1. Tardigrade
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  4. If [ textF2 textL textT text-2 ] and - text273. text15° textF then - text453. text15° textF is equal to

Q. 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

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Solution:

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 $

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