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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Units and Measurement
0 votes

Value of acceleration due to gravity is $ 9.8 m/s^2$. Find its value in $km/hr^2$

\[(a)\;127008\;km/hr^2\quad (b)\;227008\;km/hr^2 \quad (c)\;157008\;km/hr^2\quad (d)\;none\;of\;these\]

Can you answer this question?
 
 

1 Answer

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$n_1=n_2\bigg[\large\frac{L_2}{L_1}\bigg] \bigg[\large\frac{T_2}{T_1}\bigg]^{-2}$
$n_2=9.8$
Therefore $n_1=9.8\bigg[\large\frac{1m}{1km}\bigg] \bigg[\frac{1 sec}{1 hr}\bigg]^{-2}$
$\qquad=9.8 \bigg[\large\frac{1m}{1000 m}\bigg] \bigg[\large\frac{1 sec}{60 \times 60 sec}\bigg]$
$\qquad=9.8 \times \bigg[\large\frac{36 \times 36 \times 104}{1000 }\bigg]$
$n_1=127008 km/hr^2$
Hence a is the correct answer.
 

 

answered Jun 19, 2013 by meena.p
edited Jan 10, 2014 by meena.p
 

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