Question

Two mass m and 2m are attached with each other by a rope passing over a frictionless and massless pulley. If the pulley is accelerated upwards with an acceleration 'a', what is the value of T ?

• A. $\frac{g + a}{3}$
• B. $\frac{g–a}{3}$
• C. $\frac{4m(g + a)}{3}$
• D. $\frac{m(g–a)}{3}$

Answer 1

Answer: (C)

$\frac{4m(g + a)}{3}$

Answer 2

w.r.t. pulley

m :

T – (mg + ma) = ma' .....(i)

2m : (2 mg + 2 ma) – T = 2ma' .......(ii)

From (i) & (ii)

2T – 2 (mg + ma) = (2 mg + 2 ma) – T

3T = 4 (mg + ma) T = $\frac{4m}{3}$ (g + a)

Short trick

T = $\frac{2m_{1}m_{2}}{m_{1} + m_{2}}$ (g + a ) = $\frac{2(m)(2m)}{m + 2m}$ (g + a)

= $\frac{4m}{3}$ (g + a )