Physics
General
Easy

Question

Acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth, If the ratio of densities of earth rho subscript theta and moon rho subscript m is rho subscript t over rho subscript equals equals 5 divided by 3 then radius of moon Rm in terns of Re will be,

  1. 5 over 18Re
  2. 1 over 6Re
  3. 3 over 16Re
  4. fraction numerator 1 over denominator 2 square root of 3 end fractionRe

The correct answer is: 5 over 18Re

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A research satellite of mass 200 kg. circles the earth in an orbit of average radius 3R/2  where R is radius of earth. Assuming the gravitational pall 10N, the pull on the satellite will be...N

A research satellite of mass 200 kg. circles the earth in an orbit of average radius 3R/2  where R is radius of earth. Assuming the gravitational pall 10N, the pull on the satellite will be...N

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If the value of ' g ' acceleration due to gravity, at earth surface fs .10ms-2 its value in ms-2 at the center of earth, which is assumed to be a sphere of Radius ' R ' meter and uniform density is

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A man can jump to a height of 1.5 m on a planet  Α what is the height ne may be able to jump on another planet whose density and radius are respectively one- quater and one- third that of planet A

A man can jump to a height of 1.5 m on a planet  Α what is the height ne may be able to jump on another planet whose density and radius are respectively one- quater and one- third that of planet A

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If density of earth increased 4 times and its radius becomes half of then out weight will be...

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Weight of body of mass m decreases by 1% when it is raised to height h above the earth's surface. If the body is taken to a depth h in a mine. change in its weight is

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Density of the earth is doubled keeping its radius constant then acceleration, due to gravity will. be …….ms-2 (g=9.8 ms2)

Density of the earth is doubled keeping its radius constant then acceleration, due to gravity will. be …….ms-2 (g=9.8 ms2)

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The acceleration due to gravity is  at a point distance r from the center of earth . if . R.  if r<R.then

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Maths-

If u equals l o g invisible function application open parentheses x to the power of 3 end exponent plus y to the power of 3 end exponent plus z to the power of 3 end exponent minus 3 x y z close parentheses then sum x fraction numerator partial differential u over denominator partial differential x end fraction equals

The given function is u=log(x3+y3+z3-3xyz)
We have to find the value of sum x fraction numerator partial differential u over denominator partial differential x end fraction
It means we have to find the value of x fraction numerator partial differential u over denominator partial differential x end fraction plus y fraction numerator partial differential u over denominator partial differential y end fraction plus z fraction numerator partial differential u over denominator partial differential z end fraction
For simplification we will write u=logv
v = x3 + y3 + z3 -3xyz
We will take the partial derivative of u w.r.t all three variables.
fraction numerator partial differential u over denominator partial differential x end fraction equals 1 over v fraction numerator partial differential v over denominator partial differential x end fraction
v space equals x cubed plus y cubed plus z cubed minus 3 x y z
y
fraction numerator partial differential v over denominator partial differential x end fraction space equals 3 x squared minus 3 y z space space space space space space space left parenthesis 1 right parenthesis

fraction numerator partial differential u over denominator partial differential y end fraction equals 1 over v fraction numerator partial differential v over denominator partial differential y end fraction
fraction numerator partial differential v over denominator partial differential y end fraction equals 3 y squared minus 3 x z space space space space space space space space left parenthesis 2 right parenthesis

fraction numerator partial differential u over denominator partial differential z end fraction equals 1 over v fraction numerator partial differential v over denominator partial differential z space end fraction
fraction numerator partial differential v over denominator partial differential z end fraction space equals 3 z squared minus 3 x y space space space space space space space space left parenthesis 3 right parenthesis






space
Now, we will substitute all the values in the sum x fraction numerator partial differential u over denominator partial differential x end fraction.
x fraction numerator partial differential u over denominator partial differential x end fraction plus y fraction numerator partial differential u over denominator partial differential y end fraction plus z fraction numerator partial differential y over denominator partial differential z end fraction equals x left parenthesis fraction numerator 3 x squared minus 3 y z over denominator x cubed plus y cubed plus z cubed minus 3 x y z end fraction right parenthesis plus y left parenthesis fraction numerator 3 y squared minus 3 x z over denominator x cubed plus y cubed plus z cubed minus 3 x y z end fraction right parenthesis plus z left parenthesis fraction numerator 3 z squared minus 3 x y over denominator x cubed plus y cubed plus z cubed minus 3 x y z end fraction right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 3 x cubed minus 3 x y z plus 3 y cubed minus 3 x y z plus 3 z cubed minus 3 x y z over denominator x cubed plus y cubed plus z cubed minus 3 x y z end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 3 x cubed plus 3 y cubed plus 3 z cubed minus 9 x y z over denominator x cubed plus y cubed plus z cubed minus 3 x y z end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 3 fraction numerator x cubed plus y cubed plus z cubed minus 3 x y z over denominator x cubed plus y cubed plus y cubed minus 3 x y z end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 3
The required answer is 3.

If u equals l o g invisible function application open parentheses x to the power of 3 end exponent plus y to the power of 3 end exponent plus z to the power of 3 end exponent minus 3 x y z close parentheses then sum x fraction numerator partial differential u over denominator partial differential x end fraction equals

Maths-General
The given function is u=log(x3+y3+z3-3xyz)
We have to find the value of sum x fraction numerator partial differential u over denominator partial differential x end fraction
It means we have to find the value of x fraction numerator partial differential u over denominator partial differential x end fraction plus y fraction numerator partial differential u over denominator partial differential y end fraction plus z fraction numerator partial differential u over denominator partial differential z end fraction
For simplification we will write u=logv
v = x3 + y3 + z3 -3xyz
We will take the partial derivative of u w.r.t all three variables.
fraction numerator partial differential u over denominator partial differential x end fraction equals 1 over v fraction numerator partial differential v over denominator partial differential x end fraction
v space equals x cubed plus y cubed plus z cubed minus 3 x y z
y
fraction numerator partial differential v over denominator partial differential x end fraction space equals 3 x squared minus 3 y z space space space space space space space left parenthesis 1 right parenthesis

fraction numerator partial differential u over denominator partial differential y end fraction equals 1 over v fraction numerator partial differential v over denominator partial differential y end fraction
fraction numerator partial differential v over denominator partial differential y end fraction equals 3 y squared minus 3 x z space space space space space space space space left parenthesis 2 right parenthesis

fraction numerator partial differential u over denominator partial differential z end fraction equals 1 over v fraction numerator partial differential v over denominator partial differential z space end fraction
fraction numerator partial differential v over denominator partial differential z end fraction space equals 3 z squared minus 3 x y space space space space space space space space left parenthesis 3 right parenthesis






space
Now, we will substitute all the values in the sum x fraction numerator partial differential u over denominator partial differential x end fraction.
x fraction numerator partial differential u over denominator partial differential x end fraction plus y fraction numerator partial differential u over denominator partial differential y end fraction plus z fraction numerator partial differential y over denominator partial differential z end fraction equals x left parenthesis fraction numerator 3 x squared minus 3 y z over denominator x cubed plus y cubed plus z cubed minus 3 x y z end fraction right parenthesis plus y left parenthesis fraction numerator 3 y squared minus 3 x z over denominator x cubed plus y cubed plus z cubed minus 3 x y z end fraction right parenthesis plus z left parenthesis fraction numerator 3 z squared minus 3 x y over denominator x cubed plus y cubed plus z cubed minus 3 x y z end fraction right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 3 x cubed minus 3 x y z plus 3 y cubed minus 3 x y z plus 3 z cubed minus 3 x y z over denominator x cubed plus y cubed plus z cubed minus 3 x y z end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 3 x cubed plus 3 y cubed plus 3 z cubed minus 9 x y z over denominator x cubed plus y cubed plus z cubed minus 3 x y z end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 3 fraction numerator x cubed plus y cubed plus z cubed minus 3 x y z over denominator x cubed plus y cubed plus y cubed minus 3 x y z end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 3
The required answer is 3.
General
chemistry-

Which type of isomerism is observed between I and II
     and 

Which type of isomerism is observed between I and II
     and 

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How many spatial orientations are possible in following compounds-

How many spatial orientations are possible in following compounds-

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The configuration of stereo center in the compound is –

The configuration of stereo center in the compound is –

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The IUPAC name of the given compound –

The IUPAC name of the given compound –

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Shows which type of isomerism
and 

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and 

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The two compounds shown below are –

on superimposable mirror images

The two compounds shown below are –

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on superimposable mirror images
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The compound whose stereo- chemical formula is written below exhibits ‘x’ geometrical isomers and ‘y’ optical isomers :

The value of x and y are :

The compound whose stereo- chemical formula is written below exhibits ‘x’ geometrical isomers and ‘y’ optical isomers :

The value of x and y are :

chemistry-General
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