NEET Test Series from KOTA - 10 Papers In MS WORD
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Rotational Motion
149827
A solid body rotates an angle \(\theta\) about a stationary axis according to the law \(\theta=6 t-2 t^{3}\). What is the mean value of angular velocity over the time interval between \(t=0\) and the time when the body comes to rest?
1 \(1 \mathrm{rad} / \mathrm{s}\)
2 \(2 \mathrm{rad} / \mathrm{s}\)
3 \(3 \mathrm{rad} / \mathrm{s}\)
4 \(4 \mathrm{rad} / \mathrm{s}\)
Explanation:
D Given that, \(\theta=6 t-2 t^{3}\) Differentiating with respect to \(t\) on both side of the equation- \(\frac{\mathrm{d} \theta}{\mathrm{dt}}=\omega=6-6 \mathrm{t}^{2}\) \(6-6 \mathrm{t}^{2}=0\) \(\mathrm{t}^{2}=1\) \(\mathrm{t}=1 \mathrm{sec}\) Put the value of \(t\) in equation (i) we get - \(\theta =6 \times 1-2 \times 1^{3}\) \(\theta =6-2\) \(\theta =4 \mathrm{rad}\) \(\text { For } \quad \theta_{0} =6 \times 0-2 \times 0^{2} \quad\left[\because \mathrm{t}_{0}=0\right]\) \(\theta_{0}=0 \mathrm{rad}\) Mean value of angular velocity \(\omega=\frac{\theta-\theta_{0}}{t-t_{0}}=\frac{4-0}{1-0}\) \(\omega=4 \mathrm{rad} / \mathrm{sec}\)
UPSEE - 2018
Rotational Motion
149828
A wheel rotating at \(12 \mathrm{rev} / \mathrm{s}\) is brought to rest in \(6 \mathrm{~s}\). The average angular deceleration in \(\mathrm{rad} / \mathrm{s}^{2}\) of the wheel during this process is
149829
A uniform disc of mass \(M\) and radius \(R\) has a string wound around its periphery and tied to a ceiling as shown in the figure. The acceleration of the center of mass after it is released
1 \(g\)
2 \(\frac{g}{3}\)
3 \(\frac{2 g}{3}\)
4 \(\frac{g}{2}\)
Explanation:
C Let us consider Equation of motion of the disc \(\mathrm{Ma}=\mathrm{Mg}-\mathrm{T}\) Also torque about \(\mathrm{O}\). \(\mathrm{TR}=\mathrm{I} \alpha\) \(\therefore \quad \mathrm{TR}=\left(\frac{1}{2} \mathrm{MR}^{2}\right) \alpha\) For no slipping, \(a=\alpha R \Rightarrow \alpha=\frac{a}{R}\) We get, \(\mathrm{T}=\frac{\mathrm{Ma}}{2}\) From equation (i) and (ii), \(\mathrm{Ma}=\mathrm{Mg}-\frac{\mathrm{Ma}}{2}\) \(\mathrm{ma}=\mathrm{m}\left(\mathrm{g}-\frac{\mathrm{a}}{2}\right)\) \(\mathrm{a}+\frac{\mathrm{a}}{2}=\mathrm{g}\) \(\frac{2 \mathrm{a}+\mathrm{a}}{2}=\mathrm{g}\) \(\mathrm{a}=\frac{2 \mathrm{~g}}{3}\)
TS EAMCET(Medical)-2017
Rotational Motion
149830
A rod \(P Q\) of mass \(M\) and length \(L\) is hinged at end \(P\). The rod is kept horizontal by a massless string tied to point \(O\) as shown in figure. When string is cut, the initial angular acceleration of the rod is-
1 \(\frac{3 \mathrm{~g}}{2 \mathrm{~L}}\)
2 \(\frac{g}{L}\)
3 \(\frac{2 \mathrm{~g}}{\mathrm{~L}}\)
4 \(\frac{2 \mathrm{~g}}{2 \mathrm{~L}}\)
Explanation:
A \(\mathrm{P} \longrightarrow \underset{\mathrm{mg}}{\downarrow} \quad \mathrm{Q} \quad\) (Free body diagram) angular torque, \(\tau=\mathrm{I} \alpha\) Moment of inertia (I) about the point \(\mathrm{Q}\) on the rad, \(\mathrm{I}=\frac{\mathrm{mL}^{2}}{3}\) Torque due to weight, \(\tau=\operatorname{mg}\left(\frac{\mathrm{L}}{2}\right)\) So, putting the value of \(\tau\) and \(\mathrm{I}\) in \(\tau=\mathrm{I} \alpha\) - \(\mathrm{mg}\left(\frac{\mathrm{L}}{2}\right)=\frac{\mathrm{mL}^{2}}{3} \alpha\) \(\alpha=\frac{3}{2} \frac{\mathrm{g}}{\mathrm{L}}\) \(\alpha=\frac{3 \mathrm{~g}}{2 \mathrm{t}}\)
149827
A solid body rotates an angle \(\theta\) about a stationary axis according to the law \(\theta=6 t-2 t^{3}\). What is the mean value of angular velocity over the time interval between \(t=0\) and the time when the body comes to rest?
1 \(1 \mathrm{rad} / \mathrm{s}\)
2 \(2 \mathrm{rad} / \mathrm{s}\)
3 \(3 \mathrm{rad} / \mathrm{s}\)
4 \(4 \mathrm{rad} / \mathrm{s}\)
Explanation:
D Given that, \(\theta=6 t-2 t^{3}\) Differentiating with respect to \(t\) on both side of the equation- \(\frac{\mathrm{d} \theta}{\mathrm{dt}}=\omega=6-6 \mathrm{t}^{2}\) \(6-6 \mathrm{t}^{2}=0\) \(\mathrm{t}^{2}=1\) \(\mathrm{t}=1 \mathrm{sec}\) Put the value of \(t\) in equation (i) we get - \(\theta =6 \times 1-2 \times 1^{3}\) \(\theta =6-2\) \(\theta =4 \mathrm{rad}\) \(\text { For } \quad \theta_{0} =6 \times 0-2 \times 0^{2} \quad\left[\because \mathrm{t}_{0}=0\right]\) \(\theta_{0}=0 \mathrm{rad}\) Mean value of angular velocity \(\omega=\frac{\theta-\theta_{0}}{t-t_{0}}=\frac{4-0}{1-0}\) \(\omega=4 \mathrm{rad} / \mathrm{sec}\)
UPSEE - 2018
Rotational Motion
149828
A wheel rotating at \(12 \mathrm{rev} / \mathrm{s}\) is brought to rest in \(6 \mathrm{~s}\). The average angular deceleration in \(\mathrm{rad} / \mathrm{s}^{2}\) of the wheel during this process is
149829
A uniform disc of mass \(M\) and radius \(R\) has a string wound around its periphery and tied to a ceiling as shown in the figure. The acceleration of the center of mass after it is released
1 \(g\)
2 \(\frac{g}{3}\)
3 \(\frac{2 g}{3}\)
4 \(\frac{g}{2}\)
Explanation:
C Let us consider Equation of motion of the disc \(\mathrm{Ma}=\mathrm{Mg}-\mathrm{T}\) Also torque about \(\mathrm{O}\). \(\mathrm{TR}=\mathrm{I} \alpha\) \(\therefore \quad \mathrm{TR}=\left(\frac{1}{2} \mathrm{MR}^{2}\right) \alpha\) For no slipping, \(a=\alpha R \Rightarrow \alpha=\frac{a}{R}\) We get, \(\mathrm{T}=\frac{\mathrm{Ma}}{2}\) From equation (i) and (ii), \(\mathrm{Ma}=\mathrm{Mg}-\frac{\mathrm{Ma}}{2}\) \(\mathrm{ma}=\mathrm{m}\left(\mathrm{g}-\frac{\mathrm{a}}{2}\right)\) \(\mathrm{a}+\frac{\mathrm{a}}{2}=\mathrm{g}\) \(\frac{2 \mathrm{a}+\mathrm{a}}{2}=\mathrm{g}\) \(\mathrm{a}=\frac{2 \mathrm{~g}}{3}\)
TS EAMCET(Medical)-2017
Rotational Motion
149830
A rod \(P Q\) of mass \(M\) and length \(L\) is hinged at end \(P\). The rod is kept horizontal by a massless string tied to point \(O\) as shown in figure. When string is cut, the initial angular acceleration of the rod is-
1 \(\frac{3 \mathrm{~g}}{2 \mathrm{~L}}\)
2 \(\frac{g}{L}\)
3 \(\frac{2 \mathrm{~g}}{\mathrm{~L}}\)
4 \(\frac{2 \mathrm{~g}}{2 \mathrm{~L}}\)
Explanation:
A \(\mathrm{P} \longrightarrow \underset{\mathrm{mg}}{\downarrow} \quad \mathrm{Q} \quad\) (Free body diagram) angular torque, \(\tau=\mathrm{I} \alpha\) Moment of inertia (I) about the point \(\mathrm{Q}\) on the rad, \(\mathrm{I}=\frac{\mathrm{mL}^{2}}{3}\) Torque due to weight, \(\tau=\operatorname{mg}\left(\frac{\mathrm{L}}{2}\right)\) So, putting the value of \(\tau\) and \(\mathrm{I}\) in \(\tau=\mathrm{I} \alpha\) - \(\mathrm{mg}\left(\frac{\mathrm{L}}{2}\right)=\frac{\mathrm{mL}^{2}}{3} \alpha\) \(\alpha=\frac{3}{2} \frac{\mathrm{g}}{\mathrm{L}}\) \(\alpha=\frac{3 \mathrm{~g}}{2 \mathrm{t}}\)
149827
A solid body rotates an angle \(\theta\) about a stationary axis according to the law \(\theta=6 t-2 t^{3}\). What is the mean value of angular velocity over the time interval between \(t=0\) and the time when the body comes to rest?
1 \(1 \mathrm{rad} / \mathrm{s}\)
2 \(2 \mathrm{rad} / \mathrm{s}\)
3 \(3 \mathrm{rad} / \mathrm{s}\)
4 \(4 \mathrm{rad} / \mathrm{s}\)
Explanation:
D Given that, \(\theta=6 t-2 t^{3}\) Differentiating with respect to \(t\) on both side of the equation- \(\frac{\mathrm{d} \theta}{\mathrm{dt}}=\omega=6-6 \mathrm{t}^{2}\) \(6-6 \mathrm{t}^{2}=0\) \(\mathrm{t}^{2}=1\) \(\mathrm{t}=1 \mathrm{sec}\) Put the value of \(t\) in equation (i) we get - \(\theta =6 \times 1-2 \times 1^{3}\) \(\theta =6-2\) \(\theta =4 \mathrm{rad}\) \(\text { For } \quad \theta_{0} =6 \times 0-2 \times 0^{2} \quad\left[\because \mathrm{t}_{0}=0\right]\) \(\theta_{0}=0 \mathrm{rad}\) Mean value of angular velocity \(\omega=\frac{\theta-\theta_{0}}{t-t_{0}}=\frac{4-0}{1-0}\) \(\omega=4 \mathrm{rad} / \mathrm{sec}\)
UPSEE - 2018
Rotational Motion
149828
A wheel rotating at \(12 \mathrm{rev} / \mathrm{s}\) is brought to rest in \(6 \mathrm{~s}\). The average angular deceleration in \(\mathrm{rad} / \mathrm{s}^{2}\) of the wheel during this process is
149829
A uniform disc of mass \(M\) and radius \(R\) has a string wound around its periphery and tied to a ceiling as shown in the figure. The acceleration of the center of mass after it is released
1 \(g\)
2 \(\frac{g}{3}\)
3 \(\frac{2 g}{3}\)
4 \(\frac{g}{2}\)
Explanation:
C Let us consider Equation of motion of the disc \(\mathrm{Ma}=\mathrm{Mg}-\mathrm{T}\) Also torque about \(\mathrm{O}\). \(\mathrm{TR}=\mathrm{I} \alpha\) \(\therefore \quad \mathrm{TR}=\left(\frac{1}{2} \mathrm{MR}^{2}\right) \alpha\) For no slipping, \(a=\alpha R \Rightarrow \alpha=\frac{a}{R}\) We get, \(\mathrm{T}=\frac{\mathrm{Ma}}{2}\) From equation (i) and (ii), \(\mathrm{Ma}=\mathrm{Mg}-\frac{\mathrm{Ma}}{2}\) \(\mathrm{ma}=\mathrm{m}\left(\mathrm{g}-\frac{\mathrm{a}}{2}\right)\) \(\mathrm{a}+\frac{\mathrm{a}}{2}=\mathrm{g}\) \(\frac{2 \mathrm{a}+\mathrm{a}}{2}=\mathrm{g}\) \(\mathrm{a}=\frac{2 \mathrm{~g}}{3}\)
TS EAMCET(Medical)-2017
Rotational Motion
149830
A rod \(P Q\) of mass \(M\) and length \(L\) is hinged at end \(P\). The rod is kept horizontal by a massless string tied to point \(O\) as shown in figure. When string is cut, the initial angular acceleration of the rod is-
1 \(\frac{3 \mathrm{~g}}{2 \mathrm{~L}}\)
2 \(\frac{g}{L}\)
3 \(\frac{2 \mathrm{~g}}{\mathrm{~L}}\)
4 \(\frac{2 \mathrm{~g}}{2 \mathrm{~L}}\)
Explanation:
A \(\mathrm{P} \longrightarrow \underset{\mathrm{mg}}{\downarrow} \quad \mathrm{Q} \quad\) (Free body diagram) angular torque, \(\tau=\mathrm{I} \alpha\) Moment of inertia (I) about the point \(\mathrm{Q}\) on the rad, \(\mathrm{I}=\frac{\mathrm{mL}^{2}}{3}\) Torque due to weight, \(\tau=\operatorname{mg}\left(\frac{\mathrm{L}}{2}\right)\) So, putting the value of \(\tau\) and \(\mathrm{I}\) in \(\tau=\mathrm{I} \alpha\) - \(\mathrm{mg}\left(\frac{\mathrm{L}}{2}\right)=\frac{\mathrm{mL}^{2}}{3} \alpha\) \(\alpha=\frac{3}{2} \frac{\mathrm{g}}{\mathrm{L}}\) \(\alpha=\frac{3 \mathrm{~g}}{2 \mathrm{t}}\)
149827
A solid body rotates an angle \(\theta\) about a stationary axis according to the law \(\theta=6 t-2 t^{3}\). What is the mean value of angular velocity over the time interval between \(t=0\) and the time when the body comes to rest?
1 \(1 \mathrm{rad} / \mathrm{s}\)
2 \(2 \mathrm{rad} / \mathrm{s}\)
3 \(3 \mathrm{rad} / \mathrm{s}\)
4 \(4 \mathrm{rad} / \mathrm{s}\)
Explanation:
D Given that, \(\theta=6 t-2 t^{3}\) Differentiating with respect to \(t\) on both side of the equation- \(\frac{\mathrm{d} \theta}{\mathrm{dt}}=\omega=6-6 \mathrm{t}^{2}\) \(6-6 \mathrm{t}^{2}=0\) \(\mathrm{t}^{2}=1\) \(\mathrm{t}=1 \mathrm{sec}\) Put the value of \(t\) in equation (i) we get - \(\theta =6 \times 1-2 \times 1^{3}\) \(\theta =6-2\) \(\theta =4 \mathrm{rad}\) \(\text { For } \quad \theta_{0} =6 \times 0-2 \times 0^{2} \quad\left[\because \mathrm{t}_{0}=0\right]\) \(\theta_{0}=0 \mathrm{rad}\) Mean value of angular velocity \(\omega=\frac{\theta-\theta_{0}}{t-t_{0}}=\frac{4-0}{1-0}\) \(\omega=4 \mathrm{rad} / \mathrm{sec}\)
UPSEE - 2018
Rotational Motion
149828
A wheel rotating at \(12 \mathrm{rev} / \mathrm{s}\) is brought to rest in \(6 \mathrm{~s}\). The average angular deceleration in \(\mathrm{rad} / \mathrm{s}^{2}\) of the wheel during this process is
149829
A uniform disc of mass \(M\) and radius \(R\) has a string wound around its periphery and tied to a ceiling as shown in the figure. The acceleration of the center of mass after it is released
1 \(g\)
2 \(\frac{g}{3}\)
3 \(\frac{2 g}{3}\)
4 \(\frac{g}{2}\)
Explanation:
C Let us consider Equation of motion of the disc \(\mathrm{Ma}=\mathrm{Mg}-\mathrm{T}\) Also torque about \(\mathrm{O}\). \(\mathrm{TR}=\mathrm{I} \alpha\) \(\therefore \quad \mathrm{TR}=\left(\frac{1}{2} \mathrm{MR}^{2}\right) \alpha\) For no slipping, \(a=\alpha R \Rightarrow \alpha=\frac{a}{R}\) We get, \(\mathrm{T}=\frac{\mathrm{Ma}}{2}\) From equation (i) and (ii), \(\mathrm{Ma}=\mathrm{Mg}-\frac{\mathrm{Ma}}{2}\) \(\mathrm{ma}=\mathrm{m}\left(\mathrm{g}-\frac{\mathrm{a}}{2}\right)\) \(\mathrm{a}+\frac{\mathrm{a}}{2}=\mathrm{g}\) \(\frac{2 \mathrm{a}+\mathrm{a}}{2}=\mathrm{g}\) \(\mathrm{a}=\frac{2 \mathrm{~g}}{3}\)
TS EAMCET(Medical)-2017
Rotational Motion
149830
A rod \(P Q\) of mass \(M\) and length \(L\) is hinged at end \(P\). The rod is kept horizontal by a massless string tied to point \(O\) as shown in figure. When string is cut, the initial angular acceleration of the rod is-
1 \(\frac{3 \mathrm{~g}}{2 \mathrm{~L}}\)
2 \(\frac{g}{L}\)
3 \(\frac{2 \mathrm{~g}}{\mathrm{~L}}\)
4 \(\frac{2 \mathrm{~g}}{2 \mathrm{~L}}\)
Explanation:
A \(\mathrm{P} \longrightarrow \underset{\mathrm{mg}}{\downarrow} \quad \mathrm{Q} \quad\) (Free body diagram) angular torque, \(\tau=\mathrm{I} \alpha\) Moment of inertia (I) about the point \(\mathrm{Q}\) on the rad, \(\mathrm{I}=\frac{\mathrm{mL}^{2}}{3}\) Torque due to weight, \(\tau=\operatorname{mg}\left(\frac{\mathrm{L}}{2}\right)\) So, putting the value of \(\tau\) and \(\mathrm{I}\) in \(\tau=\mathrm{I} \alpha\) - \(\mathrm{mg}\left(\frac{\mathrm{L}}{2}\right)=\frac{\mathrm{mL}^{2}}{3} \alpha\) \(\alpha=\frac{3}{2} \frac{\mathrm{g}}{\mathrm{L}}\) \(\alpha=\frac{3 \mathrm{~g}}{2 \mathrm{t}}\)
