Three forces of 10 N, 20 N, and 30 N act on a body in the same direction. What is the resultant force?

The correct answer is D: 60 N. When forces act in the same direction, the resultant is the sum of all forces: 10 + 20 + 30 = 60 N.

Three forces of 5 N, 12 N, and 13 N act on a particle. If they are in equilibrium, the angle between 5 N and 12 N forces is:

The correct answer is B: 90°. Since 5² + 12² = 25 + 144 = 169 = 13², these form a right triangle. The angle between 5 N and 12 N is 90°

Which statement best describes Newton's third law of motion?

The correct answer is B: Action and reaction forces are equal in magnitude and opposite in direction. Newton's third law states that for every action, there is an equal and opposite reaction force.

Five moles of an ideal monatomic gas undergo isothermal expansion at 500 K from 10 L to 50 L. Calculate the change in internal energy and entropy change. (R = 8.314 J/mol·K)

The correct answer is A: ΔU = 0, ΔS = 67.3 J/K. # Solution: Isothermal Expansion of Ideal Monatomic Gas In an isothermal process, temperature remains constant, which has important implications for internal energy and entropy changes. **Step 1: Change in Internal Energy** For any ideal gas, internal energy depends only on temperature; since temper

Two identical gases at the same temperature and pressure occupy different volumes. Which thermodynamic property must be the same for both?

The correct answer is D: Chemical potential per molecule. At same T and P, the chemical potential per molecule is identical. Internal energy and entropy depend on amount (volume), heat capacity depends on mass.

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NEET Physics

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

72 Q 3 Topics Take Mock Test

Difficulty: All Easy Medium Hard 1–10 of 72

Topics in NEET Physics

All Mechanics & Laws of Motion 100 Thermodynamics 100 Optics 1

Q.1 Easy

Let $a_1, a_2, a_3, \dots$ be a G.P. of increasing positive terms. If $a_1 a_5 = 28$ and $a_2 + a_4 = 29$, then $a_6$ is equal to:

A 628

B 812

C 526

D 784

Correct Answer: D. 784

EXPLANATION

For a G.P. with first term $a_1$ and common ratio $r$, the n-th term is $a_n = a_1 r^{n-1}$. We use the two given conditions to find $r$, then calculate $a_6$.

Step 1: Express terms using G.P. formula

a_1 = a_1, \quad a_2 = a_1 r, \quad a_4 = a_1 r^3, \quad a_5 = a_1 r^4, \quad a_6 = a_1 r^5

Step 2: Apply condition $a_1 a_5 = 28$

a_1 \cdot a_1 r^4 = 28

\[a_1^2 r^4 = 28\]

Since all terms are positive: $a_1 r^2 = \sqrt{28} = 2\sqrt{7}$

Step 3: Apply condition $a_2 + a_4 = 29$

\[a_1 r + a_1 r^3 = 29\]

\[a_1 r(1 + r^2) = 29\]

Step 4: Solve for $r$

Divide the equation from Step 3 by the result from Step 2:

\frac{a_1 r(1 + r^2)}{a_1 r^2} = \frac{29}{2\sqrt{7}}

\frac{1 + r^2}{r} = \frac{29}{2\sqrt{7}}

2\sqrt{7}(1 + r^2) = 29r

2\sqrt{7} r^2 - 29r + 2\sqrt{7} = 0

Using the quadratic formula (or inspection): $r = \frac{29}{2\sqrt{7}} \div \frac{2\sqrt{7}}{1}$ gives $r = \frac{\sqrt{7}}{2}$ or $r = 2\sqrt{7}$.

Since the sequence is increasing: $r = 2\sqrt{7}$

Step 5: Calculate $a_6$

From $a_1 r^2 = 2\sqrt{7}$:

a_6 = a_1 r^5 = (a_1 r^2) \cdot r^3 = 2\sqrt{7} \cdot (2\sqrt{7})^3

= 2\sqrt{7} \cdot 8 \cdot 7\sqrt{7} = 2\sqrt{7} \cdot 56\sqrt{7} = 2 \cdot 56 \cdot 7 = 784

Answer: $a_6 = 784$ (Option D)

Test

Q.2 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

Test

Q.3 Easy Thermodynamics

An isolated system undergoes an irreversible process. What happens to its total entropy?

A Decreases

B Increases

C Remains constant

D First increases then decreases

Correct Answer: B. Increases

EXPLANATION

According to the second law, entropy of an isolated system always increases for irreversible processes.

Test

Q.4 Easy Thermodynamics

For a spontaneous reaction at 25°C, which condition must be satisfied?

A ΔG > 0

B ΔG < 0

C ΔH < 0

D ΔS > 0

Correct Answer: B. ΔG < 0

EXPLANATION

For spontaneous reactions, ΔG must be negative (ΔG < 0). This is the criterion for spontaneity at constant T and P.

Test

Q.5 Easy Thermodynamics

When 1000 J of heat is supplied to a system and it does 400 J of work, what is the change in internal energy?

A 400 J

B 600 J

C 1400 J

D -600 J

Correct Answer: B. 600 J

EXPLANATION

First law: ΔU = Q - W = 1000 - 400 = 600 J

Test

Q.6 Easy Thermodynamics

An ideal gas at pressure P₁ and volume V₁ undergoes heating at constant volume until pressure becomes P₂. If P₂/P₁ = 2, what is the ratio of final to initial temperature?

A 1

B 2

C 0.5

D 4

Correct Answer: B. 2

EXPLANATION

For isochoric process: P₁/T₁ = P₂/T₂, so T₂/T₁ = P₂/P₁ = 2

Test

Q.7 Easy Thermodynamics

A 2 kg block of ice at 0°C is heated until it completely melts. The latent heat of fusion of ice is 3.36 × 10⁵ J/kg. What is the heat required?

A 3.36 × 10⁵ J

B 6.72 × 10⁵ J

C 1.68 × 10⁵ J

D 2.52 × 10⁵ J

Correct Answer: B. 6.72 × 10⁵ J

EXPLANATION

Q = mL = 2 × 3.36 × 10⁵ = 6.72 × 10⁵ J

Test

Q.8 Easy Thermodynamics

For a triatomic linear gas molecule, what is the value of Cp - Cv?

A R

B 1.5R

C 2R

D 2.5R

Correct Answer: A. R

EXPLANATION

For any ideal gas, Cp - Cv = R, regardless of molecular structure. This is derived from thermodynamic relations.

Test

Q.9 Easy Thermodynamics

A reversible heat engine operates between two reservoirs at temperatures T₁ = 400 K and T₂ = 300 K. What is the maximum possible efficiency of this engine?

A 25%

B 33.33%

C 50%

D 75%

Correct Answer: A. 25%

EXPLANATION

Maximum efficiency = 1 - (T₂/T₁) = 1 - (300/400) = 1 - 0.75 = 0.25 = 25%

Test

Q.10 Easy Thermodynamics

A gas expands isothermally from volume V₁ to V₂. The work done by the gas is W and heat absorbed is Q. What is the relationship between W and Q for an ideal gas?

A W = Q

B W > Q

C W < Q

D W = 2Q

Correct Answer: A. W = Q

EXPLANATION

In an isothermal process, ΔU = 0. By first law: Q = ΔU + W, therefore Q = W for an ideal gas.

Test

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