1
AP EAPCET 2022 - 4th July Evening Shift
MCQ (Single Correct Answer)
+1
-0

An object moving along $$X$$-axis with a uniform acceleration has velocity $$\mathbf{v}=\left(12 \mathrm{cms}^{-1}\right) \hat{\mathbf{i}}$$ at $$x=3 \mathrm{~cm}$$. After 2 s if it is at $$x=-5 \mathrm{~cm}$$, then its acceleration is

A
$$\mathbf{a}=\left(-16 \mathrm{~cms}^{-2}\right) \hat{i}$$
B
$$\mathbf{a}=\left(11 \mathrm{~cms}^{-2}\right) \hat{i}$$
C
$$\mathbf{a}=\left(-11 \mathrm{~cms}^{-2}\right) \hat{i}$$
D
$$\mathbf{a}=\left(8 \mathrm{~cms}^{-2}\right) \hat{i}$$
2
AP EAPCET 2022 - 4th July Evening Shift
MCQ (Single Correct Answer)
+1
-0

A force $$\mathbf{F}_1$$ of magnitude 4 N acts on an object of mass 1 kg , at origin in a direction $$30^{\circ}$$ above the positive $$X$$-axis. A second $$F_2$$ of magnitude 4 N acts on the same object in the direction of the positive $$Y$$-axis. The magnitude of the acceleration of the object is nearly.

A
$$6.9 \mathrm{~ms}^{-2}$$
B
$$7.6 \mathrm{~ms}^{-2}$$
C
$$4.3 \mathrm{~ms}^{-2}$$
D
$$8.0 \mathrm{~ms}^{-2}$$
3
AP EAPCET 2022 - 4th July Evening Shift
MCQ (Single Correct Answer)
+1
-0

$$y=\left(P t^2-Q t^3\right) \mathrm{~m}$$ is the vertical displacement of a ball which is moving in vertical plane. Then the maximum height that the ball can reach is

A
$$\frac{27 P^3}{4 Q^2}$$
B
$$\frac{4 Q^2}{27 P^3}$$
C
$$\frac{4 P^3}{27 Q^2}$$
D
$$\frac{27 Q^2}{4 P^3}$$
4
AP EAPCET 2022 - 4th July Evening Shift
MCQ (Single Correct Answer)
+1
-0

A cricket ball of mass 50 g having velocity $$50 \mathrm{~cm} \mathrm{~s}^{-1}$$ to stopped in 0.5 s. The force applied to stop the ball is

A
0.07 N
B
0.05 N
C
5 N
D
7 N
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