A mass of 1 kg is suspended by a thread. It is(i) Lifted up with an acceleration 4.9 m/s2.(ii) Lowered with an acceleration 4.9 mls2.The ratio of the tensions is(a) 3: 1(b) 1: 2(c) 1: 3(d) 2: 1

(a)

In case (i) we have

T₁ -(1 × g) = 1 × 4.9

Þ T₁ = 9.8 + 4.9 = 14.7N

In case (ii), 1 × g - T₂ = 1 × 4.9

Þ T₂ = 9.8 - 4.9 = 4.9N

          T_1/T_2   =14.7/4.9  =3/1

Dear Student,
Please find below the solution to your problem.

Given,
a1​=a2​=4.9m/s2
g=9.8m/s2
m=1kg
Case (1):- From free body diagram,
T1​−mg=ma1​
T1​=m(a+g)
T1​=1(4.9+9.8)
T1​=17.47N
Case (2):- From free body diagram,
mg−T2​=ma2​
T2​=m(g−a)
T2​=1(9.8−4.9)
T2​=4.9N
Ratio, T2​T1​​=13​
T1​:T2​=3:1
The correct option is A

Thanks and Regards