The atomic number of an element is equal to the number of MCQ with Answer and Explanation

The atomic number of an element is equal to the number of
A. Electrons in nucleus
B. Nucleons
C. Protons
D. Neutrons
Answer: Option C
Solution (By JKSSB Mock Tests)
Atomic number Z = number of protons (and electrons in neutral atom). Mass number = protons+neutrons.

Discuss this Question (0)

No comments yet. Be the first to start the discussion!

Practice More Physics Questions

Question #1
Assertion (A): The coefficient of kinetic friction is always strictly less than the coefficient of limiting static friction. Reason (R): Once motion starts, the inertia of rest is broken and interlocking of surface irregularities is less effective.
A. Both A and R are true but R is NOT the correct explanation of A.
B. A is false but R is true.
C. Both A and R are true and R is the correct explanation of A.
D. A is true but R is false.

Correct Answer: Option C


Explanation:
Static friction opposes impending motion, reaching a maximum called limiting friction. Once the object moves, it doesn't have enough time for the microscopic irregularities (asperities) of the two surfaces to interlock strongly. Thus, the force required to keep it moving (kinetic friction) is slightly less than the force required to start the motion.

This question belongs to: Science Physics
Question #2
A physical quantity is measured as 12.345 ± 0.005 units. The relative error in this measurement is approximately:
A. 0.4%
B. 4%
C. 0.004%
D. 0.04%

Correct Answer: Option D


Explanation:
Relative error = (Absolute error) / (Measured value) = 0.005 / 12.345 ≈ 0.000405. Converting to percentage: 0.000405 × 100% ≈ 0.0405% ≈ 0.04%. Relative error quantifies precision independent of measurement scale. Absolute error (±0.005) indicates instrument precision. This concept is vital for comparing measurement reliability across different magnitudes. Memory tip: Percentage error = (Absolute error / True value) × 100. Competitive exams often test this calculation with varying significant figures to assess numerical proficiency.

This question belongs to: Science Physics
Question #3
A bullet of mass 20 g is fired from a gun of mass 2 kg with velocity 300 m/s. The recoil velocity of the gun is:
A. 3 m/s
B. 0.3 m/s
C. 6 m/s
D. 0.6 m/s

Correct Answer: Option A


Explanation:
By conservation of momentum: initial momentum = 0 (system at rest). Final momentum: m_bullet×v_bullet + m_gun×v_gun = 0. Thus (0.02 kg)(300 m/s) + (2 kg)(v_gun) = 0 ⇒ 6 + 2v_gun = 0 ⇒ v_gun = -3 m/s. Magnitude is 3 m/s; negative sign indicates opposite direction to bullet. This demonstrates momentum conservation in isolated systems. Memory tip: Recoil velocity = -(m_bullet/m_gun)×v_bullet. Such numerical problems test application of conservation laws, frequently appearing in competitive exams with varying mass ratios and velocities.

This question belongs to: Science Physics