Question

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

$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{i. Force of friction is zero in}& \text{a. Fig. (i)}\\ \text{ii. Force of friction is}2.5\text{N in}& \text{b. Fig. (ii)}\\ \text{iii. Acceleration of the block is zero in}& \text{c. Fig. (iiii)}\\ \text{iv. Normal force is not equal to}2\text{g in}& \text{d. Fig. (iv)}\end{array}$

• A. $i-b,ii-a,iii-a,b,c,iv-c$
• B. $i-b,ii-c,d,iii-d,iv-a,d$
• C. $i-a,c,ii-b,d,iii-a,b,c,d,iv-c,d$
• D. $i-d,ii-a,b,iii-d,iv-c$

Answer 1

The correct option is C $i-a,c,ii-b,d,iii-a,b,c,d,iv-c,d$

Given $\mu =0.4,\text{Limiting friction}{F}_L=\mu N$
For Fig $i\&ii$,
$N=2g$
$\Rightarrow F_L=0.4\left(20\right)=8\text{N}$
For Fig $iii$
$N=25F_L=0.4\left(25\right)=10\text{N}$
For Fig $iv$
$N=2g-5\cos {60}^{\circ }=20-4.33=15.67\text{N}$
$F_L=0.4\left(15.67\right)=6.3\text{N}$

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