371670
A body of mass \(10 \mathrm{~kg}\) is acted by two forces each of magnitude \(10 \mathrm{~N}\) making an angle of \(60^{\circ}\) with each other. Find the net acceleration of the body is
1 \(2 \sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
2 \(\sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
3 \(3 \sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
4 \(4 \sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
B Given mass \((\mathrm{m})=10 \mathrm{~kg}, \mathrm{~F}_{1}=\mathrm{F}_{2}=10 \mathrm{~N}\) and \(\theta=60^{\circ}\) \(\mathrm{F}_{\mathrm{net}} =\sqrt{\mathrm{F}_{1}^{2}+\mathrm{F}_{2}^{2}+2 \mathrm{~F}_{1} \mathrm{~F}_{2} \cos \theta}\) \(=\sqrt{(10)^{2}+(10)^{2}+2 \times 10 \times 10 \cos 60^{\circ}}\) \(=\sqrt{100+100+100}\) \(=10 \sqrt{3}\) We know, \(\quad \mathrm{F}=\mathrm{ma} \Rightarrow \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\) \(\mathrm{a}=\frac{10 \sqrt{3}}{10}\) Acceleration \((\mathrm{a})=\sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
J and K CET- 2007
LAWS OF MOTION (ADDITIONAL)
371671
Impulse is
1 a scalar
2 equal to change in the momentum of a body
3 equal to rate of change of momentum of a body
4 a force
Explanation:
B Impulse is equal to change in the momentum of a body. \(\mathrm{I}=\Delta \mathrm{P}=\mathrm{F} . \Delta \mathrm{t}\)
J and K CET- 2006
LAWS OF MOTION (ADDITIONAL)
371672
Which of the following is a self adjusting force?
1 Static friction
2 Limiting friction
3 Dynamic friction
4 Sliding friction
Explanation:
A Static friction is a force that keeps an object at rest. It is a self adjusting force because it does not have any fixed magnitude, but will adjust itself according to the applied force, till it reaches its optimum value.
J and K CET- 1999
LAWS OF MOTION (ADDITIONAL)
371673
An object is moving at constant velocity. The total force \(F\) acting on the object is given by
1 \(F=v^{2} / 2 m\)
2 \(\mathrm{F}=\mathrm{mv}\)
3 \(\mathrm{F}=0\)
4 \(F=m v^{2}\)
Explanation:
C Since the object is moving with a constant velocity. It means an acceleration of the body will be zero. So the net force on the object will be zero (i.e. \(\mathrm{F}=\) \(0)\).
371670
A body of mass \(10 \mathrm{~kg}\) is acted by two forces each of magnitude \(10 \mathrm{~N}\) making an angle of \(60^{\circ}\) with each other. Find the net acceleration of the body is
1 \(2 \sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
2 \(\sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
3 \(3 \sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
4 \(4 \sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
B Given mass \((\mathrm{m})=10 \mathrm{~kg}, \mathrm{~F}_{1}=\mathrm{F}_{2}=10 \mathrm{~N}\) and \(\theta=60^{\circ}\) \(\mathrm{F}_{\mathrm{net}} =\sqrt{\mathrm{F}_{1}^{2}+\mathrm{F}_{2}^{2}+2 \mathrm{~F}_{1} \mathrm{~F}_{2} \cos \theta}\) \(=\sqrt{(10)^{2}+(10)^{2}+2 \times 10 \times 10 \cos 60^{\circ}}\) \(=\sqrt{100+100+100}\) \(=10 \sqrt{3}\) We know, \(\quad \mathrm{F}=\mathrm{ma} \Rightarrow \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\) \(\mathrm{a}=\frac{10 \sqrt{3}}{10}\) Acceleration \((\mathrm{a})=\sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
J and K CET- 2007
LAWS OF MOTION (ADDITIONAL)
371671
Impulse is
1 a scalar
2 equal to change in the momentum of a body
3 equal to rate of change of momentum of a body
4 a force
Explanation:
B Impulse is equal to change in the momentum of a body. \(\mathrm{I}=\Delta \mathrm{P}=\mathrm{F} . \Delta \mathrm{t}\)
J and K CET- 2006
LAWS OF MOTION (ADDITIONAL)
371672
Which of the following is a self adjusting force?
1 Static friction
2 Limiting friction
3 Dynamic friction
4 Sliding friction
Explanation:
A Static friction is a force that keeps an object at rest. It is a self adjusting force because it does not have any fixed magnitude, but will adjust itself according to the applied force, till it reaches its optimum value.
J and K CET- 1999
LAWS OF MOTION (ADDITIONAL)
371673
An object is moving at constant velocity. The total force \(F\) acting on the object is given by
1 \(F=v^{2} / 2 m\)
2 \(\mathrm{F}=\mathrm{mv}\)
3 \(\mathrm{F}=0\)
4 \(F=m v^{2}\)
Explanation:
C Since the object is moving with a constant velocity. It means an acceleration of the body will be zero. So the net force on the object will be zero (i.e. \(\mathrm{F}=\) \(0)\).
371670
A body of mass \(10 \mathrm{~kg}\) is acted by two forces each of magnitude \(10 \mathrm{~N}\) making an angle of \(60^{\circ}\) with each other. Find the net acceleration of the body is
1 \(2 \sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
2 \(\sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
3 \(3 \sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
4 \(4 \sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
B Given mass \((\mathrm{m})=10 \mathrm{~kg}, \mathrm{~F}_{1}=\mathrm{F}_{2}=10 \mathrm{~N}\) and \(\theta=60^{\circ}\) \(\mathrm{F}_{\mathrm{net}} =\sqrt{\mathrm{F}_{1}^{2}+\mathrm{F}_{2}^{2}+2 \mathrm{~F}_{1} \mathrm{~F}_{2} \cos \theta}\) \(=\sqrt{(10)^{2}+(10)^{2}+2 \times 10 \times 10 \cos 60^{\circ}}\) \(=\sqrt{100+100+100}\) \(=10 \sqrt{3}\) We know, \(\quad \mathrm{F}=\mathrm{ma} \Rightarrow \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\) \(\mathrm{a}=\frac{10 \sqrt{3}}{10}\) Acceleration \((\mathrm{a})=\sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
J and K CET- 2007
LAWS OF MOTION (ADDITIONAL)
371671
Impulse is
1 a scalar
2 equal to change in the momentum of a body
3 equal to rate of change of momentum of a body
4 a force
Explanation:
B Impulse is equal to change in the momentum of a body. \(\mathrm{I}=\Delta \mathrm{P}=\mathrm{F} . \Delta \mathrm{t}\)
J and K CET- 2006
LAWS OF MOTION (ADDITIONAL)
371672
Which of the following is a self adjusting force?
1 Static friction
2 Limiting friction
3 Dynamic friction
4 Sliding friction
Explanation:
A Static friction is a force that keeps an object at rest. It is a self adjusting force because it does not have any fixed magnitude, but will adjust itself according to the applied force, till it reaches its optimum value.
J and K CET- 1999
LAWS OF MOTION (ADDITIONAL)
371673
An object is moving at constant velocity. The total force \(F\) acting on the object is given by
1 \(F=v^{2} / 2 m\)
2 \(\mathrm{F}=\mathrm{mv}\)
3 \(\mathrm{F}=0\)
4 \(F=m v^{2}\)
Explanation:
C Since the object is moving with a constant velocity. It means an acceleration of the body will be zero. So the net force on the object will be zero (i.e. \(\mathrm{F}=\) \(0)\).
NEET Test Series from KOTA - 10 Papers In MS WORD
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LAWS OF MOTION (ADDITIONAL)
371670
A body of mass \(10 \mathrm{~kg}\) is acted by two forces each of magnitude \(10 \mathrm{~N}\) making an angle of \(60^{\circ}\) with each other. Find the net acceleration of the body is
1 \(2 \sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
2 \(\sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
3 \(3 \sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
4 \(4 \sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
B Given mass \((\mathrm{m})=10 \mathrm{~kg}, \mathrm{~F}_{1}=\mathrm{F}_{2}=10 \mathrm{~N}\) and \(\theta=60^{\circ}\) \(\mathrm{F}_{\mathrm{net}} =\sqrt{\mathrm{F}_{1}^{2}+\mathrm{F}_{2}^{2}+2 \mathrm{~F}_{1} \mathrm{~F}_{2} \cos \theta}\) \(=\sqrt{(10)^{2}+(10)^{2}+2 \times 10 \times 10 \cos 60^{\circ}}\) \(=\sqrt{100+100+100}\) \(=10 \sqrt{3}\) We know, \(\quad \mathrm{F}=\mathrm{ma} \Rightarrow \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\) \(\mathrm{a}=\frac{10 \sqrt{3}}{10}\) Acceleration \((\mathrm{a})=\sqrt{3} \mathrm{~m} / \mathrm{s}^{2}\)
J and K CET- 2007
LAWS OF MOTION (ADDITIONAL)
371671
Impulse is
1 a scalar
2 equal to change in the momentum of a body
3 equal to rate of change of momentum of a body
4 a force
Explanation:
B Impulse is equal to change in the momentum of a body. \(\mathrm{I}=\Delta \mathrm{P}=\mathrm{F} . \Delta \mathrm{t}\)
J and K CET- 2006
LAWS OF MOTION (ADDITIONAL)
371672
Which of the following is a self adjusting force?
1 Static friction
2 Limiting friction
3 Dynamic friction
4 Sliding friction
Explanation:
A Static friction is a force that keeps an object at rest. It is a self adjusting force because it does not have any fixed magnitude, but will adjust itself according to the applied force, till it reaches its optimum value.
J and K CET- 1999
LAWS OF MOTION (ADDITIONAL)
371673
An object is moving at constant velocity. The total force \(F\) acting on the object is given by
1 \(F=v^{2} / 2 m\)
2 \(\mathrm{F}=\mathrm{mv}\)
3 \(\mathrm{F}=0\)
4 \(F=m v^{2}\)
Explanation:
C Since the object is moving with a constant velocity. It means an acceleration of the body will be zero. So the net force on the object will be zero (i.e. \(\mathrm{F}=\) \(0)\).
