Neet Solved Paper 2015 Question 5
Question: A block A of mass $ m _1 $ rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass $ m _2 $ is suspended. The coefficient of kinetic friction between the block and the table is $ {\mu_k} $ . When the block A is sliding on the table, the tension in the string is
Options:
A) $ \frac{(m _2+{\mu_k}m _1)g}{(m _1+m _2)} $
B) $ \frac{(m _2+{\mu_k}m _1)g}{(m _1+m _2)} $
C) $ \frac{m _1,m _2(1+{\mu_k})g}{(m _1+m _2)} $
D) $ \frac{m _1m _2(1-{\mu_k})g}{(m _1+m _2)} $
Show Answer
Answer:
Correct Answer: C
Solution:
FED of block A, $ T=m _1a=f _{k}, $ ..(i)
FBD of block B $ m _2g-T=m _2a $ …(ii)
Adding Eqs. (i) and (ii), we get $ m _2g,-m _1a,=m _2a,+f _{k} $
Þ $ m _2g,-m _1a,=m _2a,+{\mu_k},m _1g $
$ \Rightarrow ,a=\frac{( m _2-{\mu_k}m _1 )g}{m _1+m _2} $
From Eq. (ii), $ T=m _2(g-a) $ $ =m _2,[ 1-\frac{(m _2-{\mu_k}m _1)}{m _1+m _2} ]g $ $ T=\frac{m _1m _2,(1+{\mu_k})}{m _1+m _2}g $
