Physics-
General
Easy

Question

A piece of wood of mass 150g has a volume of 200 cm3 Then the density of wood in C.G.S system is

  1. 0.75 g divided by c m to the power of 3 end exponent    
  2. 7.5 g divided by c m to the power of 3 end exponent    
  3. 75 g divided by c m to the power of 3 end exponent    
  4. 750 g divided by c m to the power of 3 end exponent    

The correct answer is: 0.75 g divided by c m to the power of 3 end exponent

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c o s invisible function application e c to the power of 2 end exponent 60 to the power of ring operator end exponent s e c to the power of 2 end exponent invisible function application 30 to the power of ring operator end exponent c o s invisible function application 0 to the power of ring operator end exponent s i n invisible function application 45 to the power of ring operator end exponent c o t to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent t a n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent equals

c o s invisible function application e c to the power of 2 end exponent 60 to the power of ring operator end exponent s e c to the power of 2 end exponent invisible function application 30 to the power of ring operator end exponent c o s invisible function application 0 to the power of ring operator end exponent s i n invisible function application 45 to the power of ring operator end exponent c o t to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent t a n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent equals

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4 open parentheses s i n to the power of 4 end exponent invisible function application 30 to the power of ring operator end exponent plus c o s to the power of 4 end exponent invisible function application 60 to the power of ring operator end exponent close parentheses plus fraction numerator 2 over denominator 3 end fraction open parentheses s i n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent minus c o s to the power of 2 end exponent invisible function application 45 to the power of ring operator end exponent close parentheses plus fraction numerator 1 over denominator 2 end fraction t a n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent equals

4 open parentheses s i n to the power of 4 end exponent invisible function application 30 to the power of ring operator end exponent plus c o s to the power of 4 end exponent invisible function application 60 to the power of ring operator end exponent close parentheses plus fraction numerator 2 over denominator 3 end fraction open parentheses s i n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent minus c o s to the power of 2 end exponent invisible function application 45 to the power of ring operator end exponent close parentheses plus fraction numerator 1 over denominator 2 end fraction t a n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent equals

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cottheta + tantheta =

Hence option 2 is correct

cottheta + tantheta =

Maths-General

Hence option 2 is correct

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open parentheses 1 plus c o t to the power of 2 end exponent invisible function application theta close parentheses left parenthesis 1 minus c o s invisible function application theta right parenthesis left parenthesis 1 plus c o s invisible function application theta right parenthesis equals

Hence option 2 is correct

open parentheses 1 plus c o t to the power of 2 end exponent invisible function application theta close parentheses left parenthesis 1 minus c o s invisible function application theta right parenthesis left parenthesis 1 plus c o s invisible function application theta right parenthesis equals

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Hence option 2 is correct

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If s i n invisible function application A equals fraction numerator 1 over denominator 3 end fraction, then cos A cosecA + tan A sec A =

If s i n invisible function application A equals fraction numerator 1 over denominator 3 end fraction, then cos A cosecA + tan A sec A =

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If sin B = 1/2 then 3cos B – 4 cos3 B =

Hence option 4 is correct

If sin B = 1/2 then 3cos B – 4 cos3 B =

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Hence option 4 is correct

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A minutes hand of a table clock is 3cm long. How far does its tip move in 20 minutes

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If fraction numerator x to the power of 2 end exponent plus y to the power of 2 end exponent plus z to the power of 2 end exponent minus 64 over denominator x y minus y z minus z x end fraction equals negative 2 blank a n d blank x plus y equals 3 z comma blank t h e n blank t h e blank v a l u e blank o f blank z blank i s

Hence option 3 is correct

If fraction numerator x to the power of 2 end exponent plus y to the power of 2 end exponent plus z to the power of 2 end exponent minus 64 over denominator x y minus y z minus z x end fraction equals negative 2 blank a n d blank x plus y equals 3 z comma blank t h e n blank t h e blank v a l u e blank o f blank z blank i s

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Hence option 3 is correct

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If a to the power of 2 end exponent plus b to the power of 2 end exponent equals 117 blank a n d blank a b equals 54 comma blank t h e n blank fraction numerator a plus b over denominator a minus b end fraction blank i s

Hence option 2 is correct

If a to the power of 2 end exponent plus b to the power of 2 end exponent equals 117 blank a n d blank a b equals 54 comma blank t h e n blank fraction numerator a plus b over denominator a minus b end fraction blank i s

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Hence option 2 is correct

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The appropriate formula that can be used to find the value of open parentheses 50 fraction numerator 1 over denominator 4 end fraction cross times 49 fraction numerator 3 over denominator 4 end fraction close parentheses

Hence option 3 is correct

The appropriate formula that can be used to find the value of open parentheses 50 fraction numerator 1 over denominator 4 end fraction cross times 49 fraction numerator 3 over denominator 4 end fraction close parentheses

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Hence option 3 is correct

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Divide a x to the power of 2 end exponent minus a y to the power of 2 end exponent blank b y blank a x plus a y

Hence option 1 is correct

Divide a x to the power of 2 end exponent minus a y to the power of 2 end exponent blank b y blank a x plus a y

Maths-General

Hence option 1 is correct

parallel

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