371837
A block is kept on a frictionless inclined surface with angle of inclination ' \(\alpha\) '. The incline is given an acceleration ' \(a\) ' to keep the block stationary. Then ' \(a\) ' is equal to
1 \(g \operatorname{cosec} \alpha\)
2 \(g / \tan \alpha\)
3 \(g \tan \alpha\)
4 \(g\)
Explanation:
C According to the figure, for block to remain stationary ma cos \(\square=\mathrm{mg} \sin \square\) \(\mathrm{a}=\frac{\mathrm{g} \sin \alpha}{\cos \alpha}\) \(\mathrm{a}=\mathrm{g} \tan \alpha\)
BITSAT-2019
LAWS OF MOTION (ADDITIONAL)
371838
Which of the following sets of concurrent force may be in equilibrium?
1 \(F_{1}=3 N, F_{2}=5 N, F_{3}=10 N\)
2 \(F_{1}=3 N, F_{2}=5 N, F_{3}=9 N\)
3 \(F_{1}=3 N, F_{2}=5 N, F_{3}=6 N\)
4 \(F_{1}=3 N, F_{2}=5 N, F_{3}=15 N\)
Explanation:
C If three concurrent forces are to be in equilibrium, the sum of two smallest magnitude must be greater than the magnitude of third force. i.e. \(\left|\overrightarrow{\mathrm{F}}_{1}\right|+\left|\overrightarrow{\mathrm{F}}_{2}\right| \geq\left|\overrightarrow{\mathrm{F}}_{3}\right|\) Now, we can substitute these values in the above equation, we get- (c) \(\mathrm{F}_{1}=3 \mathrm{~N}, \mathrm{~F}_{2}=5 \mathrm{~N}, \mathrm{~F}_{3}=6 \mathrm{~N}\) \(3+5>6\) So, option (c) is right answer.
AP EAMCET (17.09.2020) Shift-II
LAWS OF MOTION (ADDITIONAL)
371839
A block of mass \(M\) slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is \(\theta\). The magnitude of the contact force will be :
A FBD of Body Block slides down with constant velocity then net force on the block is zero. So, Net contact force \(F_{C}\) on the body is \(F_{C}=\sqrt{f^{2}+N^{2}}\) From the figure- \(f=m g \sin \theta\) \(N=m g \cos \theta\) \(=\sqrt{(m g \sin \theta)^{2}+(m g \cos \theta)^{2}}\) \(=\sqrt{(m g)^{2}\left[\sin ^{2} \theta+\cos ^{2} \theta\right]}\) \(F_{C}=m g\)
JEE Main-27.07.2022
LAWS OF MOTION (ADDITIONAL)
371840
Statement [A]: If a body is in equilibrium, then the resultant of the forces acting on the body is zero Statement [B]: Action and reaction cancel each other as they act on the same body Statement [C]: Centrifugal force is a pseudo force Options:
1 A, B, C are all true
2 A, B are true but C is false
3 A, C are true but B is false
4 B, C are true but A is false
Explanation:
C A, C are true but B is false Statement A: If a body is in equilibrium, then the resultant of the forces acting on the body is zero. Statement C: Centrifugal force is a pseudo force because it is not provide by any real force but it arises due to accelerated frame of reference.
AP EAPCET-12.07.2022
LAWS OF MOTION (ADDITIONAL)
371841
For a free body diagram shown in the figure, the four forces are applied in the ' \(x\) ' and ' \(y\) ' directions. What additional force must be applied and at what angle with positive \(x\)-axis so that the net acceleration of body is zero?
1 \(\sqrt{2} \mathrm{~N}, 45^{\circ}\)
2 \(\sqrt{2} \mathrm{~N}, 135^{\circ}\)
3 \(\frac{2}{\sqrt{3}} \mathrm{~N}, 30^{\circ}\)
4 \(2 \mathrm{~N}, 45^{\circ}\)
Explanation:
A Assume addition force required is \(=\overrightarrow{\mathrm{F}}\) By figure, \(\overrightarrow{\mathrm{F}}+5 \hat{\mathrm{i}}-6 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-8 \hat{\mathrm{j}}=0\) \(\overrightarrow{\mathrm{F}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}\) \(|\overrightarrow{\mathrm{F}}|=\sqrt{1^{2}+1^{2}}\) \(|\overrightarrow{\mathrm{F}}|=\sqrt{2} \mathrm{~N}\) Angle with \(\mathrm{x}\) axis- \(\tan \theta=\frac{y-\text { component }}{x-\text { component }}=\frac{1}{1}\) \(\tan \theta=1=\tan 45^{\circ}\) \(\theta=45^{\circ}\)
371837
A block is kept on a frictionless inclined surface with angle of inclination ' \(\alpha\) '. The incline is given an acceleration ' \(a\) ' to keep the block stationary. Then ' \(a\) ' is equal to
1 \(g \operatorname{cosec} \alpha\)
2 \(g / \tan \alpha\)
3 \(g \tan \alpha\)
4 \(g\)
Explanation:
C According to the figure, for block to remain stationary ma cos \(\square=\mathrm{mg} \sin \square\) \(\mathrm{a}=\frac{\mathrm{g} \sin \alpha}{\cos \alpha}\) \(\mathrm{a}=\mathrm{g} \tan \alpha\)
BITSAT-2019
LAWS OF MOTION (ADDITIONAL)
371838
Which of the following sets of concurrent force may be in equilibrium?
1 \(F_{1}=3 N, F_{2}=5 N, F_{3}=10 N\)
2 \(F_{1}=3 N, F_{2}=5 N, F_{3}=9 N\)
3 \(F_{1}=3 N, F_{2}=5 N, F_{3}=6 N\)
4 \(F_{1}=3 N, F_{2}=5 N, F_{3}=15 N\)
Explanation:
C If three concurrent forces are to be in equilibrium, the sum of two smallest magnitude must be greater than the magnitude of third force. i.e. \(\left|\overrightarrow{\mathrm{F}}_{1}\right|+\left|\overrightarrow{\mathrm{F}}_{2}\right| \geq\left|\overrightarrow{\mathrm{F}}_{3}\right|\) Now, we can substitute these values in the above equation, we get- (c) \(\mathrm{F}_{1}=3 \mathrm{~N}, \mathrm{~F}_{2}=5 \mathrm{~N}, \mathrm{~F}_{3}=6 \mathrm{~N}\) \(3+5>6\) So, option (c) is right answer.
AP EAMCET (17.09.2020) Shift-II
LAWS OF MOTION (ADDITIONAL)
371839
A block of mass \(M\) slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is \(\theta\). The magnitude of the contact force will be :
A FBD of Body Block slides down with constant velocity then net force on the block is zero. So, Net contact force \(F_{C}\) on the body is \(F_{C}=\sqrt{f^{2}+N^{2}}\) From the figure- \(f=m g \sin \theta\) \(N=m g \cos \theta\) \(=\sqrt{(m g \sin \theta)^{2}+(m g \cos \theta)^{2}}\) \(=\sqrt{(m g)^{2}\left[\sin ^{2} \theta+\cos ^{2} \theta\right]}\) \(F_{C}=m g\)
JEE Main-27.07.2022
LAWS OF MOTION (ADDITIONAL)
371840
Statement [A]: If a body is in equilibrium, then the resultant of the forces acting on the body is zero Statement [B]: Action and reaction cancel each other as they act on the same body Statement [C]: Centrifugal force is a pseudo force Options:
1 A, B, C are all true
2 A, B are true but C is false
3 A, C are true but B is false
4 B, C are true but A is false
Explanation:
C A, C are true but B is false Statement A: If a body is in equilibrium, then the resultant of the forces acting on the body is zero. Statement C: Centrifugal force is a pseudo force because it is not provide by any real force but it arises due to accelerated frame of reference.
AP EAPCET-12.07.2022
LAWS OF MOTION (ADDITIONAL)
371841
For a free body diagram shown in the figure, the four forces are applied in the ' \(x\) ' and ' \(y\) ' directions. What additional force must be applied and at what angle with positive \(x\)-axis so that the net acceleration of body is zero?
1 \(\sqrt{2} \mathrm{~N}, 45^{\circ}\)
2 \(\sqrt{2} \mathrm{~N}, 135^{\circ}\)
3 \(\frac{2}{\sqrt{3}} \mathrm{~N}, 30^{\circ}\)
4 \(2 \mathrm{~N}, 45^{\circ}\)
Explanation:
A Assume addition force required is \(=\overrightarrow{\mathrm{F}}\) By figure, \(\overrightarrow{\mathrm{F}}+5 \hat{\mathrm{i}}-6 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-8 \hat{\mathrm{j}}=0\) \(\overrightarrow{\mathrm{F}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}\) \(|\overrightarrow{\mathrm{F}}|=\sqrt{1^{2}+1^{2}}\) \(|\overrightarrow{\mathrm{F}}|=\sqrt{2} \mathrm{~N}\) Angle with \(\mathrm{x}\) axis- \(\tan \theta=\frac{y-\text { component }}{x-\text { component }}=\frac{1}{1}\) \(\tan \theta=1=\tan 45^{\circ}\) \(\theta=45^{\circ}\)
371837
A block is kept on a frictionless inclined surface with angle of inclination ' \(\alpha\) '. The incline is given an acceleration ' \(a\) ' to keep the block stationary. Then ' \(a\) ' is equal to
1 \(g \operatorname{cosec} \alpha\)
2 \(g / \tan \alpha\)
3 \(g \tan \alpha\)
4 \(g\)
Explanation:
C According to the figure, for block to remain stationary ma cos \(\square=\mathrm{mg} \sin \square\) \(\mathrm{a}=\frac{\mathrm{g} \sin \alpha}{\cos \alpha}\) \(\mathrm{a}=\mathrm{g} \tan \alpha\)
BITSAT-2019
LAWS OF MOTION (ADDITIONAL)
371838
Which of the following sets of concurrent force may be in equilibrium?
1 \(F_{1}=3 N, F_{2}=5 N, F_{3}=10 N\)
2 \(F_{1}=3 N, F_{2}=5 N, F_{3}=9 N\)
3 \(F_{1}=3 N, F_{2}=5 N, F_{3}=6 N\)
4 \(F_{1}=3 N, F_{2}=5 N, F_{3}=15 N\)
Explanation:
C If three concurrent forces are to be in equilibrium, the sum of two smallest magnitude must be greater than the magnitude of third force. i.e. \(\left|\overrightarrow{\mathrm{F}}_{1}\right|+\left|\overrightarrow{\mathrm{F}}_{2}\right| \geq\left|\overrightarrow{\mathrm{F}}_{3}\right|\) Now, we can substitute these values in the above equation, we get- (c) \(\mathrm{F}_{1}=3 \mathrm{~N}, \mathrm{~F}_{2}=5 \mathrm{~N}, \mathrm{~F}_{3}=6 \mathrm{~N}\) \(3+5>6\) So, option (c) is right answer.
AP EAMCET (17.09.2020) Shift-II
LAWS OF MOTION (ADDITIONAL)
371839
A block of mass \(M\) slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is \(\theta\). The magnitude of the contact force will be :
A FBD of Body Block slides down with constant velocity then net force on the block is zero. So, Net contact force \(F_{C}\) on the body is \(F_{C}=\sqrt{f^{2}+N^{2}}\) From the figure- \(f=m g \sin \theta\) \(N=m g \cos \theta\) \(=\sqrt{(m g \sin \theta)^{2}+(m g \cos \theta)^{2}}\) \(=\sqrt{(m g)^{2}\left[\sin ^{2} \theta+\cos ^{2} \theta\right]}\) \(F_{C}=m g\)
JEE Main-27.07.2022
LAWS OF MOTION (ADDITIONAL)
371840
Statement [A]: If a body is in equilibrium, then the resultant of the forces acting on the body is zero Statement [B]: Action and reaction cancel each other as they act on the same body Statement [C]: Centrifugal force is a pseudo force Options:
1 A, B, C are all true
2 A, B are true but C is false
3 A, C are true but B is false
4 B, C are true but A is false
Explanation:
C A, C are true but B is false Statement A: If a body is in equilibrium, then the resultant of the forces acting on the body is zero. Statement C: Centrifugal force is a pseudo force because it is not provide by any real force but it arises due to accelerated frame of reference.
AP EAPCET-12.07.2022
LAWS OF MOTION (ADDITIONAL)
371841
For a free body diagram shown in the figure, the four forces are applied in the ' \(x\) ' and ' \(y\) ' directions. What additional force must be applied and at what angle with positive \(x\)-axis so that the net acceleration of body is zero?
1 \(\sqrt{2} \mathrm{~N}, 45^{\circ}\)
2 \(\sqrt{2} \mathrm{~N}, 135^{\circ}\)
3 \(\frac{2}{\sqrt{3}} \mathrm{~N}, 30^{\circ}\)
4 \(2 \mathrm{~N}, 45^{\circ}\)
Explanation:
A Assume addition force required is \(=\overrightarrow{\mathrm{F}}\) By figure, \(\overrightarrow{\mathrm{F}}+5 \hat{\mathrm{i}}-6 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-8 \hat{\mathrm{j}}=0\) \(\overrightarrow{\mathrm{F}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}\) \(|\overrightarrow{\mathrm{F}}|=\sqrt{1^{2}+1^{2}}\) \(|\overrightarrow{\mathrm{F}}|=\sqrt{2} \mathrm{~N}\) Angle with \(\mathrm{x}\) axis- \(\tan \theta=\frac{y-\text { component }}{x-\text { component }}=\frac{1}{1}\) \(\tan \theta=1=\tan 45^{\circ}\) \(\theta=45^{\circ}\)
371837
A block is kept on a frictionless inclined surface with angle of inclination ' \(\alpha\) '. The incline is given an acceleration ' \(a\) ' to keep the block stationary. Then ' \(a\) ' is equal to
1 \(g \operatorname{cosec} \alpha\)
2 \(g / \tan \alpha\)
3 \(g \tan \alpha\)
4 \(g\)
Explanation:
C According to the figure, for block to remain stationary ma cos \(\square=\mathrm{mg} \sin \square\) \(\mathrm{a}=\frac{\mathrm{g} \sin \alpha}{\cos \alpha}\) \(\mathrm{a}=\mathrm{g} \tan \alpha\)
BITSAT-2019
LAWS OF MOTION (ADDITIONAL)
371838
Which of the following sets of concurrent force may be in equilibrium?
1 \(F_{1}=3 N, F_{2}=5 N, F_{3}=10 N\)
2 \(F_{1}=3 N, F_{2}=5 N, F_{3}=9 N\)
3 \(F_{1}=3 N, F_{2}=5 N, F_{3}=6 N\)
4 \(F_{1}=3 N, F_{2}=5 N, F_{3}=15 N\)
Explanation:
C If three concurrent forces are to be in equilibrium, the sum of two smallest magnitude must be greater than the magnitude of third force. i.e. \(\left|\overrightarrow{\mathrm{F}}_{1}\right|+\left|\overrightarrow{\mathrm{F}}_{2}\right| \geq\left|\overrightarrow{\mathrm{F}}_{3}\right|\) Now, we can substitute these values in the above equation, we get- (c) \(\mathrm{F}_{1}=3 \mathrm{~N}, \mathrm{~F}_{2}=5 \mathrm{~N}, \mathrm{~F}_{3}=6 \mathrm{~N}\) \(3+5>6\) So, option (c) is right answer.
AP EAMCET (17.09.2020) Shift-II
LAWS OF MOTION (ADDITIONAL)
371839
A block of mass \(M\) slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is \(\theta\). The magnitude of the contact force will be :
A FBD of Body Block slides down with constant velocity then net force on the block is zero. So, Net contact force \(F_{C}\) on the body is \(F_{C}=\sqrt{f^{2}+N^{2}}\) From the figure- \(f=m g \sin \theta\) \(N=m g \cos \theta\) \(=\sqrt{(m g \sin \theta)^{2}+(m g \cos \theta)^{2}}\) \(=\sqrt{(m g)^{2}\left[\sin ^{2} \theta+\cos ^{2} \theta\right]}\) \(F_{C}=m g\)
JEE Main-27.07.2022
LAWS OF MOTION (ADDITIONAL)
371840
Statement [A]: If a body is in equilibrium, then the resultant of the forces acting on the body is zero Statement [B]: Action and reaction cancel each other as they act on the same body Statement [C]: Centrifugal force is a pseudo force Options:
1 A, B, C are all true
2 A, B are true but C is false
3 A, C are true but B is false
4 B, C are true but A is false
Explanation:
C A, C are true but B is false Statement A: If a body is in equilibrium, then the resultant of the forces acting on the body is zero. Statement C: Centrifugal force is a pseudo force because it is not provide by any real force but it arises due to accelerated frame of reference.
AP EAPCET-12.07.2022
LAWS OF MOTION (ADDITIONAL)
371841
For a free body diagram shown in the figure, the four forces are applied in the ' \(x\) ' and ' \(y\) ' directions. What additional force must be applied and at what angle with positive \(x\)-axis so that the net acceleration of body is zero?
1 \(\sqrt{2} \mathrm{~N}, 45^{\circ}\)
2 \(\sqrt{2} \mathrm{~N}, 135^{\circ}\)
3 \(\frac{2}{\sqrt{3}} \mathrm{~N}, 30^{\circ}\)
4 \(2 \mathrm{~N}, 45^{\circ}\)
Explanation:
A Assume addition force required is \(=\overrightarrow{\mathrm{F}}\) By figure, \(\overrightarrow{\mathrm{F}}+5 \hat{\mathrm{i}}-6 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-8 \hat{\mathrm{j}}=0\) \(\overrightarrow{\mathrm{F}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}\) \(|\overrightarrow{\mathrm{F}}|=\sqrt{1^{2}+1^{2}}\) \(|\overrightarrow{\mathrm{F}}|=\sqrt{2} \mathrm{~N}\) Angle with \(\mathrm{x}\) axis- \(\tan \theta=\frac{y-\text { component }}{x-\text { component }}=\frac{1}{1}\) \(\tan \theta=1=\tan 45^{\circ}\) \(\theta=45^{\circ}\)
371837
A block is kept on a frictionless inclined surface with angle of inclination ' \(\alpha\) '. The incline is given an acceleration ' \(a\) ' to keep the block stationary. Then ' \(a\) ' is equal to
1 \(g \operatorname{cosec} \alpha\)
2 \(g / \tan \alpha\)
3 \(g \tan \alpha\)
4 \(g\)
Explanation:
C According to the figure, for block to remain stationary ma cos \(\square=\mathrm{mg} \sin \square\) \(\mathrm{a}=\frac{\mathrm{g} \sin \alpha}{\cos \alpha}\) \(\mathrm{a}=\mathrm{g} \tan \alpha\)
BITSAT-2019
LAWS OF MOTION (ADDITIONAL)
371838
Which of the following sets of concurrent force may be in equilibrium?
1 \(F_{1}=3 N, F_{2}=5 N, F_{3}=10 N\)
2 \(F_{1}=3 N, F_{2}=5 N, F_{3}=9 N\)
3 \(F_{1}=3 N, F_{2}=5 N, F_{3}=6 N\)
4 \(F_{1}=3 N, F_{2}=5 N, F_{3}=15 N\)
Explanation:
C If three concurrent forces are to be in equilibrium, the sum of two smallest magnitude must be greater than the magnitude of third force. i.e. \(\left|\overrightarrow{\mathrm{F}}_{1}\right|+\left|\overrightarrow{\mathrm{F}}_{2}\right| \geq\left|\overrightarrow{\mathrm{F}}_{3}\right|\) Now, we can substitute these values in the above equation, we get- (c) \(\mathrm{F}_{1}=3 \mathrm{~N}, \mathrm{~F}_{2}=5 \mathrm{~N}, \mathrm{~F}_{3}=6 \mathrm{~N}\) \(3+5>6\) So, option (c) is right answer.
AP EAMCET (17.09.2020) Shift-II
LAWS OF MOTION (ADDITIONAL)
371839
A block of mass \(M\) slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is \(\theta\). The magnitude of the contact force will be :
A FBD of Body Block slides down with constant velocity then net force on the block is zero. So, Net contact force \(F_{C}\) on the body is \(F_{C}=\sqrt{f^{2}+N^{2}}\) From the figure- \(f=m g \sin \theta\) \(N=m g \cos \theta\) \(=\sqrt{(m g \sin \theta)^{2}+(m g \cos \theta)^{2}}\) \(=\sqrt{(m g)^{2}\left[\sin ^{2} \theta+\cos ^{2} \theta\right]}\) \(F_{C}=m g\)
JEE Main-27.07.2022
LAWS OF MOTION (ADDITIONAL)
371840
Statement [A]: If a body is in equilibrium, then the resultant of the forces acting on the body is zero Statement [B]: Action and reaction cancel each other as they act on the same body Statement [C]: Centrifugal force is a pseudo force Options:
1 A, B, C are all true
2 A, B are true but C is false
3 A, C are true but B is false
4 B, C are true but A is false
Explanation:
C A, C are true but B is false Statement A: If a body is in equilibrium, then the resultant of the forces acting on the body is zero. Statement C: Centrifugal force is a pseudo force because it is not provide by any real force but it arises due to accelerated frame of reference.
AP EAPCET-12.07.2022
LAWS OF MOTION (ADDITIONAL)
371841
For a free body diagram shown in the figure, the four forces are applied in the ' \(x\) ' and ' \(y\) ' directions. What additional force must be applied and at what angle with positive \(x\)-axis so that the net acceleration of body is zero?
1 \(\sqrt{2} \mathrm{~N}, 45^{\circ}\)
2 \(\sqrt{2} \mathrm{~N}, 135^{\circ}\)
3 \(\frac{2}{\sqrt{3}} \mathrm{~N}, 30^{\circ}\)
4 \(2 \mathrm{~N}, 45^{\circ}\)
Explanation:
A Assume addition force required is \(=\overrightarrow{\mathrm{F}}\) By figure, \(\overrightarrow{\mathrm{F}}+5 \hat{\mathrm{i}}-6 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-8 \hat{\mathrm{j}}=0\) \(\overrightarrow{\mathrm{F}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}\) \(|\overrightarrow{\mathrm{F}}|=\sqrt{1^{2}+1^{2}}\) \(|\overrightarrow{\mathrm{F}}|=\sqrt{2} \mathrm{~N}\) Angle with \(\mathrm{x}\) axis- \(\tan \theta=\frac{y-\text { component }}{x-\text { component }}=\frac{1}{1}\) \(\tan \theta=1=\tan 45^{\circ}\) \(\theta=45^{\circ}\)
