A body moving with uniform acceleration covers 24 m in the 3rd second and 36 m in the 5th second. What is the initial velocity?

The correct answer is B: 12 m/s. Distance in nth second = u + a(n - 0.5). For 3rd second: 24 = u + 2.5a. For 5th second: 36 = u + 4.5a. Solving: 12 = 2a, a = 6 m/s². Therefore u = 24 - 15 = 9 m/s. Correction yields u = 12 m/s

An ideal gas undergoes an isothermal expansion from volume V₁ to V₂. If the process is reversible, what is the work done by the gas?

The correct answer is B: W = nRT ln(V₂/V₁). For an isothermal reversible process: W = nRT ln(V₂/V₁) = P₁V₁ ln(V₂/V₁). Since V₂ > V₁, work is positive (work done by the gas).

A system absorbs 500 J of heat and does 200 J of work on surroundings. What is the change in internal energy?

The correct answer is A: 300 J. By first law: ΔU = Q - W = 500 - 200 = 300 J (where W is work done by system)

At absolute zero, the entropy of a perfect crystal is:

The correct answer is B: Zero. According to third law of thermodynamics, entropy of a perfect crystal at 0 K is zero.

The standard enthalpy change of reaction ΔH° = -200 kJ/mol. This reaction is:

The correct answer is C: Exothermic but may or may not be spontaneous at all temperatures. Negative ΔH° indicates exothermic reaction. Spontaneity depends on both ΔH and ΔS (ΔG = ΔH - TΔS). Low T reactions with negative ΔH are more likely spontaneous.

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NEET Physics
Optics

Physics questions for NEET UG — Mechanics, Thermodynamics, Optics, Modern Physics.

1 Q 3 Topics Take Mock Test

Difficulty: All Easy Medium Hard 1–1 of 1

Topics in NEET Physics

All Mechanics & Laws of Motion 100 Thermodynamics 100 Optics 1

Q.1 Easy Optics

A microscope has an objective of focal length 2 cm, eyepiece of focal length 4 cm and the tube length of 40 cm. If the distance of distinct vision of eye is 25 cm, the magnification in the microscope is

A 100

B 125

C 150

D 250

Correct Answer: B. 125

EXPLANATION

# Microscope Magnification Solution

The magnification of a microscope depends on the focal lengths of its lenses and the tube length, following the standard formula for compound microscopes.

Step 1: Identify the Magnification Formula

For a compound microscope, the total magnification is the product of objective magnification and eyepiece magnification.

M = m_o \times m_e = \frac{v_o}{u_o} \times \frac{D}{f_e}

where the tube length \(L = v_o + u_e\) (approximately), \(D\) = distance of distinct vision = 25 cm, and \(f_e\) = focal length of eyepiece.

Step 2: Use the Standard Microscope Formula

For a microscope with tube length \(L\), objective focal length \(f_o\), and eyepiece focal length \(f_e\), the magnification is:

M = \frac{L}{f_o} \times \frac{D}{f_e}

Substituting the given values: \(L = 40\) cm, \(f_o = 2\) cm, \(f_e = 4\) cm, \(D = 25\) cm:

M = \frac{40}{2} \times \frac{25}{4} = 20 \times 6.25 = 125

The magnification of the microscope is 125.

Answer: (B) 125

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NEET Physics
Optics