A mass of $1 k g$ is suspended by a thread. It is first lifted up with an acceleration $4.9 m / s^{2}$ ,and then lowered with an acceleration $4.9 m / s^{2}$ . What is the ratio of tension in the two cases?

3 : 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1 : 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 : 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 : 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 3 : 1
Given,

$a_{1} = a_{2} = 4.9 m / s^{2}$

$g = 9.8 m / s^{2}$

$m = 1 k g$

Case (1):- From free body diagram,

$T_{1} - m g = m a_{1}$

$T_{1} = m (a + g)$

$T_{1} = 1 (4.9 + 9.8)$

$T_{1} = 17.47 N$

Case (2):- From free body diagram,

$m g - T_{2} = m a_{2}$

$T_{2} = m (g - a)$

$T_{2} = 1 (9.8 - 4.9)$

$T_{2} = 4.9 N$

Ratio, $\frac{T_{1}}{T_{2}} = \frac{3}{1}$

$T_{1} : T_{2} = 3 : 1$

The correct option is A.

Suggest Corrections

Similar questions

1 k g

4.9 {m s}^{- 1}

4.9 {m s}^{- 2}

4.9

m / s^{2}

4.9

m / s^{2}

m s^{2}

m s^{- 2}

500 k g

2.0 m / s^{2}

Join BYJU'S Learning Program