144065
A cyclist covers a circular path \(34.3 \mathrm{~m}\) long in \(\sqrt{22}\) sec. The angle of inclination of the cyclist is (Given \(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\) )
1 \(50^{\circ}\)
2 \(45^{\circ}\)
3 \(30^{\circ}\)
4 \(60^{\circ}\)
Explanation:
B Let \(r\) be the radius of circular path, The length of path \(=2 \pi \mathrm{r}=34.3 \mathrm{~m}, \mathrm{t}=\sqrt{22} \mathrm{sec}\) \(\mathrm{r}=\frac{34.3}{2 \pi} \mathrm{m}\) Then, \(\quad\) Velocity \(=\frac{\text { Length of path }}{\text { Time }}=\frac{34.3}{\sqrt{22}}\) \(\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}}=\frac{\left(\frac{34.3}{\sqrt{22}}\right)^{2}}{\frac{34.3}{2 \pi} \times 9.8}=0.999\) \(\tan \theta \simeq 1=\tan 45^{\circ}\) So, \(\quad \theta=45^{\circ}\)
J and K-CET-2000
Motion in Plane
144066
A car of mass \(1000 \mathrm{~kg}\) moves on a circular road with speed of \(20 \mathrm{~m} / \mathrm{s}\). Its direction changes by \(90^{\circ}\) after travelling \(628 \mathrm{~m}\) on the road. The centripetal force acting on the car is
1 \(7500 \mathrm{~N}\)
2 \(750 \mathrm{~N}\)
3 \(1000 \mathrm{~N}\)
4 \(1500 \mathrm{~N}\)
Explanation:
C Given, \(\mathrm{v}=20 \mathrm{~m} / \mathrm{s}, \mathrm{m}=1000 \mathrm{~kg}\) Since it changes its direction after travelling \(628 \mathrm{~m}\) So, \(\quad \mathrm{r} \times \frac{\pi}{2}=628\) \(r=399.79 \simeq 400 \mathrm{~m}\) Then, Centripetal force \(=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) \(=\frac{1000 \times(20)^{2}}{400}=\frac{1000 \times 400}{400}=1000 \mathrm{~N}\)
J and K-CET-1997
Motion in Plane
144067
A particle moving in uniform circular motion has its radius and tangential speed doubled. Its centripetal acceleration is then
1 tripled
2 the same as before
3 doubled
4 become four times
Explanation:
C Given, \(\mathrm{v}_{2}=2 \mathrm{v}_{1}, \mathrm{R}_{2}=2 \mathrm{R}_{1}\) We know that, Centripetal acceleration \((a)=\frac{v^{2}}{R}\) Now, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{v}_{1}^{2}}{\mathrm{R}_{1}} \times \frac{\mathrm{R}_{2}}{\mathrm{v}_{2}^{2}}=\left(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}\right)^{2} \times \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\) \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\left(\frac{\mathrm{v}_{1}}{2 \mathrm{v}_{1}}\right)^{2} \times\left(\frac{2 \mathrm{R}_{1}}{\mathrm{R}_{1}}\right)\) \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{1}{2}\) \(\mathrm{a}_{2}=2 \mathrm{a}_{1}\) So, its acceleration will be doubled.
J and K-CET-1997
Motion in Plane
144070
A brick of mass \(\mathrm{m}\), tied to a rope, is being whirled in a vertical circle, with a uniform speed. The tension in the rope is
1 the same throughout
2 largest when the brick is at the highest point of the circular path and smallest when it is at the lowest point.
3 largest when the rope is horizontal and smallest when it is vertical
4 largest when the brick is at the lowest point and smallest when it is at the highest point
Explanation:
D When the brick is following a circular path in vertical direction then it is being acted by two forces, one due to the motion of brick that is centrifugal force and other is due to the action of gravity that is force of gravity. \(\mathrm{T}_{\mathrm{H}}+\mathrm{mg}=\mathrm{m} \omega^{2} \mathrm{r}\) \(\mathrm{T}_{\mathrm{L}}-\mathrm{mg}=\mathrm{m} \omega^{2} \mathrm{r}\) \(\mathrm{T}_{\mathrm{L}}>\mathrm{T}_{\mathrm{H}}\) Tension in largest when the brick is at the lowest point and smallest when it is at the highest point.
144065
A cyclist covers a circular path \(34.3 \mathrm{~m}\) long in \(\sqrt{22}\) sec. The angle of inclination of the cyclist is (Given \(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\) )
1 \(50^{\circ}\)
2 \(45^{\circ}\)
3 \(30^{\circ}\)
4 \(60^{\circ}\)
Explanation:
B Let \(r\) be the radius of circular path, The length of path \(=2 \pi \mathrm{r}=34.3 \mathrm{~m}, \mathrm{t}=\sqrt{22} \mathrm{sec}\) \(\mathrm{r}=\frac{34.3}{2 \pi} \mathrm{m}\) Then, \(\quad\) Velocity \(=\frac{\text { Length of path }}{\text { Time }}=\frac{34.3}{\sqrt{22}}\) \(\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}}=\frac{\left(\frac{34.3}{\sqrt{22}}\right)^{2}}{\frac{34.3}{2 \pi} \times 9.8}=0.999\) \(\tan \theta \simeq 1=\tan 45^{\circ}\) So, \(\quad \theta=45^{\circ}\)
J and K-CET-2000
Motion in Plane
144066
A car of mass \(1000 \mathrm{~kg}\) moves on a circular road with speed of \(20 \mathrm{~m} / \mathrm{s}\). Its direction changes by \(90^{\circ}\) after travelling \(628 \mathrm{~m}\) on the road. The centripetal force acting on the car is
1 \(7500 \mathrm{~N}\)
2 \(750 \mathrm{~N}\)
3 \(1000 \mathrm{~N}\)
4 \(1500 \mathrm{~N}\)
Explanation:
C Given, \(\mathrm{v}=20 \mathrm{~m} / \mathrm{s}, \mathrm{m}=1000 \mathrm{~kg}\) Since it changes its direction after travelling \(628 \mathrm{~m}\) So, \(\quad \mathrm{r} \times \frac{\pi}{2}=628\) \(r=399.79 \simeq 400 \mathrm{~m}\) Then, Centripetal force \(=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) \(=\frac{1000 \times(20)^{2}}{400}=\frac{1000 \times 400}{400}=1000 \mathrm{~N}\)
J and K-CET-1997
Motion in Plane
144067
A particle moving in uniform circular motion has its radius and tangential speed doubled. Its centripetal acceleration is then
1 tripled
2 the same as before
3 doubled
4 become four times
Explanation:
C Given, \(\mathrm{v}_{2}=2 \mathrm{v}_{1}, \mathrm{R}_{2}=2 \mathrm{R}_{1}\) We know that, Centripetal acceleration \((a)=\frac{v^{2}}{R}\) Now, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{v}_{1}^{2}}{\mathrm{R}_{1}} \times \frac{\mathrm{R}_{2}}{\mathrm{v}_{2}^{2}}=\left(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}\right)^{2} \times \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\) \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\left(\frac{\mathrm{v}_{1}}{2 \mathrm{v}_{1}}\right)^{2} \times\left(\frac{2 \mathrm{R}_{1}}{\mathrm{R}_{1}}\right)\) \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{1}{2}\) \(\mathrm{a}_{2}=2 \mathrm{a}_{1}\) So, its acceleration will be doubled.
J and K-CET-1997
Motion in Plane
144070
A brick of mass \(\mathrm{m}\), tied to a rope, is being whirled in a vertical circle, with a uniform speed. The tension in the rope is
1 the same throughout
2 largest when the brick is at the highest point of the circular path and smallest when it is at the lowest point.
3 largest when the rope is horizontal and smallest when it is vertical
4 largest when the brick is at the lowest point and smallest when it is at the highest point
Explanation:
D When the brick is following a circular path in vertical direction then it is being acted by two forces, one due to the motion of brick that is centrifugal force and other is due to the action of gravity that is force of gravity. \(\mathrm{T}_{\mathrm{H}}+\mathrm{mg}=\mathrm{m} \omega^{2} \mathrm{r}\) \(\mathrm{T}_{\mathrm{L}}-\mathrm{mg}=\mathrm{m} \omega^{2} \mathrm{r}\) \(\mathrm{T}_{\mathrm{L}}>\mathrm{T}_{\mathrm{H}}\) Tension in largest when the brick is at the lowest point and smallest when it is at the highest point.
144065
A cyclist covers a circular path \(34.3 \mathrm{~m}\) long in \(\sqrt{22}\) sec. The angle of inclination of the cyclist is (Given \(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\) )
1 \(50^{\circ}\)
2 \(45^{\circ}\)
3 \(30^{\circ}\)
4 \(60^{\circ}\)
Explanation:
B Let \(r\) be the radius of circular path, The length of path \(=2 \pi \mathrm{r}=34.3 \mathrm{~m}, \mathrm{t}=\sqrt{22} \mathrm{sec}\) \(\mathrm{r}=\frac{34.3}{2 \pi} \mathrm{m}\) Then, \(\quad\) Velocity \(=\frac{\text { Length of path }}{\text { Time }}=\frac{34.3}{\sqrt{22}}\) \(\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}}=\frac{\left(\frac{34.3}{\sqrt{22}}\right)^{2}}{\frac{34.3}{2 \pi} \times 9.8}=0.999\) \(\tan \theta \simeq 1=\tan 45^{\circ}\) So, \(\quad \theta=45^{\circ}\)
J and K-CET-2000
Motion in Plane
144066
A car of mass \(1000 \mathrm{~kg}\) moves on a circular road with speed of \(20 \mathrm{~m} / \mathrm{s}\). Its direction changes by \(90^{\circ}\) after travelling \(628 \mathrm{~m}\) on the road. The centripetal force acting on the car is
1 \(7500 \mathrm{~N}\)
2 \(750 \mathrm{~N}\)
3 \(1000 \mathrm{~N}\)
4 \(1500 \mathrm{~N}\)
Explanation:
C Given, \(\mathrm{v}=20 \mathrm{~m} / \mathrm{s}, \mathrm{m}=1000 \mathrm{~kg}\) Since it changes its direction after travelling \(628 \mathrm{~m}\) So, \(\quad \mathrm{r} \times \frac{\pi}{2}=628\) \(r=399.79 \simeq 400 \mathrm{~m}\) Then, Centripetal force \(=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) \(=\frac{1000 \times(20)^{2}}{400}=\frac{1000 \times 400}{400}=1000 \mathrm{~N}\)
J and K-CET-1997
Motion in Plane
144067
A particle moving in uniform circular motion has its radius and tangential speed doubled. Its centripetal acceleration is then
1 tripled
2 the same as before
3 doubled
4 become four times
Explanation:
C Given, \(\mathrm{v}_{2}=2 \mathrm{v}_{1}, \mathrm{R}_{2}=2 \mathrm{R}_{1}\) We know that, Centripetal acceleration \((a)=\frac{v^{2}}{R}\) Now, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{v}_{1}^{2}}{\mathrm{R}_{1}} \times \frac{\mathrm{R}_{2}}{\mathrm{v}_{2}^{2}}=\left(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}\right)^{2} \times \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\) \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\left(\frac{\mathrm{v}_{1}}{2 \mathrm{v}_{1}}\right)^{2} \times\left(\frac{2 \mathrm{R}_{1}}{\mathrm{R}_{1}}\right)\) \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{1}{2}\) \(\mathrm{a}_{2}=2 \mathrm{a}_{1}\) So, its acceleration will be doubled.
J and K-CET-1997
Motion in Plane
144070
A brick of mass \(\mathrm{m}\), tied to a rope, is being whirled in a vertical circle, with a uniform speed. The tension in the rope is
1 the same throughout
2 largest when the brick is at the highest point of the circular path and smallest when it is at the lowest point.
3 largest when the rope is horizontal and smallest when it is vertical
4 largest when the brick is at the lowest point and smallest when it is at the highest point
Explanation:
D When the brick is following a circular path in vertical direction then it is being acted by two forces, one due to the motion of brick that is centrifugal force and other is due to the action of gravity that is force of gravity. \(\mathrm{T}_{\mathrm{H}}+\mathrm{mg}=\mathrm{m} \omega^{2} \mathrm{r}\) \(\mathrm{T}_{\mathrm{L}}-\mathrm{mg}=\mathrm{m} \omega^{2} \mathrm{r}\) \(\mathrm{T}_{\mathrm{L}}>\mathrm{T}_{\mathrm{H}}\) Tension in largest when the brick is at the lowest point and smallest when it is at the highest point.
144065
A cyclist covers a circular path \(34.3 \mathrm{~m}\) long in \(\sqrt{22}\) sec. The angle of inclination of the cyclist is (Given \(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\) )
1 \(50^{\circ}\)
2 \(45^{\circ}\)
3 \(30^{\circ}\)
4 \(60^{\circ}\)
Explanation:
B Let \(r\) be the radius of circular path, The length of path \(=2 \pi \mathrm{r}=34.3 \mathrm{~m}, \mathrm{t}=\sqrt{22} \mathrm{sec}\) \(\mathrm{r}=\frac{34.3}{2 \pi} \mathrm{m}\) Then, \(\quad\) Velocity \(=\frac{\text { Length of path }}{\text { Time }}=\frac{34.3}{\sqrt{22}}\) \(\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}}=\frac{\left(\frac{34.3}{\sqrt{22}}\right)^{2}}{\frac{34.3}{2 \pi} \times 9.8}=0.999\) \(\tan \theta \simeq 1=\tan 45^{\circ}\) So, \(\quad \theta=45^{\circ}\)
J and K-CET-2000
Motion in Plane
144066
A car of mass \(1000 \mathrm{~kg}\) moves on a circular road with speed of \(20 \mathrm{~m} / \mathrm{s}\). Its direction changes by \(90^{\circ}\) after travelling \(628 \mathrm{~m}\) on the road. The centripetal force acting on the car is
1 \(7500 \mathrm{~N}\)
2 \(750 \mathrm{~N}\)
3 \(1000 \mathrm{~N}\)
4 \(1500 \mathrm{~N}\)
Explanation:
C Given, \(\mathrm{v}=20 \mathrm{~m} / \mathrm{s}, \mathrm{m}=1000 \mathrm{~kg}\) Since it changes its direction after travelling \(628 \mathrm{~m}\) So, \(\quad \mathrm{r} \times \frac{\pi}{2}=628\) \(r=399.79 \simeq 400 \mathrm{~m}\) Then, Centripetal force \(=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) \(=\frac{1000 \times(20)^{2}}{400}=\frac{1000 \times 400}{400}=1000 \mathrm{~N}\)
J and K-CET-1997
Motion in Plane
144067
A particle moving in uniform circular motion has its radius and tangential speed doubled. Its centripetal acceleration is then
1 tripled
2 the same as before
3 doubled
4 become four times
Explanation:
C Given, \(\mathrm{v}_{2}=2 \mathrm{v}_{1}, \mathrm{R}_{2}=2 \mathrm{R}_{1}\) We know that, Centripetal acceleration \((a)=\frac{v^{2}}{R}\) Now, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{v}_{1}^{2}}{\mathrm{R}_{1}} \times \frac{\mathrm{R}_{2}}{\mathrm{v}_{2}^{2}}=\left(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}\right)^{2} \times \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\) \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\left(\frac{\mathrm{v}_{1}}{2 \mathrm{v}_{1}}\right)^{2} \times\left(\frac{2 \mathrm{R}_{1}}{\mathrm{R}_{1}}\right)\) \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{1}{2}\) \(\mathrm{a}_{2}=2 \mathrm{a}_{1}\) So, its acceleration will be doubled.
J and K-CET-1997
Motion in Plane
144070
A brick of mass \(\mathrm{m}\), tied to a rope, is being whirled in a vertical circle, with a uniform speed. The tension in the rope is
1 the same throughout
2 largest when the brick is at the highest point of the circular path and smallest when it is at the lowest point.
3 largest when the rope is horizontal and smallest when it is vertical
4 largest when the brick is at the lowest point and smallest when it is at the highest point
Explanation:
D When the brick is following a circular path in vertical direction then it is being acted by two forces, one due to the motion of brick that is centrifugal force and other is due to the action of gravity that is force of gravity. \(\mathrm{T}_{\mathrm{H}}+\mathrm{mg}=\mathrm{m} \omega^{2} \mathrm{r}\) \(\mathrm{T}_{\mathrm{L}}-\mathrm{mg}=\mathrm{m} \omega^{2} \mathrm{r}\) \(\mathrm{T}_{\mathrm{L}}>\mathrm{T}_{\mathrm{H}}\) Tension in largest when the brick is at the lowest point and smallest when it is at the highest point.
