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Physics-

General

Easy

Question

A density bottle weighs 24 g when filled with water and 31 g when filled with a liquid of density $8 g divided by c. c$ Then the mass of the density bottle is

32 g
23 g
31 g
24 g

The correct answer is: 23 g

Related Questions to study

In the above problem the density of liquid is

In the above problem the density of liquid is

Physics-General

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

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

Physics-General

Which one of these is a derived physical quantity?

Which one of these is a derived physical quantity?

Physics-General

parallel

Which of the following numbers is a perfect number?

Which of the following numbers is a perfect number?

Which of the following is a perfect number?

Which of the following is a perfect number?

Which of the following is even?

Which of the following is even?

parallel

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

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

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

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

cot $theta$ + tan $theta$ =

Hence option 2 is correct

cot $theta$ + tan $theta$ =

Hence option 2 is correct

parallel

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

Hence option 2 is correct

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

Hence option 2 is correct

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 =

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 =

Hence option 4 is correct

parallel

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

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

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

Hence option 3 is correct

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

Hence option 2 is correct

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

Hence option 2 is correct

parallel

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