Question 1 of 15 Which of the following is scalar quantity? A weight B mass C magnetic moment D torque 💡 Explanation Weight, magnetic moment, and torque are all vector quantities (they have direction). Mass is a scalar quantity — it has only magnitude, with no associated direction. Question 2 of 15 Electric current is A scalar B polar vector C axial vector D none 💡 Explanation Although current has a direction of flow, it does not obey the law of vector addition (two currents meeting at a junction simply add algebraically, as per Kirchhoff's law, not vectorially). Hence current is treated as a scalar quantity. Question 3 of 15 Which of the following is odd one regarding the context of scalar and vector? A electric field B magnetic field C electric dipole D potential 💡 Explanation Electric field, magnetic field, and electric dipole moment are all vector quantities. Electric potential is a scalar quantity, making it the odd one out. Question 4 of 15 If angle between two equal forces (F) is 90°, then the magnitude of the resultant is A [latex]\dfrac{F}{\sqrt{2}}[/latex] B [latex]2F[/latex] C [latex]\sqrt{2}F[/latex] D [latex]2\sqrt{2}F[/latex] 💡 Explanation Using the law of vector addition: [latex]R = \sqrt{F^2 + F^2 + 2F \cdot F\cos 90^\circ} = \sqrt{F^2+F^2+0} = \sqrt{2F^2} = \sqrt{2},F[/latex] Question 5 of 15 If two vectors each having magnitude 'P' are acting at a point and angle between them is θ, then the resultant is given by A [latex]\sqrt{2}P\cos\dfrac{\theta}{2}[/latex] B [latex]2P\cos\dfrac{\theta}{2}[/latex] C [latex]\sqrt{2}P\sin\dfrac{\theta}{2}[/latex] D none 💡 Explanation [latex]R = \sqrt{P^2+P^2+2P^2\cos\theta} = P\sqrt{2(1+\cos\theta)}[/latex] Using the identity [latex]1+\cos\theta = 2\cos^2\dfrac{\theta}{2}[/latex]: [latex]R = P\sqrt{4\cos^2\dfrac{\theta}{2}} = 2P\cos\dfrac{\theta}{2}[/latex] Question 6 of 15 If magnitude of resultant of two forces F and F acting at a point is also F, the angle between the forces is A 0° B 60° C 120° D 90° 💡 Explanation [latex]F^2 = F^2+F^2+2F^2\cos\theta[/latex] [latex]\implies 2F^2\cos\theta = -F^2 \implies \cos\theta = -\dfrac{1}{2} \implies \theta = 120^\circ[/latex] Question 7 of 15 Small elementary area is a A scalar B vector C neither scalar nor vector D some time scalar some time vector 💡 Explanation A small elementary area is treated as a vector quantity, with magnitude equal to the area and direction taken along the outward normal to the surface. Question 8 of 15 The vector [latex]\vec{S}[/latex] is defined as [latex]\vec{S} = \vec{a} - \vec{b}[/latex], then the magnitude of [latex]\vec{S}[/latex] is A [latex]\sqrt{a+b^2+2ab\cos\theta}[/latex] B [latex]\sqrt{a^2+b^2-2ab\cos\theta}[/latex] C [latex]\sqrt{a^2+b^2+2ab\sin\theta}[/latex] D none of above 💡 Explanation For subtraction of two vectors, [latex]\vec{S} = \vec{a}-\vec{b} = \vec{a}+(-\vec{b})[/latex]. The angle between [latex]\vec{a}[/latex] and [latex]-\vec{b}[/latex] is [latex](180^\circ - \theta)[/latex], where θ is the angle between [latex]\vec{a}[/latex] and [latex]\vec{b}[/latex]. Applying the law of vector addition: [latex]S = \sqrt{a^2+b^2+2ab\cos(180^\circ-\theta)} = \sqrt{a^2+b^2-2ab\cos\theta}[/latex] since [latex]\cos(180^\circ-\theta) = -\cos\theta[/latex]. Question 9 of 15 A man is running due north at 2 m/s turns east and runs at the same speed. What is the change in velocity? A 4 m/s north east B [latex]2\sqrt{2}[/latex] m/s south west C [latex]2\sqrt{2}[/latex] m/s south east D [latex]2\sqrt{2}[/latex] m/s north west 💡 Explanation Let north be [latex]+y[/latex] and east be [latex]+x[/latex]. Initial velocity [latex]\vec{v_i} = 2\hat{j}[/latex], final velocity [latex]\vec{v_f} = 2\hat{i}[/latex]. [latex]\Delta \vec{v} = \vec{v_f}-\vec{v_i} = 2\hat{i}-2\hat{j}[/latex] [latex]|\Delta \vec{v}| = \sqrt{2^2+2^2} = 2\sqrt{2}\text{ m/s}[/latex] The vector [latex]\Delta \vec{v}[/latex] points in the east-south direction, i.e. south east. Question 10 of 15 If [latex]\vec{a}[/latex] and [latex]\vec{b}[/latex] have same magnitude 'p' and angle between [latex]\vec{a}[/latex] and [latex]\vec{b}[/latex] is θ, the magnitude of vector [latex]\vec{a}-\vec{b}[/latex] is A [latex]2p\sin\dfrac{\theta}{2}[/latex] B [latex]2p\sin\theta[/latex] C [latex]2p\cos\dfrac{\theta}{2}[/latex] D [latex]2p\cos\theta[/latex] 💡 Explanation [latex]|\vec{a}-\vec{b}| = \sqrt{p^2+p^2-2p^2\cos\theta} = p\sqrt{2(1-\cos\theta)}[/latex] Using [latex]1-\cos\theta = 2\sin^2\dfrac{\theta}{2}[/latex]: [latex]|\vec{a}-\vec{b}| = p\sqrt{4\sin^2\dfrac{\theta}{2}} = 2p\sin\dfrac{\theta}{2}[/latex] Question 11 of 15 An airplane is moving on a circular path with the speed 200 km/hr. What is the change in velocity in half revolution? A zero B 400 km/hr C 200 km/hr D [latex]200\sqrt{2}[/latex] km/hr 💡 Explanation After half a revolution, the velocity direction is exactly reversed while the speed stays the same. So [latex]\vec{v_i} = 200[/latex] (say, in one direction) and [latex]\vec{v_f} = -200[/latex] (opposite direction). [latex]\Delta v = v_f - v_i = -200-200 = -400[/latex] Magnitude of change = 400 km/hr. Question 12 of 15 If force [latex]\vec{F} = \hat{i}+\hat{j}+\hat{k}[/latex] and displacement [latex]\vec{d} = 3\hat{i}+2\hat{j}+5\hat{k}[/latex] are perpendicular to each other, then work done by force is A zero B 10 unit C 5 unit D 8 unit 💡 Explanation Work done by a force is given by [latex]W = \vec{F}\cdot\vec{d} = Fd\cos\theta[/latex]. Since [latex]\vec{F}[/latex] and [latex]\vec{d}[/latex] are stated to be perpendicular, [latex]\theta = 90^\circ[/latex], and [latex]\cos 90^\circ = 0[/latex], so: [latex]W = Fd\cos 90^\circ = 0[/latex] Question 13 of 15 If α be the angle of direction of the resultant of two equal vectors acting at an angle θ, which of the following will be the correct alternative? A [latex]\alpha = \dfrac{\theta}{2}[/latex] B [latex]\theta = \dfrac{\alpha}{2}[/latex] C [latex]\cos\alpha/2 = \cos\theta/2[/latex] D [latex]\cos\alpha/2 = \dfrac{1}{2}\cos\theta/2[/latex] 💡 Explanation When two equal vectors act at an angle θ, their resultant always bisects the angle between them, because the parallelogram formed by two equal vectors is a rhombus, and the diagonal of a rhombus bisects the angle. So the angle α that the resultant makes with either vector is exactly half of θ, i.e. [latex]\alpha = \dfrac{\theta}{2}[/latex]. Question 14 of 15 Which one will be the correct alternative for the cross product of two parallel vectors? A Parallel to one of the vectors B Normal to one of the vectors C Zero vector D A unit vector 💡 Explanation The magnitude of the cross product of two vectors is given by [latex]|\vec{A} \times \vec{B}| = AB\sin\theta[/latex], where θ is the angle between them. For two parallel vectors, [latex]\theta = 0^\circ[/latex] (or [latex]180^\circ[/latex] for antiparallel), so [latex]\sin\theta = 0[/latex]. Hence the cross product of two parallel vectors is always a zero vector. Question 15 of 15 Two forces of 4 N and 3 N act on a particle at the same time. What is the particle's net force? A 1N B 7N C 5N D Between 1N and 7N 💡 Explanation The resultant of two forces depends on the angle between them, given by [latex]R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta}[/latex]. Since the angle θ between the 4 N and 3 N forces is not specified, the resultant can take any value between the minimum, when they act in opposite directions ([latex]R = 4 - 3 = 1\text{ N}[/latex]), and the maximum, when they act in the same direction ([latex]R = 4 + 3 = 7\text{ N}[/latex]). So the net force lies between 1 N and 7 N. ← Prev Check Next → Finish & See Score 0 of 15 checked ← Back 1 / 15 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
