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Question

The coefficient of friction between the block and the surface is 0.4 in Fig. (i-iv). Match Column I with Column II.

Column IColumn IIi. Force of friction is zero ina. Fig. (i)ii. Force of friction is 2.5 N inb. Fig. (ii)iii. Acceleration of the block is zero inc. Fig. (iiii)iv. Normal force is not equal to 2g in d. Fig. (iv)

A
ib,iia,iiia,b,c,ivc
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B
ib,iic,d,iiid,iva,d
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C
ia,c,iib,d,iiia,b,c,d,ivc,d
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D
id,iia,b,iiid,ivc
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Solution

The correct option is C ia,c,iib,d,iiia,b,c,d,ivc,d
Given μ=0.4,Limiting friction FL=μN
For Fig i & ii,
N=2g
FL=0.4(20)=8 N
For Fig iii
N=25FL=0.4(25)=10 N
For Fig iv
N=2g5cos60=204.33=15.67 N
FL=0.4(15.67)=6.3 N

(i) Force of friction is zero in (a) and (c) because the block has no tendency to move.
(ii) Force of friction is 2.5 N in (b) and (d) because the applied force in horizontal direction in both cases is 2.5 N.
(iii) Acceleration is zero in all the cases, as applied force is less than limiting friction.
(iv) Normal force is not equal to 2 g in (c) and (d) because some extra vertical force is also acting.

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