372103
A small body was launched up an inclined plane set at an angle \(\theta\) against the horizontal. If the time of ascent of the body is \(k\) times less than the time of descent. What is the coefficient of friction?
A Case: (1)- When the body is launched up: Consider, \(\mu=\) coefficient of friction, \(\mathrm{u}=\) velocity of projection \(\mathrm{s}=\) distance along Now retarding force on the block - \(\mathrm{F}=\mathrm{ma}=\mathrm{mg} \sin \theta+\mu \mathrm{mg} \cos \theta\) \(\mathrm{a}=\mathrm{g}(\sin \theta+\mu \cos \theta)\) by \(2^{\text {nd }}\) equation of motion \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) Initially, \(\mathrm{u}=0\), and by equation (i), \(\mathrm{s}=0+\frac{1}{2} \mathrm{~g}(\sin \theta+\mu \cos \theta) \mathrm{t}^{2}\) \(\mathrm{~s}=\frac{1}{2} \mathrm{gt}^{2}{ }_{\text {ascent }}(\sin \theta+\mu \cos \theta)\) Case (ii) - When block comes downward direction:- \(\mathrm{f}=\mathrm{ma}=\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta\) \(\mathrm{a}=\mathrm{g} \sin \theta-\mu \mathrm{g} \cos \theta\) \(\mathrm{s}^{\prime}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) Now by equation (iii), \(s^{\prime}=\frac{1}{2} g(\sin \theta-\mu \cos \theta) t_{\text {descent---- }}^{2}\) From equation (ii) and (iv), we get - And, \(\quad \mathrm{t}_{\text {ascent }}=\frac{1}{\mathrm{k}} \mathrm{t}_{\text {descent }}\) \(\mathrm{s}^{\prime}=\mathrm{s}\) \(\frac{1}{2} g(\sin \theta+\mu \cos \theta) t_{\text {ascent }}^{2}=\frac{1}{2} g(\sin \theta-\mu \cos \theta) t_{\text {descent }}^{2}\) \((\sin \theta+\mu \cos \theta) \times \frac{1}{\mathrm{k}^{2}}=(\sin \theta-\mu \cos \theta)\) Divided by \(\cos \theta\) in both side:- \(\left(\frac{\sin \theta}{\cos \theta}+\mu\right) \cdot \frac{1}{\mathrm{k}^{2}}=\left(\frac{\sin \theta}{\cos \theta}-\mu\right)\) \((\tan \theta+\mu) \frac{1}{\mathrm{k}^{2}}=(\tan \theta-\mu)\) \(\tan \theta+\mu=\mathrm{k}^{2} \tan \theta-\mu \mathrm{k}^{2}\) \(\mu\left(1+\mathrm{k}^{2}\right)=\tan \theta\left(\mathrm{k}^{2}-1\right)\) \(\mu=\tan \theta\left(\frac{\mathrm{k}^{2}-1}{\mathrm{k}^{2}+1}\right)\) Hence, coefficient of friction \(\mu=\tan \theta\left(\frac{\mathrm{k}^{2}-1}{\mathrm{k}^{2}+1}\right)\)
SCRA-2010
LAWS OF MOTION (ADDITIONAL)
372104
A body kept on a smooth inclined plane having inclination 1 in \(l\) will remain stationary relative to the inclined plane if the plane is given a horizontal acceleration equal to
1 \(\frac{\mathrm{g}}{\sqrt{l^{2}-1}}\)
2 \(\frac{\mathrm{g} l}{\sqrt{l^{2}-1}}\)
3 \(\frac{\mathrm{g}}{2 \sqrt{l^{2}-1}}\)
4 \(\frac{2 \mathrm{~g}}{\sqrt{l^{2}-1}}\)
Explanation:
A From free body diagram (FBD) ma \(\cos \theta=m g \sin \theta\) \(a=g \frac{\sin \theta}{\cos \theta}=g \tan \theta\) Given: \(\sin \theta=\frac{1}{l}\) \(\tan \theta=\frac{1}{\sqrt{l^{2}-1}}\) \(\therefore\) From equation (i) and (ii), we get - \(\mathrm{a}=\mathrm{g} \frac{1}{\sqrt{l^{2}-1}}\) \(\mathrm{a}=\frac{\mathrm{g}}{\sqrt{l^{2}-1}}\)
SCRA-2009
LAWS OF MOTION (ADDITIONAL)
372105
Maximum acceleration of the train in which a \(50 \mathrm{~kg}\) box lying on its floor will remain stationary (Given, coefficient of static friction between the box and the train's floor is 0.3 and \(\mathbf{g}=\mathbf{1 0} \mathbf{~ m s}^{-2}\) ) :
1 \(5.0 \mathrm{~ms}^{-2}\)
2 \(3.0 \mathrm{~ms}^{-2}\)
3 \(1.5 \mathrm{~ms}^{-2}\)
4 \(15 \mathrm{~ms}^{-2}\)
Explanation:
B Given, mass of the box (m) \(=10 \mathrm{~kg}\) Coefficient of friction \((\mu)=0.3\) Acceleration due to gravity \((\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}\) Net force, Friction force \(\mathrm{f}=\mu \mathrm{N}\) Normal force \(\mathrm{N}=\mathrm{mg}\) \(\mathrm{f}=\mathrm{mg} \mu\) \(\mathrm{ma}=\) pseudo force acting opposite to motion of train to make it stationary \(\mathrm{ma}=\text { friction force }\) \(\mathrm{ma}=\mu \mathrm{mg}\) \(\mathrm{a}=\mu \mathrm{g}\) \(\mathrm{a}=0.3 \times 10\) \(\mathrm{a}=3 \mathrm{~m} / \mathrm{s}^{2}\) Hence, the maximum acceleration of the train will be \(\left(3 \mathrm{~ms}^{-2}\right)\).
Karnataka CET-2016
LAWS OF MOTION (ADDITIONAL)
372106
A cubical block rests on an inclined plane of coefficient of friction \(\mu=\frac{1}{\sqrt{3}}\). What should be the angle of inclination so that the block just slides down the inclined plane?
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
A Given, \(\mu=\frac{1}{\sqrt{3}}\) Angle of inclination \(\theta\) The block will have force \(\mathrm{mg} \sin \theta\) along the incline plane and the frictional force will have \(\mathrm{f}=\mu \mathrm{N}\) Where, \(\mathrm{N}=\mathrm{mg} \cos \theta\) For equilibrium - \(\therefore \quad \mathrm{f}=\mathrm{mg} \sin \theta\) \(\mu \mathrm{mg} \cos \theta=\mathrm{mg} \sin \theta\) \(\tan \theta=\mu\) \(\tan \theta=\frac{1}{\sqrt{3}}\) \(\theta=\tan ^{-1} \frac{1}{\sqrt{3}}\) \(\theta=30^{\circ}\) Hence, angle of inclination \(\theta=30^{\circ}\)
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LAWS OF MOTION (ADDITIONAL)
372103
A small body was launched up an inclined plane set at an angle \(\theta\) against the horizontal. If the time of ascent of the body is \(k\) times less than the time of descent. What is the coefficient of friction?
A Case: (1)- When the body is launched up: Consider, \(\mu=\) coefficient of friction, \(\mathrm{u}=\) velocity of projection \(\mathrm{s}=\) distance along Now retarding force on the block - \(\mathrm{F}=\mathrm{ma}=\mathrm{mg} \sin \theta+\mu \mathrm{mg} \cos \theta\) \(\mathrm{a}=\mathrm{g}(\sin \theta+\mu \cos \theta)\) by \(2^{\text {nd }}\) equation of motion \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) Initially, \(\mathrm{u}=0\), and by equation (i), \(\mathrm{s}=0+\frac{1}{2} \mathrm{~g}(\sin \theta+\mu \cos \theta) \mathrm{t}^{2}\) \(\mathrm{~s}=\frac{1}{2} \mathrm{gt}^{2}{ }_{\text {ascent }}(\sin \theta+\mu \cos \theta)\) Case (ii) - When block comes downward direction:- \(\mathrm{f}=\mathrm{ma}=\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta\) \(\mathrm{a}=\mathrm{g} \sin \theta-\mu \mathrm{g} \cos \theta\) \(\mathrm{s}^{\prime}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) Now by equation (iii), \(s^{\prime}=\frac{1}{2} g(\sin \theta-\mu \cos \theta) t_{\text {descent---- }}^{2}\) From equation (ii) and (iv), we get - And, \(\quad \mathrm{t}_{\text {ascent }}=\frac{1}{\mathrm{k}} \mathrm{t}_{\text {descent }}\) \(\mathrm{s}^{\prime}=\mathrm{s}\) \(\frac{1}{2} g(\sin \theta+\mu \cos \theta) t_{\text {ascent }}^{2}=\frac{1}{2} g(\sin \theta-\mu \cos \theta) t_{\text {descent }}^{2}\) \((\sin \theta+\mu \cos \theta) \times \frac{1}{\mathrm{k}^{2}}=(\sin \theta-\mu \cos \theta)\) Divided by \(\cos \theta\) in both side:- \(\left(\frac{\sin \theta}{\cos \theta}+\mu\right) \cdot \frac{1}{\mathrm{k}^{2}}=\left(\frac{\sin \theta}{\cos \theta}-\mu\right)\) \((\tan \theta+\mu) \frac{1}{\mathrm{k}^{2}}=(\tan \theta-\mu)\) \(\tan \theta+\mu=\mathrm{k}^{2} \tan \theta-\mu \mathrm{k}^{2}\) \(\mu\left(1+\mathrm{k}^{2}\right)=\tan \theta\left(\mathrm{k}^{2}-1\right)\) \(\mu=\tan \theta\left(\frac{\mathrm{k}^{2}-1}{\mathrm{k}^{2}+1}\right)\) Hence, coefficient of friction \(\mu=\tan \theta\left(\frac{\mathrm{k}^{2}-1}{\mathrm{k}^{2}+1}\right)\)
SCRA-2010
LAWS OF MOTION (ADDITIONAL)
372104
A body kept on a smooth inclined plane having inclination 1 in \(l\) will remain stationary relative to the inclined plane if the plane is given a horizontal acceleration equal to
1 \(\frac{\mathrm{g}}{\sqrt{l^{2}-1}}\)
2 \(\frac{\mathrm{g} l}{\sqrt{l^{2}-1}}\)
3 \(\frac{\mathrm{g}}{2 \sqrt{l^{2}-1}}\)
4 \(\frac{2 \mathrm{~g}}{\sqrt{l^{2}-1}}\)
Explanation:
A From free body diagram (FBD) ma \(\cos \theta=m g \sin \theta\) \(a=g \frac{\sin \theta}{\cos \theta}=g \tan \theta\) Given: \(\sin \theta=\frac{1}{l}\) \(\tan \theta=\frac{1}{\sqrt{l^{2}-1}}\) \(\therefore\) From equation (i) and (ii), we get - \(\mathrm{a}=\mathrm{g} \frac{1}{\sqrt{l^{2}-1}}\) \(\mathrm{a}=\frac{\mathrm{g}}{\sqrt{l^{2}-1}}\)
SCRA-2009
LAWS OF MOTION (ADDITIONAL)
372105
Maximum acceleration of the train in which a \(50 \mathrm{~kg}\) box lying on its floor will remain stationary (Given, coefficient of static friction between the box and the train's floor is 0.3 and \(\mathbf{g}=\mathbf{1 0} \mathbf{~ m s}^{-2}\) ) :
1 \(5.0 \mathrm{~ms}^{-2}\)
2 \(3.0 \mathrm{~ms}^{-2}\)
3 \(1.5 \mathrm{~ms}^{-2}\)
4 \(15 \mathrm{~ms}^{-2}\)
Explanation:
B Given, mass of the box (m) \(=10 \mathrm{~kg}\) Coefficient of friction \((\mu)=0.3\) Acceleration due to gravity \((\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}\) Net force, Friction force \(\mathrm{f}=\mu \mathrm{N}\) Normal force \(\mathrm{N}=\mathrm{mg}\) \(\mathrm{f}=\mathrm{mg} \mu\) \(\mathrm{ma}=\) pseudo force acting opposite to motion of train to make it stationary \(\mathrm{ma}=\text { friction force }\) \(\mathrm{ma}=\mu \mathrm{mg}\) \(\mathrm{a}=\mu \mathrm{g}\) \(\mathrm{a}=0.3 \times 10\) \(\mathrm{a}=3 \mathrm{~m} / \mathrm{s}^{2}\) Hence, the maximum acceleration of the train will be \(\left(3 \mathrm{~ms}^{-2}\right)\).
Karnataka CET-2016
LAWS OF MOTION (ADDITIONAL)
372106
A cubical block rests on an inclined plane of coefficient of friction \(\mu=\frac{1}{\sqrt{3}}\). What should be the angle of inclination so that the block just slides down the inclined plane?
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
A Given, \(\mu=\frac{1}{\sqrt{3}}\) Angle of inclination \(\theta\) The block will have force \(\mathrm{mg} \sin \theta\) along the incline plane and the frictional force will have \(\mathrm{f}=\mu \mathrm{N}\) Where, \(\mathrm{N}=\mathrm{mg} \cos \theta\) For equilibrium - \(\therefore \quad \mathrm{f}=\mathrm{mg} \sin \theta\) \(\mu \mathrm{mg} \cos \theta=\mathrm{mg} \sin \theta\) \(\tan \theta=\mu\) \(\tan \theta=\frac{1}{\sqrt{3}}\) \(\theta=\tan ^{-1} \frac{1}{\sqrt{3}}\) \(\theta=30^{\circ}\) Hence, angle of inclination \(\theta=30^{\circ}\)
372103
A small body was launched up an inclined plane set at an angle \(\theta\) against the horizontal. If the time of ascent of the body is \(k\) times less than the time of descent. What is the coefficient of friction?
A Case: (1)- When the body is launched up: Consider, \(\mu=\) coefficient of friction, \(\mathrm{u}=\) velocity of projection \(\mathrm{s}=\) distance along Now retarding force on the block - \(\mathrm{F}=\mathrm{ma}=\mathrm{mg} \sin \theta+\mu \mathrm{mg} \cos \theta\) \(\mathrm{a}=\mathrm{g}(\sin \theta+\mu \cos \theta)\) by \(2^{\text {nd }}\) equation of motion \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) Initially, \(\mathrm{u}=0\), and by equation (i), \(\mathrm{s}=0+\frac{1}{2} \mathrm{~g}(\sin \theta+\mu \cos \theta) \mathrm{t}^{2}\) \(\mathrm{~s}=\frac{1}{2} \mathrm{gt}^{2}{ }_{\text {ascent }}(\sin \theta+\mu \cos \theta)\) Case (ii) - When block comes downward direction:- \(\mathrm{f}=\mathrm{ma}=\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta\) \(\mathrm{a}=\mathrm{g} \sin \theta-\mu \mathrm{g} \cos \theta\) \(\mathrm{s}^{\prime}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) Now by equation (iii), \(s^{\prime}=\frac{1}{2} g(\sin \theta-\mu \cos \theta) t_{\text {descent---- }}^{2}\) From equation (ii) and (iv), we get - And, \(\quad \mathrm{t}_{\text {ascent }}=\frac{1}{\mathrm{k}} \mathrm{t}_{\text {descent }}\) \(\mathrm{s}^{\prime}=\mathrm{s}\) \(\frac{1}{2} g(\sin \theta+\mu \cos \theta) t_{\text {ascent }}^{2}=\frac{1}{2} g(\sin \theta-\mu \cos \theta) t_{\text {descent }}^{2}\) \((\sin \theta+\mu \cos \theta) \times \frac{1}{\mathrm{k}^{2}}=(\sin \theta-\mu \cos \theta)\) Divided by \(\cos \theta\) in both side:- \(\left(\frac{\sin \theta}{\cos \theta}+\mu\right) \cdot \frac{1}{\mathrm{k}^{2}}=\left(\frac{\sin \theta}{\cos \theta}-\mu\right)\) \((\tan \theta+\mu) \frac{1}{\mathrm{k}^{2}}=(\tan \theta-\mu)\) \(\tan \theta+\mu=\mathrm{k}^{2} \tan \theta-\mu \mathrm{k}^{2}\) \(\mu\left(1+\mathrm{k}^{2}\right)=\tan \theta\left(\mathrm{k}^{2}-1\right)\) \(\mu=\tan \theta\left(\frac{\mathrm{k}^{2}-1}{\mathrm{k}^{2}+1}\right)\) Hence, coefficient of friction \(\mu=\tan \theta\left(\frac{\mathrm{k}^{2}-1}{\mathrm{k}^{2}+1}\right)\)
SCRA-2010
LAWS OF MOTION (ADDITIONAL)
372104
A body kept on a smooth inclined plane having inclination 1 in \(l\) will remain stationary relative to the inclined plane if the plane is given a horizontal acceleration equal to
1 \(\frac{\mathrm{g}}{\sqrt{l^{2}-1}}\)
2 \(\frac{\mathrm{g} l}{\sqrt{l^{2}-1}}\)
3 \(\frac{\mathrm{g}}{2 \sqrt{l^{2}-1}}\)
4 \(\frac{2 \mathrm{~g}}{\sqrt{l^{2}-1}}\)
Explanation:
A From free body diagram (FBD) ma \(\cos \theta=m g \sin \theta\) \(a=g \frac{\sin \theta}{\cos \theta}=g \tan \theta\) Given: \(\sin \theta=\frac{1}{l}\) \(\tan \theta=\frac{1}{\sqrt{l^{2}-1}}\) \(\therefore\) From equation (i) and (ii), we get - \(\mathrm{a}=\mathrm{g} \frac{1}{\sqrt{l^{2}-1}}\) \(\mathrm{a}=\frac{\mathrm{g}}{\sqrt{l^{2}-1}}\)
SCRA-2009
LAWS OF MOTION (ADDITIONAL)
372105
Maximum acceleration of the train in which a \(50 \mathrm{~kg}\) box lying on its floor will remain stationary (Given, coefficient of static friction between the box and the train's floor is 0.3 and \(\mathbf{g}=\mathbf{1 0} \mathbf{~ m s}^{-2}\) ) :
1 \(5.0 \mathrm{~ms}^{-2}\)
2 \(3.0 \mathrm{~ms}^{-2}\)
3 \(1.5 \mathrm{~ms}^{-2}\)
4 \(15 \mathrm{~ms}^{-2}\)
Explanation:
B Given, mass of the box (m) \(=10 \mathrm{~kg}\) Coefficient of friction \((\mu)=0.3\) Acceleration due to gravity \((\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}\) Net force, Friction force \(\mathrm{f}=\mu \mathrm{N}\) Normal force \(\mathrm{N}=\mathrm{mg}\) \(\mathrm{f}=\mathrm{mg} \mu\) \(\mathrm{ma}=\) pseudo force acting opposite to motion of train to make it stationary \(\mathrm{ma}=\text { friction force }\) \(\mathrm{ma}=\mu \mathrm{mg}\) \(\mathrm{a}=\mu \mathrm{g}\) \(\mathrm{a}=0.3 \times 10\) \(\mathrm{a}=3 \mathrm{~m} / \mathrm{s}^{2}\) Hence, the maximum acceleration of the train will be \(\left(3 \mathrm{~ms}^{-2}\right)\).
Karnataka CET-2016
LAWS OF MOTION (ADDITIONAL)
372106
A cubical block rests on an inclined plane of coefficient of friction \(\mu=\frac{1}{\sqrt{3}}\). What should be the angle of inclination so that the block just slides down the inclined plane?
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
A Given, \(\mu=\frac{1}{\sqrt{3}}\) Angle of inclination \(\theta\) The block will have force \(\mathrm{mg} \sin \theta\) along the incline plane and the frictional force will have \(\mathrm{f}=\mu \mathrm{N}\) Where, \(\mathrm{N}=\mathrm{mg} \cos \theta\) For equilibrium - \(\therefore \quad \mathrm{f}=\mathrm{mg} \sin \theta\) \(\mu \mathrm{mg} \cos \theta=\mathrm{mg} \sin \theta\) \(\tan \theta=\mu\) \(\tan \theta=\frac{1}{\sqrt{3}}\) \(\theta=\tan ^{-1} \frac{1}{\sqrt{3}}\) \(\theta=30^{\circ}\) Hence, angle of inclination \(\theta=30^{\circ}\)
372103
A small body was launched up an inclined plane set at an angle \(\theta\) against the horizontal. If the time of ascent of the body is \(k\) times less than the time of descent. What is the coefficient of friction?
A Case: (1)- When the body is launched up: Consider, \(\mu=\) coefficient of friction, \(\mathrm{u}=\) velocity of projection \(\mathrm{s}=\) distance along Now retarding force on the block - \(\mathrm{F}=\mathrm{ma}=\mathrm{mg} \sin \theta+\mu \mathrm{mg} \cos \theta\) \(\mathrm{a}=\mathrm{g}(\sin \theta+\mu \cos \theta)\) by \(2^{\text {nd }}\) equation of motion \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) Initially, \(\mathrm{u}=0\), and by equation (i), \(\mathrm{s}=0+\frac{1}{2} \mathrm{~g}(\sin \theta+\mu \cos \theta) \mathrm{t}^{2}\) \(\mathrm{~s}=\frac{1}{2} \mathrm{gt}^{2}{ }_{\text {ascent }}(\sin \theta+\mu \cos \theta)\) Case (ii) - When block comes downward direction:- \(\mathrm{f}=\mathrm{ma}=\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta\) \(\mathrm{a}=\mathrm{g} \sin \theta-\mu \mathrm{g} \cos \theta\) \(\mathrm{s}^{\prime}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) Now by equation (iii), \(s^{\prime}=\frac{1}{2} g(\sin \theta-\mu \cos \theta) t_{\text {descent---- }}^{2}\) From equation (ii) and (iv), we get - And, \(\quad \mathrm{t}_{\text {ascent }}=\frac{1}{\mathrm{k}} \mathrm{t}_{\text {descent }}\) \(\mathrm{s}^{\prime}=\mathrm{s}\) \(\frac{1}{2} g(\sin \theta+\mu \cos \theta) t_{\text {ascent }}^{2}=\frac{1}{2} g(\sin \theta-\mu \cos \theta) t_{\text {descent }}^{2}\) \((\sin \theta+\mu \cos \theta) \times \frac{1}{\mathrm{k}^{2}}=(\sin \theta-\mu \cos \theta)\) Divided by \(\cos \theta\) in both side:- \(\left(\frac{\sin \theta}{\cos \theta}+\mu\right) \cdot \frac{1}{\mathrm{k}^{2}}=\left(\frac{\sin \theta}{\cos \theta}-\mu\right)\) \((\tan \theta+\mu) \frac{1}{\mathrm{k}^{2}}=(\tan \theta-\mu)\) \(\tan \theta+\mu=\mathrm{k}^{2} \tan \theta-\mu \mathrm{k}^{2}\) \(\mu\left(1+\mathrm{k}^{2}\right)=\tan \theta\left(\mathrm{k}^{2}-1\right)\) \(\mu=\tan \theta\left(\frac{\mathrm{k}^{2}-1}{\mathrm{k}^{2}+1}\right)\) Hence, coefficient of friction \(\mu=\tan \theta\left(\frac{\mathrm{k}^{2}-1}{\mathrm{k}^{2}+1}\right)\)
SCRA-2010
LAWS OF MOTION (ADDITIONAL)
372104
A body kept on a smooth inclined plane having inclination 1 in \(l\) will remain stationary relative to the inclined plane if the plane is given a horizontal acceleration equal to
1 \(\frac{\mathrm{g}}{\sqrt{l^{2}-1}}\)
2 \(\frac{\mathrm{g} l}{\sqrt{l^{2}-1}}\)
3 \(\frac{\mathrm{g}}{2 \sqrt{l^{2}-1}}\)
4 \(\frac{2 \mathrm{~g}}{\sqrt{l^{2}-1}}\)
Explanation:
A From free body diagram (FBD) ma \(\cos \theta=m g \sin \theta\) \(a=g \frac{\sin \theta}{\cos \theta}=g \tan \theta\) Given: \(\sin \theta=\frac{1}{l}\) \(\tan \theta=\frac{1}{\sqrt{l^{2}-1}}\) \(\therefore\) From equation (i) and (ii), we get - \(\mathrm{a}=\mathrm{g} \frac{1}{\sqrt{l^{2}-1}}\) \(\mathrm{a}=\frac{\mathrm{g}}{\sqrt{l^{2}-1}}\)
SCRA-2009
LAWS OF MOTION (ADDITIONAL)
372105
Maximum acceleration of the train in which a \(50 \mathrm{~kg}\) box lying on its floor will remain stationary (Given, coefficient of static friction between the box and the train's floor is 0.3 and \(\mathbf{g}=\mathbf{1 0} \mathbf{~ m s}^{-2}\) ) :
1 \(5.0 \mathrm{~ms}^{-2}\)
2 \(3.0 \mathrm{~ms}^{-2}\)
3 \(1.5 \mathrm{~ms}^{-2}\)
4 \(15 \mathrm{~ms}^{-2}\)
Explanation:
B Given, mass of the box (m) \(=10 \mathrm{~kg}\) Coefficient of friction \((\mu)=0.3\) Acceleration due to gravity \((\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}\) Net force, Friction force \(\mathrm{f}=\mu \mathrm{N}\) Normal force \(\mathrm{N}=\mathrm{mg}\) \(\mathrm{f}=\mathrm{mg} \mu\) \(\mathrm{ma}=\) pseudo force acting opposite to motion of train to make it stationary \(\mathrm{ma}=\text { friction force }\) \(\mathrm{ma}=\mu \mathrm{mg}\) \(\mathrm{a}=\mu \mathrm{g}\) \(\mathrm{a}=0.3 \times 10\) \(\mathrm{a}=3 \mathrm{~m} / \mathrm{s}^{2}\) Hence, the maximum acceleration of the train will be \(\left(3 \mathrm{~ms}^{-2}\right)\).
Karnataka CET-2016
LAWS OF MOTION (ADDITIONAL)
372106
A cubical block rests on an inclined plane of coefficient of friction \(\mu=\frac{1}{\sqrt{3}}\). What should be the angle of inclination so that the block just slides down the inclined plane?
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
A Given, \(\mu=\frac{1}{\sqrt{3}}\) Angle of inclination \(\theta\) The block will have force \(\mathrm{mg} \sin \theta\) along the incline plane and the frictional force will have \(\mathrm{f}=\mu \mathrm{N}\) Where, \(\mathrm{N}=\mathrm{mg} \cos \theta\) For equilibrium - \(\therefore \quad \mathrm{f}=\mathrm{mg} \sin \theta\) \(\mu \mathrm{mg} \cos \theta=\mathrm{mg} \sin \theta\) \(\tan \theta=\mu\) \(\tan \theta=\frac{1}{\sqrt{3}}\) \(\theta=\tan ^{-1} \frac{1}{\sqrt{3}}\) \(\theta=30^{\circ}\) Hence, angle of inclination \(\theta=30^{\circ}\)
