Chapter 25

71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers · 112 exercises

Problem 110

A particle is projected from a point \(O\) with velocity \(u\) at an angle of \(60^{\circ}\) with the horizontal. When it is moving in a direction at right angles to its direction at \(O\), its velocity then is given by (a) \(\frac{u}{3}\) (b) \(\frac{u}{2}\) (c) \(\frac{2 u}{3}\) (d) \(\frac{u}{\sqrt{3}}\)

7 step solution

Problem 112

A velocity \(\frac{1}{4} \mathrm{~m} / \mathrm{s}\) is resolved into two components along \(O A\) and \(O B\) making angles \(30^{\circ}\) and \(45^{\circ}\) respectively with the given velocity. Then the component along \(O B\) is \([2004]\) (a) \(\frac{1}{8}(\sqrt{6}-\sqrt{2}) \mathrm{m} / \mathrm{s}\) (b) \(\frac{1}{4}(\sqrt{3}-1) \mathrm{m} / \mathrm{s}\) (c) \(\frac{1}{4} \mathrm{~m} / \mathrm{s}\) (d) \(\frac{1}{8} \mathrm{~m} / \mathrm{s}\)

5 step solution

Problem 113

A paticle moves towards east from a point \(A\) to a point \(B\) at the rate of \(4 \mathrm{~km} / \mathrm{h}\) and then towards north from \(B\) to \(\mathrm{C}\) at the rate of \(5 \mathrm{~km} / \mathrm{hr}\). If \(A B=12 \mathrm{~km}\) and \(B C=5 \mathrm{~km}\), then its average speed for its journey from \(A\) to \(C\) and resultant average velocity direct from \(A\) to \(C\) are respectively [2004] (a) \(\frac{13}{9} \mathrm{~km} / \mathrm{h}\) and \(\frac{17}{9} \mathrm{~km} / \mathrm{h}\) (b) \(\frac{13}{4} \mathrm{~km} / \mathrm{h}\) and \(\frac{17}{4} \mathrm{~km} / \mathrm{h}\) (c) \(\frac{17}{9} \mathrm{~km} / \mathrm{h}\) and \(\frac{13}{9} \mathrm{~km} / \mathrm{h}\) (d) \(\frac{17}{4} \mathrm{~km} / \mathrm{h}\) and \(\frac{13}{4} \mathrm{~km} / \mathrm{h}\)

5 step solution

Problem 114

Three forces \(\vec{P}, \bar{Q}\) and \(\vec{R}\) acting along \(L A, I B\) and \(I C\), where \(I\) is the incentre of a \(\triangle A B C\) are in equilibrium. Then \(\vec{P}: \vec{Q}: \vec{R}\) is (a) \(\operatorname{cosec} \frac{A}{2}: \operatorname{cosec} \frac{B}{2}: \operatorname{cosec} \frac{C}{2}\) (b) \(\sin \frac{A}{2}: \sin \frac{B}{2}: \sin \frac{C}{2}\) (c) \(\sec \frac{A}{2}: \sec \frac{B}{2}: \sec \frac{C}{2}\) (d) \(\cos \frac{A}{2}: \cos \frac{B}{2} ; \cos \frac{C}{2}\)

4 step solution

Problem 116

With two forces acting at point, the maximum affect is obtained when their resultant is \(4 \mathrm{~N}\). If they act at right angles, then their resultant is \(3 \mathrm{~N}\). Then the forces are [2004] (a) \(\left(2+\frac{1}{2} \sqrt{3}\right) N\) and \(\left(2-\frac{1}{2} \sqrt{3}\right) N\) (b) \((2+\sqrt{3}) N\) and \((2-\sqrt{3}) N\) (c) \(\left(2+\frac{1}{2} \sqrt{2}\right) N\) and \(\left(2-\frac{1}{2} \sqrt{2}\right) N\) (d) \((2+\sqrt{2}) N\) and \((2-\sqrt{2}) N\)

6 step solution

Problem 119

Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity \(\vec{u}\) and the other from rest with uniform acceleration \(\vec{f}\). Let \(\alpha\) be the angle between their directions of motion. The relative velocity of the second particle w.r.t. the first is least after a time (a) \(\frac{u \cos \alpha}{f}\) (b) \(\frac{u \sin \alpha}{f}\) (c) \(\frac{f \cos \alpha}{u}\) (d) \(u \sin \alpha\)

6 step solution

Problem 120

Two stones are projected from the top of a cliff h metres high, with the same speed \(\mathrm{u}\), so as to hit the ground at the same spot. If one of the stones is projected horizontally and the other is projected horizontally and the other is projected at an angle \(\theta\) to the horizontal then \(\tan \theta\) equals \([2003]\) (a) \(u \sqrt{\frac{2}{g h}}\) (b) \(\sqrt{\frac{2 u}{g h}}\) (c) \(2 g \sqrt{\frac{u}{h}}\) (d) \(2 h \sqrt{\frac{u}{g}}\)

6 step solution

Problem 121

A body travels a distance s in t seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with constant acceleration \(\mathrm{f}\) and in the second part with constant retardation \(\mathrm{r}\). The value of \(\mathrm{t}\) is given by \([2003]\) (a) \(\sqrt{2 s\left(\frac{1}{f}+\frac{1}{r}\right)}\) (b) \(2 s\left(\frac{1}{f}+\frac{1}{r}\right)\) (c) \(\frac{2 s}{\frac{1}{f}+\frac{1}{r}}\) (d) \(\sqrt{2 s(f+r)}\)

8 step solution

Problem 122

The resultant of forces \(\vec{P}\) and \(\vec{Q}\) is \(\vec{R}\). If \(\vec{Q}\) is doubled then \(\vec{R}\) is doubled. If the direction of \(\vec{Q}\) is reversed, then \(\vec{R}\) is again doubled. Then \(P^{2}: Q^{2}: R^{2}\) is [2003] (a) \(2: 3: 1\) (b) \(3: 1: 1\) (c) \(2: 3: 2\) (d) \(1: 2: 3\).

4 step solution

Problem 124

A particle acted on by constant forces \(4 \hat{i}+\hat{j}-3 \hat{k}\) and \(3 \hat{i}+\hat{j}-\hat{k}\) is displaced from the point \(\hat{i}+2 \hat{j}-3 \hat{k}\) to the point \(5 \hat{i}+4 \hat{j}+\hat{k}\). The total work done by the forces is [2003] (a) 50 units (b) 20 units (c) 30 units (d) 40 units.

5 step solution

Problem 125

A bead of weight \(w\) can slide on smooth circular wire in a vertical plane. The bead is attached by a light thread to the highest point of the wire and in equilibrium, the thread is taut and make an angle \(\theta\) with the vertical then tension of the thread and reaction of the wire on the bead are (a) \(T=w \cos \theta \quad R=w \tan \theta\) (b) \(T=2 w \cos \theta \quad R=w\) (c) \(T=w R=w \sin \theta\) (d) \(T=w \sin \theta \quad R=n \cot \theta\)

6 step solution

Problem 126

The sum of two forces is \(18 \mathrm{~N}\) and resultant whose direction is at right angles to the smaller force is \(12 \mathrm{~N}\). The magnitude of the two forces are (a) 13,5 (b) 12,6 (c) 14,4 (d) 11,7

8 step solution

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