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A small block of mass m=2 kg is projected on a larger block of 20 kg and of length 20 metres, with a velocity of v m/s as shown in the figure. The coefficient of friction between the two blocks is 1.2, while between the lower block and ground is 0.1. The time taken by the small block to fall off the larger block is (Take g=10 m/s2)


A
2.5 s
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B
4 s
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C
3 s
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D
2 s
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Solution

The correct option is B 2 s
FBDs of the blocks are as shown.
Here, R1 and R2 are normal reaction forces on the respective blocks.


From the FBDs, we have,
R1=2g=20 N
f1=μ1R1=1.2×20=24 N

R2=R1+20g=20+200=220 N
f2=μ2R2=0.1×220=22 N

f1>f2 so, upper block will retard and lower block will accelerate.

Retardation of upper block is
a1=f1m=242=12 m/s2
and acceleration of lower block is
a2=f1f2m=242220=0.1 m/s2

Relative retardation of upper block w.r.t lower block is
a12=a1+a2=12.1 m/s2 (constant acceleration)

For minimum value of velocity, upper block will just come to rest as it falls from lower block
So, using 3rd law of motion, we can write
v212=u2122a12s12
0=v22×12.1×20
v2=484v=22 m/s

To find time taken, use 1st law of motion,
v12=u12a12t
0=vat=2212.1t
t=2212.1=1.82 s2 s

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