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

91) A meteor of mass ‘m’ having a speed ‘v’ at infinity reaches the surface of the earth with a speed of (ve is escape speed from the earth’s surface)

A) \sqrt{2}v_e

B) v_e

C) 2\sqrt{v^2+v_e^2}

D) \sqrt{v^2+v_e^2}

View Answer

D) \sqrt{v^2+v_e^2}
Explanation:Energy conservation:
v=\sqrt{v^2+v_e^2}
✔️ Answer: D
92) The work to be done to produce a strain of 10^{-3} in a steel wire of mass 2.96 kg and density 7.4 gcm^{-3} is (Young’s modulus of steel = 2 \times 10^{11} Nm^{-2})

A) 0.04 kJ

B) 0.04 J

C) 100 kJ

D) 400 J

View Answer

A) 0.04 kJ
Explanation:Energy:
\frac{1}{2}Y \times strain^2 × volume
Compute:
0.04 kJ
✔️ Answer: A
93) A wooden block of outer volume 1 litre and specific gravity \frac34 having a cavity floats with half of its volume immersed in water. Then the volume of the cavity is

A) 250 ml

B) 500 ml

C) 333 ml

D) 666 ml

View Answer

C) 333 ml
Explanation:Outer volume = 1 litre
Immersed volume = 1/2 litre
Weight of displaced water = weight of block
⇒ mass of block = 0.5 kg (since 1 litre water = 1 kg)
Specific gravity = 3/4 ⇒ density of wood = 0.75 kg/litre
Let actual volume of wood = V (excluding cavity)
Mass = density × volume
0.75 × V = 0.5
⇒ V = 0.5 / 0.75 = 2/3 litre
So cavity volume:
= total volume − wood volume
= 1 − 2/3 = 1/3 litre
= 333 ml
94) When ‘n’ identical mercury drops combine to form a single big drop

A) Surface area increases and heat is released

B) Surface area decreases and heat is released

C) Surface area increases and heat is absorbed

D) Surface area decreases and heat is absorbed

View Answer

B) Surface area decreases and heat is released
Explanation:Surface energy decreases → heat released
✔️ Answer: B
95) The temperature of a body shown by a faulty Celsius thermometer is 49 °C and by a correct Fahrenheit thermometer is 122 °F. The correction to be applied to the faulty thermometer is

A) -12°C

B) +1°C

C) +12°C

D) -1°C

View Answer

B) +1°C
Explanation:Convert 122°F to °C:
C = (5/9)(F − 32)
= (5/9)(122 − 32) = (5/9)(90) = 50°C
True temperature = 50°C
Faulty reading = 49°C
Correction = +1°C
96) If the radiation emitted by a perfect radiator has maximum intensity at a wavelength of 2900 Å, the intensity of radiation emitted by it is (Stefan-Boltzmann’s constant = 5.67 × 10⁻⁸ Wm⁻²K⁻⁴ and Wein’s constant = 2.9 × 10⁻³ mK)

A) 5.67×10^8 Wm⁻²

B) 5.67 Wm⁻²

C) 5670 Wm⁻²

D) 2.9 Wm⁻²

View Answer

A) 5.67×10^8 Wm⁻²
Explanation:λ_max = 2900 Å = 2.9×10⁻⁷ m
Using Wien’s law:
λT = 2.9×10⁻³
⇒ T = (2.9×10⁻³) / (2.9×10⁻⁷) = 10⁴ K
Using Stefan-Boltzmann law:
E = σT⁴
= 5.67×10⁻⁸ × (10⁴)⁴
= 5.67×10⁻⁸ × 10¹⁶
= 5.67×10⁸ W/m²
97) The ratio of the work done, change in internal energy and heat absorbed when a diatomic gas expands at constant pressure is

A) 2:3:5

B) 7:5:2

C) 5:3:2

D) 2:5:7

View Answer

D) 2:5:7
Explanation:Diatomic gas at constant pressure:
For diatomic gas:
C_v = (5/2)R
C_p = (7/2)R
Heat absorbed:
Q = C_p ΔT = (7/2)RΔT
Work done:
W = PΔV = RΔT
Change in internal energy:
ΔU = C_v ΔT = (5/2)RΔT
Ratio:
W : ΔU : Q = RΔT : (5/2)RΔT : (7/2)RΔT
Multiply by 2:
= 2 : 5 : 7
98) If the temperature of a gas is increased from 127 °C to 527 °C, then the rms speed of the gas molecules

A) increases by 4 times

B) becomes \sqrt{2} times

C) becomes half

D) decreases by \sqrt{2} times

View Answer

B) becomes \sqrt{2} times
Explanation:Temperature ratio:
400K → 800K
Speed ∝ √T ⇒ √2
✔️ Answer: B
99) An air column in a tube of length 50 cm, closed at one end is vibrating in its fifth harmoniC.The phase difference between a particle at the open end and a particle at 42 cm from the open end is

A) 90°

B) 180°

C) 0°

D) 270°

View Answer

C) 0°
Explanation:Closed pipe:
5th harmonic phase difference = 180°
✔️ Answer: B
100) A metal rod of length 125 cm is clamped at its midpoint. If the speed of the sound in the metal is 5000ms⁻¹, then the fundamental frequency of the longitudinal vibrations of the rod is

A) 2 kHz

B) 20 kHz

C) 0.2 kHz

D) 200 kHz

View Answer

A) 2 kHz
Explanation:Frequency:
f=\frac{v}{2L}
Compute:
2 kHz
✔️ Answer: A
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