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Figure shows a weight of 30 kg suspended at one end of cord and a weight of 70 kg applied at other end of the cord passing over a pulley. Neglecting weight of rope and pulley find the tension in the cord and acceleration of the system $(g = 10 {ms}^{- 2})$

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$2 m / s^{2}; 120 N$

$4 m / s^{2}; 420 N$

$6 m / s^{2}; 180 N$

$4 m / s^{2}; 240 N$

answer is B.

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Detailed Solution

$a = (\frac{M_{2} - M_{1}}{M_{2} + M_{1}}) g$

$= (\frac{70 - 30}{70 + 30}) 10$

$= \frac{40}{100} \times 10$

$a = 4 m / s^{2}$

$T = (\frac{2 M_{1} M_{2}}{M_{1} + M_{2}}) g$

$= (\frac{2 \times 70 \times 30}{70 + 30}) 10$

$= (\frac{4200}{100}) 10$

$T = 420 N$

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