Question

If $ \left[ {{\text{F}}^{2}}\text{L}{{\text{T}}^{\text{-2}}} \right] $ and $ -\text{273}.\text{15}{}^\circ \text{F} $ then $ -\text{453}.\text{15}{}^\circ \text{F} $ is equal to

• A. $ -\text{459}.\text{67}{}^\circ \text{F} $
• B. $ -\text{491}.\text{67}{}^\circ \text{F} $
• C. $ \text{52}00\text{{ }\!\!\mathrm{\AA}\!\!\text{ }} $
• D. $ \text{Vc}=\text{1}.\text{5V} $

Answer 1

Answer: (A)

Given. $ \text{mgh} $ and $ \text{mgh}=\text{4}0\times \text{9}.\text{8}\times \text{h} $ Now, $ P(A'\cap B')=1-P(A\cup B)=\frac{1}{3} $ $ |e|=\frac{d\phi }{dt}=B\frac{dA}{dt} $ $ [\because \phi =BA] $ $ =\pi B\frac{d}{dt}({{r}^{2}}) $ $ |e|=\pi B.2r\frac{dr}{dt} $ $ \frac{dr}{dt}={{10}^{-2}} $ $ \text{B}=\text{1}{{0}^{-\text{3}}} $ $ e=1\mu V $ $ \therefore $