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JEE PHYSICS MCQ: MOTION IN PLANE

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Application +4 | -1
The y-component of vector Aβƒ—\vec{A} is +3.0Β m+3.0 \mathrm{~m} if Aβƒ—\vec{A} makes an angle of 30∘30^{\circ} counter clockwise from the positive yy - axis, the magnitude of Aβƒ—\vec{A} is (assume Aβƒ—\vec{A} is in xx - y plane)
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Correct option: A
23Β m2 \sqrt{3} \mathrm{~m}
Option A (232 \sqrt{3} m) is correct because the magnitude of A⃗\vec{A} is calculated using the y-component and the given angle.

More Questions from MOTION IN PLANE

Practice All Chapter MCQs β†’
Given, Pβƒ—=Aβƒ—+B⃗  and P=A+B.\mathrm{\vec P} = \mathrm{\vec A} + \mathrm{\vec B}\,\,\mathrm{and}\,P = A + B. The angle between A⃗ and Bβƒ—\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∘0^{\circ} and 90∘90^{\circ} between them.
Reason: For two-dimensional or three-dimensional motion, velocity and acceleration of an object works in opposite direction.
If (a+3b)β‹…(7aβˆ’5b)=0\mathrm{If}\,(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5 \mathbf{b})=0 and
(aβˆ’4b)β‹…(7aβˆ’2b)=0(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b})=0. Then, the angle between a\mathbf{a} and b\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 ΞΈ\theta and another ball is thrown at an angle (90βˆ˜βˆ’ΞΈ)\left(90^{\circ}-\theta\right) with the horizontal direction from the same point with velocity 39.2Β msβˆ’139.2 \mathrm{~ms}^{-1}. The second ball reaches 50Β m50 \mathrm{~m} higher than the first ball. Find their individual heights. Take g=9.8Β msβˆ’2{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.
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