The handle of a door is at a distance $40 c m$ from axis of rotation. If a force $5 N$ is applied on the handle in a direction $30^{0}$ with plane of door, then the torque is:

0.8 N m

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

1 N m

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

1.6 N m

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

2 N m

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

Open in App

Solution

The correct option is A $1 N m$
We know that the torque is given using the relation
$τ = F R s i n θ$
Or
$τ = (5 N) (0.4 m) (s i n 30^{o}) = 1 N m$

Suggest Corrections

Similar questions

Q. A door can just be opened with

10 N

force on the handle of the door. The handle is at a distance of

50 c m

from the hinges. Then the torque applied on the door is ( in

N m

When a force is applied normally on the handle of a door, the door begins to rotate about an axis passing through the hinges on which the door resets. This is an example of

If a force is applied at the hinge of a door instead of the handle, then

Q. Force applied normally on the handle of a door is an example of:

What is the moment of force developed about the hinge of a door if a person tries to open it with a force of 100 N, the force being applied normal to the door at handle which is at a distance of 60 cm from the hinge?

Join BYJU'S Learning Program

The handle of a door is at a distance 40cm from axis of rotation. If a force 5N is applied on the handle in a direction 300 with plane of door, then the torque is:

The correct option is A 1NmWe know that the torque is given using the relation τ=FR sin θ Orτ=(5N)(0.4m)(sin30o)=1Nm

The handle of a door is at a distance $40 c m$ from axis of rotation. If a force $5 N$ is applied on the handle in a direction $30^{0}$ with plane of door, then the torque is:

The correct option is A $1 N m$
We know that the torque is given using the relation
$τ = F R s i n θ$
Or
$τ = (5 N) (0.4 m) (s i n 30^{o}) = 1 N m$