00. Scalar and Vector Quantities
Motion in Plane

143582 The position vector of a particle is \(\mathbf{r}=(\alpha \cos \omega t) \hat{\mathbf{i}}+(\alpha \sin \omega t) \hat{\mathbf{j}}\). The velocity vector of the particle is

1 Parallel to position vector
2 Perpendicular to position vector
3 Directed towards the origin
4 Directed away from the origin
Motion in Plane

143583 If a vector \(A\) having a magnitude of 8 is added to a vector \(B\) which lies along \(x\)-axis, then the resultant of two vectors lies along \(y\)-axis and has magnitude twice that of \(B\). The magnitude of \(B\) is

1 \(\frac{6}{\sqrt{5}}\)
2 \(\frac{12}{\sqrt{5}}\)
3 \(\frac{16}{\sqrt{5}}\)
4 \(\frac{8}{\sqrt{5}}\)
Motion in Plane

143584 It two forces each of \(2 \mathrm{~N}\) are inclined at \(60^{\circ}\), then resultant force is:

1 \(2 \mathrm{~N}\)
2 \(2 \sqrt{5} \mathrm{~N}\)
3 \(2 \sqrt{3} \mathrm{~N}\)
4 \(4 \sqrt{2} \mathrm{~N}\)
Motion in Plane

143586 Calculate the work done when a force \(\overrightarrow{\mathbf{F}}=\mathbf{2} \hat{\mathbf{i}}+\mathbf{3} \hat{\mathbf{j}}-\mathbf{5} \hat{\mathbf{k}}\) units acts on a body producing a displacement \(\vec{s}=\mathbf{2} \hat{\mathbf{i}}+\mathbf{4} \hat{\mathbf{j}}+\mathbf{3} \hat{\mathbf{k}}\) units :

1 1 unit
2 20 unit
3 5 unit
4 zero
Motion in Plane

143587 Three forces acting on a body are shown in the figure. To have the resultant force only along the \(y\)-direction, the magnitude of the minimum additional force needed along \(O X\) is

1 \(\frac{\sqrt{3}}{4} \mathrm{~N}\)
2 \(\sqrt{13} \mathrm{~N}\)
3 \(0.5 \mathrm{~N}\)
4 \(1.5 \mathrm{~N}\)
Motion in Plane

143582 The position vector of a particle is \(\mathbf{r}=(\alpha \cos \omega t) \hat{\mathbf{i}}+(\alpha \sin \omega t) \hat{\mathbf{j}}\). The velocity vector of the particle is

1 Parallel to position vector
2 Perpendicular to position vector
3 Directed towards the origin
4 Directed away from the origin
Motion in Plane

143583 If a vector \(A\) having a magnitude of 8 is added to a vector \(B\) which lies along \(x\)-axis, then the resultant of two vectors lies along \(y\)-axis and has magnitude twice that of \(B\). The magnitude of \(B\) is

1 \(\frac{6}{\sqrt{5}}\)
2 \(\frac{12}{\sqrt{5}}\)
3 \(\frac{16}{\sqrt{5}}\)
4 \(\frac{8}{\sqrt{5}}\)
Motion in Plane

143584 It two forces each of \(2 \mathrm{~N}\) are inclined at \(60^{\circ}\), then resultant force is:

1 \(2 \mathrm{~N}\)
2 \(2 \sqrt{5} \mathrm{~N}\)
3 \(2 \sqrt{3} \mathrm{~N}\)
4 \(4 \sqrt{2} \mathrm{~N}\)
Motion in Plane

143586 Calculate the work done when a force \(\overrightarrow{\mathbf{F}}=\mathbf{2} \hat{\mathbf{i}}+\mathbf{3} \hat{\mathbf{j}}-\mathbf{5} \hat{\mathbf{k}}\) units acts on a body producing a displacement \(\vec{s}=\mathbf{2} \hat{\mathbf{i}}+\mathbf{4} \hat{\mathbf{j}}+\mathbf{3} \hat{\mathbf{k}}\) units :

1 1 unit
2 20 unit
3 5 unit
4 zero
Motion in Plane

143587 Three forces acting on a body are shown in the figure. To have the resultant force only along the \(y\)-direction, the magnitude of the minimum additional force needed along \(O X\) is

1 \(\frac{\sqrt{3}}{4} \mathrm{~N}\)
2 \(\sqrt{13} \mathrm{~N}\)
3 \(0.5 \mathrm{~N}\)
4 \(1.5 \mathrm{~N}\)
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Motion in Plane

143582 The position vector of a particle is \(\mathbf{r}=(\alpha \cos \omega t) \hat{\mathbf{i}}+(\alpha \sin \omega t) \hat{\mathbf{j}}\). The velocity vector of the particle is

1 Parallel to position vector
2 Perpendicular to position vector
3 Directed towards the origin
4 Directed away from the origin
Motion in Plane

143583 If a vector \(A\) having a magnitude of 8 is added to a vector \(B\) which lies along \(x\)-axis, then the resultant of two vectors lies along \(y\)-axis and has magnitude twice that of \(B\). The magnitude of \(B\) is

1 \(\frac{6}{\sqrt{5}}\)
2 \(\frac{12}{\sqrt{5}}\)
3 \(\frac{16}{\sqrt{5}}\)
4 \(\frac{8}{\sqrt{5}}\)
Motion in Plane

143584 It two forces each of \(2 \mathrm{~N}\) are inclined at \(60^{\circ}\), then resultant force is:

1 \(2 \mathrm{~N}\)
2 \(2 \sqrt{5} \mathrm{~N}\)
3 \(2 \sqrt{3} \mathrm{~N}\)
4 \(4 \sqrt{2} \mathrm{~N}\)
Motion in Plane

143586 Calculate the work done when a force \(\overrightarrow{\mathbf{F}}=\mathbf{2} \hat{\mathbf{i}}+\mathbf{3} \hat{\mathbf{j}}-\mathbf{5} \hat{\mathbf{k}}\) units acts on a body producing a displacement \(\vec{s}=\mathbf{2} \hat{\mathbf{i}}+\mathbf{4} \hat{\mathbf{j}}+\mathbf{3} \hat{\mathbf{k}}\) units :

1 1 unit
2 20 unit
3 5 unit
4 zero
Motion in Plane

143587 Three forces acting on a body are shown in the figure. To have the resultant force only along the \(y\)-direction, the magnitude of the minimum additional force needed along \(O X\) is

1 \(\frac{\sqrt{3}}{4} \mathrm{~N}\)
2 \(\sqrt{13} \mathrm{~N}\)
3 \(0.5 \mathrm{~N}\)
4 \(1.5 \mathrm{~N}\)
Motion in Plane

143582 The position vector of a particle is \(\mathbf{r}=(\alpha \cos \omega t) \hat{\mathbf{i}}+(\alpha \sin \omega t) \hat{\mathbf{j}}\). The velocity vector of the particle is

1 Parallel to position vector
2 Perpendicular to position vector
3 Directed towards the origin
4 Directed away from the origin
Motion in Plane

143583 If a vector \(A\) having a magnitude of 8 is added to a vector \(B\) which lies along \(x\)-axis, then the resultant of two vectors lies along \(y\)-axis and has magnitude twice that of \(B\). The magnitude of \(B\) is

1 \(\frac{6}{\sqrt{5}}\)
2 \(\frac{12}{\sqrt{5}}\)
3 \(\frac{16}{\sqrt{5}}\)
4 \(\frac{8}{\sqrt{5}}\)
Motion in Plane

143584 It two forces each of \(2 \mathrm{~N}\) are inclined at \(60^{\circ}\), then resultant force is:

1 \(2 \mathrm{~N}\)
2 \(2 \sqrt{5} \mathrm{~N}\)
3 \(2 \sqrt{3} \mathrm{~N}\)
4 \(4 \sqrt{2} \mathrm{~N}\)
Motion in Plane

143586 Calculate the work done when a force \(\overrightarrow{\mathbf{F}}=\mathbf{2} \hat{\mathbf{i}}+\mathbf{3} \hat{\mathbf{j}}-\mathbf{5} \hat{\mathbf{k}}\) units acts on a body producing a displacement \(\vec{s}=\mathbf{2} \hat{\mathbf{i}}+\mathbf{4} \hat{\mathbf{j}}+\mathbf{3} \hat{\mathbf{k}}\) units :

1 1 unit
2 20 unit
3 5 unit
4 zero
Motion in Plane

143587 Three forces acting on a body are shown in the figure. To have the resultant force only along the \(y\)-direction, the magnitude of the minimum additional force needed along \(O X\) is

1 \(\frac{\sqrt{3}}{4} \mathrm{~N}\)
2 \(\sqrt{13} \mathrm{~N}\)
3 \(0.5 \mathrm{~N}\)
4 \(1.5 \mathrm{~N}\)
Motion in Plane

143582 The position vector of a particle is \(\mathbf{r}=(\alpha \cos \omega t) \hat{\mathbf{i}}+(\alpha \sin \omega t) \hat{\mathbf{j}}\). The velocity vector of the particle is

1 Parallel to position vector
2 Perpendicular to position vector
3 Directed towards the origin
4 Directed away from the origin
Motion in Plane

143583 If a vector \(A\) having a magnitude of 8 is added to a vector \(B\) which lies along \(x\)-axis, then the resultant of two vectors lies along \(y\)-axis and has magnitude twice that of \(B\). The magnitude of \(B\) is

1 \(\frac{6}{\sqrt{5}}\)
2 \(\frac{12}{\sqrt{5}}\)
3 \(\frac{16}{\sqrt{5}}\)
4 \(\frac{8}{\sqrt{5}}\)
Motion in Plane

143584 It two forces each of \(2 \mathrm{~N}\) are inclined at \(60^{\circ}\), then resultant force is:

1 \(2 \mathrm{~N}\)
2 \(2 \sqrt{5} \mathrm{~N}\)
3 \(2 \sqrt{3} \mathrm{~N}\)
4 \(4 \sqrt{2} \mathrm{~N}\)
Motion in Plane

143586 Calculate the work done when a force \(\overrightarrow{\mathbf{F}}=\mathbf{2} \hat{\mathbf{i}}+\mathbf{3} \hat{\mathbf{j}}-\mathbf{5} \hat{\mathbf{k}}\) units acts on a body producing a displacement \(\vec{s}=\mathbf{2} \hat{\mathbf{i}}+\mathbf{4} \hat{\mathbf{j}}+\mathbf{3} \hat{\mathbf{k}}\) units :

1 1 unit
2 20 unit
3 5 unit
4 zero
Motion in Plane

143587 Three forces acting on a body are shown in the figure. To have the resultant force only along the \(y\)-direction, the magnitude of the minimum additional force needed along \(O X\) is

1 \(\frac{\sqrt{3}}{4} \mathrm{~N}\)
2 \(\sqrt{13} \mathrm{~N}\)
3 \(0.5 \mathrm{~N}\)
4 \(1.5 \mathrm{~N}\)