Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Offsite Campus Turito Championship

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

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

$0.75 g / {c m}^{3}$
$7.5 g / {c m}^{3}$
$75 g / {c m}^{3}$
$750 g / {c m}^{3}$

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

Related Questions to study

Which one of these is a derived physical quantity?

Which one of these is a derived physical quantity?

Physics-General

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?

parallel

Which of the following is even?

Which of the following is even?

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

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

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

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

parallel

cot $theta$ + tan $theta$ =

Hence option 2 is correct

cot $theta$ + tan $theta$ =

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

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

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 =

parallel

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

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

parallel

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

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 $open parentheses 50 fraction numerator 1 over denominator 4 end fraction cross times 49 fraction numerator 3 over denominator 4 end fraction close parentheses$

Hence option 3 is correct

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

Hence option 1 is correct

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

Hence option 1 is correct

parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.