Q. No.The velocity at which \(6~\text{kg}\) mass (shown in figure) strikes the ground when it is released from a height of \(6~\text{m}\) above the ground is: (in m/s)
(Assume pulley is massless and string is light and inextensible. (Take \(g = 10~\text{m/s}^{2}\)))
1. \(7.74\)
2. \(7.20\)
3. \(6.55\)
4. \(4.50\)
(Assume pulley is massless and string is light and inextensible. (Take \(g = 10~\text{m/s}^{2}\)))
1. \(7.74\)
2. \(7.20\)
3. \(6.55\)
4. \(4.50\)
Subtopic: Application of Laws |
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A wedge \(Y\) with mass of \(10 ~\text {kg}\) and all frictionless surfaces and the inclined surface making \(37^{\circ}\) with horizontal. A block \(X\) with mass \(2 ~\text {kg}\) is placed at the highest point of the wedge as shown in figure is at rest. At \(t=0\) wedge (\(Y\)) is pulled toward right with constant force (\( f\)) of \(24~\text{N}\). Taking the block \(X\) at rest at \(t=0\) the time taken by it to slide down \(8.8~\text{m}\) on the slope, while \( Y\) is on the move, is: (in \(\text{s}\))
(take \(\tan(37^{\circ})=\dfrac{3}{4}\) and \(g= 10~\text {m/s}^2\))
1. \(2\)
2. \(4\)
3. \(\sqrt{2}\)
4. \(2 \sqrt{2}\)
(take \(\tan(37^{\circ})=\dfrac{3}{4}\) and \(g= 10~\text {m/s}^2\))
1. \(2\)
2. \(4\)
3. \(\sqrt{2}\)
4. \(2 \sqrt{2}\)
Subtopic: Application of Laws |
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The pulley shown in figure is made using a thin rim and two rods of length equal to diameter of the rim. The rim and each rod have a mass of \(M\). Two blocks of mass of \(M\) and \(m\) are attached to two ends of a light string passing over the pulley, which is hinged to rotate freely in vertical plane about its centre. The magnitudes of the acceleration experienced by the blocks is:
(assume no slipping of string on pulley.)
(assume no slipping of string on pulley.)
| 1. | \(\dfrac{({M}-{m}) {g}}{\left[\left(\dfrac{13}{6}\right) {M}+{m}\right]}\) | 2. | \(\dfrac{({M}-{m}) {g}}{{M}+{m}}\) |
| 3. | \(\dfrac{({M}-{m}){g}}{\left[\left(\dfrac{8}{3}\right) {M}+{m}\right]}\) | 4. | \(\dfrac{(M-m) g}{2 M+m}\) |
Subtopic: Application of Laws |
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A small block of mass \(m\) slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration \(a_e\). The angle between the inclined plane and ground is \(\theta\) and its base length is \(L\). Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is:
1. \(\sqrt{\dfrac{2 L}{g \sin 2 \theta-a_0(1+\cos 2 \theta)}} \)
2. \(\sqrt{\dfrac{4 {~L}}{{~g} \sin 2 \theta-{a}_0(1+\cos 2 \theta)}} \)
3. \(\sqrt{\dfrac{4 {~L}}{{~g} \cos ^2 \theta-{a}_0 \sin \theta \cos \theta}}\)
4. \(\sqrt{\dfrac{2 L}{g \sin \theta-a_0 \cos \theta}}\)
1. \(\sqrt{\dfrac{2 L}{g \sin 2 \theta-a_0(1+\cos 2 \theta)}} \)
2. \(\sqrt{\dfrac{4 {~L}}{{~g} \sin 2 \theta-{a}_0(1+\cos 2 \theta)}} \)
3. \(\sqrt{\dfrac{4 {~L}}{{~g} \cos ^2 \theta-{a}_0 \sin \theta \cos \theta}}\)
4. \(\sqrt{\dfrac{2 L}{g \sin \theta-a_0 \cos \theta}}\)
Subtopic: Application of Laws |
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A block of mass \(25~ \text{kg} \) is pulled along a horizontal surface by a force at an angle \(45^\circ\) with the horizontal. The friction coefficient between the block and the surface is \(0.25.\) The block travels at a uniform velocity. The work done by the applied force during a displacement of \(5 ~\text m\) of the block is:
1. \(970~\text{J}\)
2. \(245~\text{J}\)
3. \(735~\text{J}\)
4. \(490~\text{J}\)
1. \(970~\text{J}\)
2. \(245~\text{J}\)
3. \(735~\text{J}\)
4. \(490~\text{J}\)
Subtopic: Application of Laws |
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A body of mass m is suspended by two strings making angles \(\theta _1\) and \(\theta _2\) with the horizontal ceiling with tensions \(T_1\) and \(T_2\) simultaneously. \(T_1\) and \(T_2\) are related by \(T_1 = \sqrt{3} ~T_2,\) then the angles \(\theta _1\) and \(\theta _2\) are:
| 1. | \(\theta_1=30^{\circ}, \theta_2=60^{\circ} \text { with } T_2=\dfrac{4 mg}{5}\) |
| 2. | \(\theta_1=30^{\circ}, \theta_2=60^{\circ} \text { with } T_2=\dfrac{3 {mg}}{4}\) |
| 3. | \(\theta_1=60^{\circ}, \theta_2=30^{\circ} \text { with } T_2=\dfrac{mg}{2}\) |
| 4. | \(\theta_1=45^{\circ}, \theta_2=45^{\circ} \text { with } T_2=\dfrac{3mg}{4}\) |
Subtopic: Application of Laws |
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A body of mass \(1 ~\text{kg}\) is suspended with the help of two strings making angles as shown in figure. Magnitudes of tensions \(T_1\) and \(T_2,\) respectively, are (in \(\text{N}\)): (Take acceleration due to gravity \(10 ~\text{m/s}^2 \))
1. \(5,5\)
2. \(5,5\sqrt3 ~\)
3. \(5\sqrt3,5\sqrt3 ~\)
4. \(5\sqrt3,5 ~\)
1. \(5,5\)
2. \(5,5\sqrt3 ~\)
3. \(5\sqrt3,5\sqrt3 ~\)
4. \(5\sqrt3,5 ~\)
Subtopic: Application of Laws |
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A light unstretchable string passing over a smooth light pulley connects two blocks of masses \(m_1\) and \(m_2\). If the acceleration of the system is \(g/8\), then the ratio of the masses \(m_2\)/\(m_1\) is :
1. \(4:3\)
2. \(5:3\)
3. \(8:1\)
4. \(9:7\)
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A body of weight \(200\) N is suspended from a tree branch through a chain of mass \(10\) kg. The branch pulls the chain by a force equal to (if \(g=10 \mathrm{~m} / \mathrm{s}^2\)) :
1. \(300\) N
2. \(150\) N
3. \(100\) N
4. \(200\) N
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A light string passing over a smooth light pulley connects two blocks of masses \(\text{m}_1\) and \(\text{m}_2\) (where \(\text{m}_2>\text{m}_1\)). If the acceleration of the system is \(\text g/\sqrt 2\) , then the ratio of the masses \(\text{m}_1/\text{m}_2\) is:
1. \(\frac{\sqrt{3}+1}{\sqrt{2}-1}\)
2. \(\frac{1+\sqrt{5}}{\sqrt{2}-1}\)
3. \(\frac{1+\sqrt{5}}{\sqrt{5}-1}\)
4. \(\frac{\sqrt{2}-1}{\sqrt{2}+1}\)
1. \(\frac{\sqrt{3}+1}{\sqrt{2}-1}\)
2. \(\frac{1+\sqrt{5}}{\sqrt{2}-1}\)
3. \(\frac{1+\sqrt{5}}{\sqrt{5}-1}\)
4. \(\frac{\sqrt{2}-1}{\sqrt{2}+1}\)
Subtopic: Application of Laws |
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