SEE Science Preparation-2077

Group - D  Solution of Q.N.21

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 Unit-1                                           Force               ( class:-10)

     NUMEICAL SOLUTION

1.     The distance between earth and moon is 36x10⁸m. If the mass of the earth is 6x10²⁴kg and mass of the moon is 7.2x10²²kg. What is the gravitational force between them?

Solution:-

Given:-

Distance between them(d) = 36x10⁸m = 3.6x10⁹m

Mass of the earth (M) = 6x10²⁴kg

Mass of the moon (m) = 7.2x10²²kg

Gravitational force (F) = ?

Gravitational constant (G) = 6.67 x10^-11 Nm²/kg².

Now, By using formula

 F = GMm/d² =6.67 x10^-11 x 6x10²⁴ x 7.2x10²² /(3.6x10⁹)²

Or , F = 288.144 x 10³⁵ / 12.96 x10¹⁸ = 22.23 x 10¹⁷

\ F = 2.223 x 10¹⁸ N

2.     The mass and radius of Jupiter is 1.9x10²⁷ kg and 71x10⁶m respectively. Find the acceleration due to gravity of it. What will be the weight  of an object having mass 75 kg on that planet?

Solution:-

Given:-

Mass of the Jupiter (M) = 1.9x10²⁷

Mass of an object (m) = 75 kg

Radius of the Jupiter( R ) =71x10⁶m = 7.1x10⁷m

Acceleration due to gravity on Jupiter surface (g) = ?

Weight of an object (w) = ?

Now , By using formula

  g = GM/R² /

or, g = 6.67 x10^-11 x 1.9x10²⁷  / (7.1x10⁷ )²

or, g = 12.67 x10¹⁶ / 50.41 x 10¹⁴

or, g= 0.2513 x 10²

\ g = 25.13 m/s²

Again,

\W = mxg = 75 x 25.13 = 1884.75 N

3.     The mass of Jupiter is 319 times greater than the mass of earth and the radius is 11 times greater than radius of earth. If the acceleration due to gravity on the earth surface is 9.8 m/s². Calculate the acceleration due to gravity on the surface of Jupiter.

Solution:-

Given:-

Mass of the Jupiter (M_J) = 319 M_e.

Radius of the Jupiter(R_J) = 11 R_e.

Acceleration due to gravity on the earth surface (g_e) = 9.8 m/s².

Acceleration due to gravity on the Jupiter surface (g_j) = ?

Now,

Acceleration due to gravity on the earth surface is

  g_e  =  GM_e / R_e²  or   9.8 =  GM_e / R_e²  …………………..eqⁿ (i)

Acceleration due to gravity on the earth surface is

     g_j  =  GM_j / R_j² 

or, g_j = G 319 M_e / (11R_e)²

or, g_j = 319 GM_e / 121R_e²

or, g_j = 319 / 121 x 9.8                                       [ GM_e / R_e²  = 9.8]

\ g_j  = 25.83 m/s².

4.     The earth is compressed to the size of the moon. Calculate the acceleration due to gravity of newly formed earth based on the data given below:-

Mass of the earth (M_e) = 6x10²⁴ kg

Mass of the moon (M_m) = 7.2x10²² kg.

Radius of the earth (R_e) = 63800 km

Radius of the moon (R_m) = 1.7x10³km

Solution:-

Given:-

Mass of the earth (M_e) = 6x10²⁴ kg

Radius of the moon (R_m) = 1.7x10³km =1.7x10⁶m

Acceleration due to gravity of new earth(g_n) = ?

g_n = GM_e / R_e²  = 6.67x10^-11 x 6x10²⁴ / (1.7x10⁶)² = 40.02 x 10¹³/ 2.89x10¹²

  or, g_n = 13.847x10

\ g_n  = 138.47  m/s².

                                            # CARBON DIOXIDE GAS:-

Ø  Molecular formula is CO₂  and it’s molecular mass is 44 amu.

·        Occurrences:-

Ø CO₂ is found in free as well as in combined state in nature.

Ø It’s volume in atmosphere is found to be 0.03%.

Unit-2                                                                                                  PRESSURE                                                       

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                                                       Gravitational force