361908
A train 1 moves from east to west and another train 2 moves from west to east on the equator with equal speeds relative to ground. The ratio of their centripetal accelerations \(\frac{{{a_1}}}{{{a_2}}}\) relative to centre of earth is:
1 \( > 1\)
2 \( = 1\)
3 \( < 1\)
4 \( \ge 1\)
Explanation:
The earth rotates west to east. So the velocity of train 2 increases \({a_c} = {v^2}{\rm{/R}}\) . Hence centripetal acceleration of train 2 is more.
PHXI04:MOTION IN A PLANE
361909
Which of the following statements is false for a particle moving in a circle with a constant angular speed?
1 The velocity vector is tangent to the circle.
2 The acceleration vector is tangent to the circle.
3 The acceleration vector points to the centre of the circle.
4 The velocity and acceleration vectors are perpendicular to each other.
Explanation:
Conceptual Question
PHXI04:MOTION IN A PLANE
361910
A particle is moving with constant speed in a circular path. When the particle turns by angle \(90^{\circ}\), the ratio of instantaneous velocity to its average velocity is \(\pi: x \sqrt{2}\). The value of \(x\) will be
1 7
2 1
3 5
4 2
Explanation:
Instantaneous velocity \(=\omega R\) Time taken by particle \(=\dfrac{\pi}{2 \omega}\) Displacement \(=R \sqrt{2}\) Average velocity \(=\dfrac{R \sqrt{2} \times 2 \omega}{\pi}=\dfrac{2 \sqrt{2}}{\pi} \omega R\) \(\therefore \dfrac{v_{\text {ins }}}{v_{\text {avg }}}=\dfrac{\omega R \pi}{2 \sqrt{2} \omega R}=\dfrac{\pi}{2 \sqrt{2}}\) On comparing with given value, we get, \(x=2\)
JEE - 2023
PHXI04:MOTION IN A PLANE
361911
A body moving along a circular path of radius \(R\) with velocity \(v\), has centripetal acceleration \(a\). If its velocity is made equal to \(2 v\). What will be the centripetal acceleration?
1 \(4 a\)
2 \(2 a\)
3 \(\dfrac{a}{4}\)
4 \(\dfrac{a}{2}\)
Explanation:
We know that centripetal acceleration is given by \(a=\dfrac{V^{2}}{R}\) For constant \(R . \quad a \propto V^{2}\) or \(\dfrac{a_{1}}{a_{2}}=\dfrac{V_{1}^{2}}{V_{2}^{2}}\) \(\therefore \dfrac{a_{1}}{a_{2}}=\dfrac{v^{2}}{(2 v)^{2}}=\dfrac{1}{4}\) \(a_{2}=4 a_{1}\) or \(a_{2}=4 a \quad\left(\because a_{2}=a\right)\)
PHXI04:MOTION IN A PLANE
361912
A point moves along a circle with a velocity \(v=a t\), where \(a = 0.50\;m{\rm{/}}{s^2}\). Find the net acceleration of the point at the moment when it has covered the \(n^{\text {th }}(n=0.10)\) fraction of the circle after beginning of the motion.
361908
A train 1 moves from east to west and another train 2 moves from west to east on the equator with equal speeds relative to ground. The ratio of their centripetal accelerations \(\frac{{{a_1}}}{{{a_2}}}\) relative to centre of earth is:
1 \( > 1\)
2 \( = 1\)
3 \( < 1\)
4 \( \ge 1\)
Explanation:
The earth rotates west to east. So the velocity of train 2 increases \({a_c} = {v^2}{\rm{/R}}\) . Hence centripetal acceleration of train 2 is more.
PHXI04:MOTION IN A PLANE
361909
Which of the following statements is false for a particle moving in a circle with a constant angular speed?
1 The velocity vector is tangent to the circle.
2 The acceleration vector is tangent to the circle.
3 The acceleration vector points to the centre of the circle.
4 The velocity and acceleration vectors are perpendicular to each other.
Explanation:
Conceptual Question
PHXI04:MOTION IN A PLANE
361910
A particle is moving with constant speed in a circular path. When the particle turns by angle \(90^{\circ}\), the ratio of instantaneous velocity to its average velocity is \(\pi: x \sqrt{2}\). The value of \(x\) will be
1 7
2 1
3 5
4 2
Explanation:
Instantaneous velocity \(=\omega R\) Time taken by particle \(=\dfrac{\pi}{2 \omega}\) Displacement \(=R \sqrt{2}\) Average velocity \(=\dfrac{R \sqrt{2} \times 2 \omega}{\pi}=\dfrac{2 \sqrt{2}}{\pi} \omega R\) \(\therefore \dfrac{v_{\text {ins }}}{v_{\text {avg }}}=\dfrac{\omega R \pi}{2 \sqrt{2} \omega R}=\dfrac{\pi}{2 \sqrt{2}}\) On comparing with given value, we get, \(x=2\)
JEE - 2023
PHXI04:MOTION IN A PLANE
361911
A body moving along a circular path of radius \(R\) with velocity \(v\), has centripetal acceleration \(a\). If its velocity is made equal to \(2 v\). What will be the centripetal acceleration?
1 \(4 a\)
2 \(2 a\)
3 \(\dfrac{a}{4}\)
4 \(\dfrac{a}{2}\)
Explanation:
We know that centripetal acceleration is given by \(a=\dfrac{V^{2}}{R}\) For constant \(R . \quad a \propto V^{2}\) or \(\dfrac{a_{1}}{a_{2}}=\dfrac{V_{1}^{2}}{V_{2}^{2}}\) \(\therefore \dfrac{a_{1}}{a_{2}}=\dfrac{v^{2}}{(2 v)^{2}}=\dfrac{1}{4}\) \(a_{2}=4 a_{1}\) or \(a_{2}=4 a \quad\left(\because a_{2}=a\right)\)
PHXI04:MOTION IN A PLANE
361912
A point moves along a circle with a velocity \(v=a t\), where \(a = 0.50\;m{\rm{/}}{s^2}\). Find the net acceleration of the point at the moment when it has covered the \(n^{\text {th }}(n=0.10)\) fraction of the circle after beginning of the motion.
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI04:MOTION IN A PLANE
361908
A train 1 moves from east to west and another train 2 moves from west to east on the equator with equal speeds relative to ground. The ratio of their centripetal accelerations \(\frac{{{a_1}}}{{{a_2}}}\) relative to centre of earth is:
1 \( > 1\)
2 \( = 1\)
3 \( < 1\)
4 \( \ge 1\)
Explanation:
The earth rotates west to east. So the velocity of train 2 increases \({a_c} = {v^2}{\rm{/R}}\) . Hence centripetal acceleration of train 2 is more.
PHXI04:MOTION IN A PLANE
361909
Which of the following statements is false for a particle moving in a circle with a constant angular speed?
1 The velocity vector is tangent to the circle.
2 The acceleration vector is tangent to the circle.
3 The acceleration vector points to the centre of the circle.
4 The velocity and acceleration vectors are perpendicular to each other.
Explanation:
Conceptual Question
PHXI04:MOTION IN A PLANE
361910
A particle is moving with constant speed in a circular path. When the particle turns by angle \(90^{\circ}\), the ratio of instantaneous velocity to its average velocity is \(\pi: x \sqrt{2}\). The value of \(x\) will be
1 7
2 1
3 5
4 2
Explanation:
Instantaneous velocity \(=\omega R\) Time taken by particle \(=\dfrac{\pi}{2 \omega}\) Displacement \(=R \sqrt{2}\) Average velocity \(=\dfrac{R \sqrt{2} \times 2 \omega}{\pi}=\dfrac{2 \sqrt{2}}{\pi} \omega R\) \(\therefore \dfrac{v_{\text {ins }}}{v_{\text {avg }}}=\dfrac{\omega R \pi}{2 \sqrt{2} \omega R}=\dfrac{\pi}{2 \sqrt{2}}\) On comparing with given value, we get, \(x=2\)
JEE - 2023
PHXI04:MOTION IN A PLANE
361911
A body moving along a circular path of radius \(R\) with velocity \(v\), has centripetal acceleration \(a\). If its velocity is made equal to \(2 v\). What will be the centripetal acceleration?
1 \(4 a\)
2 \(2 a\)
3 \(\dfrac{a}{4}\)
4 \(\dfrac{a}{2}\)
Explanation:
We know that centripetal acceleration is given by \(a=\dfrac{V^{2}}{R}\) For constant \(R . \quad a \propto V^{2}\) or \(\dfrac{a_{1}}{a_{2}}=\dfrac{V_{1}^{2}}{V_{2}^{2}}\) \(\therefore \dfrac{a_{1}}{a_{2}}=\dfrac{v^{2}}{(2 v)^{2}}=\dfrac{1}{4}\) \(a_{2}=4 a_{1}\) or \(a_{2}=4 a \quad\left(\because a_{2}=a\right)\)
PHXI04:MOTION IN A PLANE
361912
A point moves along a circle with a velocity \(v=a t\), where \(a = 0.50\;m{\rm{/}}{s^2}\). Find the net acceleration of the point at the moment when it has covered the \(n^{\text {th }}(n=0.10)\) fraction of the circle after beginning of the motion.
361908
A train 1 moves from east to west and another train 2 moves from west to east on the equator with equal speeds relative to ground. The ratio of their centripetal accelerations \(\frac{{{a_1}}}{{{a_2}}}\) relative to centre of earth is:
1 \( > 1\)
2 \( = 1\)
3 \( < 1\)
4 \( \ge 1\)
Explanation:
The earth rotates west to east. So the velocity of train 2 increases \({a_c} = {v^2}{\rm{/R}}\) . Hence centripetal acceleration of train 2 is more.
PHXI04:MOTION IN A PLANE
361909
Which of the following statements is false for a particle moving in a circle with a constant angular speed?
1 The velocity vector is tangent to the circle.
2 The acceleration vector is tangent to the circle.
3 The acceleration vector points to the centre of the circle.
4 The velocity and acceleration vectors are perpendicular to each other.
Explanation:
Conceptual Question
PHXI04:MOTION IN A PLANE
361910
A particle is moving with constant speed in a circular path. When the particle turns by angle \(90^{\circ}\), the ratio of instantaneous velocity to its average velocity is \(\pi: x \sqrt{2}\). The value of \(x\) will be
1 7
2 1
3 5
4 2
Explanation:
Instantaneous velocity \(=\omega R\) Time taken by particle \(=\dfrac{\pi}{2 \omega}\) Displacement \(=R \sqrt{2}\) Average velocity \(=\dfrac{R \sqrt{2} \times 2 \omega}{\pi}=\dfrac{2 \sqrt{2}}{\pi} \omega R\) \(\therefore \dfrac{v_{\text {ins }}}{v_{\text {avg }}}=\dfrac{\omega R \pi}{2 \sqrt{2} \omega R}=\dfrac{\pi}{2 \sqrt{2}}\) On comparing with given value, we get, \(x=2\)
JEE - 2023
PHXI04:MOTION IN A PLANE
361911
A body moving along a circular path of radius \(R\) with velocity \(v\), has centripetal acceleration \(a\). If its velocity is made equal to \(2 v\). What will be the centripetal acceleration?
1 \(4 a\)
2 \(2 a\)
3 \(\dfrac{a}{4}\)
4 \(\dfrac{a}{2}\)
Explanation:
We know that centripetal acceleration is given by \(a=\dfrac{V^{2}}{R}\) For constant \(R . \quad a \propto V^{2}\) or \(\dfrac{a_{1}}{a_{2}}=\dfrac{V_{1}^{2}}{V_{2}^{2}}\) \(\therefore \dfrac{a_{1}}{a_{2}}=\dfrac{v^{2}}{(2 v)^{2}}=\dfrac{1}{4}\) \(a_{2}=4 a_{1}\) or \(a_{2}=4 a \quad\left(\because a_{2}=a\right)\)
PHXI04:MOTION IN A PLANE
361912
A point moves along a circle with a velocity \(v=a t\), where \(a = 0.50\;m{\rm{/}}{s^2}\). Find the net acceleration of the point at the moment when it has covered the \(n^{\text {th }}(n=0.10)\) fraction of the circle after beginning of the motion.
361908
A train 1 moves from east to west and another train 2 moves from west to east on the equator with equal speeds relative to ground. The ratio of their centripetal accelerations \(\frac{{{a_1}}}{{{a_2}}}\) relative to centre of earth is:
1 \( > 1\)
2 \( = 1\)
3 \( < 1\)
4 \( \ge 1\)
Explanation:
The earth rotates west to east. So the velocity of train 2 increases \({a_c} = {v^2}{\rm{/R}}\) . Hence centripetal acceleration of train 2 is more.
PHXI04:MOTION IN A PLANE
361909
Which of the following statements is false for a particle moving in a circle with a constant angular speed?
1 The velocity vector is tangent to the circle.
2 The acceleration vector is tangent to the circle.
3 The acceleration vector points to the centre of the circle.
4 The velocity and acceleration vectors are perpendicular to each other.
Explanation:
Conceptual Question
PHXI04:MOTION IN A PLANE
361910
A particle is moving with constant speed in a circular path. When the particle turns by angle \(90^{\circ}\), the ratio of instantaneous velocity to its average velocity is \(\pi: x \sqrt{2}\). The value of \(x\) will be
1 7
2 1
3 5
4 2
Explanation:
Instantaneous velocity \(=\omega R\) Time taken by particle \(=\dfrac{\pi}{2 \omega}\) Displacement \(=R \sqrt{2}\) Average velocity \(=\dfrac{R \sqrt{2} \times 2 \omega}{\pi}=\dfrac{2 \sqrt{2}}{\pi} \omega R\) \(\therefore \dfrac{v_{\text {ins }}}{v_{\text {avg }}}=\dfrac{\omega R \pi}{2 \sqrt{2} \omega R}=\dfrac{\pi}{2 \sqrt{2}}\) On comparing with given value, we get, \(x=2\)
JEE - 2023
PHXI04:MOTION IN A PLANE
361911
A body moving along a circular path of radius \(R\) with velocity \(v\), has centripetal acceleration \(a\). If its velocity is made equal to \(2 v\). What will be the centripetal acceleration?
1 \(4 a\)
2 \(2 a\)
3 \(\dfrac{a}{4}\)
4 \(\dfrac{a}{2}\)
Explanation:
We know that centripetal acceleration is given by \(a=\dfrac{V^{2}}{R}\) For constant \(R . \quad a \propto V^{2}\) or \(\dfrac{a_{1}}{a_{2}}=\dfrac{V_{1}^{2}}{V_{2}^{2}}\) \(\therefore \dfrac{a_{1}}{a_{2}}=\dfrac{v^{2}}{(2 v)^{2}}=\dfrac{1}{4}\) \(a_{2}=4 a_{1}\) or \(a_{2}=4 a \quad\left(\because a_{2}=a\right)\)
PHXI04:MOTION IN A PLANE
361912
A point moves along a circle with a velocity \(v=a t\), where \(a = 0.50\;m{\rm{/}}{s^2}\). Find the net acceleration of the point at the moment when it has covered the \(n^{\text {th }}(n=0.10)\) fraction of the circle after beginning of the motion.