366196
Let \(\vec{F}\) be the force acting on a particle having position vector \(\vec{r}\) and \(\vec{T}\) be the torque of this force about the origin.
1 \(\vec r \cdot \vec T = 0\,\,and\,\,\vec F \cdot \vec T = 0\)
2 \(\vec r \cdot \vec T \ne 0\,\,and\,\,\vec F \cdot \vec T = 0\)
3 \(\vec r \cdot \vec T = 0\,\,and\,\,\vec F \cdot \vec T \ne 0\)
4 \(\vec r \cdot \vec T \ne 0\,\,and\,\,\vec F \cdot \vec T \ne 0\)
Explanation:
\(\vec{T}=\vec{r} \times \vec{F}\) \(\vec{T}\) is \(\perp\) to both \(\vec{r}\) and \(\vec{F}\) so \(\vec{r} \cdot \vec{T}=0\) and \(\vec{F} \cdot \vec{T}=0\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366197
If the external torque acting on a system \(\vec{\tau}=0\), then
1 \(\alpha=0\)
2 \(F=0\)
3 \(\omega=0\)
4 \(J=0\)
Explanation:
\(\tau=I \alpha\), if \(\tau=0\) then \(\alpha=0\) (since \(I \neq 0\) )
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366198
A rod of \(l\;m\) long is acted upon by a couple as shown in figure. The moment of couple is \(\tau Nm\). If the force at each end of the rod is \(F\), then magnitude of each force is \(\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5\right)\)
1 \(\dfrac{3 l}{\tau}\)
2 \(\dfrac{l}{\tau}\)
3 \(\dfrac{2 \tau}{l}\)
4 \(\dfrac{2 l}{\tau}\)
Explanation:
Moment of couple, \(\tau=F l \sin \theta\) \(\therefore\) Force acting on each end \(F=\dfrac{\tau}{l \sin \theta}=\dfrac{\tau}{l \sin 30^{\circ}}=\dfrac{\tau}{l \times \dfrac{1}{2}}=\dfrac{2 \tau}{l}\)
MHTCET - 2019
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366199
Turning effect is produced by
1 Radial component of force
2 Transverse component of force
3 Both radial & transverse components of force
4 None of the above
Explanation:
Consider a force \(F\) that acts at point \(P\). The torque produced by the force about \(O\) is \(\tau=r F \sin \phi\) \(F \sin \phi\) is the transverse component of force correct option is (2)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366196
Let \(\vec{F}\) be the force acting on a particle having position vector \(\vec{r}\) and \(\vec{T}\) be the torque of this force about the origin.
1 \(\vec r \cdot \vec T = 0\,\,and\,\,\vec F \cdot \vec T = 0\)
2 \(\vec r \cdot \vec T \ne 0\,\,and\,\,\vec F \cdot \vec T = 0\)
3 \(\vec r \cdot \vec T = 0\,\,and\,\,\vec F \cdot \vec T \ne 0\)
4 \(\vec r \cdot \vec T \ne 0\,\,and\,\,\vec F \cdot \vec T \ne 0\)
Explanation:
\(\vec{T}=\vec{r} \times \vec{F}\) \(\vec{T}\) is \(\perp\) to both \(\vec{r}\) and \(\vec{F}\) so \(\vec{r} \cdot \vec{T}=0\) and \(\vec{F} \cdot \vec{T}=0\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366197
If the external torque acting on a system \(\vec{\tau}=0\), then
1 \(\alpha=0\)
2 \(F=0\)
3 \(\omega=0\)
4 \(J=0\)
Explanation:
\(\tau=I \alpha\), if \(\tau=0\) then \(\alpha=0\) (since \(I \neq 0\) )
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366198
A rod of \(l\;m\) long is acted upon by a couple as shown in figure. The moment of couple is \(\tau Nm\). If the force at each end of the rod is \(F\), then magnitude of each force is \(\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5\right)\)
1 \(\dfrac{3 l}{\tau}\)
2 \(\dfrac{l}{\tau}\)
3 \(\dfrac{2 \tau}{l}\)
4 \(\dfrac{2 l}{\tau}\)
Explanation:
Moment of couple, \(\tau=F l \sin \theta\) \(\therefore\) Force acting on each end \(F=\dfrac{\tau}{l \sin \theta}=\dfrac{\tau}{l \sin 30^{\circ}}=\dfrac{\tau}{l \times \dfrac{1}{2}}=\dfrac{2 \tau}{l}\)
MHTCET - 2019
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366199
Turning effect is produced by
1 Radial component of force
2 Transverse component of force
3 Both radial & transverse components of force
4 None of the above
Explanation:
Consider a force \(F\) that acts at point \(P\). The torque produced by the force about \(O\) is \(\tau=r F \sin \phi\) \(F \sin \phi\) is the transverse component of force correct option is (2)
366196
Let \(\vec{F}\) be the force acting on a particle having position vector \(\vec{r}\) and \(\vec{T}\) be the torque of this force about the origin.
1 \(\vec r \cdot \vec T = 0\,\,and\,\,\vec F \cdot \vec T = 0\)
2 \(\vec r \cdot \vec T \ne 0\,\,and\,\,\vec F \cdot \vec T = 0\)
3 \(\vec r \cdot \vec T = 0\,\,and\,\,\vec F \cdot \vec T \ne 0\)
4 \(\vec r \cdot \vec T \ne 0\,\,and\,\,\vec F \cdot \vec T \ne 0\)
Explanation:
\(\vec{T}=\vec{r} \times \vec{F}\) \(\vec{T}\) is \(\perp\) to both \(\vec{r}\) and \(\vec{F}\) so \(\vec{r} \cdot \vec{T}=0\) and \(\vec{F} \cdot \vec{T}=0\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366197
If the external torque acting on a system \(\vec{\tau}=0\), then
1 \(\alpha=0\)
2 \(F=0\)
3 \(\omega=0\)
4 \(J=0\)
Explanation:
\(\tau=I \alpha\), if \(\tau=0\) then \(\alpha=0\) (since \(I \neq 0\) )
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366198
A rod of \(l\;m\) long is acted upon by a couple as shown in figure. The moment of couple is \(\tau Nm\). If the force at each end of the rod is \(F\), then magnitude of each force is \(\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5\right)\)
1 \(\dfrac{3 l}{\tau}\)
2 \(\dfrac{l}{\tau}\)
3 \(\dfrac{2 \tau}{l}\)
4 \(\dfrac{2 l}{\tau}\)
Explanation:
Moment of couple, \(\tau=F l \sin \theta\) \(\therefore\) Force acting on each end \(F=\dfrac{\tau}{l \sin \theta}=\dfrac{\tau}{l \sin 30^{\circ}}=\dfrac{\tau}{l \times \dfrac{1}{2}}=\dfrac{2 \tau}{l}\)
MHTCET - 2019
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366199
Turning effect is produced by
1 Radial component of force
2 Transverse component of force
3 Both radial & transverse components of force
4 None of the above
Explanation:
Consider a force \(F\) that acts at point \(P\). The torque produced by the force about \(O\) is \(\tau=r F \sin \phi\) \(F \sin \phi\) is the transverse component of force correct option is (2)
366196
Let \(\vec{F}\) be the force acting on a particle having position vector \(\vec{r}\) and \(\vec{T}\) be the torque of this force about the origin.
1 \(\vec r \cdot \vec T = 0\,\,and\,\,\vec F \cdot \vec T = 0\)
2 \(\vec r \cdot \vec T \ne 0\,\,and\,\,\vec F \cdot \vec T = 0\)
3 \(\vec r \cdot \vec T = 0\,\,and\,\,\vec F \cdot \vec T \ne 0\)
4 \(\vec r \cdot \vec T \ne 0\,\,and\,\,\vec F \cdot \vec T \ne 0\)
Explanation:
\(\vec{T}=\vec{r} \times \vec{F}\) \(\vec{T}\) is \(\perp\) to both \(\vec{r}\) and \(\vec{F}\) so \(\vec{r} \cdot \vec{T}=0\) and \(\vec{F} \cdot \vec{T}=0\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366197
If the external torque acting on a system \(\vec{\tau}=0\), then
1 \(\alpha=0\)
2 \(F=0\)
3 \(\omega=0\)
4 \(J=0\)
Explanation:
\(\tau=I \alpha\), if \(\tau=0\) then \(\alpha=0\) (since \(I \neq 0\) )
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366198
A rod of \(l\;m\) long is acted upon by a couple as shown in figure. The moment of couple is \(\tau Nm\). If the force at each end of the rod is \(F\), then magnitude of each force is \(\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5\right)\)
1 \(\dfrac{3 l}{\tau}\)
2 \(\dfrac{l}{\tau}\)
3 \(\dfrac{2 \tau}{l}\)
4 \(\dfrac{2 l}{\tau}\)
Explanation:
Moment of couple, \(\tau=F l \sin \theta\) \(\therefore\) Force acting on each end \(F=\dfrac{\tau}{l \sin \theta}=\dfrac{\tau}{l \sin 30^{\circ}}=\dfrac{\tau}{l \times \dfrac{1}{2}}=\dfrac{2 \tau}{l}\)
MHTCET - 2019
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366199
Turning effect is produced by
1 Radial component of force
2 Transverse component of force
3 Both radial & transverse components of force
4 None of the above
Explanation:
Consider a force \(F\) that acts at point \(P\). The torque produced by the force about \(O\) is \(\tau=r F \sin \phi\) \(F \sin \phi\) is the transverse component of force correct option is (2)
