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A Diwali rocket is ejected $0.05 kg$ of gases per second at a velocity of $400 m / s$ . The accelerating force on the rocket is

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20

dyne

20

Newton

20

None of the above

answer is B.

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Detailed Solution

A Diwali rocket is ejected

0.05 kg

of gases per second at a velocity of

400 m / s

. The accelerating force on the rocket is

20

Newton.
Given;
Mass

m = 0.05 kg

Velocity

v = 400 m / s

Time

t = 1 \sec

Using the formula

v = u + at

, since the initial velocity of the rocket is zero, then

u = 0

We have

a = \frac{v}{t}

\Rightarrow a = \frac{400}{1} m / s^{2}

[

m / s^{2} = (m / s) / s

]
Using Newton's second law, we have

F = ma

\Rightarrow F = 0.05 \times 400

\Rightarrow F = 20 N

Dyne is the centimetre-gram-seconds (CGS) unit of force. Since the rest of the options are given in SI units. Therefore,

20

Newton is the correct answer.

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