363497
An intense stream of water of cross-sectional area \(A\) strikes a wall at an angle \(\theta \) with the normal to the wall and returns back elastically. If the density of water is \(\rho \) and its velocity is \(\upsilon ,\) then the force exerted in the wall will be
1 \(2\,A{\upsilon ^2}\rho \cos \theta \)
2 \(2\,A\upsilon \rho \cos \theta \)
3 \(2\,A\upsilon \rho \)
4 \(2\,A{\upsilon ^2}\rho \)
Explanation:
Linear momentum of water striking per second to the wall \({P_i} = m\upsilon = A\upsilon \rho \,\upsilon = A{\upsilon ^2}\rho \,,\) similarly linear momentum of reflected water per second \({P_r} = A{\upsilon ^2}\rho \,.\) Change in momentum of water per second \( = {P_i}\cos \theta + {P_r}\cos \theta \) \( = 2A{\upsilon ^2}\rho \cos \theta \) Force exerted on the wall \( = 2A{\upsilon ^2}\rho \cos \theta .\)
PHXI05:LAWS OF MOTION
363498
A ball of mass 0.15 \(kg\) is dropped from a height 10 \(m\), strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (\(g = 10\,m/{s^2}\)) nearly
1 \(4.2\,kg\,m/s\)
2 \(2.1\,kg\,m/s\)
3 \(1.4\,kg\,m/s\)
4 \(0\,kg\,m/s\)
Explanation:
Ball rebounds to the same height so velocity will be same after striking So impulse imparted is 2\(mv\) \(\Delta P = 2mv = 2 \times m \times \sqrt {2gh} \) \( = 2 \times 0.15 \times \sqrt {2 \times 10 \times 10} = 4.2\)
NEET - 2021
PHXI05:LAWS OF MOTION
363499
A force exerts an impulse \(I\) on a particle changing its speed from \(u\) to 2 \(u\). The applied force and the initial velocity are oppositely directed along the same line. The work done by the force is:
1 \(\frac{1}{2}I\;u\)
2 \(\frac{3}{2}I\;u\)
3 2 \(Iu\)
4 \(Iu\)
Explanation:
Let \(m\) is the mass of the particle. \(\vec I = {{\vec P}_f} - {{\vec P}_i}\) \(I\,\hat i = m2\,u\hat i + mu\hat i\) \(I = 3\,mu\) From work energy theorem \(W = \Delta K = \frac{1}{2}m[4\,{u^2} - {u^2}]\) \(W = \frac{1}{2}Iu\)
PHXI05:LAWS OF MOTION
363500
Assertion : The slope of momentum versus time curve give us the acceleration. Reason : Acceleration is given by the rate of change of velocity.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Rate of change of momentum \( = \) Slope of momentum - time graph \( = \) force.
PHXI05:LAWS OF MOTION
363501
The magnitude of the impulse developed by a mass of 0.2\(kg\) which changes its velocity from \(5\widehat i - 3\widehat j + 7\widehat k\,m/s\) to \(2\widehat i + 3\widehat j + \widehat k\,m/s\) is
363497
An intense stream of water of cross-sectional area \(A\) strikes a wall at an angle \(\theta \) with the normal to the wall and returns back elastically. If the density of water is \(\rho \) and its velocity is \(\upsilon ,\) then the force exerted in the wall will be
1 \(2\,A{\upsilon ^2}\rho \cos \theta \)
2 \(2\,A\upsilon \rho \cos \theta \)
3 \(2\,A\upsilon \rho \)
4 \(2\,A{\upsilon ^2}\rho \)
Explanation:
Linear momentum of water striking per second to the wall \({P_i} = m\upsilon = A\upsilon \rho \,\upsilon = A{\upsilon ^2}\rho \,,\) similarly linear momentum of reflected water per second \({P_r} = A{\upsilon ^2}\rho \,.\) Change in momentum of water per second \( = {P_i}\cos \theta + {P_r}\cos \theta \) \( = 2A{\upsilon ^2}\rho \cos \theta \) Force exerted on the wall \( = 2A{\upsilon ^2}\rho \cos \theta .\)
PHXI05:LAWS OF MOTION
363498
A ball of mass 0.15 \(kg\) is dropped from a height 10 \(m\), strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (\(g = 10\,m/{s^2}\)) nearly
1 \(4.2\,kg\,m/s\)
2 \(2.1\,kg\,m/s\)
3 \(1.4\,kg\,m/s\)
4 \(0\,kg\,m/s\)
Explanation:
Ball rebounds to the same height so velocity will be same after striking So impulse imparted is 2\(mv\) \(\Delta P = 2mv = 2 \times m \times \sqrt {2gh} \) \( = 2 \times 0.15 \times \sqrt {2 \times 10 \times 10} = 4.2\)
NEET - 2021
PHXI05:LAWS OF MOTION
363499
A force exerts an impulse \(I\) on a particle changing its speed from \(u\) to 2 \(u\). The applied force and the initial velocity are oppositely directed along the same line. The work done by the force is:
1 \(\frac{1}{2}I\;u\)
2 \(\frac{3}{2}I\;u\)
3 2 \(Iu\)
4 \(Iu\)
Explanation:
Let \(m\) is the mass of the particle. \(\vec I = {{\vec P}_f} - {{\vec P}_i}\) \(I\,\hat i = m2\,u\hat i + mu\hat i\) \(I = 3\,mu\) From work energy theorem \(W = \Delta K = \frac{1}{2}m[4\,{u^2} - {u^2}]\) \(W = \frac{1}{2}Iu\)
PHXI05:LAWS OF MOTION
363500
Assertion : The slope of momentum versus time curve give us the acceleration. Reason : Acceleration is given by the rate of change of velocity.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Rate of change of momentum \( = \) Slope of momentum - time graph \( = \) force.
PHXI05:LAWS OF MOTION
363501
The magnitude of the impulse developed by a mass of 0.2\(kg\) which changes its velocity from \(5\widehat i - 3\widehat j + 7\widehat k\,m/s\) to \(2\widehat i + 3\widehat j + \widehat k\,m/s\) is
363497
An intense stream of water of cross-sectional area \(A\) strikes a wall at an angle \(\theta \) with the normal to the wall and returns back elastically. If the density of water is \(\rho \) and its velocity is \(\upsilon ,\) then the force exerted in the wall will be
1 \(2\,A{\upsilon ^2}\rho \cos \theta \)
2 \(2\,A\upsilon \rho \cos \theta \)
3 \(2\,A\upsilon \rho \)
4 \(2\,A{\upsilon ^2}\rho \)
Explanation:
Linear momentum of water striking per second to the wall \({P_i} = m\upsilon = A\upsilon \rho \,\upsilon = A{\upsilon ^2}\rho \,,\) similarly linear momentum of reflected water per second \({P_r} = A{\upsilon ^2}\rho \,.\) Change in momentum of water per second \( = {P_i}\cos \theta + {P_r}\cos \theta \) \( = 2A{\upsilon ^2}\rho \cos \theta \) Force exerted on the wall \( = 2A{\upsilon ^2}\rho \cos \theta .\)
PHXI05:LAWS OF MOTION
363498
A ball of mass 0.15 \(kg\) is dropped from a height 10 \(m\), strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (\(g = 10\,m/{s^2}\)) nearly
1 \(4.2\,kg\,m/s\)
2 \(2.1\,kg\,m/s\)
3 \(1.4\,kg\,m/s\)
4 \(0\,kg\,m/s\)
Explanation:
Ball rebounds to the same height so velocity will be same after striking So impulse imparted is 2\(mv\) \(\Delta P = 2mv = 2 \times m \times \sqrt {2gh} \) \( = 2 \times 0.15 \times \sqrt {2 \times 10 \times 10} = 4.2\)
NEET - 2021
PHXI05:LAWS OF MOTION
363499
A force exerts an impulse \(I\) on a particle changing its speed from \(u\) to 2 \(u\). The applied force and the initial velocity are oppositely directed along the same line. The work done by the force is:
1 \(\frac{1}{2}I\;u\)
2 \(\frac{3}{2}I\;u\)
3 2 \(Iu\)
4 \(Iu\)
Explanation:
Let \(m\) is the mass of the particle. \(\vec I = {{\vec P}_f} - {{\vec P}_i}\) \(I\,\hat i = m2\,u\hat i + mu\hat i\) \(I = 3\,mu\) From work energy theorem \(W = \Delta K = \frac{1}{2}m[4\,{u^2} - {u^2}]\) \(W = \frac{1}{2}Iu\)
PHXI05:LAWS OF MOTION
363500
Assertion : The slope of momentum versus time curve give us the acceleration. Reason : Acceleration is given by the rate of change of velocity.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Rate of change of momentum \( = \) Slope of momentum - time graph \( = \) force.
PHXI05:LAWS OF MOTION
363501
The magnitude of the impulse developed by a mass of 0.2\(kg\) which changes its velocity from \(5\widehat i - 3\widehat j + 7\widehat k\,m/s\) to \(2\widehat i + 3\widehat j + \widehat k\,m/s\) is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI05:LAWS OF MOTION
363497
An intense stream of water of cross-sectional area \(A\) strikes a wall at an angle \(\theta \) with the normal to the wall and returns back elastically. If the density of water is \(\rho \) and its velocity is \(\upsilon ,\) then the force exerted in the wall will be
1 \(2\,A{\upsilon ^2}\rho \cos \theta \)
2 \(2\,A\upsilon \rho \cos \theta \)
3 \(2\,A\upsilon \rho \)
4 \(2\,A{\upsilon ^2}\rho \)
Explanation:
Linear momentum of water striking per second to the wall \({P_i} = m\upsilon = A\upsilon \rho \,\upsilon = A{\upsilon ^2}\rho \,,\) similarly linear momentum of reflected water per second \({P_r} = A{\upsilon ^2}\rho \,.\) Change in momentum of water per second \( = {P_i}\cos \theta + {P_r}\cos \theta \) \( = 2A{\upsilon ^2}\rho \cos \theta \) Force exerted on the wall \( = 2A{\upsilon ^2}\rho \cos \theta .\)
PHXI05:LAWS OF MOTION
363498
A ball of mass 0.15 \(kg\) is dropped from a height 10 \(m\), strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (\(g = 10\,m/{s^2}\)) nearly
1 \(4.2\,kg\,m/s\)
2 \(2.1\,kg\,m/s\)
3 \(1.4\,kg\,m/s\)
4 \(0\,kg\,m/s\)
Explanation:
Ball rebounds to the same height so velocity will be same after striking So impulse imparted is 2\(mv\) \(\Delta P = 2mv = 2 \times m \times \sqrt {2gh} \) \( = 2 \times 0.15 \times \sqrt {2 \times 10 \times 10} = 4.2\)
NEET - 2021
PHXI05:LAWS OF MOTION
363499
A force exerts an impulse \(I\) on a particle changing its speed from \(u\) to 2 \(u\). The applied force and the initial velocity are oppositely directed along the same line. The work done by the force is:
1 \(\frac{1}{2}I\;u\)
2 \(\frac{3}{2}I\;u\)
3 2 \(Iu\)
4 \(Iu\)
Explanation:
Let \(m\) is the mass of the particle. \(\vec I = {{\vec P}_f} - {{\vec P}_i}\) \(I\,\hat i = m2\,u\hat i + mu\hat i\) \(I = 3\,mu\) From work energy theorem \(W = \Delta K = \frac{1}{2}m[4\,{u^2} - {u^2}]\) \(W = \frac{1}{2}Iu\)
PHXI05:LAWS OF MOTION
363500
Assertion : The slope of momentum versus time curve give us the acceleration. Reason : Acceleration is given by the rate of change of velocity.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Rate of change of momentum \( = \) Slope of momentum - time graph \( = \) force.
PHXI05:LAWS OF MOTION
363501
The magnitude of the impulse developed by a mass of 0.2\(kg\) which changes its velocity from \(5\widehat i - 3\widehat j + 7\widehat k\,m/s\) to \(2\widehat i + 3\widehat j + \widehat k\,m/s\) is
363497
An intense stream of water of cross-sectional area \(A\) strikes a wall at an angle \(\theta \) with the normal to the wall and returns back elastically. If the density of water is \(\rho \) and its velocity is \(\upsilon ,\) then the force exerted in the wall will be
1 \(2\,A{\upsilon ^2}\rho \cos \theta \)
2 \(2\,A\upsilon \rho \cos \theta \)
3 \(2\,A\upsilon \rho \)
4 \(2\,A{\upsilon ^2}\rho \)
Explanation:
Linear momentum of water striking per second to the wall \({P_i} = m\upsilon = A\upsilon \rho \,\upsilon = A{\upsilon ^2}\rho \,,\) similarly linear momentum of reflected water per second \({P_r} = A{\upsilon ^2}\rho \,.\) Change in momentum of water per second \( = {P_i}\cos \theta + {P_r}\cos \theta \) \( = 2A{\upsilon ^2}\rho \cos \theta \) Force exerted on the wall \( = 2A{\upsilon ^2}\rho \cos \theta .\)
PHXI05:LAWS OF MOTION
363498
A ball of mass 0.15 \(kg\) is dropped from a height 10 \(m\), strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (\(g = 10\,m/{s^2}\)) nearly
1 \(4.2\,kg\,m/s\)
2 \(2.1\,kg\,m/s\)
3 \(1.4\,kg\,m/s\)
4 \(0\,kg\,m/s\)
Explanation:
Ball rebounds to the same height so velocity will be same after striking So impulse imparted is 2\(mv\) \(\Delta P = 2mv = 2 \times m \times \sqrt {2gh} \) \( = 2 \times 0.15 \times \sqrt {2 \times 10 \times 10} = 4.2\)
NEET - 2021
PHXI05:LAWS OF MOTION
363499
A force exerts an impulse \(I\) on a particle changing its speed from \(u\) to 2 \(u\). The applied force and the initial velocity are oppositely directed along the same line. The work done by the force is:
1 \(\frac{1}{2}I\;u\)
2 \(\frac{3}{2}I\;u\)
3 2 \(Iu\)
4 \(Iu\)
Explanation:
Let \(m\) is the mass of the particle. \(\vec I = {{\vec P}_f} - {{\vec P}_i}\) \(I\,\hat i = m2\,u\hat i + mu\hat i\) \(I = 3\,mu\) From work energy theorem \(W = \Delta K = \frac{1}{2}m[4\,{u^2} - {u^2}]\) \(W = \frac{1}{2}Iu\)
PHXI05:LAWS OF MOTION
363500
Assertion : The slope of momentum versus time curve give us the acceleration. Reason : Acceleration is given by the rate of change of velocity.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Rate of change of momentum \( = \) Slope of momentum - time graph \( = \) force.
PHXI05:LAWS OF MOTION
363501
The magnitude of the impulse developed by a mass of 0.2\(kg\) which changes its velocity from \(5\widehat i - 3\widehat j + 7\widehat k\,m/s\) to \(2\widehat i + 3\widehat j + \widehat k\,m/s\) is
