Derivatives & Integrals in Physics — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Derivatives & Integrals in Physics MCQs with step-by-step solutions (20 questions). Part of Mathematical Methods (Std 11). Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Derivatives & Integrals in Physics · easy · theory
The derivative of a quantity with respect to time represents its:
A. Rate of change ✓ Correct
B. Average value
C. Maximum value
D. Total accumulated value
Solution: Differentiation answers how fast something is changing at an instant.
Q2 — Derivatives & Integrals in Physics · easy · theory
Instantaneous velocity is defined as:
A. $\dfrac{x}{t}$
B. $\int x\,dt$
C. $\dfrac{dv}{dt}$
D. $\dfrac{dx}{dt}$ ✓ Correct
Solution: It is the limit of the average velocity as the time interval tends to zero.
Q3 — Derivatives & Integrals in Physics · easy · theory
Instantaneous acceleration is defined as:
A. $\dfrac{dx}{dt}$
B. $\int v\,dt$
C. $\dfrac{v}{t}$
D. $\dfrac{dv}{dt}$ ✓ Correct
Solution: It may also be written $\dfrac{d^2x}{dt^2}$.
Q4 — Derivatives & Integrals in Physics · easy · theory
The slope of a displacement-time graph at any point gives the:
A. Instantaneous acceleration
B. Total distance
C. Instantaneous velocity ✓ Correct
D. Average speed
Solution: A steeper graph means a faster motion.
Q5 — Derivatives & Integrals in Physics · easy · theory
The slope of a velocity-time graph at any point gives the:
A. Displacement
B. Instantaneous velocity
C. Distance travelled
D. Instantaneous acceleration ✓ Correct
Solution: A horizontal velocity-time graph therefore indicates zero acceleration.
Q6 — Derivatives & Integrals in Physics · medium · theory
The area under a velocity-time graph gives the:
A. Speed
B. Force
C. Acceleration
D. Displacement ✓ Correct
Solution: Integration accumulates the small displacements $v\,dt$ over the interval.
Q7 — Derivatives & Integrals in Physics · medium · theory
The work done by a variable force along a straight line is given by:
A. $Fx$ always
B. $\dfrac{F}{x}$
C. $\int F\,dx$ ✓ Correct
D. $\dfrac{dF}{dx}$
Solution: For a constant force this reduces to the familiar $W = Fx$.
Q8 — Derivatives & Integrals in Physics · medium · theory
Integration is the mathematical operation that:
A. Finds the rate of change of a quantity
B. Converts a vector into a scalar
C. Accumulates a quantity over an interval, being the reverse of differentiation ✓ Correct
D. Finds the maximum of a function
Solution: The definite integral corresponds to the area under a curve.
Q9 — Derivatives & Integrals in Physics · medium · numerical
The displacement of a particle varies as $x = 5t^2$ metre. Its velocity at $t = 2\text{ s}$ is:
A. $10\text{ m/s}$
B. $20\text{ m/s}$ ✓ Correct
C. $5\text{ m/s}$
D. $40\text{ m/s}$
Solution: $v = \dfrac{dx}{dt} = 10t$, so at $t = 2\text{ s}$, $v = 20\text{ m/s}$.
Q10 — Derivatives & Integrals in Physics · medium · numerical
The velocity of a particle varies as $v = 3t^2 + 2t$. Its acceleration at $t = 1\text{ s}$ is:
A. $5\text{ m/s}^2$
B. $2\text{ m/s}^2$
C. $8\text{ m/s}^2$ ✓ Correct
D. $6\text{ m/s}^2$
Solution: $a = \dfrac{dv}{dt} = 6t + 2$, so at $t = 1\text{ s}$, $a = 8\text{ m/s}^2$.
Q11 — Derivatives & Integrals in Physics · medium · numerical
The displacement of a particle varies as $x = t^3$ metre. Its velocity at $t = 2\text{ s}$ is:
A. $6\text{ m/s}$
B. $12\text{ m/s}$ ✓ Correct
C. $4\text{ m/s}$
D. $8\text{ m/s}$
Solution: $v = 3t^2$, so at $t = 2\text{ s}$, $v = 12\text{ m/s}$.
Q12 — Derivatives & Integrals in Physics · hard · numerical
The displacement of a particle varies as $x = 2t^3 - 3t^2$. Its velocity at $t = 2\text{ s}$ is:
A. $6\text{ m/s}$
B. $4\text{ m/s}$
C. $12\text{ m/s}$ ✓ Correct
D. $24\text{ m/s}$
Solution: $v = 6t^2 - 6t$, so at $t = 2\text{ s}$, $v = 24 - 12 = 12\text{ m/s}$.
Q13 — Derivatives & Integrals in Physics · hard · numerical
A particle moves with velocity $v = 4t\text{ m/s}$. Its displacement between $t = 0$ and $t = 2\text{ s}$ is:
A. $16\text{ m}$
B. $2\text{ m}$
C. $8\text{ m}$ ✓ Correct
D. $4\text{ m}$
Solution: $s = \int_0^2 4t\,dt = \left[2t^2\right]_0^2 = 8\text{ m}$.
Q14 — Derivatives & Integrals in Physics · hard · numerical
A variable force $F = 2x\text{ N}$ acts on a body moving from $x = 0$ to $x = 3\text{ m}$. The work done is:
A. $9\text{ J}$ ✓ Correct
B. $6\text{ J}$
C. $3\text{ J}$
D. $18\text{ J}$
Solution: $W = \int_0^3 2x\,dx = \left[x^2\right]_0^3 = 9\text{ J}$.
Q15 — Derivatives & Integrals in Physics · medium · numerical
The velocity of a particle is $v = (10 - 2t)\text{ m/s}$. It comes momentarily to rest at:
A. $t = 5\text{ s}$ ✓ Correct
B. $t = 10\text{ s}$
C. $t = 20\text{ s}$
D. $t = 2\text{ s}$
Solution: Setting $10 - 2t = 0$ gives $t = 5\text{ s}$.
Q16 — Derivatives & Integrals in Physics · medium · numerical
A particle moves with velocity $v = 5t\text{ m/s}$. Its acceleration is:
A. $5\text{ m/s}^2$, constant ✓ Correct
B. Zero
C. $5t\text{ m/s}^2$
D. $10\text{ m/s}^2$
Solution: $a = \dfrac{dv}{dt} = 5\text{ m/s}^2$, independent of time.
Q17 — Derivatives & Integrals in Physics · medium · numerical
The displacement of a particle varies as $x = 4t^2 + 2t$. Its velocity at $t = 3\text{ s}$ is:
A. $24\text{ m/s}$
B. $42\text{ m/s}$
C. $14\text{ m/s}$
D. $26\text{ m/s}$ ✓ Correct
Solution: $v = 8t + 2$, so at $t = 3\text{ s}$, $v = 24 + 2 = 26\text{ m/s}$.
Q18 — Derivatives & Integrals in Physics · easy · numerical
A body moves at a constant velocity. The slope of its velocity-time graph is:
A. Constant and non-zero
B. Increasing with time
C. Zero ✓ Correct
D. Negative
Solution: A constant velocity means zero acceleration, and acceleration is that slope.
Q19 — Derivatives & Integrals in Physics · medium · numerical
A velocity-time graph is a straight line through the origin. The motion has:
A. Zero acceleration
B. Constant velocity
C. Constant acceleration ✓ Correct
D. Increasing acceleration
Solution: A constant slope means a uniform rate of change of velocity.
Q20 — Derivatives & Integrals in Physics · medium · numerical
The displacement of a particle is $x = 6t$ metre. Its acceleration is:
A. $3\text{ m/s}^2$
B. $6\text{ m/s}^2$
C. Zero ✓ Correct
D. $12\text{ m/s}^2$
Solution: $v = 6\text{ m/s}$ is constant, so $a = \dfrac{dv}{dt} = 0$.