During photosynthesis, the lumenal pH of thylakoids drops to approximately:

The correct answer is B: 4-5. Protons accumulate in the thylakoid lumen (pH 4-5) due to water photolysis and electron transport, creating a chemiosmotic gradient for ATP synthesis. The stroma maintains pH ~8. This proton gradient is essential for understanding light reactions.

A car accelerates from rest at 2 m/s². How long does it take to reach 20 m/s?

The correct answer is B: 10 seconds. **This problem uses the kinematic equation relating velocity, acceleration, and time for motion starting from rest.** **Step 1: Identify the Given Values** We are given the initial velocity (starting from rest), final velocity, and acceleration. We need to find the time taken. $$v_i = 0 \text{ m/s},

Which of the following statements about stomata is correct?

The correct answer is B: Stomata close at night to conserve water. Stomata generally close at night (except CAM plants) to reduce transpiration. Guard cells are rich in mitochondria and chloroplasts to provide ATP and regulate osmotic potential. This circadian regulation is fundamental to plant physiology.

A wire of resistance R is stretched to double its length. What is the new resistance?

The correct answer is C: 4R. Resistance R = ρL/A. When length doubles, L becomes 2L and area becomes A/2 (volume constant). New R = ρ(2L)/(A/2) = 4ρL/A = 4R

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Difficulty: All Easy Medium Hard 21–30 of 96

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Q.21 Medium Physics

A spring with spring constant k = 100 N/m is compressed by 0.1 m. What is the elastic potential energy stored?

A 0.5 J

B 1 J

C 5 J

D 10 J

Correct Answer: A. 0.5 J

EXPLANATION

Elastic PE = ½kx² = ½ × 100 × (0.1)² = ½ × 100 × 0.01 = 0.5 J.

Test

Q.22 Medium Physics

A block slides down a frictionless incline of angle 30°. What is its acceleration down the incline?

A 4.9 m/s²

B 8.66 m/s²

C 9.8 m/s²

D 10 m/s²

Correct Answer: A. 4.9 m/s²

EXPLANATION

When a block slides down a frictionless incline, gravity's component along the slope determines the acceleration.

Step 1: [Identify the Force Component Along the Incline]

On an incline, gravity acts vertically downward, but only its component parallel to the slope causes acceleration down the incline.

F_{\parallel} = mg\sin(\theta)

Step 2: [Apply Newton's Second Law]

Using F = ma, the net force along the incline equals mass times acceleration. The mass cancels out, giving us a pure kinematic result.

a = g\sin(\theta) = 9.8 \times \sin(30°) = 9.8 \times 0.5 = 4.9 \text{ m/s}^2

The acceleration down the incline is 4.9 m/s² (Answer: A)

Test

Q.23 Medium Physics

Two identical objects moving with velocities v and 2v collide and stick together. What fraction of kinetic energy is lost?

A 1/3

B 2/3

C 1/2

D 2/5

Correct Answer: B. 2/3

EXPLANATION

Initial KE = ½m(v)² + ½m(2v)² = 2.5mv². Final velocity = 3v/2 (by momentum conservation). Final KE = ½(2m)(3v/2)² = 2.25mv²/4 = 0.84mv². Loss = 2.5 - 0.84 = 1.66, which is 2/3 of initial KE.

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Q.24 Medium Physics

A conducting sphere is given a charge of 8 μC. What is the electric field at its surface if the radius is 2 cm?

A 1.8 × 10⁶ N/C

B 3.6 × 10⁷ N/C

C 1.8 × 10⁷ N/C

D 7.2 × 10⁷ N/C

Correct Answer: C. 1.8 × 10⁷ N/C

EXPLANATION

Electric field at surface: E = kQ/r² = (9 × 10⁹ × 8 × 10⁻⁶)/(0.02)² = (72)/(4 × 10⁻⁴) = 1.8 × 10⁷ N/C.

Test

Q.25 Medium Physics

A submarine sonar uses ultrasound with frequency 50 kHz and speed 1500 m/s in water. What is the wavelength?

A 0.03 m

B 0.3 m

C 3 m

D 30 m

Correct Answer: A. 0.03 m

EXPLANATION

Wavelength λ = v/f = 1500/(50 × 10³) = 1500/50000 = 0.03 m = 3 cm. Higher frequencies in denser mediums produce shorter wavelengths.

Test

Q.26 Medium Physics

An object undergoes simple harmonic motion with amplitude 5 cm and frequency 2 Hz. What is the maximum velocity?

A 10π cm/s

B 20π cm/s

C 5π cm/s

D 2.5π cm/s

Correct Answer: B. 20π cm/s

EXPLANATION

Maximum velocity in SHM: v_max = ωA = 2πfA = 2π × 2 × 5 = 20π cm/s ≈ 62.8 cm/s.

Test

Q.27 Medium Physics

A radioactive element has a half-life of 10 years. What fraction of the original sample remains after 30 years?

A 1/2

B 1/4

C 1/8

D 3/4

Correct Answer: C. 1/8

EXPLANATION

After 30 years = 3 half-lives: N = N₀ × (1/2)³ = N₀/8. The sample reduces by half three times successively.

Test

Q.28 Medium Physics

A 10 m long ladder leans against a vertical wall at 60° to the ground. How high does it reach on the wall?

A 5 m

B 5√3 m or 8.66 m

C 10√3 m

D 10/√3 m

Correct Answer: B. 5√3 m or 8.66 m

EXPLANATION

This problem involves using trigonometry to find the height of a ladder against a wall.

Step 1: [Identify the Right Triangle]

The ladder, wall, and ground form a right triangle where the ladder is the hypotenuse, the wall height is the opposite side to the 60° angle, and the ground distance is the adjacent side.

\text{Ladder (hypotenuse)} = 10 \text{ m, Angle with ground} = 60°

Step 2: [Apply Sine Ratio]

Since we need the height (opposite side) and have the hypotenuse and angle, we use the sine function: sine of the angle equals opposite divided by hypotenuse.

\sin(60°) = \frac{\text{Height}}{10}

Step 3: [Calculate the Height]

Substitute the value of sin(60°) = √3/2 and solve for height.

\text{Height} = 10 \times \sin(60°) = 10 \times \frac{\sqrt{3}}{2} = \frac{10\sqrt{3}}{2} = 5\sqrt{3} \text{ m}

Step 4: [Convert to Decimal]

Calculate the approximate decimal value: 5√3 ≈ 5 × 1.732 = 8.66 m

The ladder reaches a height of 5√3 m or approximately 8.66 m on the wall.

Answer: (B) 5√3 m or 8.66 m

Test

Q.29 Medium Physics

A concave mirror has a focal length of 20 cm. An object is placed at 40 cm from the mirror. What type of image is formed?

A Virtual, erect, magnified

B Real, inverted, same size

C Virtual, erect, diminished

D Real, inverted, diminished

Correct Answer: B. Real, inverted, same size

EXPLANATION

Object at 2f (40 cm = 2 × 20 cm) forms image at 2f. Using mirror equation confirms v = 40 cm, m = -1 (real, inverted, same size as object).

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Q.30 Medium Physics

A wire of length L and cross-sectional area A has resistance R. If the wire is cut into two equal halves and connected in parallel, what is the new resistance?

A R/4

B R/2

C R

D 4R

Correct Answer: A. R/4

EXPLANATION

Each half has resistance R/2. When connected in parallel: 1/R_new = 1/(R/2) + 1/(R/2) = 4/R, so R_new = R/4.

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