372150
A body at rest slides down a \(30^{\circ}\) inclined plane. The time taken by it to slide down is twice the time it takes when it slides down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is
1 0.43
2 0.37
3 0.64
4 0.75
Explanation:
A Given, \(\theta=30^{\circ}\) From the figure \(\mathrm{R}=\mathrm{mg} \cos \theta\) Net force on the body when slides down on inclined plane \(\mathrm{F}=\mathrm{mg} \sin \theta-\mathrm{f} .\) And, \(\quad \mathrm{F}=\mathrm{ma}\) Then, \(\quad m a=m g \sin \theta-\mu R[\because f=\mu R]\) \(\therefore \quad \mathrm{ma}=\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta \quad[\because \mathrm{R}=\mathrm{mg} \cos \theta]\) \(\mathrm{ma}=\mathrm{mg}(\sin \theta-\mu \cos \theta)\) \(\mathrm{a}=\mathrm{g}(\sin \theta-\mu \cos \theta)\) Time taken by the body to slide down the plane \(t_{1}=\sqrt{\frac{2 S}{a}}=\sqrt{\frac{2 S}{g \sin \theta-\mu g \cos \theta}}\) And time taken by the body in the absence of friction \(\mathrm{t}_{2}=\sqrt{\frac{2 \mathrm{~S}}{\mathrm{~g} \sin \theta}}\) Given, \(t_{1}=2 t_{2}\) \(\sqrt{\frac{2 S}{g(\sin \theta-\mu \cos \theta)}}=2 \sqrt{\frac{2 S}{g \sin \theta}}\) \(\frac{2 S}{g(\sin \theta-\mu \cos \theta)}=4 \times \frac{2 S}{g \sin \theta}\) \(\sin \theta=4 \sin \theta-4 \mu \cos \theta\) \(3 \sin \theta=4 \mu \cos \theta\) \(3 \tan \theta=4 \mu\) \(\mu=\frac{3}{4} \tan \theta=\frac{3}{4} \tan 30^{\circ}=0.43\)
JIPMER-2016
LAWS OF MOTION (ADDITIONAL)
372151
A rod \(P Q\) of length \(l\) is moving with ends remaining in contact with frictionless wall and floor. If at the instant, shown the velocity of end \(Q\) is \(2 \mathrm{~m} / \mathrm{s}\) towards negative direction of \(x\). The speed of end \(P\) will be
1 \(\sqrt{3} \mathrm{~ms}^{-1}\)
2 \(\frac{2}{\sqrt{3}} \mathrm{~ms}^{-1}\)
3 \(\sqrt{2} \mathrm{~ms}^{-1}\)
4 \(\frac{\sqrt{5}}{2} \mathrm{~ms}^{-1}\)
Explanation:
B Given that velocity at \(\mathrm{Q}\) \(\Delta\) POQ \(\mathrm{v}_{\mathrm{Q}}=-2 \mathrm{~m} / \mathrm{sec}\) \(\mathrm{X}^{2}+\mathrm{Y}^{2}=l^{2}\) On differentiating equation (i) w.r to \(t\) \(2 X \frac{d X}{d t}+2 Y \frac{d Y}{d t}=0\) From the figure \(\mathrm{X}=l \cos 60^{\circ}=\frac{l}{2} \text { and } \mathrm{Y}=l \sin 60^{\circ}=\frac{\sqrt{3} l}{2}\) \(2 \times\left(\frac{l}{2}\right) \times(-2)+2 \times\left(\frac{\sqrt{3} l}{2}\right) \frac{\mathrm{dY}}{\mathrm{dt}}=0\) \(-2+(\sqrt{3}) \frac{\mathrm{dY}}{\mathrm{dt}}=0\) \(-2=-\frac{\sqrt{3} \mathrm{dY}}{\mathrm{dt}}\) \(\frac{\mathrm{dY}}{\mathrm{dt}}=\mathrm{v}_{\mathrm{P}}=\frac{2}{\sqrt{3}} \mathrm{~m} / \mathrm{sec}\) Hence, the speed of end \(P\) will be \(v_{p}=\frac{2}{\sqrt{3}} \mathrm{~m} / \mathrm{sec}\).
JIPMER-2015
LAWS OF MOTION (ADDITIONAL)
372152
A block slides down on an incline of angle \(30^{\circ}\) with acceleration \(\frac{\mathrm{g}}{4}\). Find the kinetic friction coefficient.
1 \(\frac{1}{2 \sqrt{2}}\)
2 0.6
3 \(\frac{1}{2 \sqrt{3}}\)
4 \(\frac{1}{\sqrt{2}}\)
Explanation:
C Given, \(\mathrm{a}=\frac{\mathrm{g}}{4}, \theta=30^{\circ}\) Let block has mass \(m\) and \(\mu\) is kinetic friction. from the figure \(\mathrm{N}=\mathrm{mg} \cos 30^{\circ}\) We know \(\mathrm{f}=\mu \mathrm{N}\) \(\mathrm{f}=\mu \mathrm{mg} \cos 30^{\circ}\) And \(\quad \mathrm{mg} \sin 30^{\circ}-\mathrm{f}=\mathrm{ma}\) \(\mathrm{mg} \sin 30^{\circ}-\mu \mathrm{mg} \cos 30^{\circ}=\mathrm{m} \cdot \mathrm{g} / 4 \quad\left[\right.\) from \(\mathrm{eq}^{\mathrm{n}}\) (i)] \(\sin 30^{\circ}-\mu \cos 30^{\circ}=\frac{1}{4}\) \(\sin 30^{\circ}-\mu \cos 30^{\circ}=\frac{1}{4}\) \(\frac{1}{2}-\frac{\sqrt{3}}{2} \mu=\frac{1}{4}\) \(\frac{1}{2}-\frac{1}{4}=\frac{\sqrt{3}}{2} \mu\) \(\frac{2-1}{4}=\frac{\sqrt{3}}{2} \mu\) \(\frac{1}{4}=\frac{\sqrt{3}}{2} \mu\) \(\mu=\frac{1}{2 \sqrt{3}}\) \(\mu=\frac{1}{2 \sqrt{3}}\)
JIPMER-2015
LAWS OF MOTION (ADDITIONAL)
372153
A tangential force acting on the top of sphere of mass \(m\) kept on a rough horizontal place as shown in figure If the sphere rolls without slipping, then the acceleration with which the centre of sphere moves, is
1 \(\frac{10 \mathrm{~F}}{7 \mathrm{~m}}\)
2 \(\frac{\mathrm{F}}{2 \mathrm{~m}}\)
3 \(\frac{3 \mathrm{~F}}{3 \mathrm{~m}}\)
4 \(\frac{7 \mathrm{~F}}{2 \mathrm{~m}}\)
Explanation:
A Let \(a\) is the acceleration of the centre of sphere. Then, We know, \(\text { Angular acceleration }(\alpha)=\frac{a}{r}\) For linear motion of the centre \(\mathrm{f}+\mathrm{F}=\mathrm{ma}\) And for rotational motion about the centre. We know Then, \(\text { Torque }(\tau)=\mathrm{I} \alpha\) \(\mathrm{Fr}-\mathrm{fr}=\mathrm{I} \alpha\) \(\mathrm{Fr}-\mathrm{fr}=\left(\frac{2}{5} \mathrm{mr}^{2}\right)\left(\frac{\mathrm{a}}{\mathrm{r}}\right)\) [moment of Inertia of solid sphere (I) \(=\frac{2}{5} \mathrm{mr}^{2}\) ] \(\mathrm{F}-\mathrm{f}==\frac{2}{5} \mathrm{ma}\) From equation (i) and (ii) \(2 \mathrm{~F}=\frac{7}{5} \mathrm{ma}\) \(\mathrm{a}=\frac{10 \mathrm{~F}}{7 \mathrm{~m}}\)
372150
A body at rest slides down a \(30^{\circ}\) inclined plane. The time taken by it to slide down is twice the time it takes when it slides down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is
1 0.43
2 0.37
3 0.64
4 0.75
Explanation:
A Given, \(\theta=30^{\circ}\) From the figure \(\mathrm{R}=\mathrm{mg} \cos \theta\) Net force on the body when slides down on inclined plane \(\mathrm{F}=\mathrm{mg} \sin \theta-\mathrm{f} .\) And, \(\quad \mathrm{F}=\mathrm{ma}\) Then, \(\quad m a=m g \sin \theta-\mu R[\because f=\mu R]\) \(\therefore \quad \mathrm{ma}=\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta \quad[\because \mathrm{R}=\mathrm{mg} \cos \theta]\) \(\mathrm{ma}=\mathrm{mg}(\sin \theta-\mu \cos \theta)\) \(\mathrm{a}=\mathrm{g}(\sin \theta-\mu \cos \theta)\) Time taken by the body to slide down the plane \(t_{1}=\sqrt{\frac{2 S}{a}}=\sqrt{\frac{2 S}{g \sin \theta-\mu g \cos \theta}}\) And time taken by the body in the absence of friction \(\mathrm{t}_{2}=\sqrt{\frac{2 \mathrm{~S}}{\mathrm{~g} \sin \theta}}\) Given, \(t_{1}=2 t_{2}\) \(\sqrt{\frac{2 S}{g(\sin \theta-\mu \cos \theta)}}=2 \sqrt{\frac{2 S}{g \sin \theta}}\) \(\frac{2 S}{g(\sin \theta-\mu \cos \theta)}=4 \times \frac{2 S}{g \sin \theta}\) \(\sin \theta=4 \sin \theta-4 \mu \cos \theta\) \(3 \sin \theta=4 \mu \cos \theta\) \(3 \tan \theta=4 \mu\) \(\mu=\frac{3}{4} \tan \theta=\frac{3}{4} \tan 30^{\circ}=0.43\)
JIPMER-2016
LAWS OF MOTION (ADDITIONAL)
372151
A rod \(P Q\) of length \(l\) is moving with ends remaining in contact with frictionless wall and floor. If at the instant, shown the velocity of end \(Q\) is \(2 \mathrm{~m} / \mathrm{s}\) towards negative direction of \(x\). The speed of end \(P\) will be
1 \(\sqrt{3} \mathrm{~ms}^{-1}\)
2 \(\frac{2}{\sqrt{3}} \mathrm{~ms}^{-1}\)
3 \(\sqrt{2} \mathrm{~ms}^{-1}\)
4 \(\frac{\sqrt{5}}{2} \mathrm{~ms}^{-1}\)
Explanation:
B Given that velocity at \(\mathrm{Q}\) \(\Delta\) POQ \(\mathrm{v}_{\mathrm{Q}}=-2 \mathrm{~m} / \mathrm{sec}\) \(\mathrm{X}^{2}+\mathrm{Y}^{2}=l^{2}\) On differentiating equation (i) w.r to \(t\) \(2 X \frac{d X}{d t}+2 Y \frac{d Y}{d t}=0\) From the figure \(\mathrm{X}=l \cos 60^{\circ}=\frac{l}{2} \text { and } \mathrm{Y}=l \sin 60^{\circ}=\frac{\sqrt{3} l}{2}\) \(2 \times\left(\frac{l}{2}\right) \times(-2)+2 \times\left(\frac{\sqrt{3} l}{2}\right) \frac{\mathrm{dY}}{\mathrm{dt}}=0\) \(-2+(\sqrt{3}) \frac{\mathrm{dY}}{\mathrm{dt}}=0\) \(-2=-\frac{\sqrt{3} \mathrm{dY}}{\mathrm{dt}}\) \(\frac{\mathrm{dY}}{\mathrm{dt}}=\mathrm{v}_{\mathrm{P}}=\frac{2}{\sqrt{3}} \mathrm{~m} / \mathrm{sec}\) Hence, the speed of end \(P\) will be \(v_{p}=\frac{2}{\sqrt{3}} \mathrm{~m} / \mathrm{sec}\).
JIPMER-2015
LAWS OF MOTION (ADDITIONAL)
372152
A block slides down on an incline of angle \(30^{\circ}\) with acceleration \(\frac{\mathrm{g}}{4}\). Find the kinetic friction coefficient.
1 \(\frac{1}{2 \sqrt{2}}\)
2 0.6
3 \(\frac{1}{2 \sqrt{3}}\)
4 \(\frac{1}{\sqrt{2}}\)
Explanation:
C Given, \(\mathrm{a}=\frac{\mathrm{g}}{4}, \theta=30^{\circ}\) Let block has mass \(m\) and \(\mu\) is kinetic friction. from the figure \(\mathrm{N}=\mathrm{mg} \cos 30^{\circ}\) We know \(\mathrm{f}=\mu \mathrm{N}\) \(\mathrm{f}=\mu \mathrm{mg} \cos 30^{\circ}\) And \(\quad \mathrm{mg} \sin 30^{\circ}-\mathrm{f}=\mathrm{ma}\) \(\mathrm{mg} \sin 30^{\circ}-\mu \mathrm{mg} \cos 30^{\circ}=\mathrm{m} \cdot \mathrm{g} / 4 \quad\left[\right.\) from \(\mathrm{eq}^{\mathrm{n}}\) (i)] \(\sin 30^{\circ}-\mu \cos 30^{\circ}=\frac{1}{4}\) \(\sin 30^{\circ}-\mu \cos 30^{\circ}=\frac{1}{4}\) \(\frac{1}{2}-\frac{\sqrt{3}}{2} \mu=\frac{1}{4}\) \(\frac{1}{2}-\frac{1}{4}=\frac{\sqrt{3}}{2} \mu\) \(\frac{2-1}{4}=\frac{\sqrt{3}}{2} \mu\) \(\frac{1}{4}=\frac{\sqrt{3}}{2} \mu\) \(\mu=\frac{1}{2 \sqrt{3}}\) \(\mu=\frac{1}{2 \sqrt{3}}\)
JIPMER-2015
LAWS OF MOTION (ADDITIONAL)
372153
A tangential force acting on the top of sphere of mass \(m\) kept on a rough horizontal place as shown in figure If the sphere rolls without slipping, then the acceleration with which the centre of sphere moves, is
1 \(\frac{10 \mathrm{~F}}{7 \mathrm{~m}}\)
2 \(\frac{\mathrm{F}}{2 \mathrm{~m}}\)
3 \(\frac{3 \mathrm{~F}}{3 \mathrm{~m}}\)
4 \(\frac{7 \mathrm{~F}}{2 \mathrm{~m}}\)
Explanation:
A Let \(a\) is the acceleration of the centre of sphere. Then, We know, \(\text { Angular acceleration }(\alpha)=\frac{a}{r}\) For linear motion of the centre \(\mathrm{f}+\mathrm{F}=\mathrm{ma}\) And for rotational motion about the centre. We know Then, \(\text { Torque }(\tau)=\mathrm{I} \alpha\) \(\mathrm{Fr}-\mathrm{fr}=\mathrm{I} \alpha\) \(\mathrm{Fr}-\mathrm{fr}=\left(\frac{2}{5} \mathrm{mr}^{2}\right)\left(\frac{\mathrm{a}}{\mathrm{r}}\right)\) [moment of Inertia of solid sphere (I) \(=\frac{2}{5} \mathrm{mr}^{2}\) ] \(\mathrm{F}-\mathrm{f}==\frac{2}{5} \mathrm{ma}\) From equation (i) and (ii) \(2 \mathrm{~F}=\frac{7}{5} \mathrm{ma}\) \(\mathrm{a}=\frac{10 \mathrm{~F}}{7 \mathrm{~m}}\)
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LAWS OF MOTION (ADDITIONAL)
372150
A body at rest slides down a \(30^{\circ}\) inclined plane. The time taken by it to slide down is twice the time it takes when it slides down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is
1 0.43
2 0.37
3 0.64
4 0.75
Explanation:
A Given, \(\theta=30^{\circ}\) From the figure \(\mathrm{R}=\mathrm{mg} \cos \theta\) Net force on the body when slides down on inclined plane \(\mathrm{F}=\mathrm{mg} \sin \theta-\mathrm{f} .\) And, \(\quad \mathrm{F}=\mathrm{ma}\) Then, \(\quad m a=m g \sin \theta-\mu R[\because f=\mu R]\) \(\therefore \quad \mathrm{ma}=\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta \quad[\because \mathrm{R}=\mathrm{mg} \cos \theta]\) \(\mathrm{ma}=\mathrm{mg}(\sin \theta-\mu \cos \theta)\) \(\mathrm{a}=\mathrm{g}(\sin \theta-\mu \cos \theta)\) Time taken by the body to slide down the plane \(t_{1}=\sqrt{\frac{2 S}{a}}=\sqrt{\frac{2 S}{g \sin \theta-\mu g \cos \theta}}\) And time taken by the body in the absence of friction \(\mathrm{t}_{2}=\sqrt{\frac{2 \mathrm{~S}}{\mathrm{~g} \sin \theta}}\) Given, \(t_{1}=2 t_{2}\) \(\sqrt{\frac{2 S}{g(\sin \theta-\mu \cos \theta)}}=2 \sqrt{\frac{2 S}{g \sin \theta}}\) \(\frac{2 S}{g(\sin \theta-\mu \cos \theta)}=4 \times \frac{2 S}{g \sin \theta}\) \(\sin \theta=4 \sin \theta-4 \mu \cos \theta\) \(3 \sin \theta=4 \mu \cos \theta\) \(3 \tan \theta=4 \mu\) \(\mu=\frac{3}{4} \tan \theta=\frac{3}{4} \tan 30^{\circ}=0.43\)
JIPMER-2016
LAWS OF MOTION (ADDITIONAL)
372151
A rod \(P Q\) of length \(l\) is moving with ends remaining in contact with frictionless wall and floor. If at the instant, shown the velocity of end \(Q\) is \(2 \mathrm{~m} / \mathrm{s}\) towards negative direction of \(x\). The speed of end \(P\) will be
1 \(\sqrt{3} \mathrm{~ms}^{-1}\)
2 \(\frac{2}{\sqrt{3}} \mathrm{~ms}^{-1}\)
3 \(\sqrt{2} \mathrm{~ms}^{-1}\)
4 \(\frac{\sqrt{5}}{2} \mathrm{~ms}^{-1}\)
Explanation:
B Given that velocity at \(\mathrm{Q}\) \(\Delta\) POQ \(\mathrm{v}_{\mathrm{Q}}=-2 \mathrm{~m} / \mathrm{sec}\) \(\mathrm{X}^{2}+\mathrm{Y}^{2}=l^{2}\) On differentiating equation (i) w.r to \(t\) \(2 X \frac{d X}{d t}+2 Y \frac{d Y}{d t}=0\) From the figure \(\mathrm{X}=l \cos 60^{\circ}=\frac{l}{2} \text { and } \mathrm{Y}=l \sin 60^{\circ}=\frac{\sqrt{3} l}{2}\) \(2 \times\left(\frac{l}{2}\right) \times(-2)+2 \times\left(\frac{\sqrt{3} l}{2}\right) \frac{\mathrm{dY}}{\mathrm{dt}}=0\) \(-2+(\sqrt{3}) \frac{\mathrm{dY}}{\mathrm{dt}}=0\) \(-2=-\frac{\sqrt{3} \mathrm{dY}}{\mathrm{dt}}\) \(\frac{\mathrm{dY}}{\mathrm{dt}}=\mathrm{v}_{\mathrm{P}}=\frac{2}{\sqrt{3}} \mathrm{~m} / \mathrm{sec}\) Hence, the speed of end \(P\) will be \(v_{p}=\frac{2}{\sqrt{3}} \mathrm{~m} / \mathrm{sec}\).
JIPMER-2015
LAWS OF MOTION (ADDITIONAL)
372152
A block slides down on an incline of angle \(30^{\circ}\) with acceleration \(\frac{\mathrm{g}}{4}\). Find the kinetic friction coefficient.
1 \(\frac{1}{2 \sqrt{2}}\)
2 0.6
3 \(\frac{1}{2 \sqrt{3}}\)
4 \(\frac{1}{\sqrt{2}}\)
Explanation:
C Given, \(\mathrm{a}=\frac{\mathrm{g}}{4}, \theta=30^{\circ}\) Let block has mass \(m\) and \(\mu\) is kinetic friction. from the figure \(\mathrm{N}=\mathrm{mg} \cos 30^{\circ}\) We know \(\mathrm{f}=\mu \mathrm{N}\) \(\mathrm{f}=\mu \mathrm{mg} \cos 30^{\circ}\) And \(\quad \mathrm{mg} \sin 30^{\circ}-\mathrm{f}=\mathrm{ma}\) \(\mathrm{mg} \sin 30^{\circ}-\mu \mathrm{mg} \cos 30^{\circ}=\mathrm{m} \cdot \mathrm{g} / 4 \quad\left[\right.\) from \(\mathrm{eq}^{\mathrm{n}}\) (i)] \(\sin 30^{\circ}-\mu \cos 30^{\circ}=\frac{1}{4}\) \(\sin 30^{\circ}-\mu \cos 30^{\circ}=\frac{1}{4}\) \(\frac{1}{2}-\frac{\sqrt{3}}{2} \mu=\frac{1}{4}\) \(\frac{1}{2}-\frac{1}{4}=\frac{\sqrt{3}}{2} \mu\) \(\frac{2-1}{4}=\frac{\sqrt{3}}{2} \mu\) \(\frac{1}{4}=\frac{\sqrt{3}}{2} \mu\) \(\mu=\frac{1}{2 \sqrt{3}}\) \(\mu=\frac{1}{2 \sqrt{3}}\)
JIPMER-2015
LAWS OF MOTION (ADDITIONAL)
372153
A tangential force acting on the top of sphere of mass \(m\) kept on a rough horizontal place as shown in figure If the sphere rolls without slipping, then the acceleration with which the centre of sphere moves, is
1 \(\frac{10 \mathrm{~F}}{7 \mathrm{~m}}\)
2 \(\frac{\mathrm{F}}{2 \mathrm{~m}}\)
3 \(\frac{3 \mathrm{~F}}{3 \mathrm{~m}}\)
4 \(\frac{7 \mathrm{~F}}{2 \mathrm{~m}}\)
Explanation:
A Let \(a\) is the acceleration of the centre of sphere. Then, We know, \(\text { Angular acceleration }(\alpha)=\frac{a}{r}\) For linear motion of the centre \(\mathrm{f}+\mathrm{F}=\mathrm{ma}\) And for rotational motion about the centre. We know Then, \(\text { Torque }(\tau)=\mathrm{I} \alpha\) \(\mathrm{Fr}-\mathrm{fr}=\mathrm{I} \alpha\) \(\mathrm{Fr}-\mathrm{fr}=\left(\frac{2}{5} \mathrm{mr}^{2}\right)\left(\frac{\mathrm{a}}{\mathrm{r}}\right)\) [moment of Inertia of solid sphere (I) \(=\frac{2}{5} \mathrm{mr}^{2}\) ] \(\mathrm{F}-\mathrm{f}==\frac{2}{5} \mathrm{ma}\) From equation (i) and (ii) \(2 \mathrm{~F}=\frac{7}{5} \mathrm{ma}\) \(\mathrm{a}=\frac{10 \mathrm{~F}}{7 \mathrm{~m}}\)
372150
A body at rest slides down a \(30^{\circ}\) inclined plane. The time taken by it to slide down is twice the time it takes when it slides down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is
1 0.43
2 0.37
3 0.64
4 0.75
Explanation:
A Given, \(\theta=30^{\circ}\) From the figure \(\mathrm{R}=\mathrm{mg} \cos \theta\) Net force on the body when slides down on inclined plane \(\mathrm{F}=\mathrm{mg} \sin \theta-\mathrm{f} .\) And, \(\quad \mathrm{F}=\mathrm{ma}\) Then, \(\quad m a=m g \sin \theta-\mu R[\because f=\mu R]\) \(\therefore \quad \mathrm{ma}=\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta \quad[\because \mathrm{R}=\mathrm{mg} \cos \theta]\) \(\mathrm{ma}=\mathrm{mg}(\sin \theta-\mu \cos \theta)\) \(\mathrm{a}=\mathrm{g}(\sin \theta-\mu \cos \theta)\) Time taken by the body to slide down the plane \(t_{1}=\sqrt{\frac{2 S}{a}}=\sqrt{\frac{2 S}{g \sin \theta-\mu g \cos \theta}}\) And time taken by the body in the absence of friction \(\mathrm{t}_{2}=\sqrt{\frac{2 \mathrm{~S}}{\mathrm{~g} \sin \theta}}\) Given, \(t_{1}=2 t_{2}\) \(\sqrt{\frac{2 S}{g(\sin \theta-\mu \cos \theta)}}=2 \sqrt{\frac{2 S}{g \sin \theta}}\) \(\frac{2 S}{g(\sin \theta-\mu \cos \theta)}=4 \times \frac{2 S}{g \sin \theta}\) \(\sin \theta=4 \sin \theta-4 \mu \cos \theta\) \(3 \sin \theta=4 \mu \cos \theta\) \(3 \tan \theta=4 \mu\) \(\mu=\frac{3}{4} \tan \theta=\frac{3}{4} \tan 30^{\circ}=0.43\)
JIPMER-2016
LAWS OF MOTION (ADDITIONAL)
372151
A rod \(P Q\) of length \(l\) is moving with ends remaining in contact with frictionless wall and floor. If at the instant, shown the velocity of end \(Q\) is \(2 \mathrm{~m} / \mathrm{s}\) towards negative direction of \(x\). The speed of end \(P\) will be
1 \(\sqrt{3} \mathrm{~ms}^{-1}\)
2 \(\frac{2}{\sqrt{3}} \mathrm{~ms}^{-1}\)
3 \(\sqrt{2} \mathrm{~ms}^{-1}\)
4 \(\frac{\sqrt{5}}{2} \mathrm{~ms}^{-1}\)
Explanation:
B Given that velocity at \(\mathrm{Q}\) \(\Delta\) POQ \(\mathrm{v}_{\mathrm{Q}}=-2 \mathrm{~m} / \mathrm{sec}\) \(\mathrm{X}^{2}+\mathrm{Y}^{2}=l^{2}\) On differentiating equation (i) w.r to \(t\) \(2 X \frac{d X}{d t}+2 Y \frac{d Y}{d t}=0\) From the figure \(\mathrm{X}=l \cos 60^{\circ}=\frac{l}{2} \text { and } \mathrm{Y}=l \sin 60^{\circ}=\frac{\sqrt{3} l}{2}\) \(2 \times\left(\frac{l}{2}\right) \times(-2)+2 \times\left(\frac{\sqrt{3} l}{2}\right) \frac{\mathrm{dY}}{\mathrm{dt}}=0\) \(-2+(\sqrt{3}) \frac{\mathrm{dY}}{\mathrm{dt}}=0\) \(-2=-\frac{\sqrt{3} \mathrm{dY}}{\mathrm{dt}}\) \(\frac{\mathrm{dY}}{\mathrm{dt}}=\mathrm{v}_{\mathrm{P}}=\frac{2}{\sqrt{3}} \mathrm{~m} / \mathrm{sec}\) Hence, the speed of end \(P\) will be \(v_{p}=\frac{2}{\sqrt{3}} \mathrm{~m} / \mathrm{sec}\).
JIPMER-2015
LAWS OF MOTION (ADDITIONAL)
372152
A block slides down on an incline of angle \(30^{\circ}\) with acceleration \(\frac{\mathrm{g}}{4}\). Find the kinetic friction coefficient.
1 \(\frac{1}{2 \sqrt{2}}\)
2 0.6
3 \(\frac{1}{2 \sqrt{3}}\)
4 \(\frac{1}{\sqrt{2}}\)
Explanation:
C Given, \(\mathrm{a}=\frac{\mathrm{g}}{4}, \theta=30^{\circ}\) Let block has mass \(m\) and \(\mu\) is kinetic friction. from the figure \(\mathrm{N}=\mathrm{mg} \cos 30^{\circ}\) We know \(\mathrm{f}=\mu \mathrm{N}\) \(\mathrm{f}=\mu \mathrm{mg} \cos 30^{\circ}\) And \(\quad \mathrm{mg} \sin 30^{\circ}-\mathrm{f}=\mathrm{ma}\) \(\mathrm{mg} \sin 30^{\circ}-\mu \mathrm{mg} \cos 30^{\circ}=\mathrm{m} \cdot \mathrm{g} / 4 \quad\left[\right.\) from \(\mathrm{eq}^{\mathrm{n}}\) (i)] \(\sin 30^{\circ}-\mu \cos 30^{\circ}=\frac{1}{4}\) \(\sin 30^{\circ}-\mu \cos 30^{\circ}=\frac{1}{4}\) \(\frac{1}{2}-\frac{\sqrt{3}}{2} \mu=\frac{1}{4}\) \(\frac{1}{2}-\frac{1}{4}=\frac{\sqrt{3}}{2} \mu\) \(\frac{2-1}{4}=\frac{\sqrt{3}}{2} \mu\) \(\frac{1}{4}=\frac{\sqrt{3}}{2} \mu\) \(\mu=\frac{1}{2 \sqrt{3}}\) \(\mu=\frac{1}{2 \sqrt{3}}\)
JIPMER-2015
LAWS OF MOTION (ADDITIONAL)
372153
A tangential force acting on the top of sphere of mass \(m\) kept on a rough horizontal place as shown in figure If the sphere rolls without slipping, then the acceleration with which the centre of sphere moves, is
1 \(\frac{10 \mathrm{~F}}{7 \mathrm{~m}}\)
2 \(\frac{\mathrm{F}}{2 \mathrm{~m}}\)
3 \(\frac{3 \mathrm{~F}}{3 \mathrm{~m}}\)
4 \(\frac{7 \mathrm{~F}}{2 \mathrm{~m}}\)
Explanation:
A Let \(a\) is the acceleration of the centre of sphere. Then, We know, \(\text { Angular acceleration }(\alpha)=\frac{a}{r}\) For linear motion of the centre \(\mathrm{f}+\mathrm{F}=\mathrm{ma}\) And for rotational motion about the centre. We know Then, \(\text { Torque }(\tau)=\mathrm{I} \alpha\) \(\mathrm{Fr}-\mathrm{fr}=\mathrm{I} \alpha\) \(\mathrm{Fr}-\mathrm{fr}=\left(\frac{2}{5} \mathrm{mr}^{2}\right)\left(\frac{\mathrm{a}}{\mathrm{r}}\right)\) [moment of Inertia of solid sphere (I) \(=\frac{2}{5} \mathrm{mr}^{2}\) ] \(\mathrm{F}-\mathrm{f}==\frac{2}{5} \mathrm{ma}\) From equation (i) and (ii) \(2 \mathrm{~F}=\frac{7}{5} \mathrm{ma}\) \(\mathrm{a}=\frac{10 \mathrm{~F}}{7 \mathrm{~m}}\)
