A person cannot see objects clearly beyond 2 m. This defect of vision is: MCQ with Answer and Explanation

A person cannot see objects clearly beyond 2 m. This defect of vision is:
A. Hypermetropia
B. Astigmatism
C. Presbyopia
D. Myopia
Answer: Option D
Solution (By JKSSB Mock Tests)
Myopia (nearsightedness) is the inability to see distant objects clearly, with far point less than infinity. Here, far point is 2 m. Corrected using concave lens. Hypermetropia (farsightedness) affects near vision; presbyopia is age-related loss of accommodation; astigmatism involves uneven corneal curvature. Memory aid: 'Myopia = near-sighted; far point reduced'. This application question tests knowledge of vision defects and corrections, frequently appearing in competitive exams. Always link symptom (blurred distant vision) to defect (myopia) and correction (diverging lens).

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Practice More Physics Questions

Question #1
In a potentiometer experiment, the balancing length for a cell is 60 cm. When the cell is shunted by a 2Ω resistor, the balancing length becomes 40 cm. The internal resistance of the cell is:
A. 3Ω
B. 4Ω
C. 2Ω
D. 1Ω

Correct Answer: Option D


Explanation:
Potentiometer measures EMF and internal resistance. Formula: r = R(L₁ - L₂)/L₂, where R is shunt resistance, L₁ initial balancing length, L₂ after shunting. Here R=2Ω, L₁=60 cm, L₂=40 cm. Thus r = 2×(60-40)/40 = 2×20/40 = 1Ω. This derives from terminal voltage V = E - Ir and potentiometer principle V ∝ balancing length. Memory tip: 'r = R(L₁/L₂ - 1)'. This advanced numerical tests potentiometer applications, common in competitive exams for experimental physics. Always verify the formula derivation to avoid misapplication.

This question belongs to: Science Physics
Question #2
A body of volume 100 cm³ and density 0.8 g/cm³ floats in water. Volume submerged is
A. 20 cm³
B. 100 cm³
C. 80 cm³
D. 50 cm³

Correct Answer: Option C


Explanation:
Weight = buoyancy => ρ_obj V_total g = ρ_water V_sub g => 0.8×100 = 1×V_sub => V_sub = 80 cm³.

This question belongs to: Science Physics
Question #3
Atmospheric pressure at sea level is 1.013×10⁵ Pa. Equivalent height of mercury column (density 13600 kg/m³, g=9.8) is
A. 0.76 m
B. 7.6 m
C. 0.76 cm
D. 13.6 m

Correct Answer: Option A


Explanation:
P = ρgh => h = P/(ρg) = 1.013×10⁵/(13600×9.8) ≈ 0.76 m = 76 cm.

This question belongs to: Science Physics