163770
If \(I\) is the moment of inertia and \(E\) is the kinetic energy of ratation of a body, then its angular momentum will be :
1 \(\sqrt{\mathrm{EI}}\)
2 \(2 \mathrm{E}\) I
3 \(E / I\)
4 \(\sqrt{(2 \mathrm{El})}\)
Explanation:
Here \(\quad E=\frac{1}{2} \mid \omega^2\) \( \mathrm{L}=\mathrm{I} \omega \quad \therefore \quad \mathrm{L}^2=\mathrm{I} \times \mathrm{I} \omega^2 \) or \( L^2=I \times 2 E \text { or } L=\sqrt{2 E I} \)
NCERT-XI- I -106
4 RBTS PAPER
163771
The angular momentum of a system of particles is not conserved :
1 When net external force acts upon the system
2 When a net external torque acts upon the system
3 When a net external impulse acts upon the system
4 None of these.
Explanation:
If \(\tau \neq 0\), then \(L\) is not constant
NCERT-XI- I -108
4 RBTS PAPER
163772
The centre of an equilateral triangle is 0 . Three forces \(F_1, F_2\) and \(F_3\) are appointed along \(A B, B C\) and \(A C\) respectively. How much the magnitude of \(F_3\) be so that the total torque about \(O\) should be zero:
1 \(\left(F_1-F_2\right)\)
2 \(\left(F_1+F_2\right)\)
3 \(F_1+F_2 / 2\)
4 \(2\left(F_1+F_2\right)\)
Explanation:
Let perpendicular distance of any side of the triangle form \(O\) is \(x\). Taking moment about \(O\), we get \( F_1 x+F_2 x-F_3 x=0 \) \( \Rightarrow F_1+F_2=F_3 \)
NCERT-XI- I -105
4 RBTS PAPER
163773
A disc of mass \(\mathbf{M}\) and radius \(\mathbf{R}\) is free to rotate in vertical plane about an axis \(x x\) ' shown in fig. It is released from topmost position, its velocity of centre of mass at lowest position is:
1 \(\sqrt{g R}\)
2 \(\sqrt{\frac{16}{5} g R}\)
3 \(\sqrt{\frac{8}{3} g R}\)
4 \(\sqrt{8 g R}\)
Explanation:
Applying Conservation of energy \(\mathrm{mg}(2 \mathrm{R})=\frac{1}{2} \mathrm{I} \omega^2\) \(2 \mathrm{mgR}=\frac{1}{2}\left(\frac{5}{4} m R^2\right) \times \frac{v^2}{R^2}\) \(2 \mathrm{mgR}=\frac{5}{8} m v^2\) \(\mathrm{v}^2=\frac{16 g R}{5}\) \( \Rightarrow v=\sqrt{\frac{16 g R}{5}} . \)
NEET Test Series from KOTA - 10 Papers In MS WORD
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4 RBTS PAPER
163770
If \(I\) is the moment of inertia and \(E\) is the kinetic energy of ratation of a body, then its angular momentum will be :
1 \(\sqrt{\mathrm{EI}}\)
2 \(2 \mathrm{E}\) I
3 \(E / I\)
4 \(\sqrt{(2 \mathrm{El})}\)
Explanation:
Here \(\quad E=\frac{1}{2} \mid \omega^2\) \( \mathrm{L}=\mathrm{I} \omega \quad \therefore \quad \mathrm{L}^2=\mathrm{I} \times \mathrm{I} \omega^2 \) or \( L^2=I \times 2 E \text { or } L=\sqrt{2 E I} \)
NCERT-XI- I -106
4 RBTS PAPER
163771
The angular momentum of a system of particles is not conserved :
1 When net external force acts upon the system
2 When a net external torque acts upon the system
3 When a net external impulse acts upon the system
4 None of these.
Explanation:
If \(\tau \neq 0\), then \(L\) is not constant
NCERT-XI- I -108
4 RBTS PAPER
163772
The centre of an equilateral triangle is 0 . Three forces \(F_1, F_2\) and \(F_3\) are appointed along \(A B, B C\) and \(A C\) respectively. How much the magnitude of \(F_3\) be so that the total torque about \(O\) should be zero:
1 \(\left(F_1-F_2\right)\)
2 \(\left(F_1+F_2\right)\)
3 \(F_1+F_2 / 2\)
4 \(2\left(F_1+F_2\right)\)
Explanation:
Let perpendicular distance of any side of the triangle form \(O\) is \(x\). Taking moment about \(O\), we get \( F_1 x+F_2 x-F_3 x=0 \) \( \Rightarrow F_1+F_2=F_3 \)
NCERT-XI- I -105
4 RBTS PAPER
163773
A disc of mass \(\mathbf{M}\) and radius \(\mathbf{R}\) is free to rotate in vertical plane about an axis \(x x\) ' shown in fig. It is released from topmost position, its velocity of centre of mass at lowest position is:
1 \(\sqrt{g R}\)
2 \(\sqrt{\frac{16}{5} g R}\)
3 \(\sqrt{\frac{8}{3} g R}\)
4 \(\sqrt{8 g R}\)
Explanation:
Applying Conservation of energy \(\mathrm{mg}(2 \mathrm{R})=\frac{1}{2} \mathrm{I} \omega^2\) \(2 \mathrm{mgR}=\frac{1}{2}\left(\frac{5}{4} m R^2\right) \times \frac{v^2}{R^2}\) \(2 \mathrm{mgR}=\frac{5}{8} m v^2\) \(\mathrm{v}^2=\frac{16 g R}{5}\) \( \Rightarrow v=\sqrt{\frac{16 g R}{5}} . \)
163770
If \(I\) is the moment of inertia and \(E\) is the kinetic energy of ratation of a body, then its angular momentum will be :
1 \(\sqrt{\mathrm{EI}}\)
2 \(2 \mathrm{E}\) I
3 \(E / I\)
4 \(\sqrt{(2 \mathrm{El})}\)
Explanation:
Here \(\quad E=\frac{1}{2} \mid \omega^2\) \( \mathrm{L}=\mathrm{I} \omega \quad \therefore \quad \mathrm{L}^2=\mathrm{I} \times \mathrm{I} \omega^2 \) or \( L^2=I \times 2 E \text { or } L=\sqrt{2 E I} \)
NCERT-XI- I -106
4 RBTS PAPER
163771
The angular momentum of a system of particles is not conserved :
1 When net external force acts upon the system
2 When a net external torque acts upon the system
3 When a net external impulse acts upon the system
4 None of these.
Explanation:
If \(\tau \neq 0\), then \(L\) is not constant
NCERT-XI- I -108
4 RBTS PAPER
163772
The centre of an equilateral triangle is 0 . Three forces \(F_1, F_2\) and \(F_3\) are appointed along \(A B, B C\) and \(A C\) respectively. How much the magnitude of \(F_3\) be so that the total torque about \(O\) should be zero:
1 \(\left(F_1-F_2\right)\)
2 \(\left(F_1+F_2\right)\)
3 \(F_1+F_2 / 2\)
4 \(2\left(F_1+F_2\right)\)
Explanation:
Let perpendicular distance of any side of the triangle form \(O\) is \(x\). Taking moment about \(O\), we get \( F_1 x+F_2 x-F_3 x=0 \) \( \Rightarrow F_1+F_2=F_3 \)
NCERT-XI- I -105
4 RBTS PAPER
163773
A disc of mass \(\mathbf{M}\) and radius \(\mathbf{R}\) is free to rotate in vertical plane about an axis \(x x\) ' shown in fig. It is released from topmost position, its velocity of centre of mass at lowest position is:
1 \(\sqrt{g R}\)
2 \(\sqrt{\frac{16}{5} g R}\)
3 \(\sqrt{\frac{8}{3} g R}\)
4 \(\sqrt{8 g R}\)
Explanation:
Applying Conservation of energy \(\mathrm{mg}(2 \mathrm{R})=\frac{1}{2} \mathrm{I} \omega^2\) \(2 \mathrm{mgR}=\frac{1}{2}\left(\frac{5}{4} m R^2\right) \times \frac{v^2}{R^2}\) \(2 \mathrm{mgR}=\frac{5}{8} m v^2\) \(\mathrm{v}^2=\frac{16 g R}{5}\) \( \Rightarrow v=\sqrt{\frac{16 g R}{5}} . \)
163770
If \(I\) is the moment of inertia and \(E\) is the kinetic energy of ratation of a body, then its angular momentum will be :
1 \(\sqrt{\mathrm{EI}}\)
2 \(2 \mathrm{E}\) I
3 \(E / I\)
4 \(\sqrt{(2 \mathrm{El})}\)
Explanation:
Here \(\quad E=\frac{1}{2} \mid \omega^2\) \( \mathrm{L}=\mathrm{I} \omega \quad \therefore \quad \mathrm{L}^2=\mathrm{I} \times \mathrm{I} \omega^2 \) or \( L^2=I \times 2 E \text { or } L=\sqrt{2 E I} \)
NCERT-XI- I -106
4 RBTS PAPER
163771
The angular momentum of a system of particles is not conserved :
1 When net external force acts upon the system
2 When a net external torque acts upon the system
3 When a net external impulse acts upon the system
4 None of these.
Explanation:
If \(\tau \neq 0\), then \(L\) is not constant
NCERT-XI- I -108
4 RBTS PAPER
163772
The centre of an equilateral triangle is 0 . Three forces \(F_1, F_2\) and \(F_3\) are appointed along \(A B, B C\) and \(A C\) respectively. How much the magnitude of \(F_3\) be so that the total torque about \(O\) should be zero:
1 \(\left(F_1-F_2\right)\)
2 \(\left(F_1+F_2\right)\)
3 \(F_1+F_2 / 2\)
4 \(2\left(F_1+F_2\right)\)
Explanation:
Let perpendicular distance of any side of the triangle form \(O\) is \(x\). Taking moment about \(O\), we get \( F_1 x+F_2 x-F_3 x=0 \) \( \Rightarrow F_1+F_2=F_3 \)
NCERT-XI- I -105
4 RBTS PAPER
163773
A disc of mass \(\mathbf{M}\) and radius \(\mathbf{R}\) is free to rotate in vertical plane about an axis \(x x\) ' shown in fig. It is released from topmost position, its velocity of centre of mass at lowest position is:
1 \(\sqrt{g R}\)
2 \(\sqrt{\frac{16}{5} g R}\)
3 \(\sqrt{\frac{8}{3} g R}\)
4 \(\sqrt{8 g R}\)
Explanation:
Applying Conservation of energy \(\mathrm{mg}(2 \mathrm{R})=\frac{1}{2} \mathrm{I} \omega^2\) \(2 \mathrm{mgR}=\frac{1}{2}\left(\frac{5}{4} m R^2\right) \times \frac{v^2}{R^2}\) \(2 \mathrm{mgR}=\frac{5}{8} m v^2\) \(\mathrm{v}^2=\frac{16 g R}{5}\) \( \Rightarrow v=\sqrt{\frac{16 g R}{5}} . \)
