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Three blocks S, R and P connected by string are arranged as shown in the figure. The masses of S, R and P are 5 kg, 5 kg and 10 kg respectively. The tension in the string connecting R and P is (ignore friction)

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20 N

30 N

25 N

10 N

answer is B.

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Detailed Solution

$m_{p} g - T = m_{p} a \dots . (1)$

$T + m_{R} g - T^{1} = m_{R} a \dots . (2)$

$T = m_{s} a$

$a = \frac{(m_{R} + m_{P}) g}{m_{P} + m_{R} + m_{S}}$

$= \frac{5 g + 10 g}{20} = \frac{15 \times 10}{20} = 7.5 m / s^{2}$

$m_{p} g - T = m_{p} a$

100 - T = 10(7.5)
T = 100 - 75
= 25N

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