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JEE PHYSICS MCQ: MOTION IN PLANE

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In a tetrahedron OABCO A B C, the edges are of lengths, ∣OA∣=∣BC∣=a,∣OB∣=∣AC∣=b,∣OC∣=∣AB∣=c|O A|=|B C|=a,|O B|=|A C|=b,|O C|=|A B|=c. Let G1G_1 and C2C_2 be the centroids of the triangle ABCA B C and AOCA O C such that OG1⊥BG2O G_1 \perp B G_2, then the value of a2+c2b2\frac{a^2+c^2}{b^2} is
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Correct option: B
3
Option B (3) is correct because by using the given condition OG1⋅BG2=0OG_1 \cdot BG_2 = 0 and the properties of centroids, we can derive the relationship a2+c2b2=3\frac{{a^2 + c^2}}{{b^2}} = 3.

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Given, P⃗=A⃗+B⃗  and P=A+B.\mathrm{\vec P} = \mathrm{\vec A} + \mathrm{\vec B}\,\,\mathrm{and}\,P = A + B. The angle between A⃗ and B⃗\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)⋅(7a−5b)=0\mathrm{If}\,(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5 \mathbf{b})=0 and
(a−4b)⋅(7a−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 θ\theta and another ball is thrown at an angle (90∘−θ)\left(90^{\circ}-\theta\right) with the horizontal direction from the same point with velocity 39.2 ms−139.2 \mathrm{~ms}^{-1}. The second ball reaches 50 m50 \mathrm{~m} higher than the first ball. Find their individual heights. Take g=9.8 ms−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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