366209
Assertion : It is harder to open and shut the door if we apply force near the hinge. Reason : Torque is maximum at hinge of the door.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
For rotation \(\tau \neq 0\) is the condition At hinges \(r=0 \Rightarrow \tau=0\) \(\Rightarrow\) Door cannot be rotated if \(F\) is applied at hinges. \(\tau=r F \sin \theta\) For \(\theta=90^{\circ}\) and \(r\) is maximum at handle, \(\Rightarrow\) We have \(\tau=\tau_{\text {max }}\) at handle \(\Rightarrow\) Reason is false The torque required to open or close a door is highest when the force is applied at the handle which is the point farthest from the axis of rotation. So correct option is (3).
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366210
Two blocks each of the mass \(m\) are attached to the ends of a massless rod which pivotes as shown in the figure. Initially the rod is held in the horizontal position and then released. Calculate the net torque on this system above pivot.
1 \(\left(m l_{1} g-m l_{2} g\right) \hat{k}\)
2 \(-\left(m l_{1} g+m l_{2} g\right) \hat{k}\)
3 \(\left( {m{l_2}g + m{l_1}g} \right)\hat k\)
4 \(\left(m l_{2} g-m l_{1} g\right) \hat{k}\)
Explanation:
Torque due to weight of left \(\mathrm{m}\) : \(\vec{\tau}_{1}=m g l_{1} \hat{k}\) Torque due to weight of right \(\mathrm{m}\) : \(\vec{\tau}_{2}=-m g l_{2} \hat{k}\) So net torque: \(\vec{\tau}=\vec{\tau}_{1}+\vec{\tau}_{2}=\left(m g l_{1}-m g l_{2}\right) \hat{k}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366211
If there is a change of angular momentum from \(1\,Js\,\,to\,\,4\,Js\) in \(4\;s\), then the torque applied is
1 \(\left(\dfrac{5}{4}\right) J\)
2 \(\left(\dfrac{3}{4}\right) J\)
3 \(1\;J\)
4 \(\left(\dfrac{4}{3}\right) J\)
Explanation:
The relation between torque \(\tau\) and angular momentum \(L\) is \(\tau=\dfrac{d L}{d t} \approx \dfrac{\Delta L}{\Delta t}\) Here, \(\Delta L = (4 - 1) = 3Js\) and \(\Delta L = (4 - 1) = 3Js\) \(\therefore \tau=\left(\dfrac{3}{4}\right) J\)
MHTCET - 2020
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366212
Three forces \({F, 2 F}\), and \({3 F}\) act on a rod \({A B}\), which is pivoted at \({O}\). The net torque produced by the forces about \({O}\) is
1 \(3\,Fa\)
2 \(4\,Fa\)
3 \(2\,Fa\)
4 \(6\,Fa\)
Explanation:
The forces that act on the rod are shown in the figure. Hence we consider torque about point \({O}\). \({\tau_{F}=-F(2 a), \tau_{2 F}=2(F a)}\) \({\tau_{3 F}=3 F(2 a) \quad(2 a}\)-is a the lever arm \({)}\) The net torque by the three forces about \({O}\) is \({\tau_{o}=\tau_{2 F}+\tau_{F}+\tau_{3 F}=6 F a}\)
366209
Assertion : It is harder to open and shut the door if we apply force near the hinge. Reason : Torque is maximum at hinge of the door.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
For rotation \(\tau \neq 0\) is the condition At hinges \(r=0 \Rightarrow \tau=0\) \(\Rightarrow\) Door cannot be rotated if \(F\) is applied at hinges. \(\tau=r F \sin \theta\) For \(\theta=90^{\circ}\) and \(r\) is maximum at handle, \(\Rightarrow\) We have \(\tau=\tau_{\text {max }}\) at handle \(\Rightarrow\) Reason is false The torque required to open or close a door is highest when the force is applied at the handle which is the point farthest from the axis of rotation. So correct option is (3).
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366210
Two blocks each of the mass \(m\) are attached to the ends of a massless rod which pivotes as shown in the figure. Initially the rod is held in the horizontal position and then released. Calculate the net torque on this system above pivot.
1 \(\left(m l_{1} g-m l_{2} g\right) \hat{k}\)
2 \(-\left(m l_{1} g+m l_{2} g\right) \hat{k}\)
3 \(\left( {m{l_2}g + m{l_1}g} \right)\hat k\)
4 \(\left(m l_{2} g-m l_{1} g\right) \hat{k}\)
Explanation:
Torque due to weight of left \(\mathrm{m}\) : \(\vec{\tau}_{1}=m g l_{1} \hat{k}\) Torque due to weight of right \(\mathrm{m}\) : \(\vec{\tau}_{2}=-m g l_{2} \hat{k}\) So net torque: \(\vec{\tau}=\vec{\tau}_{1}+\vec{\tau}_{2}=\left(m g l_{1}-m g l_{2}\right) \hat{k}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366211
If there is a change of angular momentum from \(1\,Js\,\,to\,\,4\,Js\) in \(4\;s\), then the torque applied is
1 \(\left(\dfrac{5}{4}\right) J\)
2 \(\left(\dfrac{3}{4}\right) J\)
3 \(1\;J\)
4 \(\left(\dfrac{4}{3}\right) J\)
Explanation:
The relation between torque \(\tau\) and angular momentum \(L\) is \(\tau=\dfrac{d L}{d t} \approx \dfrac{\Delta L}{\Delta t}\) Here, \(\Delta L = (4 - 1) = 3Js\) and \(\Delta L = (4 - 1) = 3Js\) \(\therefore \tau=\left(\dfrac{3}{4}\right) J\)
MHTCET - 2020
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366212
Three forces \({F, 2 F}\), and \({3 F}\) act on a rod \({A B}\), which is pivoted at \({O}\). The net torque produced by the forces about \({O}\) is
1 \(3\,Fa\)
2 \(4\,Fa\)
3 \(2\,Fa\)
4 \(6\,Fa\)
Explanation:
The forces that act on the rod are shown in the figure. Hence we consider torque about point \({O}\). \({\tau_{F}=-F(2 a), \tau_{2 F}=2(F a)}\) \({\tau_{3 F}=3 F(2 a) \quad(2 a}\)-is a the lever arm \({)}\) The net torque by the three forces about \({O}\) is \({\tau_{o}=\tau_{2 F}+\tau_{F}+\tau_{3 F}=6 F a}\)
366209
Assertion : It is harder to open and shut the door if we apply force near the hinge. Reason : Torque is maximum at hinge of the door.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
For rotation \(\tau \neq 0\) is the condition At hinges \(r=0 \Rightarrow \tau=0\) \(\Rightarrow\) Door cannot be rotated if \(F\) is applied at hinges. \(\tau=r F \sin \theta\) For \(\theta=90^{\circ}\) and \(r\) is maximum at handle, \(\Rightarrow\) We have \(\tau=\tau_{\text {max }}\) at handle \(\Rightarrow\) Reason is false The torque required to open or close a door is highest when the force is applied at the handle which is the point farthest from the axis of rotation. So correct option is (3).
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366210
Two blocks each of the mass \(m\) are attached to the ends of a massless rod which pivotes as shown in the figure. Initially the rod is held in the horizontal position and then released. Calculate the net torque on this system above pivot.
1 \(\left(m l_{1} g-m l_{2} g\right) \hat{k}\)
2 \(-\left(m l_{1} g+m l_{2} g\right) \hat{k}\)
3 \(\left( {m{l_2}g + m{l_1}g} \right)\hat k\)
4 \(\left(m l_{2} g-m l_{1} g\right) \hat{k}\)
Explanation:
Torque due to weight of left \(\mathrm{m}\) : \(\vec{\tau}_{1}=m g l_{1} \hat{k}\) Torque due to weight of right \(\mathrm{m}\) : \(\vec{\tau}_{2}=-m g l_{2} \hat{k}\) So net torque: \(\vec{\tau}=\vec{\tau}_{1}+\vec{\tau}_{2}=\left(m g l_{1}-m g l_{2}\right) \hat{k}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366211
If there is a change of angular momentum from \(1\,Js\,\,to\,\,4\,Js\) in \(4\;s\), then the torque applied is
1 \(\left(\dfrac{5}{4}\right) J\)
2 \(\left(\dfrac{3}{4}\right) J\)
3 \(1\;J\)
4 \(\left(\dfrac{4}{3}\right) J\)
Explanation:
The relation between torque \(\tau\) and angular momentum \(L\) is \(\tau=\dfrac{d L}{d t} \approx \dfrac{\Delta L}{\Delta t}\) Here, \(\Delta L = (4 - 1) = 3Js\) and \(\Delta L = (4 - 1) = 3Js\) \(\therefore \tau=\left(\dfrac{3}{4}\right) J\)
MHTCET - 2020
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366212
Three forces \({F, 2 F}\), and \({3 F}\) act on a rod \({A B}\), which is pivoted at \({O}\). The net torque produced by the forces about \({O}\) is
1 \(3\,Fa\)
2 \(4\,Fa\)
3 \(2\,Fa\)
4 \(6\,Fa\)
Explanation:
The forces that act on the rod are shown in the figure. Hence we consider torque about point \({O}\). \({\tau_{F}=-F(2 a), \tau_{2 F}=2(F a)}\) \({\tau_{3 F}=3 F(2 a) \quad(2 a}\)-is a the lever arm \({)}\) The net torque by the three forces about \({O}\) is \({\tau_{o}=\tau_{2 F}+\tau_{F}+\tau_{3 F}=6 F a}\)
366209
Assertion : It is harder to open and shut the door if we apply force near the hinge. Reason : Torque is maximum at hinge of the door.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
For rotation \(\tau \neq 0\) is the condition At hinges \(r=0 \Rightarrow \tau=0\) \(\Rightarrow\) Door cannot be rotated if \(F\) is applied at hinges. \(\tau=r F \sin \theta\) For \(\theta=90^{\circ}\) and \(r\) is maximum at handle, \(\Rightarrow\) We have \(\tau=\tau_{\text {max }}\) at handle \(\Rightarrow\) Reason is false The torque required to open or close a door is highest when the force is applied at the handle which is the point farthest from the axis of rotation. So correct option is (3).
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366210
Two blocks each of the mass \(m\) are attached to the ends of a massless rod which pivotes as shown in the figure. Initially the rod is held in the horizontal position and then released. Calculate the net torque on this system above pivot.
1 \(\left(m l_{1} g-m l_{2} g\right) \hat{k}\)
2 \(-\left(m l_{1} g+m l_{2} g\right) \hat{k}\)
3 \(\left( {m{l_2}g + m{l_1}g} \right)\hat k\)
4 \(\left(m l_{2} g-m l_{1} g\right) \hat{k}\)
Explanation:
Torque due to weight of left \(\mathrm{m}\) : \(\vec{\tau}_{1}=m g l_{1} \hat{k}\) Torque due to weight of right \(\mathrm{m}\) : \(\vec{\tau}_{2}=-m g l_{2} \hat{k}\) So net torque: \(\vec{\tau}=\vec{\tau}_{1}+\vec{\tau}_{2}=\left(m g l_{1}-m g l_{2}\right) \hat{k}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366211
If there is a change of angular momentum from \(1\,Js\,\,to\,\,4\,Js\) in \(4\;s\), then the torque applied is
1 \(\left(\dfrac{5}{4}\right) J\)
2 \(\left(\dfrac{3}{4}\right) J\)
3 \(1\;J\)
4 \(\left(\dfrac{4}{3}\right) J\)
Explanation:
The relation between torque \(\tau\) and angular momentum \(L\) is \(\tau=\dfrac{d L}{d t} \approx \dfrac{\Delta L}{\Delta t}\) Here, \(\Delta L = (4 - 1) = 3Js\) and \(\Delta L = (4 - 1) = 3Js\) \(\therefore \tau=\left(\dfrac{3}{4}\right) J\)
MHTCET - 2020
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366212
Three forces \({F, 2 F}\), and \({3 F}\) act on a rod \({A B}\), which is pivoted at \({O}\). The net torque produced by the forces about \({O}\) is
1 \(3\,Fa\)
2 \(4\,Fa\)
3 \(2\,Fa\)
4 \(6\,Fa\)
Explanation:
The forces that act on the rod are shown in the figure. Hence we consider torque about point \({O}\). \({\tau_{F}=-F(2 a), \tau_{2 F}=2(F a)}\) \({\tau_{3 F}=3 F(2 a) \quad(2 a}\)-is a the lever arm \({)}\) The net torque by the three forces about \({O}\) is \({\tau_{o}=\tau_{2 F}+\tau_{F}+\tau_{3 F}=6 F a}\)
