Consider the situation shown in fig.
Column –I | Column-II |
(i) The ratio of acceleration of block of 1 kg to that of 4 kg is | [A] 0.5 |
(ii) The ratio of velocity of 1 kg to that of 4 kg is | [B] 1 |
(iii) The coefficient of kinetic friction between the block and table is. For 1 kg block having a speed of 0.3 ms–1 after descending 1m | [C] 0.6 |
(iv) The velocity of 4 kg block at this instant mentioned in (C) is ms–1 | [D] 0.12 |
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i)-[A], (ii)-[A], (iii)-[D], (iv)-[C]
Sol.
2a A = a B ⇒ 
= 
= 0.5
2V A = V B ⇒ 
= 
= 0.5
m A g x A = 
× 1 × (0.3) 2 + 
× 4 × (0.6) 2 + μ B m B g x B
μ = 0.12
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