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NEET Practice Questions (MCQs) with Answers & Solutions

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If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is

90°

45°

180°

0°

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Explanation

(a)

$Suppose two vectors are P and Q . It is given that - |P + Q| = |P - Q| Let the angle between P and Q is θ . \Rightarrow P^{2} + Q^{2} + 2 PQcosθ = P^{2} + Q^{2} - 2 PQcosθ \Rightarrow cosθ = 0 = \cos 90^{0} \Rightarrow θ = 90^{0}$

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A force of 6 N and another of 8 N can be applied together to produce the effect of a single force of -

1 N

11 N

15 N

20 N

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Which of the sets given below may represent the magnitude of resultant of three vectors adding to zero?

2, 4, 8

4, 8, 16

1, 2, 1

0.5, 1, 2

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Explanation

(C)

$Let there be 3 vectors \vec{A}, \vec{B} & \vec{C} . Suppose magnitude of C is highest . For zero resultant, C \leq A + B Where A, B & C represents magnitude of \vec{A}, \vec{B} & \vec{C} For pair 1, 2, 1 C = 2 A = 1 B = 1 Also, 2 \leq 1 + 1$

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A blind person after walking 10 steps in one direction, each of length 80 cm, turns randomly to the left or to the right by $90 ° .$ After walking a total of 40 steps the maximum possible displacement of the person from his starting position could be -

$320 m$

$32 m$

$16 / \sqrt{2} m$

$16 \sqrt{2} m$

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Explanation

Each step is 80 cm, and the person takes 40 steps. If the person walks in a straight line after the initial turn, the maximum displacement is 40 Ã— 0.8 m = 32 m. However, since the person can turn at 90Â° after every 10 steps, the maximum displacement is âˆš(32^2 + 32^2) = 16âˆš2 m.

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A truck travelling due north with 20m/s turns towards west and travels at the same speed. Then the change in velocity is -

40 m/s north-west

$20 \sqrt{2}$ m/s north-west

40 m/s south-west

$20 \sqrt{2}$ m/s south-west

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A particle is moving in a circle of radius r centered at O with constant speed v. The change in velocity moving from A to B $(∠ AOB = 40 °)$ is -

$2 v \cos 40 °$

$2 v \sin 40 °$

$2 v \cos 20 °$

$2 v \sin 20 °$

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If $\vec{A} = \vec{B} + \vec{C}$ and magnitudes of A, B and C are 5, 4 and 3 units respectively, the angle between A and C is -

$\sin^{- 1} (3 / 4)$

$\cos^{- 1} (4 / 5)$

$\cos^{- 1} (3 / 5)$

$π / 2$

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What is the resultant of three coplanar forces: $300 N$ at $0 °$ , $400 N$ at $30 °$ and $400 N$ at $150 °$ ?

$500 N$

$700 N$

$1100 N$

$300 N$

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Two forces, $F_{1}$ and $F_{2}$ are acting on a body. One force is doubled of the other force and the resultant is equal to the greater force. Then the angle between the two forces is -

$\cos^{- 1} (1 / 2)$

$\cos^{- 1} (- 1 / 2)$

$\cos^{- 1} (- 1 / 4)$

$\cos^{- 1} (1 / 4)$

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Explanation

(C)

$F_{2} = 2 F_{1} and R = F_{2} R = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cosθ} Squaring both sides {(F_{2})}^{2} = F_{1}^{2} + {(2 F_{1})}^{2} + 2 F_{1} (2 F_{1}) cosθ {(2 F_{1})}^{2} = 5 F_{1}^{2} + 4 F_{1}^{2} cosθ Solving, - F_{1}^{2} = 4 F_{1}^{2} cosθ cosθ = \frac{- 1}{4} θ = \cos^{- 1} (\frac{- 1}{4})$

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What displacement must be added to the displacement $25 \hat{i} - 6 \hat{j} m$ to give a displacement of $7.0 m$ pointing in the x-direction?

$18 \hat{i} - 6 \hat{j}$

$32 \hat{i} - 13 \hat{j}$

$- 18 \hat{i} + 6 \hat{j}$

$- 25 \hat{i} + 13 \hat{j}$

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Explanation

$L e t t h e r e q u i r e d v e c t o r b e x \hat{i} + y \hat{j} S o, 25 \hat{i} + - 6 \hat{j} + (x \hat{i} + y \hat{j}) = 7 \hat{i} x \hat{i} + y \hat{j} = - 18 \hat{i} + 6 \hat{j} C o m p a r i n g, x = - 18, y = 6$

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