Two blocks of masses m1 and m2 are connected as shown in the figure. The acceleration of the block m2 is
A
m2gm1+m2
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B
m1gm1+m2
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C
4m2g−m1gm1+m2
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D
m2gm1+4m2
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Solution
The correct option is Am2gm1+m2 Let acceleration of m1 be a1 and m2 be a2. a2=Fnetm2
Considering force body diagram of m2 a2=m2g−Tm2a2=g−Tm2
Now considering the pulley connected to mass m2
Since pulley is massless, so net force on iy would be zero.
Hence by force balance T′=2T
where T′ is the tension in the string joining two pulleys
Now taking the pulley above the previous pulley we get T′=2T"
where T" is the tension in strings which are joining the uppermost pulley with middle one.
From this we get T"=T
Taking m1 into consideration and balancing force T=m1a1
Now putting this into force balance equation of m2 a2=g−m1a1m2
Now as the middle pulley is the exact mirror symmetry of the bottom pulley we can say a1=a2=a ⇒a=g−m1am2⇒a=m2gm1+m2