Two blocks masses '3m' and '2m' are in contact on a smooth table. A force P is first applied horizontally on block of mass '3m' and then on mass '2m'. The contact force between the two block sin the two cases are in the ratio.

1 : 2

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

2 : 3

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

3 : 2

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

5 : 3

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

Open in App

Solution

The correct option is B 2 : 3
$f_{1} = (2 m) (a c c .)$

$= 2 m \times \frac{P}{5 m} = \frac{2}{5} P$

$f_{2} = (3 m) (a c c .)$
$= 3 m \times \frac{P}{5 m} = \frac{3}{5} P$
$\frac{f_{1}}{f_{2}} = \frac{2 P / 5}{3 P / 5} = \frac{2}{3}$

Suggest Corrections

Similar questions

Q. Two blocks are in contact on a frictionless table. One has a mass

m

and the other

2 m

. A force

F

is applied on

m

.
In the two cases respectively the ratio of force of contact between the two blocks will be:

Two blocks are in contact on a frictionless table one has a mass m and the other 2 m as shown in figure. Force F is applied on mass 2m then system moves towards right. Now the same force F is applied on m. The ratio of force of contact between the two blocks will be in the two cases respectively.

Two blocks are in contact on a frictionless table masses $2 m$ and $m$ respectively as shown in figure. Force F is applied on mass $2 m$ then system moves towards right. Now the same force F is applied on $m$ . The ratio of force of contact between the two blocks will be in the two cases respectively.

Q. Two blocks of masses 2kg and 1 kg are placed on a smooth horizontal table i contact with each other.A horizontal force of 3 newton is applied on the first so that the block moves with a constant acceleration. The force between the blocks would be:-

Q. Two blocks of masses

2 k g

and

1 k g

are in contact with each other on a frictionless table, when a horizontal force of

3.0 N

is applied to the block of mass

2 k g

the value of the force of contact between the two blocks is:

Join BYJU'S Learning Program

Two blocks masses '3m' and '2m' are in contact on a smooth table. A force P is first applied horizontally on block of mass '3m' and then on mass '2m'. The contact force between the two block sin the two cases are in the ratio.

The correct option is B 2 : 3f1=(2m)(acc.) =2m×P5m=25P f2=(3m)(acc.) =3m×P5m=35P f1f2=2P/53P/5=23

The correct option is B 2 : 3
$f_{1} = (2 m) (a c c .)$

$= 2 m \times \frac{P}{5 m} = \frac{2}{5} P$

$f_{2} = (3 m) (a c c .)$
$= 3 m \times \frac{P}{5 m} = \frac{3}{5} P$
$\frac{f_{1}}{f_{2}} = \frac{2 P / 5}{3 P / 5} = \frac{2}{3}$