269441
The handle of a door is at a distance\(40 \mathrm{~cm}\) from axis of rotation. If a force \(5 \mathrm{~N}\) is applied on the handle in a direction \(30^{\circ}\) with plane of door, then the torque is
1 \(0.8 \mathrm{Nm}\)
2 \(1 \mathrm{Nm}\)
3 \(1.6 \mathrm{Nm}\)
4 \(2 \mathrm{Nm}\)
Explanation:
\(\tau=r F \sin \theta\)
Rotational Motion
269442
A door can just be opened with\(10 \mathrm{~N}\) force on the handle of the door. The handle is at a distance of \(50 \mathrm{~cm}\) from the hinges. Then, the torque applied on the door (in \(\mathrm{Nm}\) ) is
1 5
2 10
3 15
4 20
Explanation:
\(\tau=r F\)
Rotational Motion
269443
A particle of mass \(m\) is projected with an initial velocity \(u\) at an angle \(\theta\) tohorizontal.The torque of gravity on projectile at maximum height about the point of projection is
1 \(\frac{m g u^{2} \sin 2 \theta}{2}\)
2 \(m g u^{2} \sin 2 \theta\)
3 \(\frac{m g u^{2} \sin \theta}{2}\)
4 \(\frac{1}{2} m u^{2} \sin 2 \theta\)
Explanation:
\(\vec{\tau}=\square_{\square} 2\) in \(\hat{i}+H \hat{j} \square \times m g \hat{j}\)
Rotational Motion
269444
A uniform rod is\(4 \mathrm{~m}\) long and weights \(10 \mathrm{~kg}\). If it is supported on a knife edge at one meter from the end, what weight placed at that end keeps the rod horizontal.
269441
The handle of a door is at a distance\(40 \mathrm{~cm}\) from axis of rotation. If a force \(5 \mathrm{~N}\) is applied on the handle in a direction \(30^{\circ}\) with plane of door, then the torque is
1 \(0.8 \mathrm{Nm}\)
2 \(1 \mathrm{Nm}\)
3 \(1.6 \mathrm{Nm}\)
4 \(2 \mathrm{Nm}\)
Explanation:
\(\tau=r F \sin \theta\)
Rotational Motion
269442
A door can just be opened with\(10 \mathrm{~N}\) force on the handle of the door. The handle is at a distance of \(50 \mathrm{~cm}\) from the hinges. Then, the torque applied on the door (in \(\mathrm{Nm}\) ) is
1 5
2 10
3 15
4 20
Explanation:
\(\tau=r F\)
Rotational Motion
269443
A particle of mass \(m\) is projected with an initial velocity \(u\) at an angle \(\theta\) tohorizontal.The torque of gravity on projectile at maximum height about the point of projection is
1 \(\frac{m g u^{2} \sin 2 \theta}{2}\)
2 \(m g u^{2} \sin 2 \theta\)
3 \(\frac{m g u^{2} \sin \theta}{2}\)
4 \(\frac{1}{2} m u^{2} \sin 2 \theta\)
Explanation:
\(\vec{\tau}=\square_{\square} 2\) in \(\hat{i}+H \hat{j} \square \times m g \hat{j}\)
Rotational Motion
269444
A uniform rod is\(4 \mathrm{~m}\) long and weights \(10 \mathrm{~kg}\). If it is supported on a knife edge at one meter from the end, what weight placed at that end keeps the rod horizontal.
269441
The handle of a door is at a distance\(40 \mathrm{~cm}\) from axis of rotation. If a force \(5 \mathrm{~N}\) is applied on the handle in a direction \(30^{\circ}\) with plane of door, then the torque is
1 \(0.8 \mathrm{Nm}\)
2 \(1 \mathrm{Nm}\)
3 \(1.6 \mathrm{Nm}\)
4 \(2 \mathrm{Nm}\)
Explanation:
\(\tau=r F \sin \theta\)
Rotational Motion
269442
A door can just be opened with\(10 \mathrm{~N}\) force on the handle of the door. The handle is at a distance of \(50 \mathrm{~cm}\) from the hinges. Then, the torque applied on the door (in \(\mathrm{Nm}\) ) is
1 5
2 10
3 15
4 20
Explanation:
\(\tau=r F\)
Rotational Motion
269443
A particle of mass \(m\) is projected with an initial velocity \(u\) at an angle \(\theta\) tohorizontal.The torque of gravity on projectile at maximum height about the point of projection is
1 \(\frac{m g u^{2} \sin 2 \theta}{2}\)
2 \(m g u^{2} \sin 2 \theta\)
3 \(\frac{m g u^{2} \sin \theta}{2}\)
4 \(\frac{1}{2} m u^{2} \sin 2 \theta\)
Explanation:
\(\vec{\tau}=\square_{\square} 2\) in \(\hat{i}+H \hat{j} \square \times m g \hat{j}\)
Rotational Motion
269444
A uniform rod is\(4 \mathrm{~m}\) long and weights \(10 \mathrm{~kg}\). If it is supported on a knife edge at one meter from the end, what weight placed at that end keeps the rod horizontal.
269441
The handle of a door is at a distance\(40 \mathrm{~cm}\) from axis of rotation. If a force \(5 \mathrm{~N}\) is applied on the handle in a direction \(30^{\circ}\) with plane of door, then the torque is
1 \(0.8 \mathrm{Nm}\)
2 \(1 \mathrm{Nm}\)
3 \(1.6 \mathrm{Nm}\)
4 \(2 \mathrm{Nm}\)
Explanation:
\(\tau=r F \sin \theta\)
Rotational Motion
269442
A door can just be opened with\(10 \mathrm{~N}\) force on the handle of the door. The handle is at a distance of \(50 \mathrm{~cm}\) from the hinges. Then, the torque applied on the door (in \(\mathrm{Nm}\) ) is
1 5
2 10
3 15
4 20
Explanation:
\(\tau=r F\)
Rotational Motion
269443
A particle of mass \(m\) is projected with an initial velocity \(u\) at an angle \(\theta\) tohorizontal.The torque of gravity on projectile at maximum height about the point of projection is
1 \(\frac{m g u^{2} \sin 2 \theta}{2}\)
2 \(m g u^{2} \sin 2 \theta\)
3 \(\frac{m g u^{2} \sin \theta}{2}\)
4 \(\frac{1}{2} m u^{2} \sin 2 \theta\)
Explanation:
\(\vec{\tau}=\square_{\square} 2\) in \(\hat{i}+H \hat{j} \square \times m g \hat{j}\)
Rotational Motion
269444
A uniform rod is\(4 \mathrm{~m}\) long and weights \(10 \mathrm{~kg}\). If it is supported on a knife edge at one meter from the end, what weight placed at that end keeps the rod horizontal.
