144044
What is the least radius of curve on a horizontal road, at which a vehicle can travel with a speed of \(36 \mathrm{~km} / \mathrm{hr}\) at an angle of inclination \(45^{\circ}\) ? [g \(\left.=10 \mathrm{~m} / \mathrm{s}^{2}, \tan 45^{\circ}=1\right]\)
144045
A particle is performing vertical circular motion. The difference in tension at lowest and highest point is
1 \(8 \mathrm{mg}\)
2 \(2 \mathrm{mg}\)
3 \(6 \mathrm{mg}\)
4 \(4 \mathrm{mg}\)
Explanation:
C According to the question Let, \(\quad \mathrm{T}_{1}=\) Tension at lowest point \(\mathrm{T}_{2}=\) Tension at highest point \(\mathrm{v}=\) Velocity at highest point \(\mathrm{u}=\) Velocity at lowest point At the lowest point- \(\mathrm{T}_{1}=\frac{\mathrm{mu}^{2}}{\mathrm{R}}+\mathrm{mg}\) At the highest point- Substracting equation (i) from (ii) \(\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)=\left(\frac{\mathrm{mu}^{2}}{\mathrm{R}}+\mathrm{mg}\right)-\left(\frac{\mathrm{mv}^{2}}{\mathrm{R}}-\mathrm{mg}\right)\) \(\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)=\frac{\mathrm{m}}{\mathrm{R}}\left(\mathrm{u}^{2}-\mathrm{v}^{2}\right)+2 \mathrm{mg}\) From conservation of energy, \((\mathrm{P} . \mathrm{E}+\mathrm{K} . \mathrm{E})_{\text {Lowest }}=(\mathrm{P} . \mathrm{E}+\mathrm{K} . \mathrm{E})_{\text {Highest }}\) \(0+\frac{1}{2} \mathrm{mu}^{2}=\mathrm{mg} 2 \mathrm{R}+\frac{1}{2} \mathrm{mv}^{2}\) \(\frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}-\mathrm{v}^{2}\right)=\mathrm{mg} 2 \mathrm{R}\) \(\mathrm{u}^{2}-\mathrm{v}^{2}=4 \mathrm{gR}\) Putting these value in equation (iii) \(\mathrm{T}_{1}-\mathrm{T}_{2} =\frac{\mathrm{m}}{\mathrm{R}} \times 4 \mathrm{gR}+2 \mathrm{mg}\) \(=4 \mathrm{mg}+2 \mathrm{mg}=6 \mathrm{mg}\)
MHT-CET 2020
Motion in Plane
144046
A particle is moving along the circular path with constant speed and centripetal acceleration ' \(a\) '. If the speed is doubled, the ratio of its acceleration after and before the change is
1 \(4: 1\)
2 \(2: 1\)
3 \(3: 1\)
4 \(1: 4\)
Explanation:
A We know, Centripetal acceleration \((a)=\frac{v^{2}}{r}\) Acceleration before change, \(a=\frac{v^{2}}{r}\) and acceleration after change, \(a_{2}=\frac{(2 v)^{2}}{r}=\frac{4 v^{2}}{r}\) Ratio of both acceleration, \(\frac{\mathrm{a}_{2}}{\mathrm{a}}=\frac{4 \mathrm{v}^{2} / \mathrm{r}}{\mathrm{v}^{2} / \mathrm{r}} \Rightarrow \frac{\mathrm{a}_{2}}{\mathrm{a}}=\frac{4}{1}\)
MHT-CET 2019
Motion in Plane
144047
A rod of length ' \(L\) ' is hung from its one end and a mass ' \(m\) ' is attached to its free end. What tangential velocity must be imparted to ' \(m\) '. So that it reaches the top of the vertical circle? (g = acceleration due to gravity)
1 \(4 \sqrt{\mathrm{gL}}\)
2 \(2 \sqrt{\mathrm{gL}}\)
3 \(5 \sqrt{\mathrm{gL}}\)
4 \(3 \sqrt{\mathrm{gL}}\)
Explanation:
B According to the question From conservation of energy, \((\mathrm{K} \cdot \mathrm{E})_{\mathrm{A}}+(\mathrm{P} . \mathrm{E})_{\mathrm{A}}=(\mathrm{K} \cdot \mathrm{E})_{\mathrm{B}}+(\mathrm{P} . \mathrm{E})_{\mathrm{B}}\) \(\frac{1}{2} \mathrm{mv}^{2}+0=0+\mathrm{mg} \times(2 \mathrm{~L})\) \(\frac{1}{2} \mathrm{mv} \mathrm{v}^{2}=\mathrm{mg} \times 2 \mathrm{~L}\) \(\frac{1}{2} \mathrm{v}^{2}=2 \mathrm{gL}\) \(\mathrm{v}^{2}=4 \mathrm{gL}\) \(\mathrm{v}=2 \sqrt{\mathrm{gL}}\)
144044
What is the least radius of curve on a horizontal road, at which a vehicle can travel with a speed of \(36 \mathrm{~km} / \mathrm{hr}\) at an angle of inclination \(45^{\circ}\) ? [g \(\left.=10 \mathrm{~m} / \mathrm{s}^{2}, \tan 45^{\circ}=1\right]\)
144045
A particle is performing vertical circular motion. The difference in tension at lowest and highest point is
1 \(8 \mathrm{mg}\)
2 \(2 \mathrm{mg}\)
3 \(6 \mathrm{mg}\)
4 \(4 \mathrm{mg}\)
Explanation:
C According to the question Let, \(\quad \mathrm{T}_{1}=\) Tension at lowest point \(\mathrm{T}_{2}=\) Tension at highest point \(\mathrm{v}=\) Velocity at highest point \(\mathrm{u}=\) Velocity at lowest point At the lowest point- \(\mathrm{T}_{1}=\frac{\mathrm{mu}^{2}}{\mathrm{R}}+\mathrm{mg}\) At the highest point- Substracting equation (i) from (ii) \(\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)=\left(\frac{\mathrm{mu}^{2}}{\mathrm{R}}+\mathrm{mg}\right)-\left(\frac{\mathrm{mv}^{2}}{\mathrm{R}}-\mathrm{mg}\right)\) \(\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)=\frac{\mathrm{m}}{\mathrm{R}}\left(\mathrm{u}^{2}-\mathrm{v}^{2}\right)+2 \mathrm{mg}\) From conservation of energy, \((\mathrm{P} . \mathrm{E}+\mathrm{K} . \mathrm{E})_{\text {Lowest }}=(\mathrm{P} . \mathrm{E}+\mathrm{K} . \mathrm{E})_{\text {Highest }}\) \(0+\frac{1}{2} \mathrm{mu}^{2}=\mathrm{mg} 2 \mathrm{R}+\frac{1}{2} \mathrm{mv}^{2}\) \(\frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}-\mathrm{v}^{2}\right)=\mathrm{mg} 2 \mathrm{R}\) \(\mathrm{u}^{2}-\mathrm{v}^{2}=4 \mathrm{gR}\) Putting these value in equation (iii) \(\mathrm{T}_{1}-\mathrm{T}_{2} =\frac{\mathrm{m}}{\mathrm{R}} \times 4 \mathrm{gR}+2 \mathrm{mg}\) \(=4 \mathrm{mg}+2 \mathrm{mg}=6 \mathrm{mg}\)
MHT-CET 2020
Motion in Plane
144046
A particle is moving along the circular path with constant speed and centripetal acceleration ' \(a\) '. If the speed is doubled, the ratio of its acceleration after and before the change is
1 \(4: 1\)
2 \(2: 1\)
3 \(3: 1\)
4 \(1: 4\)
Explanation:
A We know, Centripetal acceleration \((a)=\frac{v^{2}}{r}\) Acceleration before change, \(a=\frac{v^{2}}{r}\) and acceleration after change, \(a_{2}=\frac{(2 v)^{2}}{r}=\frac{4 v^{2}}{r}\) Ratio of both acceleration, \(\frac{\mathrm{a}_{2}}{\mathrm{a}}=\frac{4 \mathrm{v}^{2} / \mathrm{r}}{\mathrm{v}^{2} / \mathrm{r}} \Rightarrow \frac{\mathrm{a}_{2}}{\mathrm{a}}=\frac{4}{1}\)
MHT-CET 2019
Motion in Plane
144047
A rod of length ' \(L\) ' is hung from its one end and a mass ' \(m\) ' is attached to its free end. What tangential velocity must be imparted to ' \(m\) '. So that it reaches the top of the vertical circle? (g = acceleration due to gravity)
1 \(4 \sqrt{\mathrm{gL}}\)
2 \(2 \sqrt{\mathrm{gL}}\)
3 \(5 \sqrt{\mathrm{gL}}\)
4 \(3 \sqrt{\mathrm{gL}}\)
Explanation:
B According to the question From conservation of energy, \((\mathrm{K} \cdot \mathrm{E})_{\mathrm{A}}+(\mathrm{P} . \mathrm{E})_{\mathrm{A}}=(\mathrm{K} \cdot \mathrm{E})_{\mathrm{B}}+(\mathrm{P} . \mathrm{E})_{\mathrm{B}}\) \(\frac{1}{2} \mathrm{mv}^{2}+0=0+\mathrm{mg} \times(2 \mathrm{~L})\) \(\frac{1}{2} \mathrm{mv} \mathrm{v}^{2}=\mathrm{mg} \times 2 \mathrm{~L}\) \(\frac{1}{2} \mathrm{v}^{2}=2 \mathrm{gL}\) \(\mathrm{v}^{2}=4 \mathrm{gL}\) \(\mathrm{v}=2 \sqrt{\mathrm{gL}}\)
144044
What is the least radius of curve on a horizontal road, at which a vehicle can travel with a speed of \(36 \mathrm{~km} / \mathrm{hr}\) at an angle of inclination \(45^{\circ}\) ? [g \(\left.=10 \mathrm{~m} / \mathrm{s}^{2}, \tan 45^{\circ}=1\right]\)
144045
A particle is performing vertical circular motion. The difference in tension at lowest and highest point is
1 \(8 \mathrm{mg}\)
2 \(2 \mathrm{mg}\)
3 \(6 \mathrm{mg}\)
4 \(4 \mathrm{mg}\)
Explanation:
C According to the question Let, \(\quad \mathrm{T}_{1}=\) Tension at lowest point \(\mathrm{T}_{2}=\) Tension at highest point \(\mathrm{v}=\) Velocity at highest point \(\mathrm{u}=\) Velocity at lowest point At the lowest point- \(\mathrm{T}_{1}=\frac{\mathrm{mu}^{2}}{\mathrm{R}}+\mathrm{mg}\) At the highest point- Substracting equation (i) from (ii) \(\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)=\left(\frac{\mathrm{mu}^{2}}{\mathrm{R}}+\mathrm{mg}\right)-\left(\frac{\mathrm{mv}^{2}}{\mathrm{R}}-\mathrm{mg}\right)\) \(\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)=\frac{\mathrm{m}}{\mathrm{R}}\left(\mathrm{u}^{2}-\mathrm{v}^{2}\right)+2 \mathrm{mg}\) From conservation of energy, \((\mathrm{P} . \mathrm{E}+\mathrm{K} . \mathrm{E})_{\text {Lowest }}=(\mathrm{P} . \mathrm{E}+\mathrm{K} . \mathrm{E})_{\text {Highest }}\) \(0+\frac{1}{2} \mathrm{mu}^{2}=\mathrm{mg} 2 \mathrm{R}+\frac{1}{2} \mathrm{mv}^{2}\) \(\frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}-\mathrm{v}^{2}\right)=\mathrm{mg} 2 \mathrm{R}\) \(\mathrm{u}^{2}-\mathrm{v}^{2}=4 \mathrm{gR}\) Putting these value in equation (iii) \(\mathrm{T}_{1}-\mathrm{T}_{2} =\frac{\mathrm{m}}{\mathrm{R}} \times 4 \mathrm{gR}+2 \mathrm{mg}\) \(=4 \mathrm{mg}+2 \mathrm{mg}=6 \mathrm{mg}\)
MHT-CET 2020
Motion in Plane
144046
A particle is moving along the circular path with constant speed and centripetal acceleration ' \(a\) '. If the speed is doubled, the ratio of its acceleration after and before the change is
1 \(4: 1\)
2 \(2: 1\)
3 \(3: 1\)
4 \(1: 4\)
Explanation:
A We know, Centripetal acceleration \((a)=\frac{v^{2}}{r}\) Acceleration before change, \(a=\frac{v^{2}}{r}\) and acceleration after change, \(a_{2}=\frac{(2 v)^{2}}{r}=\frac{4 v^{2}}{r}\) Ratio of both acceleration, \(\frac{\mathrm{a}_{2}}{\mathrm{a}}=\frac{4 \mathrm{v}^{2} / \mathrm{r}}{\mathrm{v}^{2} / \mathrm{r}} \Rightarrow \frac{\mathrm{a}_{2}}{\mathrm{a}}=\frac{4}{1}\)
MHT-CET 2019
Motion in Plane
144047
A rod of length ' \(L\) ' is hung from its one end and a mass ' \(m\) ' is attached to its free end. What tangential velocity must be imparted to ' \(m\) '. So that it reaches the top of the vertical circle? (g = acceleration due to gravity)
1 \(4 \sqrt{\mathrm{gL}}\)
2 \(2 \sqrt{\mathrm{gL}}\)
3 \(5 \sqrt{\mathrm{gL}}\)
4 \(3 \sqrt{\mathrm{gL}}\)
Explanation:
B According to the question From conservation of energy, \((\mathrm{K} \cdot \mathrm{E})_{\mathrm{A}}+(\mathrm{P} . \mathrm{E})_{\mathrm{A}}=(\mathrm{K} \cdot \mathrm{E})_{\mathrm{B}}+(\mathrm{P} . \mathrm{E})_{\mathrm{B}}\) \(\frac{1}{2} \mathrm{mv}^{2}+0=0+\mathrm{mg} \times(2 \mathrm{~L})\) \(\frac{1}{2} \mathrm{mv} \mathrm{v}^{2}=\mathrm{mg} \times 2 \mathrm{~L}\) \(\frac{1}{2} \mathrm{v}^{2}=2 \mathrm{gL}\) \(\mathrm{v}^{2}=4 \mathrm{gL}\) \(\mathrm{v}=2 \sqrt{\mathrm{gL}}\)
144044
What is the least radius of curve on a horizontal road, at which a vehicle can travel with a speed of \(36 \mathrm{~km} / \mathrm{hr}\) at an angle of inclination \(45^{\circ}\) ? [g \(\left.=10 \mathrm{~m} / \mathrm{s}^{2}, \tan 45^{\circ}=1\right]\)
144045
A particle is performing vertical circular motion. The difference in tension at lowest and highest point is
1 \(8 \mathrm{mg}\)
2 \(2 \mathrm{mg}\)
3 \(6 \mathrm{mg}\)
4 \(4 \mathrm{mg}\)
Explanation:
C According to the question Let, \(\quad \mathrm{T}_{1}=\) Tension at lowest point \(\mathrm{T}_{2}=\) Tension at highest point \(\mathrm{v}=\) Velocity at highest point \(\mathrm{u}=\) Velocity at lowest point At the lowest point- \(\mathrm{T}_{1}=\frac{\mathrm{mu}^{2}}{\mathrm{R}}+\mathrm{mg}\) At the highest point- Substracting equation (i) from (ii) \(\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)=\left(\frac{\mathrm{mu}^{2}}{\mathrm{R}}+\mathrm{mg}\right)-\left(\frac{\mathrm{mv}^{2}}{\mathrm{R}}-\mathrm{mg}\right)\) \(\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)=\frac{\mathrm{m}}{\mathrm{R}}\left(\mathrm{u}^{2}-\mathrm{v}^{2}\right)+2 \mathrm{mg}\) From conservation of energy, \((\mathrm{P} . \mathrm{E}+\mathrm{K} . \mathrm{E})_{\text {Lowest }}=(\mathrm{P} . \mathrm{E}+\mathrm{K} . \mathrm{E})_{\text {Highest }}\) \(0+\frac{1}{2} \mathrm{mu}^{2}=\mathrm{mg} 2 \mathrm{R}+\frac{1}{2} \mathrm{mv}^{2}\) \(\frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}-\mathrm{v}^{2}\right)=\mathrm{mg} 2 \mathrm{R}\) \(\mathrm{u}^{2}-\mathrm{v}^{2}=4 \mathrm{gR}\) Putting these value in equation (iii) \(\mathrm{T}_{1}-\mathrm{T}_{2} =\frac{\mathrm{m}}{\mathrm{R}} \times 4 \mathrm{gR}+2 \mathrm{mg}\) \(=4 \mathrm{mg}+2 \mathrm{mg}=6 \mathrm{mg}\)
MHT-CET 2020
Motion in Plane
144046
A particle is moving along the circular path with constant speed and centripetal acceleration ' \(a\) '. If the speed is doubled, the ratio of its acceleration after and before the change is
1 \(4: 1\)
2 \(2: 1\)
3 \(3: 1\)
4 \(1: 4\)
Explanation:
A We know, Centripetal acceleration \((a)=\frac{v^{2}}{r}\) Acceleration before change, \(a=\frac{v^{2}}{r}\) and acceleration after change, \(a_{2}=\frac{(2 v)^{2}}{r}=\frac{4 v^{2}}{r}\) Ratio of both acceleration, \(\frac{\mathrm{a}_{2}}{\mathrm{a}}=\frac{4 \mathrm{v}^{2} / \mathrm{r}}{\mathrm{v}^{2} / \mathrm{r}} \Rightarrow \frac{\mathrm{a}_{2}}{\mathrm{a}}=\frac{4}{1}\)
MHT-CET 2019
Motion in Plane
144047
A rod of length ' \(L\) ' is hung from its one end and a mass ' \(m\) ' is attached to its free end. What tangential velocity must be imparted to ' \(m\) '. So that it reaches the top of the vertical circle? (g = acceleration due to gravity)
1 \(4 \sqrt{\mathrm{gL}}\)
2 \(2 \sqrt{\mathrm{gL}}\)
3 \(5 \sqrt{\mathrm{gL}}\)
4 \(3 \sqrt{\mathrm{gL}}\)
Explanation:
B According to the question From conservation of energy, \((\mathrm{K} \cdot \mathrm{E})_{\mathrm{A}}+(\mathrm{P} . \mathrm{E})_{\mathrm{A}}=(\mathrm{K} \cdot \mathrm{E})_{\mathrm{B}}+(\mathrm{P} . \mathrm{E})_{\mathrm{B}}\) \(\frac{1}{2} \mathrm{mv}^{2}+0=0+\mathrm{mg} \times(2 \mathrm{~L})\) \(\frac{1}{2} \mathrm{mv} \mathrm{v}^{2}=\mathrm{mg} \times 2 \mathrm{~L}\) \(\frac{1}{2} \mathrm{v}^{2}=2 \mathrm{gL}\) \(\mathrm{v}^{2}=4 \mathrm{gL}\) \(\mathrm{v}=2 \sqrt{\mathrm{gL}}\)
