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4-1.Newton's Laws of Motion
hard
$ m_1 = 4m_2$ છે. $m_1 $ નો પ્રવેગ $a$ છે. $m_1$ ને સ્થિર થતાં ........ $\sec$ લાગે.
A
$0.2$
B
$0.4$
C
$0.6$
D
$0.8$
Solution
${m_1}a = {m_1}g – 2T$ …..(i)
${m_2}\left( {2a} \right) = T – {m_2}g$…..(ii)
$a = g/4$
$t = \sqrt {\frac{{2h}}{a}} \, = \,\sqrt {\frac{{2 \times 0.2}}{{g/4}}} \, = \,\sqrt {\frac{{2 \times 0.2}}{{2.5}}} $$ = 0.4\,\sec $
Standard 11
Physics
