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TS EAMCET 2025 May 4 Shift 1 Question Paper with Answers | Engineering Previous Paper

TS EAMCET 2025 May 4 Shift 1
Physics

81) Match the ”Technology” given in List-1 with the ”Principle of Physics” given in List-2.

List – 1 (Technology)		List – 2 (Principle of Physics)
A	Steam engine	I	Magnetic confinement of plasma
B	Electron microscope	II	Laws of thermodynamics
C	Non-reflecting coatings	III	Wave nature of electrons
D	Tokamak	IV	Interference of light

A) A-I, B-II, C-III, D-IV

B) A-II, B-III, C-IV, D-I

C) A-II, B-IV, C-III, D-I

D) A-II, B-I, C-III, D-IV

View Answer

B) A-II, B-III, C-IV, D-I
Explanation:

Technology	Correct Match
A (Steam engine)	II
B (Electron microscope)	III
C (Non-reflecting coatings)	IV
D (Tokamak)	I

82) In an experiment, the coefficient of viscosity (in mPa s) of a liquid was determined as 2.62, 2.68, 2.58, 2.57, 2.54 and 2.55. The mean absolute error in the determination of the coefficient of viscosity of the liquid is

A) 0.08 mPa s

B) 0.12 mPa s

C) 0.06 mPa s

D) 0.04 mPa s

View Answer

D) 0.04 mPa s
Explanation:Mean value ≈ 2.59
Absolute deviations:
Average = 0.04
✔️ Answer:

83) The relation between the displacement ‘x’ (in metre) and the time ‘t’ (in second) of a particle is t = 2x² +3x. If the displacement of the particle is 25 cm from the origin (x=0), then the acceleration of the particle is

A) + $\frac{1}{16}$

B) $-\frac{1}{16}$

C) + $\frac{1}{8}$

D) $-\frac{1}{8}$

View Answer

B) $-\frac{1}{16}$

Explanation:Given:
t = 2x² + 3x
Differentiate w.r.t x:
dt/dx = 4x + 3
Velocity:
v = dx/dt = 1 / (4x + 3)
Acceleration:
a = dv/dt = (dv/dx)(dx/dt)
First find dv/dx:
v = (4x+3)⁻¹
⇒ dv/dx = −4 / (4x+3)²
Now:
a = (−4/(4x+3)²) × (1/(4x+3))
⇒ a = −4 / (4x+3)³
Now x = 25 cm = 1/4 m
⇒ 4x + 3 = 1 + 3 = 4
⇒ a = −4 / 4³ = −4/64 = −1/16
Final Answer:
(B) −1/16

84) A body projected at certain angle (̸= 90◦) from the ground crosses a point in its path at a time of 2.3 s and from there it reaches the ground after a time of 5.7 s. The maximum height reached by the body is (Acceleration due to gravity = 10ms−2

A) 80 m

B) 120 m

C) 40 m

D) 160 m

View Answer

A) 80 m
Explanation:Total time = 2.3 + 5.7 = 8 s
Max height at t=4s:
$H=\frac{1}{2}gt^2 = 80m$

✔️ Answer: A

85) A circular path of radius 75 m is banked at an angle of $\tan^{-1}(0.2)$ . If the coefficient of static friction between the tyres of the car and the circular path is 0.1, then the maximum permissible speed of the car to avoid slipping is

A) 10 $ms^{-1}$

B) 20 $ms^{-1}$

C) 15 $ms^{-1}$

D) 30 $ms^{-1}$

View Answer

C) 15 $ms^{-1}$

Explanation:Given:
r = 75 m
tanθ = 0.2 ⇒ θ ≈ small angle
μ = 0.1
g ≈ 10 m/s²
Formula (maximum speed on banked road with friction):
v_max = √[ r g ( (tanθ + μ) / (1 − μ tanθ) ) ] Substitute:
tanθ = 0.2
⇒ v² = 75 × 10 × (0.2 + 0.1)/(1 − 0.1×0.2)
= 750 × (0.3 / 0.98)
≈ 750 × 0.306 ≈ 229.5
⇒ v ≈ √229.5 ≈ 15 m/s
Final Answer:
(C) 15 ms⁻¹

86) A horizontal force of 10 N is applied on a block of mass 1.5 kg which is initially at rest on a rough horizontal surface. The work done by the applied force in a time of 6 s from the beginning of the motion is (Acceleration due to gravity = 10 ms⁻²; the coefficient of kinetic friction between the block and the surface is 0.2)

A) 588 J

B) 360 J

C) 840 J

D) 420 J

View Answer

C) 840 J
Explanation:Given:
F = 10 N
m = 1.5 kg
μ = 0.2
g = 10 m/s²
Friction:
f = μmg = 0.2 × 1.5 × 10 = 3 N
Net force:
F_net = 10 − 3 = 7 N
Acceleration:
a = F_net / m = 7 / 1.5 = 14/3 m/s²
Initial velocity u = 0
Displacement in 6 s:
s = ut + (1/2)at²
= 0 + (1/2)(14/3)(36)
= (7/3) × 36 = 84 m
Work done by applied force:
W = F × s = 10 × 84 = 840 J
Final Answer:
(C) 840 J

87) A ball is allowed to fall freely from a height of 42 m from the grounD.If the coefficient of restitution between the ball and the ground is 0.4, then the total distance travelled by the ball before it comes to rest is

A) 84 m

B) 87 m

C) 72 m

D) 58 m

View Answer

D) 58 m
Explanation:Total distance:
$H\frac{1+e^2}{1-e^2}$

Substitute:
≈ 87 m
✔️ Answer:

88) A thin uniform wire of mass ‘m’ and linear density ‘ρ’ is bent in the form of a circular ring. The moment of inertia of the ring about a tangent parallel to its diameter is

A) $\frac{3m^3}{8\pi^2\rho^2}$

B) $\frac{8m^3}{3\pi^2\rho^2}$

C) $\frac{8\pi^2 m^3}{3\rho^2}$

D) $\frac{3\pi^2 m^3}{8\rho^2}$

View Answer

A) $\frac{3m^3}{8\pi^2\rho^2}$

Explanation:Length of wire = circumference:
2πR = m / ρ
⇒ R = m / (2πρ)
Moment of inertia of ring about diameter:
I_d = (1/2) mR²
Moment of inertia about tangent parallel to diameter:
Use parallel axis theorem:
I = I_d + mR²
= (1/2)mR² + mR²
= (3/2)mR²
Now substitute R:
R² = m² / (4π²ρ²)
⇒ I = (3/2)m × (m² / 4π²ρ²)
= (3m³) / (8π²ρ²)
Final Answer:
(A) 3m³ / (8π²ρ²)

89) A solid sphere and a thin uniform circular disc of same radius are rolling down an inclined plane without slipping. If the acceleration of the sphere is 3 ms⁻², then the acceleration of the disc is

A) 4 ms⁻²

B) 2.8 ms⁻²

C) 3 ms⁻²

D) 3.2 ms⁻²

View Answer

B) 2.8 ms⁻²
Explanation:Rolling motion:
Disc acceleration:
$a=\frac{g\sin\theta}{1+k}$

Sphere given 3 ⇒ disc =2.8
✔️ Answer: B

90) If the amplitudes of a damped harmonic oscillator at times t=0, $t_1$ and $t_2$ are $A_0$ , $A_1$ and $A_2$ respectively, then the amplitude of the oscillator at a time of $(t_1+t_2)$ is $\frac{A_1 A_2}{A_0}$

A) $\frac{A_0+A_1+A_2}{3}$

B) $\frac{A_2A_0}{A_1}$

C) $\frac{A_1A_0}{A_2}$

D) $\frac{A_1A_2}{A_0}$

View Answer

D) $\frac{A_1A_2}{A_0}$

Explanation:For damped oscillator:
A(t) = A₀ e^(−kt)
So:
A₁ = A₀ e^(−k t₁)
A₂ = A₀ e^(−k t₂)
Multiply:
A₁ A₂ = A₀² e^(−k(t₁ + t₂))
⇒ A₀ e^(−k(t₁ + t₂)) = (A₁ A₂)/A₀
But:
A(t₁ + t₂) = A₀ e^(−k(t₁ + t₂))
⇒ A(t₁ + t₂) = (A₁ A₂)/A₀
Final Answer:
(D) A₁A₂ / A₀

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TS EAMCET 2025 May 4 Shift 1 Physics

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TS EAMCET 2025 May 4 Shift 1
Physics