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A rocket motor consumes $100 \mathrm{~kg}$ of fuel per second exhausting it with a speed of $5 \mathrm{~km} / \mathrm{s}$. The speed of the rocket when its mass is reduced to $\frac{1^{\text {th }}}{20}$ of its initial mass, is (Assume initial speed to be zero and ignored gravitational and viscous forces.)
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Verified Answer
The correct answer is:
$5 \ln (20) \mathrm{km} / \mathrm{s}$
Key Idea Velocity of a rocket at any timet, $v=u\left(\frac{m_0}{m}\right)-g t$
where, $u=$ speed of exhausted gases,
$m_0=$ initial mass of the rocket
and $m=$ mass of the rocket at time $t$
Given, fuel burned rate, $\left(\frac{d m}{d t}\right)=100 \mathrm{~kg} / \mathrm{s}, u=5$
$\mathrm{km} / \mathrm{s}$
Then, $v=5 \ln \left(\frac{m_0}{m}\right)-g t$
As, it is given that the gravitational force is ignored and mass of rocket is reduced to $\frac{1}{20}$ th of it's initial mass i.e, $m=\frac{1}{20} m_0$.
or
$$
\frac{m_0}{m}=20
$$
So, $v=5 \ln (20) \mathrm{km} / \mathrm{s}$
Hence, the correct option is (3).
where, $u=$ speed of exhausted gases,
$m_0=$ initial mass of the rocket
and $m=$ mass of the rocket at time $t$
Given, fuel burned rate, $\left(\frac{d m}{d t}\right)=100 \mathrm{~kg} / \mathrm{s}, u=5$
$\mathrm{km} / \mathrm{s}$
Then, $v=5 \ln \left(\frac{m_0}{m}\right)-g t$
As, it is given that the gravitational force is ignored and mass of rocket is reduced to $\frac{1}{20}$ th of it's initial mass i.e, $m=\frac{1}{20} m_0$.
or
$$
\frac{m_0}{m}=20
$$
So, $v=5 \ln (20) \mathrm{km} / \mathrm{s}$
Hence, the correct option is (3).
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