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A fluid container is containing a liquid of density rho is accelerating upward with acceleration an along the inclined place of inclination alpha as shown. Then the angle of inclination theta of free surface is

  1. t a n to the power of negative 1 end exponent invisible function application open square brackets fraction numerator a over denominator g c o s invisible function application alpha end fraction close square brackets    
  2. t a n to the power of negative 1 end exponent invisible function application open square brackets fraction numerator a plus g s i n invisible function application alpha over denominator g c o s invisible function application alpha end fraction close square brackets    
  3. t a n to the power of negative 1 end exponent invisible function application open square brackets fraction numerator a minus g s i n invisible function application alpha over denominator g left parenthesis 1 plus c o s invisible function application alpha right parenthesis end fraction close square brackets    
  4. t a n to the power of negative 1 end exponent invisible function application open square brackets fraction numerator a minus g s i n invisible function application alpha over denominator g left parenthesis 1 minus c o s invisible function application alpha right parenthesis end fraction close square brackets    

The correct answer is: t a n to the power of negative 1 end exponent invisible function application open square brackets fraction numerator a plus g s i n invisible function application alpha over denominator g c o s invisible function application alpha end fraction close square brackets

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