MOTION OF A BODY ON THE INCLINED PLANE
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Laws of Motion

270286 The upper half of an inclined plane of inclination '\(\theta\) ' is perfectly smooth while the lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom. The coefficient of friction between the block and the lower half of the plane is given by
(2013E)

1 \(\mu=2 \tan \theta\)
2 \(\mu=\frac{2}{\tan \theta}\)
3 \(\mu=\tan \theta\)
4 \(\mu=\frac{1}{\tan \theta}\)
Laws of Motion

270287 A\(30 \mathrm{~kg}\) box has to move up an inclined plane of slope \(30^{\circ}\) to the horizontal with a uniform velocity of \(5 \mathrm{~ms}^{-1}\).If the frictional force retarding the motion is \(150 \mathrm{~N}\), the horizontal force required to move the box up is \(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\)

1 \(300 \times \frac{2}{\sqrt{3}} \mathrm{~N}\)
2 \(300 \times \frac{\sqrt{3}}{2} \mathrm{~N}\)
3 \(300 \mathrm{~N}\)
4 \(150 \mathrm{~N}\)
Laws of Motion

270337 A body is allowed to slide from the top along a smooth inclined plane of length\(5 \mathrm{~m}\) at an angle of inclination \(30^{0}\). If \(\mathbf{g}=10 \mathrm{~ms}^{-2}\), time taken by the body to reach the bottom of the plane is

1 \(\frac{\sqrt{3}}{2} s\)
2 \(1.414 \mathrm{~s}\)
3 \(\frac{1}{\sqrt{2}} s\)
4 \(2 \mathrm{~s}\)
Laws of Motion

270338 A body slides down a smooth inclined plane of height \(h\) and angle of inclination \(30^{\circ}\) reaching the bottom with a velocity \(v\). Without changing the height, if the angle of inclination is doubled, the velocity with which it reaches the bottom of the plane is

1 \(\mathrm{v}\)
2 \(v / 2\)
3 \(2 \mathrm{v}\)
4 \(\sqrt{2} \mathrm{v}\)
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Laws of Motion

270286 The upper half of an inclined plane of inclination '\(\theta\) ' is perfectly smooth while the lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom. The coefficient of friction between the block and the lower half of the plane is given by
(2013E)

1 \(\mu=2 \tan \theta\)
2 \(\mu=\frac{2}{\tan \theta}\)
3 \(\mu=\tan \theta\)
4 \(\mu=\frac{1}{\tan \theta}\)
Laws of Motion

270287 A\(30 \mathrm{~kg}\) box has to move up an inclined plane of slope \(30^{\circ}\) to the horizontal with a uniform velocity of \(5 \mathrm{~ms}^{-1}\).If the frictional force retarding the motion is \(150 \mathrm{~N}\), the horizontal force required to move the box up is \(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\)

1 \(300 \times \frac{2}{\sqrt{3}} \mathrm{~N}\)
2 \(300 \times \frac{\sqrt{3}}{2} \mathrm{~N}\)
3 \(300 \mathrm{~N}\)
4 \(150 \mathrm{~N}\)
Laws of Motion

270337 A body is allowed to slide from the top along a smooth inclined plane of length\(5 \mathrm{~m}\) at an angle of inclination \(30^{0}\). If \(\mathbf{g}=10 \mathrm{~ms}^{-2}\), time taken by the body to reach the bottom of the plane is

1 \(\frac{\sqrt{3}}{2} s\)
2 \(1.414 \mathrm{~s}\)
3 \(\frac{1}{\sqrt{2}} s\)
4 \(2 \mathrm{~s}\)
Laws of Motion

270338 A body slides down a smooth inclined plane of height \(h\) and angle of inclination \(30^{\circ}\) reaching the bottom with a velocity \(v\). Without changing the height, if the angle of inclination is doubled, the velocity with which it reaches the bottom of the plane is

1 \(\mathrm{v}\)
2 \(v / 2\)
3 \(2 \mathrm{v}\)
4 \(\sqrt{2} \mathrm{v}\)
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Laws of Motion

270286 The upper half of an inclined plane of inclination '\(\theta\) ' is perfectly smooth while the lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom. The coefficient of friction between the block and the lower half of the plane is given by
(2013E)

1 \(\mu=2 \tan \theta\)
2 \(\mu=\frac{2}{\tan \theta}\)
3 \(\mu=\tan \theta\)
4 \(\mu=\frac{1}{\tan \theta}\)
Laws of Motion

270287 A\(30 \mathrm{~kg}\) box has to move up an inclined plane of slope \(30^{\circ}\) to the horizontal with a uniform velocity of \(5 \mathrm{~ms}^{-1}\).If the frictional force retarding the motion is \(150 \mathrm{~N}\), the horizontal force required to move the box up is \(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\)

1 \(300 \times \frac{2}{\sqrt{3}} \mathrm{~N}\)
2 \(300 \times \frac{\sqrt{3}}{2} \mathrm{~N}\)
3 \(300 \mathrm{~N}\)
4 \(150 \mathrm{~N}\)
Laws of Motion

270337 A body is allowed to slide from the top along a smooth inclined plane of length\(5 \mathrm{~m}\) at an angle of inclination \(30^{0}\). If \(\mathbf{g}=10 \mathrm{~ms}^{-2}\), time taken by the body to reach the bottom of the plane is

1 \(\frac{\sqrt{3}}{2} s\)
2 \(1.414 \mathrm{~s}\)
3 \(\frac{1}{\sqrt{2}} s\)
4 \(2 \mathrm{~s}\)
Laws of Motion

270338 A body slides down a smooth inclined plane of height \(h\) and angle of inclination \(30^{\circ}\) reaching the bottom with a velocity \(v\). Without changing the height, if the angle of inclination is doubled, the velocity with which it reaches the bottom of the plane is

1 \(\mathrm{v}\)
2 \(v / 2\)
3 \(2 \mathrm{v}\)
4 \(\sqrt{2} \mathrm{v}\)
For More Updates join with Us
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
Laws of Motion

270286 The upper half of an inclined plane of inclination '\(\theta\) ' is perfectly smooth while the lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom. The coefficient of friction between the block and the lower half of the plane is given by
(2013E)

1 \(\mu=2 \tan \theta\)
2 \(\mu=\frac{2}{\tan \theta}\)
3 \(\mu=\tan \theta\)
4 \(\mu=\frac{1}{\tan \theta}\)
Laws of Motion

270287 A\(30 \mathrm{~kg}\) box has to move up an inclined plane of slope \(30^{\circ}\) to the horizontal with a uniform velocity of \(5 \mathrm{~ms}^{-1}\).If the frictional force retarding the motion is \(150 \mathrm{~N}\), the horizontal force required to move the box up is \(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\)

1 \(300 \times \frac{2}{\sqrt{3}} \mathrm{~N}\)
2 \(300 \times \frac{\sqrt{3}}{2} \mathrm{~N}\)
3 \(300 \mathrm{~N}\)
4 \(150 \mathrm{~N}\)
Laws of Motion

270337 A body is allowed to slide from the top along a smooth inclined plane of length\(5 \mathrm{~m}\) at an angle of inclination \(30^{0}\). If \(\mathbf{g}=10 \mathrm{~ms}^{-2}\), time taken by the body to reach the bottom of the plane is

1 \(\frac{\sqrt{3}}{2} s\)
2 \(1.414 \mathrm{~s}\)
3 \(\frac{1}{\sqrt{2}} s\)
4 \(2 \mathrm{~s}\)
Laws of Motion

270338 A body slides down a smooth inclined plane of height \(h\) and angle of inclination \(30^{\circ}\) reaching the bottom with a velocity \(v\). Without changing the height, if the angle of inclination is doubled, the velocity with which it reaches the bottom of the plane is

1 \(\mathrm{v}\)
2 \(v / 2\)
3 \(2 \mathrm{v}\)
4 \(\sqrt{2} \mathrm{v}\)