Question

Match statements in column $I$ with results in column $II$ for the figure shown.$(g=10{\text{m/s}}^2)$

$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{(a)}& \text{Acceleration of}2\text{kg block}& \text{(p)}& 8\text{SI unit}\\ \text{(b)}& \text{Net force on}3\text{kg block}& \text{(q)}& 25\text{SI unit}\\ \text{(c)}& \text{Normal reaction between}& \text{(r)}& 2\text{SI unit}\\ & 2\text{kg and}1\text{kg}& & \\ \text{(d)}& \text{Normal reaction between}& \text{(s)}& 45\text{SI unit}\\ & 3\text{kg and}2\text{kg}& & \\ & & \text{(t)}& \text{None}\end{array}$

• A. a - s, b - r, c - q, d - p
• B. a - r, b - t, c - q, d - t
• C. a - s, b - p, c - q, d - r
• D. a - r, b - q, c - r, d - p

Answer 1

The correct option is B a - r, b - t, c - q, d - t

(a) → (r); (b) → (t); (c) → (q); (d) → (t)



Let $a$ be acceleration of masses together as a system along the inclined plane
$42-mg\sin {30}^{\circ }=ma$
$\Rightarrow -6\times 10\times \frac{1}{2}+42=6a$
$\Rightarrow 12=6a$
$\therefore a=2{\text{m/s}}^2$

Net force on $3kg$ mass is $3a=6N$

From FBD of $1kg$ mass
$R-18-g\sin {30}^{\circ }=a$
$R-18-5=a$
$R-23=2$
$R=25\text{N}$

From FBD of $3kg$ mass
$60-R^{\prime }-3g\sin {30}^{\circ }=3a$
$60-R^{\prime }-15=6$
$45-R^{\prime }=6$
$R^{\prime }=39\text{N}$