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

101) If the distances of the object and its real image from the principal focus of a concave mirror are 16 cm and 9 cm respectively, then the focal length of the mirror is

A) 30 cm

B) 12 cm

C) 18 cm

D) 24 cm

View Answer

B) 12 cm
Explanation:Using Newton’s formula:
x · y = f²
Given:
x = 16 cm, y = 9 cm
⇒ f² = 16 × 9 = 144
⇒ f = 12 cm

102) If the angle of minimum deviation produced by an equilateral prism is equal to the angle of the prism, then the refractive index of the material of the prism is nearly

A) 1.515

B) 1.414

C) 1.732

D) 1.625

View Answer

C) 1.732
Explanation:For prism:
μ = sin[(A + δ)/2] / sin(A/2)
Given:
A = 60°, δ = 60°
⇒ μ = sin(120/2) / sin(60/2)
= sin60° / sin30°
= (√3/2) / (1/2)
= √3 ≈ 1.732

103) When two light waves of equal intensity superimpose, the maximum intensity obtained is I. If the intensity of one of the waves is quadrupled, then the maximum intensity obtained is

A) $\frac{4I}{9}$

B) $\frac{9I}{4}$

C) $\frac{2I}{3}$

D) $\frac{3I}{2}$

View Answer

B) $\frac{9I}{4}$

Explanation:Intensity ∝ amplitude²
New amplitude doubles:
Intensity = $(2+1)^2 = 9/4 I$

✔️ Answer: B

104) The electric field due to an infinitely long thin straight wire with uniform linear charge density of 2.5 × 10⁻⁷ Cm⁻¹ at a radial distance of x from the wire is 7.5 × 10⁴ NC⁻¹. Then x =

A) 2 cm

B) 3 cm

C) 4 cm

D) 6 cm

View Answer

D) 6 cm
Explanation:Formula:
E = λ / (2π ε₀ r)
Given:
λ = 2.5 × 10⁻⁷ C/m
E = 7.5 × 10⁴ N/C
ε₀ = 8.85 × 10⁻¹²
Solve for r:
r = λ / (2π ε₀ E)
≈ (2.5×10⁻⁷) / [2π × 8.85×10⁻¹² × 7.5×10⁴] ≈ (2.5×10⁻⁷) / (4.17×10⁻⁶)
≈ 0.06 m = 6 cm

105) A parallel plate capacitor of capacitance 10 µF is charged by a 220 V supply. The capacitor is then disconnected from the supply and is connected to another uncharged parallel plate capacitor of capacitance 12 µF. The loss of electrostatic energy in this process is

A) 132 mJ

B) 220 mJ

C) 66 mJ

D) 110 mJ

View Answer

A) 132 mJ
Explanation:Given:
C₁ = 10 µF, V = 220 V
C₂ = 12 µF
Initial energy:
U₁ = (1/2) C₁ V²
= (1/2)(10×10⁻⁶)(220²)
= 5×10⁻⁶ × 48400 = 0.242 J
Final voltage after connection:
V’ = (C₁V)/(C₁ + C₂)
= (10×220)/(22) = 100 V
Final energy:
U₂ = (1/2)(C₁ + C₂)V’²
= (1/2)(22×10⁻⁶)(100²)
= 11×10⁻⁶ × 10000 = 0.11 J
Energy loss:
= U₁ − U₂ = 0.242 − 0.11 = 0.132 J
= 132 mJ

106) The lengths of two wires made of the same material are in the ratio 2:3 and their radii are in the ratio 1:2.If the two wires are connected in parallel to a battery, then the ratio of the drift velocities of free electrons in the two wires is

A) 2:1

B) 3:1

C) 3:2

D) 3:4

View Answer

C) 3:2
Explanation:Drift velocity ∝ current density
Ratio simplifies:
3:2
✔️ Answer: C

107) In a potentiometer experiment for the determination of the internal resis tance of a cell, when an external resistance of R is connected parallel to the cell, the balancing length decreases by 10%. The internal resistance of the cell is

A) $\frac{R}{9}$

B) $\frac{R}{7}$

C) $\frac{R}{5}$

D) $\frac{R}{11}$

View Answer

A) $\frac{R}{9}$

Explanation:Balancing length proportional:
$r = \frac{R}{9}$

✔️ Answer: A

108) The number of turns of two circular coils A and B are 300 and 200 respectively. The magnetic moments of the two coils A and B are in the ratio 1:2.If the two coils carry equal currents, then the ratio of radii of coils A and B is

A) 2:√3

B) 2:3

C) 1:2

D) 1:√3

View Answer

D) 1:√3
Explanation:Magnetic moment:
M = N I A = N I (π r²)
Since currents are equal:
M ∝ N r²
Given:
M_A : M_B = 1 : 2
N_A = 300, N_B = 200
⇒ (300 r_A²) / (200 r_B²) = 1/2
⇒ (3 r_A²) / (2 r_B²) = 1/2
Cross multiply:
6 r_A² = 2 r_B²
⇒ r_B² = 3 r_A²
⇒ r_B = √3 r_A
⇒ r_A : r_B = 1 : √3

109) Two long straight parallel wires carry currents of 8 A and 10 A in opposite directions. If the distance of separation between the wires is 9 cm, then the net magnetic field at a point between the two wires, which is at a perpendicular distance of 4 cm from the wire carrying 8 A current is

A) Zero

B) $4×10^{-5}$

C) $8×10^{-5}$

D) $12×10^{-5}$

View Answer

C) $8×10^{-5}$

Explanation:Field between wires:
Add fields:
$8×10^{-5} T$

✔️ Answer: C

110) A short bar magnet of magnetic moment 2.5 $Am^2$ is kept in a uniform magnetic field of $4 \times 10^{-5}$ T. The work done in moving the magnet from its most stable position to most unstable position is