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Physics MCQs for NEET — Practice Questions with Answers

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Given: $\vec{C} = \vec{A} + \vec{B}$ . Also, the magnitude of $\vec{A}, \vec{B}$ and $\vec{C}$ are 12, 5 and 13 units respectively. The angle between $\vec{A}$ and $\vec{B}$ is

$0 °$

$\frac{π}{4}$

$\frac{π}{2}$

$π$

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Explanation

$A s \vec{C} = \vec{A} + \vec{B}$

Let angle between A and B is $θ$

$\Rightarrow C^{2} = A^{2} + B^{2} + 2 A B C o s θ \Rightarrow {(13)}^{2} = {(12)}^{2} + {(5)}^{2} + 2 \times 12 + 5 + \cos θ \Rightarrow \cos θ = 0 \Rightarrow θ = 90 °$

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If $\vec{P} + \vec{Q} = \vec{P} - \vec{Q}$ and $θ$ is the angle between $\vec{P}$ and $\vec{Q}$ , then

$θ = 0 °$

$θ = 90 °$

$P = 0$

$Q = 0$

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Explanation

$\vec{P} + \vec{Q} = \vec{P} - \vec{Q}$

$2 \vec{Q} = 0$

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Three non zero vectors $\vec{A}, \vec{B} & \vec{C}$ satisfy the relation $\vec{A} \cdot \vec{B} = 0 & \vec{A} \cdot \vec{C} = 0$ . Then $\vec{A}$ can be parallel to:

$\vec{B}$

$\vec{C}$

$\vec{B} \cdot \vec{C}$

$\vec{B} \times \vec{C}$

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Explanation

$(4) As \vec{A} . \vec{B} = 0 and \vec{A} . \vec{C} = 0, so \vec{A} is ⊥ ar to both \vec{B} and \vec{C} . Also, \vec{B} \times \vec{C} is also ⊥ ar to both \vec{B} and \vec{C} . So, \vec{A} can be parallel to \vec{B} \times \vec{C} .$

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Given: $\vec{a} + \vec{b} + \vec{c} = 0$ . Out of the three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ two are equal in magnitude. The magnitude of the third vector is $\sqrt{2}$ times that of either of the two having equal magnitude. The angles between the vectors are:

90 $°$ , 135 $°$ , 135 $°$

30 $°$ , 60 $°$ , 90 $°$

45 $°$ , 45 $°$ , 90 $°$

45 $°$ , 60 $°$ , 90 $°$

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Vector $\vec{A}$ is of length 2 cm and is 60 $°$ above the x-axis in the first quadrant. Vector $\vec{B}$ is of length 2 cm and 60 $°$ below the x-axis in the fourth quadrant. The sum $\vec{A}$ + $\vec{B}$ is a vector of magnitude -

2 along + y-axis

2 along + x-axis

1 along - x-axis

2 along - x-axis

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Explanation

Vector A has a magnitude of 2 cm at an angle of 60Â° above the x-axis. Vector B has the same magnitude of 2 cm but at 60Â° below the x-axis. Their sum A + B will be a vector along the positive x-axis with a magnitude of 2âˆš3 â‰ˆ 3.46 cm.

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Six forces, 9.81 N each, acting at a point are coplanar. If the angle between neighboring forces are equal, then the resultant is

0 N

9.81 N

2 $\times$ 9.81 N

3 $\times$ 9.81 N

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Explanation

When six equal forces of 9.81 N each are acting at a point in the same plane, and the angles between neighboring forces are equal, the forces will form a closed polygon. The resultant of such a system of coplanar forces is always zero, as the vector sum of all forces is zero.

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Temperature of a body varies with time as $T = (T_{0} + {at}^{2} + b sint) K$ , where $T_{0}$ is the temperature in Kelvin at $t = 0 \sec & a = 2 / π K / s^{2} & b = - 4 K$ , then the rate of change of temperature $(\frac{dT}{dt})$ at $t = π \sec$ is-

$8 K$

$8^{0} K$

$8 K / \sec$

$8^{0} K / \sec$

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Explanation

$T = (T_{0} + {at}^{2} + b \sin t) K \frac{dT}{dt} = 2 at + b \cos t At t = π \sec \frac{dT}{dt} = 2 (\frac{2}{π}) π + (- 4) cosπ = 8 K / \sec$

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If the distance 's' travelled by a body in time 't' is given by $s = \frac{a}{t} + {bt}^{2}$ then the acceleration equals

$\frac{2 a}{t^{3}} + 2 b$

$\frac{2 s}{t^{3}}$

$2 b - \frac{2 a}{t^{3}}$

$\frac{s}{t^{2}}$

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Explanation

(A)

$s = \frac{a}{t} + {bt}^{2} v = \frac{ds}{dt} = \frac{- a}{t^{2}} + 2 bt a = \frac{dv}{dt} = \frac{2 a}{t^{3}} + 2 b$

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The velocity of a particle moving on the x-axis is given by $v = x^{2} + x$ where v is in m/s and x is in m. Find its acceleration in $m / s^{2}$ when passing through the point x=2m.

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Explanation

$v = x^{2} + x Differentiating both sides w . r . t t \frac{dv}{dt} = \frac{d (x^{2})}{dt} + \frac{dx}{dt} a = 2 x (\frac{dx}{dt}) + \frac{dx}{dt} a = (2 x + 1) v [∵ v = \frac{dx}{dt}] = (2 x + 1) (x^{2} + x) At x = 2 m a = [2 (2) + 1] [2^{2} + 2] = 30 {ms}^{- 2}$

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A particle moves in the XY plane and at time t is at the point whose coordinates are $(t^{2}, t^{3} - 2 t)$ . Then at what instant of time, will its velocity and acceleration vectors be perpendicular to each other?

1/3 sec

2/3 sec

3/2 sec

never

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Explanation

(B)

$\vec{r} = t^{2} \hat{i} + (t^{3} - 2 t) \hat{j} \vec{v} = \frac{d \vec{r}}{dt} = 2 t \hat{i} + (3 t^{2} - 2) \hat{j} \vec{a} = 2 \hat{i} + (6 t) \hat{j} When \vec{v} ⊥ \vec{a}, \vec{v} . \vec{a} = 0 [2 t \hat{i} + (3 t^{2} - 2) \hat{j}] . [2 \hat{i} + 6 t \hat{j}] = 0 4 t + 6 t (3 t^{2} - 2) = 0 2 t [2 + 3 (3 t^{2} - 2)] = 0 t = 0 or 2 + 9 t^{2} - 6 = 0 t^{2} = \frac{4}{9} or t = \frac{2}{3} \sec$

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