150290
A door \(1.6 \mathrm{~m}\) wide requires a force of \(1 \mathrm{~N}\) to be applied at the free end to open or close it. The force that is required at a point \(0.4 \mathrm{~m}\) distance from the hinges for opening or closing the door is
1 ) \(1.2 \mathrm{~N}\)
2 ) \(3.6 \mathrm{~N}\)
3 ) \(2.4 \mathrm{~N}\)
4 ) \(4 \mathrm{~N}\)
Explanation:
D Exp: (D) : Given that, \(\mathrm{F}=1 \mathrm{~N}\) \(\mathrm{d}=1.6 \mathrm{~m}\) So, torque, \(\tau=\mathrm{F} \times \mathrm{d}\) \(\tau=1 \times 1.6 \Rightarrow 1.6 \mathrm{Nm}\) Now force is required to open the door at distance 0.4 from hinges point, \(\tau =\mathrm{F}_{1} \times \mathrm{d}_{1}\) \(\mathrm{~F}_{1} =\frac{\tau}{\mathrm{d}_{1}}\) \(=\frac{1.6}{0.4}\) \(\mathrm{~F}_{1} =4 \mathrm{~N}\)
Manipal UGET-2009
Rotational Motion
150291
If a street light of mass \(M\) is suspended from the end of a uniform rod of length \(L\) in different possible patterns as shown in figure, then: Cable
1 ) Pattern \(\mathrm{A}\) is more sturdy
2 ) Pattern B is more sturdy
3 ) Pattern C is more sturdy
4 ) All will have same sturdiness
Explanation:
A Exp: (A) : If \(\tau\) be the torque and \(T\) be tension in the string and \(\mathrm{L}\) be perpendicular distance of cable from the axis of rotation then, \(\tau=\mathrm{T} \cdot \mathrm{L}\) Tension is same in all three cases, so the perpendicular distance between the line of action of tension to the axis of rotation. Now, from the figure we can see that in case A the perpendicular distance is maximum but in other two cases it is less than case A. Therefore the correct option is (A) Pattern A is more sturdy.
AIIMS-2006
Rotational Motion
150292
Three uniform circular discs, each of mass \(M\) and radius \(R\) are kept in contact with each other as shown in the figure. Moment of inertia of the system about the axis \(A B\) is
1 ) \(\frac{7}{4} \mathrm{MR}^{2}\)
2 ) \(\frac{11}{4} \mathrm{MR}^{2}\)
3 ) \(\frac{11}{2} \mathrm{MR}^{2}\)
4 ) \(\frac{M R^{2}}{4}\)
Explanation:
B Exp: (B) : According to question - Moment of inertia of system - \(=(\mathrm{M} . \mathrm{I})_{\mathrm{P}}+(\mathrm{M} . \mathrm{I})_{\mathrm{Q}}+(\mathrm{M} . \mathrm{I} .)_{\mathrm{R}}\) \(=\frac{5}{4} \mathrm{MR}^{2}+\frac{5}{4} \mathrm{MR}^{2}+\frac{1}{4} \mathrm{MR}^{2}\) \(\mathrm{M} . \mathrm{I})_{\text {system }}=\frac{11}{4} \mathrm{MR}^{2}\)
AP EAMCET(Medical)-2014
Rotational Motion
150293
Four particles each of mass \(m\) are placed at the corners of a square of side length \(l\). The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is:
1 ) \(\sqrt{2} \ell\)
2 ) \(\frac{\ell}{2}\)
3 ) \(l\)
4 ) \(\frac{l}{\sqrt{2}}\)
Explanation:
D Exp: (D) : The distance of mass from center, \(\mathrm{r}=\frac{1}{\sqrt{2}}\) Moment of inertia, \(\mathrm{I}=4 \mathrm{mr}^{2}\) Ans: b Exp: (B) : In pure rolling, mechanical energy remains conserved. When heights of incline are equals speed of sphere will be same in both the cases. \(\therefore 2 \mathrm{~m} l^{2}=4 \mathrm{mk}^{2}\) \(\mathrm{I}=2 \mathrm{ml}^{2}\) \(\mathrm{I}=4 \mathrm{mk}^{2}\) \(\mathrm{k}=\frac{l}{\sqrt{2}} \quad\) (Radius of gyration \(\mathrm{k}\) As acceleration down to the plane, \(\mathrm{a} \propto \sin \theta\) Therefore, acceleration and time of descent will be different.
150290
A door \(1.6 \mathrm{~m}\) wide requires a force of \(1 \mathrm{~N}\) to be applied at the free end to open or close it. The force that is required at a point \(0.4 \mathrm{~m}\) distance from the hinges for opening or closing the door is
1 ) \(1.2 \mathrm{~N}\)
2 ) \(3.6 \mathrm{~N}\)
3 ) \(2.4 \mathrm{~N}\)
4 ) \(4 \mathrm{~N}\)
Explanation:
D Exp: (D) : Given that, \(\mathrm{F}=1 \mathrm{~N}\) \(\mathrm{d}=1.6 \mathrm{~m}\) So, torque, \(\tau=\mathrm{F} \times \mathrm{d}\) \(\tau=1 \times 1.6 \Rightarrow 1.6 \mathrm{Nm}\) Now force is required to open the door at distance 0.4 from hinges point, \(\tau =\mathrm{F}_{1} \times \mathrm{d}_{1}\) \(\mathrm{~F}_{1} =\frac{\tau}{\mathrm{d}_{1}}\) \(=\frac{1.6}{0.4}\) \(\mathrm{~F}_{1} =4 \mathrm{~N}\)
Manipal UGET-2009
Rotational Motion
150291
If a street light of mass \(M\) is suspended from the end of a uniform rod of length \(L\) in different possible patterns as shown in figure, then: Cable
1 ) Pattern \(\mathrm{A}\) is more sturdy
2 ) Pattern B is more sturdy
3 ) Pattern C is more sturdy
4 ) All will have same sturdiness
Explanation:
A Exp: (A) : If \(\tau\) be the torque and \(T\) be tension in the string and \(\mathrm{L}\) be perpendicular distance of cable from the axis of rotation then, \(\tau=\mathrm{T} \cdot \mathrm{L}\) Tension is same in all three cases, so the perpendicular distance between the line of action of tension to the axis of rotation. Now, from the figure we can see that in case A the perpendicular distance is maximum but in other two cases it is less than case A. Therefore the correct option is (A) Pattern A is more sturdy.
AIIMS-2006
Rotational Motion
150292
Three uniform circular discs, each of mass \(M\) and radius \(R\) are kept in contact with each other as shown in the figure. Moment of inertia of the system about the axis \(A B\) is
1 ) \(\frac{7}{4} \mathrm{MR}^{2}\)
2 ) \(\frac{11}{4} \mathrm{MR}^{2}\)
3 ) \(\frac{11}{2} \mathrm{MR}^{2}\)
4 ) \(\frac{M R^{2}}{4}\)
Explanation:
B Exp: (B) : According to question - Moment of inertia of system - \(=(\mathrm{M} . \mathrm{I})_{\mathrm{P}}+(\mathrm{M} . \mathrm{I})_{\mathrm{Q}}+(\mathrm{M} . \mathrm{I} .)_{\mathrm{R}}\) \(=\frac{5}{4} \mathrm{MR}^{2}+\frac{5}{4} \mathrm{MR}^{2}+\frac{1}{4} \mathrm{MR}^{2}\) \(\mathrm{M} . \mathrm{I})_{\text {system }}=\frac{11}{4} \mathrm{MR}^{2}\)
AP EAMCET(Medical)-2014
Rotational Motion
150293
Four particles each of mass \(m\) are placed at the corners of a square of side length \(l\). The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is:
1 ) \(\sqrt{2} \ell\)
2 ) \(\frac{\ell}{2}\)
3 ) \(l\)
4 ) \(\frac{l}{\sqrt{2}}\)
Explanation:
D Exp: (D) : The distance of mass from center, \(\mathrm{r}=\frac{1}{\sqrt{2}}\) Moment of inertia, \(\mathrm{I}=4 \mathrm{mr}^{2}\) Ans: b Exp: (B) : In pure rolling, mechanical energy remains conserved. When heights of incline are equals speed of sphere will be same in both the cases. \(\therefore 2 \mathrm{~m} l^{2}=4 \mathrm{mk}^{2}\) \(\mathrm{I}=2 \mathrm{ml}^{2}\) \(\mathrm{I}=4 \mathrm{mk}^{2}\) \(\mathrm{k}=\frac{l}{\sqrt{2}} \quad\) (Radius of gyration \(\mathrm{k}\) As acceleration down to the plane, \(\mathrm{a} \propto \sin \theta\) Therefore, acceleration and time of descent will be different.
150290
A door \(1.6 \mathrm{~m}\) wide requires a force of \(1 \mathrm{~N}\) to be applied at the free end to open or close it. The force that is required at a point \(0.4 \mathrm{~m}\) distance from the hinges for opening or closing the door is
1 ) \(1.2 \mathrm{~N}\)
2 ) \(3.6 \mathrm{~N}\)
3 ) \(2.4 \mathrm{~N}\)
4 ) \(4 \mathrm{~N}\)
Explanation:
D Exp: (D) : Given that, \(\mathrm{F}=1 \mathrm{~N}\) \(\mathrm{d}=1.6 \mathrm{~m}\) So, torque, \(\tau=\mathrm{F} \times \mathrm{d}\) \(\tau=1 \times 1.6 \Rightarrow 1.6 \mathrm{Nm}\) Now force is required to open the door at distance 0.4 from hinges point, \(\tau =\mathrm{F}_{1} \times \mathrm{d}_{1}\) \(\mathrm{~F}_{1} =\frac{\tau}{\mathrm{d}_{1}}\) \(=\frac{1.6}{0.4}\) \(\mathrm{~F}_{1} =4 \mathrm{~N}\)
Manipal UGET-2009
Rotational Motion
150291
If a street light of mass \(M\) is suspended from the end of a uniform rod of length \(L\) in different possible patterns as shown in figure, then: Cable
1 ) Pattern \(\mathrm{A}\) is more sturdy
2 ) Pattern B is more sturdy
3 ) Pattern C is more sturdy
4 ) All will have same sturdiness
Explanation:
A Exp: (A) : If \(\tau\) be the torque and \(T\) be tension in the string and \(\mathrm{L}\) be perpendicular distance of cable from the axis of rotation then, \(\tau=\mathrm{T} \cdot \mathrm{L}\) Tension is same in all three cases, so the perpendicular distance between the line of action of tension to the axis of rotation. Now, from the figure we can see that in case A the perpendicular distance is maximum but in other two cases it is less than case A. Therefore the correct option is (A) Pattern A is more sturdy.
AIIMS-2006
Rotational Motion
150292
Three uniform circular discs, each of mass \(M\) and radius \(R\) are kept in contact with each other as shown in the figure. Moment of inertia of the system about the axis \(A B\) is
1 ) \(\frac{7}{4} \mathrm{MR}^{2}\)
2 ) \(\frac{11}{4} \mathrm{MR}^{2}\)
3 ) \(\frac{11}{2} \mathrm{MR}^{2}\)
4 ) \(\frac{M R^{2}}{4}\)
Explanation:
B Exp: (B) : According to question - Moment of inertia of system - \(=(\mathrm{M} . \mathrm{I})_{\mathrm{P}}+(\mathrm{M} . \mathrm{I})_{\mathrm{Q}}+(\mathrm{M} . \mathrm{I} .)_{\mathrm{R}}\) \(=\frac{5}{4} \mathrm{MR}^{2}+\frac{5}{4} \mathrm{MR}^{2}+\frac{1}{4} \mathrm{MR}^{2}\) \(\mathrm{M} . \mathrm{I})_{\text {system }}=\frac{11}{4} \mathrm{MR}^{2}\)
AP EAMCET(Medical)-2014
Rotational Motion
150293
Four particles each of mass \(m\) are placed at the corners of a square of side length \(l\). The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is:
1 ) \(\sqrt{2} \ell\)
2 ) \(\frac{\ell}{2}\)
3 ) \(l\)
4 ) \(\frac{l}{\sqrt{2}}\)
Explanation:
D Exp: (D) : The distance of mass from center, \(\mathrm{r}=\frac{1}{\sqrt{2}}\) Moment of inertia, \(\mathrm{I}=4 \mathrm{mr}^{2}\) Ans: b Exp: (B) : In pure rolling, mechanical energy remains conserved. When heights of incline are equals speed of sphere will be same in both the cases. \(\therefore 2 \mathrm{~m} l^{2}=4 \mathrm{mk}^{2}\) \(\mathrm{I}=2 \mathrm{ml}^{2}\) \(\mathrm{I}=4 \mathrm{mk}^{2}\) \(\mathrm{k}=\frac{l}{\sqrt{2}} \quad\) (Radius of gyration \(\mathrm{k}\) As acceleration down to the plane, \(\mathrm{a} \propto \sin \theta\) Therefore, acceleration and time of descent will be different.
150290
A door \(1.6 \mathrm{~m}\) wide requires a force of \(1 \mathrm{~N}\) to be applied at the free end to open or close it. The force that is required at a point \(0.4 \mathrm{~m}\) distance from the hinges for opening or closing the door is
1 ) \(1.2 \mathrm{~N}\)
2 ) \(3.6 \mathrm{~N}\)
3 ) \(2.4 \mathrm{~N}\)
4 ) \(4 \mathrm{~N}\)
Explanation:
D Exp: (D) : Given that, \(\mathrm{F}=1 \mathrm{~N}\) \(\mathrm{d}=1.6 \mathrm{~m}\) So, torque, \(\tau=\mathrm{F} \times \mathrm{d}\) \(\tau=1 \times 1.6 \Rightarrow 1.6 \mathrm{Nm}\) Now force is required to open the door at distance 0.4 from hinges point, \(\tau =\mathrm{F}_{1} \times \mathrm{d}_{1}\) \(\mathrm{~F}_{1} =\frac{\tau}{\mathrm{d}_{1}}\) \(=\frac{1.6}{0.4}\) \(\mathrm{~F}_{1} =4 \mathrm{~N}\)
Manipal UGET-2009
Rotational Motion
150291
If a street light of mass \(M\) is suspended from the end of a uniform rod of length \(L\) in different possible patterns as shown in figure, then: Cable
1 ) Pattern \(\mathrm{A}\) is more sturdy
2 ) Pattern B is more sturdy
3 ) Pattern C is more sturdy
4 ) All will have same sturdiness
Explanation:
A Exp: (A) : If \(\tau\) be the torque and \(T\) be tension in the string and \(\mathrm{L}\) be perpendicular distance of cable from the axis of rotation then, \(\tau=\mathrm{T} \cdot \mathrm{L}\) Tension is same in all three cases, so the perpendicular distance between the line of action of tension to the axis of rotation. Now, from the figure we can see that in case A the perpendicular distance is maximum but in other two cases it is less than case A. Therefore the correct option is (A) Pattern A is more sturdy.
AIIMS-2006
Rotational Motion
150292
Three uniform circular discs, each of mass \(M\) and radius \(R\) are kept in contact with each other as shown in the figure. Moment of inertia of the system about the axis \(A B\) is
1 ) \(\frac{7}{4} \mathrm{MR}^{2}\)
2 ) \(\frac{11}{4} \mathrm{MR}^{2}\)
3 ) \(\frac{11}{2} \mathrm{MR}^{2}\)
4 ) \(\frac{M R^{2}}{4}\)
Explanation:
B Exp: (B) : According to question - Moment of inertia of system - \(=(\mathrm{M} . \mathrm{I})_{\mathrm{P}}+(\mathrm{M} . \mathrm{I})_{\mathrm{Q}}+(\mathrm{M} . \mathrm{I} .)_{\mathrm{R}}\) \(=\frac{5}{4} \mathrm{MR}^{2}+\frac{5}{4} \mathrm{MR}^{2}+\frac{1}{4} \mathrm{MR}^{2}\) \(\mathrm{M} . \mathrm{I})_{\text {system }}=\frac{11}{4} \mathrm{MR}^{2}\)
AP EAMCET(Medical)-2014
Rotational Motion
150293
Four particles each of mass \(m\) are placed at the corners of a square of side length \(l\). The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is:
1 ) \(\sqrt{2} \ell\)
2 ) \(\frac{\ell}{2}\)
3 ) \(l\)
4 ) \(\frac{l}{\sqrt{2}}\)
Explanation:
D Exp: (D) : The distance of mass from center, \(\mathrm{r}=\frac{1}{\sqrt{2}}\) Moment of inertia, \(\mathrm{I}=4 \mathrm{mr}^{2}\) Ans: b Exp: (B) : In pure rolling, mechanical energy remains conserved. When heights of incline are equals speed of sphere will be same in both the cases. \(\therefore 2 \mathrm{~m} l^{2}=4 \mathrm{mk}^{2}\) \(\mathrm{I}=2 \mathrm{ml}^{2}\) \(\mathrm{I}=4 \mathrm{mk}^{2}\) \(\mathrm{k}=\frac{l}{\sqrt{2}} \quad\) (Radius of gyration \(\mathrm{k}\) As acceleration down to the plane, \(\mathrm{a} \propto \sin \theta\) Therefore, acceleration and time of descent will be different.
