149798
A coin placed on a rotating turn table just slips if it is placed at a distance of \(4 \mathrm{~cm}\) from the centre. If the angular velocity of the turntable is doubled, it will just slip at a distance of
1 \(1 \mathrm{~cm}\)
2 \(2 \mathrm{~cm}\)
3 \(4 \mathrm{~cm}\)
4 \(8 \mathrm{~cm}\)
Explanation:
A Given, \(\omega_{2}=2 \omega_{1}, r_{1}=4 \mathrm{~cm}\) A body placed on a non-inertial frame of reference which is rotating about its axis, experiences a centrifugal force. \(\because \quad \mathrm{F}=\mathrm{Mr} \omega^{2}\) According to the question, \(\mathrm{F}_{1_{1}}=\mathrm{F}_{2}\) \(\mathrm{Mr}_{1} \omega_{1}=\mathrm{Mr}_{2} \omega_{2}{ }^{2}\) \(4 \times \omega_{1}^{2}=\mathrm{r}_{2}\left(2 \omega_{1}\right)^{2}\) \(4 \times \omega_{1}^{2}=4 \omega_{1}{ }^{2} \mathrm{r}_{2}\) \(\mathrm{r}_{2}=\frac{4}{4}\) \(\mathrm{r}_{2}=1 \mathrm{~cm}\)
AP EAMCET-25.09.2020
Rotational Motion
149799
A disc spinning at the rate \(27.5 \mathrm{rad} \mathrm{s}^{-1}\) is slowed at the rate 10 rad \(\mathrm{s}^{-2}\). The time after which it will come to rest is
1 \(2.75 \mathrm{~s}\)
2 \(5.5 \mathrm{~s}\)
3 \(1.25 \mathrm{~s}\)
4 \(3.5 \mathrm{~s}\)
5 \(6.2 \mathrm{~s}\)
Explanation:
A Given, initial angular velocity \(\left(\omega_{i}\right)=27.5\) \(\mathrm{rad} / \mathrm{sec}\), angular acceleration \((\alpha)=-10 \mathrm{rad} / \mathrm{sec}^{2}\), final angular velocity \(\left(\omega_{\mathrm{f}}\right)=0\) According to kinematics law of rotational motion- \(\omega_{\mathrm{f}}=\omega_{\mathrm{i}}+\alpha \mathrm{t}\) \(\alpha \mathrm{t}=\omega_{\mathrm{f}}-\omega_{\mathrm{i}}\) \(\mathrm{t}=\frac{\omega_{\mathrm{f}}-\omega_{\mathrm{i}}}{\alpha}\) \(\mathrm{t}=\frac{0-27.5}{-10} \Rightarrow \mathrm{t}=\frac{27.5}{10}\) \(\mathrm{t}=2.75 \mathrm{sec}\)
Kerala CEE 2020
Rotational Motion
149800
Assume proton is rotating along a circular path of radius \(1 \mathrm{~m}\) under a centrifugal force of \(4 \times 10^{-12} \mathrm{~N}\). If the mass of proton is \(1.6 \times 10^{-27} \mathrm{~kg}\), then its angular velocity of rotation is
149801
A wheel starting from rest gains an angular velocity of \(10 \mathrm{rad} / \mathrm{s}\) after uniformly accelerated for \(5 \mathrm{~s}\). The total angle through which it has turned is :
1 \(25 \mathrm{rad}\)
2 \(100 \mathrm{rad}\)
3 \(25 \pi \mathrm{rad}\)
4 \(50 \pi \mathrm{rad}\) and a vertical axis
Explanation:
A Given that, \(\omega_{\mathrm{i}}=\mathrm{rad} / \mathrm{sec}, \omega_{\mathrm{f}}=10 \mathrm{rad} / \mathrm{sec}, \mathrm{t}=\) \(5 \mathrm{sec}\) According to kinematics law of rotational motion- \(\omega_{\mathrm{f}}=\omega_{\mathrm{i}}+\alpha \mathrm{t}\) \(10=0+\alpha \times 5\) \(\alpha=2\) Now, \(\quad \theta=\omega_{\mathrm{i}} \mathrm{t}+1 / 2 \alpha \mathrm{t}^{2}\) \(\theta=(0 \times 5)+1 / 2 \times 2 \times 5^{2}\) \(\theta=25 \mathrm{rad}\).
149798
A coin placed on a rotating turn table just slips if it is placed at a distance of \(4 \mathrm{~cm}\) from the centre. If the angular velocity of the turntable is doubled, it will just slip at a distance of
1 \(1 \mathrm{~cm}\)
2 \(2 \mathrm{~cm}\)
3 \(4 \mathrm{~cm}\)
4 \(8 \mathrm{~cm}\)
Explanation:
A Given, \(\omega_{2}=2 \omega_{1}, r_{1}=4 \mathrm{~cm}\) A body placed on a non-inertial frame of reference which is rotating about its axis, experiences a centrifugal force. \(\because \quad \mathrm{F}=\mathrm{Mr} \omega^{2}\) According to the question, \(\mathrm{F}_{1_{1}}=\mathrm{F}_{2}\) \(\mathrm{Mr}_{1} \omega_{1}=\mathrm{Mr}_{2} \omega_{2}{ }^{2}\) \(4 \times \omega_{1}^{2}=\mathrm{r}_{2}\left(2 \omega_{1}\right)^{2}\) \(4 \times \omega_{1}^{2}=4 \omega_{1}{ }^{2} \mathrm{r}_{2}\) \(\mathrm{r}_{2}=\frac{4}{4}\) \(\mathrm{r}_{2}=1 \mathrm{~cm}\)
AP EAMCET-25.09.2020
Rotational Motion
149799
A disc spinning at the rate \(27.5 \mathrm{rad} \mathrm{s}^{-1}\) is slowed at the rate 10 rad \(\mathrm{s}^{-2}\). The time after which it will come to rest is
1 \(2.75 \mathrm{~s}\)
2 \(5.5 \mathrm{~s}\)
3 \(1.25 \mathrm{~s}\)
4 \(3.5 \mathrm{~s}\)
5 \(6.2 \mathrm{~s}\)
Explanation:
A Given, initial angular velocity \(\left(\omega_{i}\right)=27.5\) \(\mathrm{rad} / \mathrm{sec}\), angular acceleration \((\alpha)=-10 \mathrm{rad} / \mathrm{sec}^{2}\), final angular velocity \(\left(\omega_{\mathrm{f}}\right)=0\) According to kinematics law of rotational motion- \(\omega_{\mathrm{f}}=\omega_{\mathrm{i}}+\alpha \mathrm{t}\) \(\alpha \mathrm{t}=\omega_{\mathrm{f}}-\omega_{\mathrm{i}}\) \(\mathrm{t}=\frac{\omega_{\mathrm{f}}-\omega_{\mathrm{i}}}{\alpha}\) \(\mathrm{t}=\frac{0-27.5}{-10} \Rightarrow \mathrm{t}=\frac{27.5}{10}\) \(\mathrm{t}=2.75 \mathrm{sec}\)
Kerala CEE 2020
Rotational Motion
149800
Assume proton is rotating along a circular path of radius \(1 \mathrm{~m}\) under a centrifugal force of \(4 \times 10^{-12} \mathrm{~N}\). If the mass of proton is \(1.6 \times 10^{-27} \mathrm{~kg}\), then its angular velocity of rotation is
149801
A wheel starting from rest gains an angular velocity of \(10 \mathrm{rad} / \mathrm{s}\) after uniformly accelerated for \(5 \mathrm{~s}\). The total angle through which it has turned is :
1 \(25 \mathrm{rad}\)
2 \(100 \mathrm{rad}\)
3 \(25 \pi \mathrm{rad}\)
4 \(50 \pi \mathrm{rad}\) and a vertical axis
Explanation:
A Given that, \(\omega_{\mathrm{i}}=\mathrm{rad} / \mathrm{sec}, \omega_{\mathrm{f}}=10 \mathrm{rad} / \mathrm{sec}, \mathrm{t}=\) \(5 \mathrm{sec}\) According to kinematics law of rotational motion- \(\omega_{\mathrm{f}}=\omega_{\mathrm{i}}+\alpha \mathrm{t}\) \(10=0+\alpha \times 5\) \(\alpha=2\) Now, \(\quad \theta=\omega_{\mathrm{i}} \mathrm{t}+1 / 2 \alpha \mathrm{t}^{2}\) \(\theta=(0 \times 5)+1 / 2 \times 2 \times 5^{2}\) \(\theta=25 \mathrm{rad}\).
149798
A coin placed on a rotating turn table just slips if it is placed at a distance of \(4 \mathrm{~cm}\) from the centre. If the angular velocity of the turntable is doubled, it will just slip at a distance of
1 \(1 \mathrm{~cm}\)
2 \(2 \mathrm{~cm}\)
3 \(4 \mathrm{~cm}\)
4 \(8 \mathrm{~cm}\)
Explanation:
A Given, \(\omega_{2}=2 \omega_{1}, r_{1}=4 \mathrm{~cm}\) A body placed on a non-inertial frame of reference which is rotating about its axis, experiences a centrifugal force. \(\because \quad \mathrm{F}=\mathrm{Mr} \omega^{2}\) According to the question, \(\mathrm{F}_{1_{1}}=\mathrm{F}_{2}\) \(\mathrm{Mr}_{1} \omega_{1}=\mathrm{Mr}_{2} \omega_{2}{ }^{2}\) \(4 \times \omega_{1}^{2}=\mathrm{r}_{2}\left(2 \omega_{1}\right)^{2}\) \(4 \times \omega_{1}^{2}=4 \omega_{1}{ }^{2} \mathrm{r}_{2}\) \(\mathrm{r}_{2}=\frac{4}{4}\) \(\mathrm{r}_{2}=1 \mathrm{~cm}\)
AP EAMCET-25.09.2020
Rotational Motion
149799
A disc spinning at the rate \(27.5 \mathrm{rad} \mathrm{s}^{-1}\) is slowed at the rate 10 rad \(\mathrm{s}^{-2}\). The time after which it will come to rest is
1 \(2.75 \mathrm{~s}\)
2 \(5.5 \mathrm{~s}\)
3 \(1.25 \mathrm{~s}\)
4 \(3.5 \mathrm{~s}\)
5 \(6.2 \mathrm{~s}\)
Explanation:
A Given, initial angular velocity \(\left(\omega_{i}\right)=27.5\) \(\mathrm{rad} / \mathrm{sec}\), angular acceleration \((\alpha)=-10 \mathrm{rad} / \mathrm{sec}^{2}\), final angular velocity \(\left(\omega_{\mathrm{f}}\right)=0\) According to kinematics law of rotational motion- \(\omega_{\mathrm{f}}=\omega_{\mathrm{i}}+\alpha \mathrm{t}\) \(\alpha \mathrm{t}=\omega_{\mathrm{f}}-\omega_{\mathrm{i}}\) \(\mathrm{t}=\frac{\omega_{\mathrm{f}}-\omega_{\mathrm{i}}}{\alpha}\) \(\mathrm{t}=\frac{0-27.5}{-10} \Rightarrow \mathrm{t}=\frac{27.5}{10}\) \(\mathrm{t}=2.75 \mathrm{sec}\)
Kerala CEE 2020
Rotational Motion
149800
Assume proton is rotating along a circular path of radius \(1 \mathrm{~m}\) under a centrifugal force of \(4 \times 10^{-12} \mathrm{~N}\). If the mass of proton is \(1.6 \times 10^{-27} \mathrm{~kg}\), then its angular velocity of rotation is
149801
A wheel starting from rest gains an angular velocity of \(10 \mathrm{rad} / \mathrm{s}\) after uniformly accelerated for \(5 \mathrm{~s}\). The total angle through which it has turned is :
1 \(25 \mathrm{rad}\)
2 \(100 \mathrm{rad}\)
3 \(25 \pi \mathrm{rad}\)
4 \(50 \pi \mathrm{rad}\) and a vertical axis
Explanation:
A Given that, \(\omega_{\mathrm{i}}=\mathrm{rad} / \mathrm{sec}, \omega_{\mathrm{f}}=10 \mathrm{rad} / \mathrm{sec}, \mathrm{t}=\) \(5 \mathrm{sec}\) According to kinematics law of rotational motion- \(\omega_{\mathrm{f}}=\omega_{\mathrm{i}}+\alpha \mathrm{t}\) \(10=0+\alpha \times 5\) \(\alpha=2\) Now, \(\quad \theta=\omega_{\mathrm{i}} \mathrm{t}+1 / 2 \alpha \mathrm{t}^{2}\) \(\theta=(0 \times 5)+1 / 2 \times 2 \times 5^{2}\) \(\theta=25 \mathrm{rad}\).
149798
A coin placed on a rotating turn table just slips if it is placed at a distance of \(4 \mathrm{~cm}\) from the centre. If the angular velocity of the turntable is doubled, it will just slip at a distance of
1 \(1 \mathrm{~cm}\)
2 \(2 \mathrm{~cm}\)
3 \(4 \mathrm{~cm}\)
4 \(8 \mathrm{~cm}\)
Explanation:
A Given, \(\omega_{2}=2 \omega_{1}, r_{1}=4 \mathrm{~cm}\) A body placed on a non-inertial frame of reference which is rotating about its axis, experiences a centrifugal force. \(\because \quad \mathrm{F}=\mathrm{Mr} \omega^{2}\) According to the question, \(\mathrm{F}_{1_{1}}=\mathrm{F}_{2}\) \(\mathrm{Mr}_{1} \omega_{1}=\mathrm{Mr}_{2} \omega_{2}{ }^{2}\) \(4 \times \omega_{1}^{2}=\mathrm{r}_{2}\left(2 \omega_{1}\right)^{2}\) \(4 \times \omega_{1}^{2}=4 \omega_{1}{ }^{2} \mathrm{r}_{2}\) \(\mathrm{r}_{2}=\frac{4}{4}\) \(\mathrm{r}_{2}=1 \mathrm{~cm}\)
AP EAMCET-25.09.2020
Rotational Motion
149799
A disc spinning at the rate \(27.5 \mathrm{rad} \mathrm{s}^{-1}\) is slowed at the rate 10 rad \(\mathrm{s}^{-2}\). The time after which it will come to rest is
1 \(2.75 \mathrm{~s}\)
2 \(5.5 \mathrm{~s}\)
3 \(1.25 \mathrm{~s}\)
4 \(3.5 \mathrm{~s}\)
5 \(6.2 \mathrm{~s}\)
Explanation:
A Given, initial angular velocity \(\left(\omega_{i}\right)=27.5\) \(\mathrm{rad} / \mathrm{sec}\), angular acceleration \((\alpha)=-10 \mathrm{rad} / \mathrm{sec}^{2}\), final angular velocity \(\left(\omega_{\mathrm{f}}\right)=0\) According to kinematics law of rotational motion- \(\omega_{\mathrm{f}}=\omega_{\mathrm{i}}+\alpha \mathrm{t}\) \(\alpha \mathrm{t}=\omega_{\mathrm{f}}-\omega_{\mathrm{i}}\) \(\mathrm{t}=\frac{\omega_{\mathrm{f}}-\omega_{\mathrm{i}}}{\alpha}\) \(\mathrm{t}=\frac{0-27.5}{-10} \Rightarrow \mathrm{t}=\frac{27.5}{10}\) \(\mathrm{t}=2.75 \mathrm{sec}\)
Kerala CEE 2020
Rotational Motion
149800
Assume proton is rotating along a circular path of radius \(1 \mathrm{~m}\) under a centrifugal force of \(4 \times 10^{-12} \mathrm{~N}\). If the mass of proton is \(1.6 \times 10^{-27} \mathrm{~kg}\), then its angular velocity of rotation is
149801
A wheel starting from rest gains an angular velocity of \(10 \mathrm{rad} / \mathrm{s}\) after uniformly accelerated for \(5 \mathrm{~s}\). The total angle through which it has turned is :
1 \(25 \mathrm{rad}\)
2 \(100 \mathrm{rad}\)
3 \(25 \pi \mathrm{rad}\)
4 \(50 \pi \mathrm{rad}\) and a vertical axis
Explanation:
A Given that, \(\omega_{\mathrm{i}}=\mathrm{rad} / \mathrm{sec}, \omega_{\mathrm{f}}=10 \mathrm{rad} / \mathrm{sec}, \mathrm{t}=\) \(5 \mathrm{sec}\) According to kinematics law of rotational motion- \(\omega_{\mathrm{f}}=\omega_{\mathrm{i}}+\alpha \mathrm{t}\) \(10=0+\alpha \times 5\) \(\alpha=2\) Now, \(\quad \theta=\omega_{\mathrm{i}} \mathrm{t}+1 / 2 \alpha \mathrm{t}^{2}\) \(\theta=(0 \times 5)+1 / 2 \times 2 \times 5^{2}\) \(\theta=25 \mathrm{rad}\).
