1
AP EAPCET 2025 - 26th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

If force $=\frac{\alpha}{\operatorname{density}+\beta^3}$, then the dimensional formulae of $\alpha$ and $\beta$ are respectively

A

$\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right],\left[\mathrm{ML}^{-1 / 3} \mathrm{~T}^0\right]$

B

$\left[M^2 L^4 T^{-2}\right],\left[M^{1 / 3} L^{-1} T^0\right]$

C

$\left[\mathrm{M}^2 \mathrm{~L}^{-2} \mathrm{~T}^{-2}\right],\left[\mathrm{M}^{1 / 3} \mathrm{~L}^{-1} mathrm{~T}^0\right]$

D

$\left[\mathrm{M}^2 \mathrm{~L}^{-2} \mathrm{~T}^{-2}\right],\left[\mathrm{ML}^{-3} \mathrm{~T}^0\right]$

2
AP EAPCET 2025 - 26th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The displacement $(x)$ and time $(t)$ graph of a particle moving along a straight line is shown in the figure. The average velocity of the particle in the time of 10 s is

AP EAPCET 2025 - 26th May Evening Shift Physics - Motion in a Straight Line Question 2 English

A

$2 \mathrm{~ms}^{-1}$

B

$4 \mathrm{~ms}^{-1}$

C

$6 \mathrm{~ms}^{-1}$

D

$8 \mathrm{~ms}^{-1}$

3
AP EAPCET 2025 - 26th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

If the horizontal range of a body projected with a velocity ' $u$ ' is 3 times the maximum height reached by it, then the range of the body is

( $g=$ Acceleration due to gravity)

A

$\frac{2 u^2}{3 g}$

B

$\frac{4 u^2}{5 g}$

C

$\frac{12 u^2}{13 g}$

D

$\frac{24 u^2}{25 g}$

4
AP EAPCET 2025 - 26th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

If the velocity at the maximum height of a projectile projected at an angle of $45^{\circ}$ is $20 \mathrm{~ms}^{-1}$, then the maximum height reached by the projectile is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

A

10 m

B

20 m

C

30 m

D

40 m