In the diagram shown in figure- match the following- g -10 m - s 2a - text Acceeration of 2 kg block - p - t e x t 8 SI unit hlinec - ltext Normal reaction between 2 kg and 1 kg - r - text 2 SI unit hlined - 1 text Normal reaction between 3 kg and 2 kg - s - text45 N hline -text t - -endarray A- a - r - b - t - c - q - d - p B- a - r - b - t - c - t - d - q C- a - p- b - t- c - q- d - t

The correct option is D (a) – (r); (b) – (t); (c) – (q); (d) – (t)(a) a = 42N 6 × g × sin 30^∘/6 = 2ms^ 2 (b) F_net on 3kg = 3 × a = 6N (c) R_1 1 × g sin ...

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Byju's Answer

Standard XII

Physics

Second Law of Motion

In the diagra...

Question

In the diagram shown in figure, match the following: $(g = 10 m / s^{2})$

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

(a) – (r); (b) – (t); (c) – (q); (d) – (p)

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

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(a) – (p); (b) – (t); (c) – (q); (d) – (t)

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

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Solution

The correct option is D (a) – (r); (b) – (t); (c) – (q); (d) – (t)
$(a) a = \frac{42 N - 6 \times g \times s i n 30^{\circ}}{6} = 2 m s^{- 2}$

(b) $F_{n e t}$ on 3kg = $3 \times a = 6 N$

(c) $R_{1} - 1 \times g s i n 30^{\circ} - 18 = 1 \times a$

$R_{1} = 18 + 5 + 2 = 25 N$

(d) $60 N - R_{2} - 3 \times g s i n 30^{\circ} = 3 \times 2$

$R_{2} = 60 - 15 - 6 = 39 N$

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In the diagram shown in figure, match the following: (g=10m/s2) Column -IColumn-II(a)Acceeration of 2 kg block(p)8 SI unit(b)Net force on 3 kg block(q)25 SI unit(c)Normal reaction between 2 kg and 1 kg(r)2 SI unit(d)Normal reaction between 3 kg and 2 kg(s)45 N (t)None

The correct option is D (a) – (r); (b) – (t); (c) – (q); (d) – (t)(a) a=42N−6×g×sin30∘6=2ms−2 (b) Fnet on 3kg = 3×a=6N (c) R1−1× g sin 30∘−18=1×a R1=18+5+2=25N (d) 60N−R2−3×gsin30∘=3×2 R2=60−15−6=39N