The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be:

(Radius of earth will, be:

Option 1 -

1%

Option 2 -

3%

Option 3 -

4%

Option 4 -

0.5%

0 2 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    2 months ago
    Correct Option - 1


    Detailed Solution:

     g1=g (12hR)=g (12×326400)

    g1=99g100=0.99g

    % decrease is wt = gg1g×100=1%

Similar Questions for you

V
Vishal Baghel

g = A x x 2 + a 2 3 / 2  

v 0 ? d V = - x ? g d x

O - V = - A x a 2 + x 2 3 / 2

Let, a 2 + x 2 = t 2

2 x d x = 2 t d t

x d x = t d t

V = A t d t t 3 - A t - A a 2 + x 2 x

V = A a 2 + x 2

A
alok kumar singh

F = Gm2d2 - (i)

Now,  F'=Gm1m2d2=G2m3×4m3d2

k'=89Gm2d2

F'=8GF

 

V
Vishal Baghel

 g'=gω2Rsinθ

g' (atθ=0pole)=g

g' (atθ=90°, equator)=gω2R

g decreases as we move bole to equator

So, A is false statement

But ‘R’ statement is true

A
alok kumar singh

M A = ρ A × 4 3 π R A 3 ρ B = 4 ρ A

M B = ρ B × 4 3 π R B 3 R B = R A 2

M A M B = 2 , R A R B = 2 V E A V E B = 2 G 1 M A R A × R B 2 G 1 M B

v E A = v E B = 1 2 k m s e c 1

P
Payal Gupta

GM (3R/2)2=GMR3×r

OA=4R9=r

AB=R4R9=5R9OA:AB=4:5=x:yx=4

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