    Question

# 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
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4 s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3 s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2 s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is B 2 sFBDs 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=f1−f2m=24−2220=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=u212−2a12s12 ⇒0=v2−2×12.1×20 ⇒v2=484⇒v=22 m/s To find time taken, use 1st law of motion, v12=u12−a12t ⇒0=v−at=22−12.1t ⇒t=2212.1=1.82 s≈2 s  Suggest Corrections  0      Explore more