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Byju's Answer

Standard XII

Physics

Reynold's Number

In a rocket, ...

Question

In a rocket, fuel burns at the rate of ${1 k g s}^{- 1}$ . This fuel is ejected from the rocket with the velocity of ${60 k m s}^{- 1}$ . This exerts a force, in newton, on the rocket of:

60 N

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600 N

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6000 N

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60000 N

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Solution

The correct option is D. 60000 N.
Given,
$v = 60 k m / s = 60000 m / s$
Also, force is defined as the rate of change of momentum, here mass is the rate of fuel supply or m/t=1\ kg/s
So $F = \frac{m v}{t} = \frac{1 \times 60000}{1} = 60000 N$ .

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In a rocket, fuel burns at the rate of 1kgs−1. This fuel is ejected from the rocket with the velocity of 60kms−1. This exerts a force, in newton, on the rocket of:

The correct option is D. 60000 N.Given,v=60 km/s=60000 m/sAlso, force is defined as the rate of change of momentum, here mass is the rate of fuel supply or m/t=1\ kg/sSo F=mvt=1×600001=60000 N.

In a rocket, fuel burns at the rate of ${1 k g s}^{- 1}$ . This fuel is ejected from the rocket with the velocity of ${60 k m s}^{- 1}$ . This exerts a force, in newton, on the rocket of:

The correct option is D. 60000 N.
Given,
$v = 60 k m / s = 60000 m / s$
Also, force is defined as the rate of change of momentum, here mass is the rate of fuel supply or m/t=1\ kg/s
So $F = \frac{m v}{t} = \frac{1 \times 60000}{1} = 60000 N$ .