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CASE – I: A, B, C are three boys climbing up/down on a light ladder supported by a light string. $m_{A} = 20 k g, m_{B} = 30 k g, m_{C} = 40 k g .$ A is going up with $1.3 m s^{- 2}$ and B with $7.5 m s^{- 2}$ down.

CASE – II: 1 & 2 are two blocks of equal mass kept on a rough cubical wedge having coefficient of friction $(μ)$ and connected by an ideal pulley and ideal string as shown. Wedge is accelerated to right so that the blocks are NOT slipping on it. $μ = 0.3$ between the blocks and wedge and mass of each block is 5kg. $(g = 9.8 m / s^{2})$

All the quantities in Column – II are in S.I units. Match the following two columns.
Column-I Column-II
a) Numerical value of acceleration of “C” so that the tension in the supporting string is zero in case I. P) 18.2
b) Acceleration of C.M of system for the condition in case (a). Q) 5.8
c) Maximum acceleration of wedge so that blocks don’t slip in case-II. R) 9.8
d) Tension in the string in multiples of 10N in case-(c). S) 7.6
T) 17.1

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$a - P, b - R, c - T, d - Q$

$a - T, b - Q, c - R, d - P$

$a - T, b - R, c - P, d - S$

$a - P, b - R, c - T, d - S$

answer is C.

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Detailed Solution

$\begin{array}{l} a) F o r T = 0 \\ \bar{F_{A}} + \bar{F_{B}} + \bar{F_{C}} = \bar{0} \\ \Rightarrow 20 (9.8 + 1.3) + 30 (9.8 - 7.5) + 40 (9.8 + a_{C}) = 0 \\ \Rightarrow 222 + 69 + 40 (9.8 + a_{C}) = 0 \\ a_{C} = - 17.1 \\ b) a_{C M} = g a s F_{e x t} = Weight only \\ c) m a_{0} = μ m g + T \\ T = m g + μ m a_{0} s o l v i n g a_{0} = \frac{g (1 + μ)}{1 - μ} \\ d) T = m [g + μ a_{0}] \end{array}$

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