JEE PHYSICS MCQ: MOTION IN PLANE

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Let

\vec{a}

and

\vec{b}

be two vectors inclined at an angle

\theta

, where

\frac{\pi}{2} < \theta < \pi

.
Statement-I:

|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}| < |\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|

Statement-II:

|\vec{a}+\vec{b}|^2=a^2+b^2+2\vec{a} \cdot \vec{b}

View answer and explanation

Correct option: A

Statement-I is true and statement-II is true

Option A (Statement-I is true and statement-II is true) is correct because when the angle is obtuse, the magnitude of the difference is greater than the magnitude of the sum, and the expansion of |a+b|^2 is correct.

More Questions from MOTION IN PLANE

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Given,

\mathrm{\vec P} = \mathrm{\vec A} + \mathrm{\vec B}\,\,\mathrm{and}\,P = A + B.

The angle between

\mathrm{\vec A}\,and\,\mathrm{\vec B}

is Assertion: The necessary condition for motion in two dimensions, velocity and acceleration vectors should have any angle between

0^{\circ}

and

90^{\circ}

between them.
Reason: For two-dimensional or three-dimensional motion, velocity and acceleration of an object works in opposite direction.

\mathrm{If}\,(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5 \mathbf{b})=0

and

(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b})=0

. Then, the angle between

\mathbf{a}

and

\mathbf{b}

is At the top of the trajectory of a projectile, the direction of its velocity and acceleration are A ball is thrown at an angle

\theta

and another ball is thrown at an angle

\left(90^{\circ}-\theta\right)

with the horizontal direction from the same point with velocity

39.2 \mathrm{~ms}^{-1}

. The second ball reaches

50 \mathrm{~m}

higher than the first ball. Find their individual heights. Take

{g}=9.8 \mathrm{~ms}^{-2}

. Assertion: The drover in a vehicle moving with a constant speed on a straight road is in a non-inertial frame of reference.

Reason: A reference frame in which Newton's laws of motion are applicable is non-inertial.