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In the diagram shown in figure, match the following columns (Take, g = 10 ms^-2)
Column I
(A) Normal reaction
(8) Force of friction
(C) Acceleration of block
Column II
(p) 12 SI unit
(q) 20 SI unit
(r ) zero
(s) 2 SI unit

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A B C

p q r

A B C

q p r

A B C

q p s

A B C

p s r

answer is C.

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Detailed Solution

$N = m g - 20 \sqrt{2} \sin 45 ° = 20 N, f = μ_{k} N = 12 N$

Since, $20 \sqrt{2} \cos 45 ° > μ N$ block will move and its acceleration will be

$a = \frac{20 \sqrt{2} \cos 45 ° - μ_{k} N}{m} = \frac{20 - 12}{4} = 2 {ms}^{- 2}$

Hence, $A \to q, B \to p, C \to s$ .

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In the diagram shown in figure, match the following columns (Take, g = 10 ms-2)Column I(A) Normal reaction(8) Force of friction(C) Acceleration of blockColumn II(p) 12 SI unit(q) 20 SI unit(r ) zero(s) 2 SI unit

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In the diagram shown in figure, match the following columns (Take, g = 10 ms^-2)
Column I
(A) Normal reaction
(8) Force of friction
(C) Acceleration of block
Column II
(p) 12 SI unit
(q) 20 SI unit
(r ) zero
(s) 2 SI unit