366213
Average torque on a projectile of mass \(m\) (initial speed \(u\) and angle of projection \(\theta\) ) between initial and final positions \(P\) and \(Q\) about the point of projection is:
1 \(m u^{2} \cos \theta\)
2 \(m u^{2} \sin \theta\)
3 \(\dfrac{m u^{2} \sin 2 \theta}{2}\)
4 \(\dfrac{m u^{2} \cos \theta}{2}\)
Explanation:
\(\vec{\tau}_{a v} \cdot \Delta t=\Delta \vec{L}\) Here \(\Delta t=\) time of flight \(==\dfrac{2 u \sin \theta}{g}\) Change in angular momentum about point of projection (initially it is zero) \(|\overrightarrow{\Delta L}|=\left|\vec{L}_{f}-\vec{L}_{i}\right|=\mid(m u \sin \theta)\) Range \(-0 \mid\) \(=\dfrac{(m u \sin \theta)\left(u^{2} \sin 2 \theta\right)}{g}=\dfrac{m u^{3} \sin \theta \sin 2 \theta}{g}\) Now \(\left|\vec{\tau}_{a v}\right|=\left|\dfrac{\overrightarrow{\Delta L}}{\Delta t}\right|=\dfrac{m u^{3} \sin \theta \sin 2 \theta}{g} \times \dfrac{g}{2 u \sin \theta}\) \(=\dfrac{m u^{2} \sin 2 \theta}{2}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366214
A door \(1.6\;m\) wide requires a force of \(1\;N\) to be applied at the free end to open or close it. The force that is required at a point \(0.4\;m\) distant from the hinges for opening or closing the door is
366213
Average torque on a projectile of mass \(m\) (initial speed \(u\) and angle of projection \(\theta\) ) between initial and final positions \(P\) and \(Q\) about the point of projection is:
1 \(m u^{2} \cos \theta\)
2 \(m u^{2} \sin \theta\)
3 \(\dfrac{m u^{2} \sin 2 \theta}{2}\)
4 \(\dfrac{m u^{2} \cos \theta}{2}\)
Explanation:
\(\vec{\tau}_{a v} \cdot \Delta t=\Delta \vec{L}\) Here \(\Delta t=\) time of flight \(==\dfrac{2 u \sin \theta}{g}\) Change in angular momentum about point of projection (initially it is zero) \(|\overrightarrow{\Delta L}|=\left|\vec{L}_{f}-\vec{L}_{i}\right|=\mid(m u \sin \theta)\) Range \(-0 \mid\) \(=\dfrac{(m u \sin \theta)\left(u^{2} \sin 2 \theta\right)}{g}=\dfrac{m u^{3} \sin \theta \sin 2 \theta}{g}\) Now \(\left|\vec{\tau}_{a v}\right|=\left|\dfrac{\overrightarrow{\Delta L}}{\Delta t}\right|=\dfrac{m u^{3} \sin \theta \sin 2 \theta}{g} \times \dfrac{g}{2 u \sin \theta}\) \(=\dfrac{m u^{2} \sin 2 \theta}{2}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366214
A door \(1.6\;m\) wide requires a force of \(1\;N\) to be applied at the free end to open or close it. The force that is required at a point \(0.4\;m\) distant from the hinges for opening or closing the door is
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366213
Average torque on a projectile of mass \(m\) (initial speed \(u\) and angle of projection \(\theta\) ) between initial and final positions \(P\) and \(Q\) about the point of projection is:
1 \(m u^{2} \cos \theta\)
2 \(m u^{2} \sin \theta\)
3 \(\dfrac{m u^{2} \sin 2 \theta}{2}\)
4 \(\dfrac{m u^{2} \cos \theta}{2}\)
Explanation:
\(\vec{\tau}_{a v} \cdot \Delta t=\Delta \vec{L}\) Here \(\Delta t=\) time of flight \(==\dfrac{2 u \sin \theta}{g}\) Change in angular momentum about point of projection (initially it is zero) \(|\overrightarrow{\Delta L}|=\left|\vec{L}_{f}-\vec{L}_{i}\right|=\mid(m u \sin \theta)\) Range \(-0 \mid\) \(=\dfrac{(m u \sin \theta)\left(u^{2} \sin 2 \theta\right)}{g}=\dfrac{m u^{3} \sin \theta \sin 2 \theta}{g}\) Now \(\left|\vec{\tau}_{a v}\right|=\left|\dfrac{\overrightarrow{\Delta L}}{\Delta t}\right|=\dfrac{m u^{3} \sin \theta \sin 2 \theta}{g} \times \dfrac{g}{2 u \sin \theta}\) \(=\dfrac{m u^{2} \sin 2 \theta}{2}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366214
A door \(1.6\;m\) wide requires a force of \(1\;N\) to be applied at the free end to open or close it. The force that is required at a point \(0.4\;m\) distant from the hinges for opening or closing the door is
366213
Average torque on a projectile of mass \(m\) (initial speed \(u\) and angle of projection \(\theta\) ) between initial and final positions \(P\) and \(Q\) about the point of projection is:
1 \(m u^{2} \cos \theta\)
2 \(m u^{2} \sin \theta\)
3 \(\dfrac{m u^{2} \sin 2 \theta}{2}\)
4 \(\dfrac{m u^{2} \cos \theta}{2}\)
Explanation:
\(\vec{\tau}_{a v} \cdot \Delta t=\Delta \vec{L}\) Here \(\Delta t=\) time of flight \(==\dfrac{2 u \sin \theta}{g}\) Change in angular momentum about point of projection (initially it is zero) \(|\overrightarrow{\Delta L}|=\left|\vec{L}_{f}-\vec{L}_{i}\right|=\mid(m u \sin \theta)\) Range \(-0 \mid\) \(=\dfrac{(m u \sin \theta)\left(u^{2} \sin 2 \theta\right)}{g}=\dfrac{m u^{3} \sin \theta \sin 2 \theta}{g}\) Now \(\left|\vec{\tau}_{a v}\right|=\left|\dfrac{\overrightarrow{\Delta L}}{\Delta t}\right|=\dfrac{m u^{3} \sin \theta \sin 2 \theta}{g} \times \dfrac{g}{2 u \sin \theta}\) \(=\dfrac{m u^{2} \sin 2 \theta}{2}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366214
A door \(1.6\;m\) wide requires a force of \(1\;N\) to be applied at the free end to open or close it. The force that is required at a point \(0.4\;m\) distant from the hinges for opening or closing the door is
