04. Friction, and Inclined Plane Friction Motion
Laws of Motion

146167 One end of a massless spring of spring constant
\(100 \mathrm{~N} \mathrm{~m}^{-1}\) and natural length \(0.49 \mathrm{~m}\) is fixed and other end is connected to a body of mass \(0.5 \mathrm{~kg}\) lying on a frictionless horizontal table. The spring remains horizontal. If the body is made to rotate at an angular velocity of 2 rad \(\sec ^{-1}\), then the elongation of the spring will be

1 \(2 \mathrm{~cm}\)
2 \(1 \mathrm{~cm}\)
3 \(0.5 \mathrm{~cm}\)
4 \(0.25 \mathrm{~cm}\)
Laws of Motion

146168 The time taken by a block to slide down a smooth inclined plane of inclination \(45^{\circ}\) is \(t_{1}\) and the time taken by the same block to slide down a rough inclined plane of coefficient of kinetic friction \(\frac{2}{3}\) is \(t_{2}\). If the length and angle of inclination of the rough plane are same as those of the smooth plane, then \(\frac{t_{1}}{t_{2}}\) is

1 \(\frac{1}{\sqrt{3}}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\frac{1}{2}\)
4 \(\frac{2}{\sqrt{3}}\)
Laws of Motion

146171 A block of mass \(m\) is placed on a smooth inclined wedge \(\mathrm{ABC}\) of inclination \(\theta\) as shown in the figure. The wedge is given an acceleration a towards the right. The relation between \(a\) and \(\theta\) for the block to remain stationary on the wedge is

1 \(a=g \cos \theta\)
2 \(a=\frac{g}{\sin \theta}\)
3 \(\mathrm{a}=\frac{\mathrm{g}}{\operatorname{cosec} \theta}\)
4 \(\mathrm{a}=\mathrm{g} \tan \theta\)
Laws of Motion

146172 A body initially at rest and sliding along a frictionless track from a height \(h\) (as shown in the figure) just completes a vertical circle of diameter \(A B=D\). The height \(h\) is equal to

1 \(\frac{7}{5} \mathrm{D}\)
2 \(\mathrm{D}\)
3 \(\frac{3}{2} \mathrm{D}\)
4 \(\frac{5}{4} \mathrm{D}\)
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Laws of Motion

146167 One end of a massless spring of spring constant
\(100 \mathrm{~N} \mathrm{~m}^{-1}\) and natural length \(0.49 \mathrm{~m}\) is fixed and other end is connected to a body of mass \(0.5 \mathrm{~kg}\) lying on a frictionless horizontal table. The spring remains horizontal. If the body is made to rotate at an angular velocity of 2 rad \(\sec ^{-1}\), then the elongation of the spring will be

1 \(2 \mathrm{~cm}\)
2 \(1 \mathrm{~cm}\)
3 \(0.5 \mathrm{~cm}\)
4 \(0.25 \mathrm{~cm}\)
Laws of Motion

146168 The time taken by a block to slide down a smooth inclined plane of inclination \(45^{\circ}\) is \(t_{1}\) and the time taken by the same block to slide down a rough inclined plane of coefficient of kinetic friction \(\frac{2}{3}\) is \(t_{2}\). If the length and angle of inclination of the rough plane are same as those of the smooth plane, then \(\frac{t_{1}}{t_{2}}\) is

1 \(\frac{1}{\sqrt{3}}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\frac{1}{2}\)
4 \(\frac{2}{\sqrt{3}}\)
Laws of Motion

146171 A block of mass \(m\) is placed on a smooth inclined wedge \(\mathrm{ABC}\) of inclination \(\theta\) as shown in the figure. The wedge is given an acceleration a towards the right. The relation between \(a\) and \(\theta\) for the block to remain stationary on the wedge is

1 \(a=g \cos \theta\)
2 \(a=\frac{g}{\sin \theta}\)
3 \(\mathrm{a}=\frac{\mathrm{g}}{\operatorname{cosec} \theta}\)
4 \(\mathrm{a}=\mathrm{g} \tan \theta\)
Laws of Motion

146172 A body initially at rest and sliding along a frictionless track from a height \(h\) (as shown in the figure) just completes a vertical circle of diameter \(A B=D\). The height \(h\) is equal to

1 \(\frac{7}{5} \mathrm{D}\)
2 \(\mathrm{D}\)
3 \(\frac{3}{2} \mathrm{D}\)
4 \(\frac{5}{4} \mathrm{D}\)
Laws of Motion

146167 One end of a massless spring of spring constant
\(100 \mathrm{~N} \mathrm{~m}^{-1}\) and natural length \(0.49 \mathrm{~m}\) is fixed and other end is connected to a body of mass \(0.5 \mathrm{~kg}\) lying on a frictionless horizontal table. The spring remains horizontal. If the body is made to rotate at an angular velocity of 2 rad \(\sec ^{-1}\), then the elongation of the spring will be

1 \(2 \mathrm{~cm}\)
2 \(1 \mathrm{~cm}\)
3 \(0.5 \mathrm{~cm}\)
4 \(0.25 \mathrm{~cm}\)
Laws of Motion

146168 The time taken by a block to slide down a smooth inclined plane of inclination \(45^{\circ}\) is \(t_{1}\) and the time taken by the same block to slide down a rough inclined plane of coefficient of kinetic friction \(\frac{2}{3}\) is \(t_{2}\). If the length and angle of inclination of the rough plane are same as those of the smooth plane, then \(\frac{t_{1}}{t_{2}}\) is

1 \(\frac{1}{\sqrt{3}}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\frac{1}{2}\)
4 \(\frac{2}{\sqrt{3}}\)
Laws of Motion

146171 A block of mass \(m\) is placed on a smooth inclined wedge \(\mathrm{ABC}\) of inclination \(\theta\) as shown in the figure. The wedge is given an acceleration a towards the right. The relation between \(a\) and \(\theta\) for the block to remain stationary on the wedge is

1 \(a=g \cos \theta\)
2 \(a=\frac{g}{\sin \theta}\)
3 \(\mathrm{a}=\frac{\mathrm{g}}{\operatorname{cosec} \theta}\)
4 \(\mathrm{a}=\mathrm{g} \tan \theta\)
Laws of Motion

146172 A body initially at rest and sliding along a frictionless track from a height \(h\) (as shown in the figure) just completes a vertical circle of diameter \(A B=D\). The height \(h\) is equal to

1 \(\frac{7}{5} \mathrm{D}\)
2 \(\mathrm{D}\)
3 \(\frac{3}{2} \mathrm{D}\)
4 \(\frac{5}{4} \mathrm{D}\)
Laws of Motion

146167 One end of a massless spring of spring constant
\(100 \mathrm{~N} \mathrm{~m}^{-1}\) and natural length \(0.49 \mathrm{~m}\) is fixed and other end is connected to a body of mass \(0.5 \mathrm{~kg}\) lying on a frictionless horizontal table. The spring remains horizontal. If the body is made to rotate at an angular velocity of 2 rad \(\sec ^{-1}\), then the elongation of the spring will be

1 \(2 \mathrm{~cm}\)
2 \(1 \mathrm{~cm}\)
3 \(0.5 \mathrm{~cm}\)
4 \(0.25 \mathrm{~cm}\)
Laws of Motion

146168 The time taken by a block to slide down a smooth inclined plane of inclination \(45^{\circ}\) is \(t_{1}\) and the time taken by the same block to slide down a rough inclined plane of coefficient of kinetic friction \(\frac{2}{3}\) is \(t_{2}\). If the length and angle of inclination of the rough plane are same as those of the smooth plane, then \(\frac{t_{1}}{t_{2}}\) is

1 \(\frac{1}{\sqrt{3}}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\frac{1}{2}\)
4 \(\frac{2}{\sqrt{3}}\)
Laws of Motion

146171 A block of mass \(m\) is placed on a smooth inclined wedge \(\mathrm{ABC}\) of inclination \(\theta\) as shown in the figure. The wedge is given an acceleration a towards the right. The relation between \(a\) and \(\theta\) for the block to remain stationary on the wedge is

1 \(a=g \cos \theta\)
2 \(a=\frac{g}{\sin \theta}\)
3 \(\mathrm{a}=\frac{\mathrm{g}}{\operatorname{cosec} \theta}\)
4 \(\mathrm{a}=\mathrm{g} \tan \theta\)
Laws of Motion

146172 A body initially at rest and sliding along a frictionless track from a height \(h\) (as shown in the figure) just completes a vertical circle of diameter \(A B=D\). The height \(h\) is equal to

1 \(\frac{7}{5} \mathrm{D}\)
2 \(\mathrm{D}\)
3 \(\frac{3}{2} \mathrm{D}\)
4 \(\frac{5}{4} \mathrm{D}\)