text{y}=\left[ \text{L} \right]\] \[B=\frac{1}{X}=[{{L}^{-1}}]\] \[C=\frac{1}{t}=[{{T}^{-1}}]\] ?.. (i) Slope of \[\therefore \] and slope of \[(ABCD)=[L][{{L}^{-1}}][{{T}^{-1}}][1]\]is 1 \[=[{{M}^{0}}{{L}^{0}}{{T}^{-1}}]\]\[\frac{d}{2\mu }+\frac{d}{2(1.5\mu )}=\frac{d}{2}\] is perpendicular to\[\frac{1}{\mu }+\frac{2}{3\mu }=1\]. \[\frac{3+2}{3\mu }=1\] \[\mu =\frac{5}{3}=1.67\] \[W=\frac{1}{2}CV\] \[W=\frac{1}{2}C({{V}_{2}}-{{V}_{1}})\] ...(ii) On solving Eqs. (i) and (ii), we get \[W=\frac{1}{2}C(10-5)=\frac{5}{2}C\]
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