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  • If the maximum possible exhaust velocity of a rocket be 2 Km/s  , calculate the ratio <img alt="\frac{m_0}{m_r}" src="http://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7Bm_0%7D%7Bm_r

If the maximum possible exhaust velocity of a rocket be 2 Km/s  , calculate the ratio \frac{m_0}{m_r}  for it is to require the escape velocity of 11.2 km/s after starting from rest (approx ) [ {m_0}\rightarrow initial velocity , {m_r}\rightarrow  mass of emptied rocket ]  

Option: 1

270 


Option: 2

300 


Option: 3

350 


Option: 4

200 


Answers (1)

best_answer

Given-

Maximum relative velocity of fuel, u=2000m/s,

Escape velocity of rocket vb=11.2km/hr=11200m/s,

Neglecting the effect of gravity, the expression for the speed of rocket-

V_{b}=V_{max}=u\:log_e\left ( \frac{m_\circ }{m_r} \right )

m_{r}\rightarrow residual\:mass\:of\:empty\:container

V_{b} = 2.303 u log \left ( \frac{m_0}{m_r } \right )

11.2 = 2.303 \times 2 \times log _{10}\left ( \frac{m_0}{m_r } \right )

log _{10}\left ( \frac{m_0}{m_r } \right ) = \frac{11.2 }{2.303\times 2} = 2.432

\left ( \frac{m_0}{m } \right ) = antilog (2.432 ) = 270.4

 

 

 

 

Posted by

HARSH KANKARIA

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