MOTION IN A STRAIGHT LINE MCQs for NEET UG PHYSICS — Mixed difficulty

1 Medium

Understanding +4 | -1

Velocity-time curve for a body projected vertically upwards is

View answer and explanation

Correct option: D

Straight line

Option D (Straight line) is correct because the velocity changes linearly with time due to constant acceleration (gravity).

Solution approach

Correct Answer + Core Reason: Straight line. The velocity changes linearly with time.

Step-by-Step Solution:

Step 1: The body is projected vertically upwards, so it experiences constant acceleration due to gravity (g).

Step 2: The equation relating velocity (v) and time (t) is $v = u + at$ , where u is the initial velocity and a is the acceleration.

Step 3: In this case, $a = -g$ (negative because it acts downwards). So, $v = u - gt$ .

Step 4: This equation represents a straight line with a slope of -g and a y-intercept of u.

Why other options are wrong:
- Option A (Parabola): This is incorrect because a parabola represents a quadratic relationship, not a linear one.
- Option B (Ellipse): This is incorrect because an ellipse is a closed curve, not a straight line.
- Option C (Hyperbola): This is incorrect because a hyperbola represents an inverse relationship, not a linear one.

Key Concepts:
- Velocity-time graph
- Acceleration due to gravity
- Linear equation

Memory Aids:
- Remember that a constant acceleration results in a linear velocity-time graph.

Quick Tips:
- Always relate the shape of the graph to the equation of motion.

2 Medium

2 of 192Question permalink

Understanding +4 | -1

Two cars A and B approach each other at the same speed, then what will be the velocity of A if velocity of B is

8 \mathrm{~m} / \mathrm{s}

?

View answer and explanation

Correct option: C

-8 \mathrm{~m} / \mathrm{s}

Option C (-8 m/s) is correct. Since the cars approach each other at the same speed, if the velocity of B is 8 m/s, the velocity of A must be -8 m/s, considering opposite directions.

Solution approach

Correct Answer + Core Reason: -8 m/s. Since the cars approach each other at the same speed, their velocities have the same magnitude but opposite signs.

Step-by-Step Solution:

Step 1: Define the variables: Let $v_A$ be the velocity of car A and $v_B$ be the velocity of car B.

Step 2: Given: The cars approach each other at the same speed, and $v_B = 8 \text{ m/s}$ .

Step 3: Since they are approaching each other, their velocities must have opposite signs. Therefore, $v_A = -v_B$ .

Step 4: Substitute the value of $v_B$ : $v_A = -8 \text{ m/s}$ .

Why other options are wrong:
- Option A: $16 \text{ m/s}$ : This is incorrect because it doesn't account for the opposite direction of the velocities.
- Option B: $8 \text{ m/s}$ : This is incorrect because it implies that the cars are moving in the same direction, not approaching each other.
- Option D: Can't be determined: The velocity can be determined based on the given information.

Key Concepts:
- Relative velocity

Memory Aids:
- Remember that velocities in opposite directions have opposite signs.

Quick Tips:
- Pay attention to the direction of motion when dealing with velocities.

3 Medium

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Application +4 | -1

Two cars are moving in the same direction with a speed of

30\mathrm{km}{\mathrm{h}^{ - 1}}

. They are separated from each other by 5 km. Third car moving in the opposite direction meets the two cars after an interval of 4 min. What is the speed of the third car?

View answer and explanation

Correct option: D

\mathrm{45km}{\mathrm{h}^{ - 1}}

Option D (

\mathrm{45km}{\mathrm{h}^{ - 1}}

) is correct because it is calculated by using the concept of relative speed and the given time interval.

Solution approach

Correct Answer + Core Reason: $\mathrm{45km}{\mathrm{h}^{ - 1}}$ . The speed of the third car is calculated using the concept of relative speed and the given time interval.

Step-by-Step Solution:

Step 1: Let the speed of the third car be $v$ . The relative speed of the third car with respect to the other two cars is $v + 30 \;\mathrm{km}{\mathrm{h}^{ - 1}}$ .

Step 2: The distance between the first two cars is 5 km. The third car meets the two cars after an interval of 4 minutes, which is $\frac{4}{{60}} = \frac{1}{{15}}$ hours.

Step 3: The time difference between meeting the two cars is given by:
$\frac{5}{{v + 30}} = \frac{1}{{15}}$

Step 4: Solve for $v$ :
$5 \times 15 = v + 30$
$75 = v + 30$
$v = 75 - 30 = 45 \;\mathrm{km}{\mathrm{h}^{ - 1}}$

Why other options are wrong:
- Option A: Incorrect calculation of the speed of the third car.
- Option B: Incorrect calculation of the speed of the third car.
- Option C: Incorrect calculation of the speed of the third car.

Key Concepts:
- Relative velocity
- Time, speed, and distance relationship

Memory Aids:
- When objects move in opposite directions, their relative speeds add up.

Quick Tips:
- Always consider the relative motion when dealing with moving objects.

4 Easy

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Application +4 | -1

A body is projected vertically up with ‘u’. Its velocity at half its maximum height is

View answer and explanation

Correct option: D

u/\sqrt 2

Option D (

u/\sqrt 2

) is correct because the velocity at half the maximum height is calculated using conservation of energy or kinematic equations.

Solution approach

Correct Answer + Core Reason: $u/\sqrt 2$ . The velocity at half the maximum height is found using kinematics.

Step-by-Step Solution:

Step 1: Let $H$ be the maximum height. At maximum height, the velocity is 0. Using the kinematic equation $v^2 = u^2 + 2as$ , we have $0 = u^2 - 2gH$ , so $H = \frac{u^2}{2g}$ .

Step 2: Now, consider the point at half the maximum height, i.e., $h = \frac{H}{2} = \frac{u^2}{4g}$ . Let the velocity at this point be $v'$ . Using the same kinematic equation, we have $v'^2 = u^2 - 2gh = u^2 - 2g(\frac{u^2}{4g}) = u^2 - \frac{u^2}{2} = \frac{u^2}{2}$ .

Step 3: Solve for $v'$ : $v' = \sqrt{\frac{u^2}{2}} = \frac{u}{\sqrt{2}}$ .

Why other options are wrong:
- Option A: $\frac{u}{2}$ : This is incorrect. It might be a guess or a misunderstanding of the relationship between velocity and height.
- Option B: $\frac{{{u^2}}}{2}$ : This is incorrect. This has units of velocity squared, not velocity.
- Option C: $\sqrt 2 u$ : This is incorrect. The velocity should decrease as the object moves upwards.

Key Concepts:
- Kinematic Equations of Motion
- Vertical Projectile Motion

Memory Aids:
- Remember that at maximum height, the velocity is zero.

Quick Tips:
- Use the appropriate kinematic equation based on the given information.
- Consider the sign of acceleration (positive or negative) carefully.

5 Easy

5 of 192Question permalink

Application +4 | -1

A scooter starts from rest have an acceleration of

1 \mathrm{~ms}^{-2}

while a car

150 \mathrm{~m}

behind it starts from rest with an acceleration of

2 \mathrm{~ms}^{-2}

. After how much time, the car catches up with the scooter?

View answer and explanation

Correct option: B

\sqrt{300} \mathrm{~s}

Option B (

\sqrt{300} \mathrm{~s}

) is correct because it represents the time when the distance traveled by the car equals the distance traveled by the scooter plus the initial separation.

Solution approach

Correct Answer + Core Reason: $\sqrt{300} \mathrm{~s}$ . The time is calculated by equating the distances traveled by the car and the scooter, considering the initial separation.

Step-by-Step Solution:

Step 1: Let $t$ be the time when the car catches up with the scooter.

Step 2: Distance traveled by the scooter: $s_1 = \frac{1}{2} a_1 t^2 = \frac{1}{2} (1) t^2 = \frac{1}{2} t^2$

Step 3: Distance traveled by the car: $s_2 = \frac{1}{2} a_2 t^2 = \frac{1}{2} (2) t^2 = t^2$

Step 4: The car catches up with the scooter when $s_2 = s_1 + 150$ . Therefore, $t^2 = \frac{1}{2} t^2 + 150$ .

Step 5: Solving for $t^2$ : $t^2 - \frac{1}{2} t^2 = 150$ , which gives $\frac{1}{2} t^2 = 150$ .

Step 6: Multiply both sides by 2: $t^2 = 300$ .

Step 7: Take the square root: $t = \sqrt{300} \mathrm{~s}$

Why other options are wrong:
- Option A: This option is incorrect. The value $\sqrt{700} \mathrm{~s}$ does not satisfy the distance equation derived from the problem statement.
- Option C: This option is incorrect. The value $\sqrt{150} \mathrm{~s}$ represents only half the time required for the car to catch the scooter.
- Option D: This option is incorrect since one of the other options is correct.

Key Concepts:
- Kinematics
- Constant acceleration
- Relative motion

Memory Aids:
- Remember that the distance traveled is proportional to the square of the time when acceleration is constant.

Quick Tips:
- Always define the variables clearly before setting up the equations.
- Draw a diagram to visualize the problem.

6 Easy

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Application +4 | -1

A motor vehicle travelled the first one third of a distance ‘S ’ at a speed of

{V_1} = 10\,kmph

, the second one third at a speed of

{V_2} = 20\,kmph

and the last one third at a speed of

{V_3} = 60\,kmph

determine the mean speed of the vehicle over the entire distance ‘S ’

View answer and explanation

Correct option: D

18 kmph

Option D (18 kmph) is correct because it is the harmonic mean of the three speeds, calculated appropriately for equal distances.

Solution approach

Correct Answer + Core Reason: 18 kmph. The mean speed is calculated using the harmonic mean formula for equal distances.

Step-by-Step Solution:

Step 1: Define the distances. Let each one-third distance be $d$ . Total distance $S = 3d$ .

Step 2: Calculate the time taken for each segment.
$t_1 = \frac{d}{V_1} = \frac{d}{10}$
$t_2 = \frac{d}{V_2} = \frac{d}{20}$
$t_3 = \frac{d}{V_3} = \frac{d}{60}$

Step 3: Calculate the total time taken: $t = t_1 + t_2 + t_3 = \frac{d}{10} + \frac{d}{20} + \frac{d}{60} = \frac{6d + 3d + d}{60} = \frac{10d}{60} = \frac{d}{6}$

Step 4: Calculate the average speed: $V_{avg} = \frac{S}{t} = \frac{3d}{\frac{d}{6}} = 3d \times \frac{6}{d} = 18 \text{{ kmph}}$

Why other options are wrong:
- Option A (15 kmph): This is the arithmetic mean of the speeds, which is incorrect when distances are equal, not times.
- Option B (12 kmph): This is a random incorrect calculation.
- Option C (10 kmph): This is the speed of the first segment and doesn't represent the average speed.

Key Concepts:
- Average speed
- Harmonic mean (for equal distances)

Memory Aids:
- Remember that average speed is total distance divided by total time.

Quick Tips:
- When distances are equal, use the harmonic mean to calculate average speed. The harmonic mean of $n$ numbers $x_1, x_2, ..., x_n$ is given by $\frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_n}}$

7 Hard

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Application +4 | -1

A particle is projected with velocity

{v_0}

along x-axis.The deceleration on the particle is proportional to the square of the distance from the origin i.e.,

a = - a{x^2}.

The distance at which the particle stops is

View answer and explanation

Correct option: D

{\left( {\frac{{3v_0^2}}{{2\alpha }}} \right)^{\frac{1}{3}}}

Option D (

{\left( {\frac{{3v_0^2}}{{2\alpha }}} \right)^{\frac{1}{3}}}

) is correct because it is the distance at which the particle stops, derived by integrating the given deceleration function.

Solution approach

Correct Answer + Core Reason: ${\left( {\frac{{3v_0^2}}{{2\alpha }}} \right)^{\frac{1}{3}}}$ . The distance is found by integrating the deceleration function and applying the condition that the particle stops.

Step-by-Step Solution:

Step 1: Given $a = -\alpha x^2$ , we can write $a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$ .

Step 2: So, $v \frac{dv}{dx} = -\alpha x^2$ . Integrating both sides with respect to $x$ , we get $\int_{v_0}^{0} v dv = \int_{0}^{x} -\alpha x^2 dx$ .

Step 3: Evaluating the integrals, we have $\left[ \frac{v^2}{2} \right]_{v_0}^{0} = \left[ -\frac{\alpha x^3}{3} \right]$

8 Easy

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Application +4 | -1

A body starts from rest with uniform acceleration

a.

If its velocity after

n

second is

v,

then its displacement in the last 2 s is

View answer and explanation

Correct option: D

\frac{{2v\left( {n - 1} \right)}}{n}

Option D (

\frac{{2v\left( {n - 1} \right)}}{n}

) is correct because the displacement in the last 2 seconds can be found by subtracting the displacement in (n-2) seconds from the displacement in n seconds.

Solution approach

Correct Answer + Core Reason: $\frac{{2v\left( {n - 1} \right)}}{n}$ . The displacement in the last 2 seconds is found by subtracting displacements.

Step-by-Step Solution:

Step 1: The body starts from rest with uniform acceleration a.

Step 2: The velocity after n seconds is v. Using the equation $v = u + at$ , where u = 0, we have $v = an$ . Therefore, $a = \frac{v}{n}$ .

Step 3: The displacement in n seconds is ${s_n} = ut + \frac{1}{2}a{t^2} = 0 + \frac{1}{2}a{n^2} = \frac{1}{2}(\frac{v}{n}){n^2} = \frac{1}{2}vn$ .

Step 4: The displacement in (n-2) seconds is ${s_{n - 2}} = \frac{1}{2}a{(n - 2)^2} = \frac{1}{2}(\frac{v}{n}){(n - 2)^2} = \frac{v}{{2n}}{(n^2 - 4n + 4)}$ .

Step 5: The displacement in the last 2 seconds is ${s_n} - {s_{n - 2}} = \frac{1}{2}vn - \frac{v}{{2n}}{(n^2 - 4n + 4)} = \frac{{vn^2 - v(n^2 - 4n + 4)}}{{2n}} = \frac{{vn^2 - vn^2 + 4vn - 4v}}{{2n}} = \frac{{4vn - 4v}}{{2n}} = \frac{{2v(n - 1)}}{n}$ .

Why other options are wrong:
- **Option A ( $\frac{{2v\left( {n + 1} \right)}}{n}$ ): This is incorrect because it is not the correct calculation for the displacement in the last 2 seconds.
- Option B ( $\frac{{v\left( {n + 1} \right)}}{n}$ ): This is incorrect because it is not the correct calculation for the displacement in the last 2 seconds.
- Option C ( $\frac{{v\left( {n - 1} \right)}}{n}$ ): This is incorrect because it is not the correct calculation for the displacement in the last 2 seconds.

Key Concepts:
- Uniform acceleration
- Displacement

Memory Aids:
- Remember the equations of motion for uniform acceleration.

Quick Tips:**
- Always break down the problem into smaller steps.

9 Medium

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Analyzing +4 | -1

Which of the following 4 statements is false

View answer and explanation

Correct option: B

A body can have a constant velocity and still have a varying speed

Option B (A body can have a constant velocity and still have a varying speed) is false because constant velocity implies constant speed and direction.

Solution approach

Correct Answer + Core Reason: A body can have a constant velocity and still have a varying speed. Constant velocity implies both constant speed and constant direction.

Step-by-Step Solution:

Step 1: Analyze each statement:

Step 2: A: A body can have zero velocity and still be accelerated (e.g., at the highest point of a vertical throw).

Step 3: B: A body cannot have a constant velocity and still have a varying speed. Constant velocity means both speed and direction are constant.

Step 4: C: A body can have a constant speed and still have a varying velocity (e.g., uniform circular motion).

Step 5: D: The direction of the velocity of a body can change when its acceleration is constant (e.g., projectile motion).

Step 6: Identify the false statement: B.

Why other options are wrong:
- Option A (A body can have zero velocity and still be accelerated): This is true; consider the instant a ball thrown upwards reaches its peak.
- Option C (A body can have a constant speed and still have a varying velocity): This is true; consider uniform circular motion.
- Option D (The direction of the velocity of a body can change when its acceleration is constant): This is true; consider projectile motion.

Key Concepts:
- Velocity and speed
- Acceleration
- Constant velocity
- Uniform circular motion
- Projectile motion

Memory Aids:
- Velocity is a vector (magnitude and direction).
- Speed is a scalar (magnitude only).

Quick Tips:
- Constant velocity implies constant speed and direction.

10 Easy

10 of 192Question permalink

Understanding +4 | -1

A ball of mass

{m_1}

and another ball of mass

{m_2}

are dropped from equal height. If time taken by the balls are

{t_1}

and

{t_2}

respectively, then

View answer and explanation

Correct option: B

{t_1} = {t_2}

Option B (

{t_1} = {t_2}

) is correct because the time taken for an object to fall from a height is independent of its mass when air resistance is negligible.

Solution approach

Correct Answer + Core Reason: ${t_1} = {t_2}$ . The time taken is independent of mass.

Step-by-Step Solution:

Step 1: The balls are dropped from the same height, so the initial conditions are the same.

Step 2: The equation for the distance fallen under constant acceleration (gravity) is $h = ut + \frac{1}{2}g{t^2}$ , where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time.

Step 3: Since both balls are dropped, their initial velocity u = 0. Therefore, the equation simplifies to $h = \frac{1}{2}g{t^2}$ .

Step 4: Solving for t, we get $t = \sqrt{\frac{{2h}}{g}}$ .

Step 5: Since h and g are the same for both balls, the time taken is the same, regardless of their masses. Therefore, ${t_1} = {t_2}$ .

Why other options are wrong:
- **Option A ( ${t_1} = \frac{{{t_2}}}{2}$ ): This is incorrect because the time taken is independent of the mass of the balls.
- Option C ( ${t_1} = 4{t_2}$ ): This is incorrect because the time taken is independent of the mass of the balls.
- Option D ( ${t_1} = \frac{{{t_2}}}{4}$ ): This is incorrect because the time taken is independent of the mass of the balls.

Key Concepts:
- Free fall
- Acceleration due to gravity
- Kinematics

Memory Aids:
- Remember that in the absence of air resistance, all objects fall with the same acceleration due to gravity.

Quick Tips:**
- When dealing with free fall problems, remember that the acceleration is constant and equal to g.

11 Easy

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Application +4 | -1

A stone is dropped from the top of a cliff. It is seen to hit the ground below after 4.2 s. How high is the cliff?

View answer and explanation

Correct option: A

86.44 m

Option A (86.44 m) is correct because it is calculated using the equation of motion for free fall,

h = \frac{1}{2}gt^2

.

Solution approach

Correct Answer + Core Reason: 86.44 m. The height is calculated using the free fall equation.

Step-by-Step Solution:

Step 1: Identify the given information: time $t = 4.2 \text{ s}$ , acceleration due to gravity $g = 9.8 \text{ m/s}^2$ .

Step 2: The stone is dropped, so the initial velocity $u = 0 \text{ m/s}$ .

Step 3: Use the equation of motion for free fall: $h = ut + \frac{1}{2}gt^2$ .

Step 4: Substitute the values: $h = (0 \text{ m/s})(4.2 \text{ s}) + \frac{1}{2}(9.8 \text{ m/s}^2)(4.2 \text{ s})^2$ .

Step 5: Simplify: $h = 0 + \frac{1}{2}(9.8)(17.64) = 4.9 \times 17.64 = 86.436 \text{ m}$ .

Step 6: Round to two decimal places: $h = 86.44 \text{ m}$ .

Why other options are wrong:
- Option B: Incorrect, likely a miscalculation or misunderstanding of the formula.
- Option C: Incorrect, likely a miscalculation or misunderstanding of the formula.
- Option D: Incorrect, likely a miscalculation or misunderstanding of the formula.

Key Concepts:
- Free fall
- Equations of motion
- Acceleration due to gravity

Memory Aids:
- $h = ut + \frac{1}{2}gt^2$ (equation for free fall)

Quick Tips:
- Remember the equation for free fall and ensure correct substitution of values.

12 Easy

12 of 192Question permalink

Application +4 | -1

A particle is moving in a straight line such that its velocity is given by

v = {t^4} + 3{t^2} + 8\,\,m/s

. Find acceleration at time

t = 1\,s

View answer and explanation

Correct option: A

10\,m/{s^2}

Option A (

10\,m/{s^2}

) is correct because it is the value obtained by differentiating the velocity function with respect to time and evaluating at t=1s.

Solution approach

Correct Answer + Core Reason: $10\,m/{s^2}$ . The acceleration is the derivative of the velocity function evaluated at t=1s.

Step-by-Step Solution:

Step 1: Identify the given information: Velocity function $v = t^4 + 3t^2 + 8$ m/s, and time $t = 1\,s$ .

Step 2: Recall the relationship between velocity and acceleration: Acceleration is the derivative of velocity with respect to time, $a = \frac{{dv}}{{dt}}$ .

Step 3: Differentiate the velocity function with respect to time: $a = \frac{{d}}{{dt}}(t^4 + 3t^2 + 8) = 4t^3 + 6t$ .

Step 4: Substitute $t = 1\,s$ into the acceleration function: $a(1) = 4(1)^3 + 6(1) = 4 + 6 = 10\,m/{s^2}$ .

Step 5: Final Answer: The acceleration at $t = 1\,s$ is $10\,m/{s^2}$ .

Why other options are wrong:
- **Option B ( $10.5\,m/{s^2}$ ): This is incorrect due to a calculation error in the differentiation or substitution process.
- Option C ( $11\,m/{s^2}$ ): This is incorrect due to a calculation error in the differentiation or substitution process.
- Option D ( $9\,m/{s^2}$ ): This is incorrect due to a calculation error in the differentiation or substitution process.

Key Concepts:**
- Differentiation
- Velocity and Acceleration relationship: $a = \frac{{dv}}{{dt}}$

Memory Aids:
- Remember that acceleration is the rate of change of velocity.

Quick Tips:
- Always remember to differentiate the velocity function correctly before substituting the value of time.

13 Easy

13 of 192Question permalink

Application +4 | -1

If

t=\sqrt{x}+4

, then

\left(\frac{d x}{d t}\right)_{t=4}

is

View answer and explanation

Correct option: B

zero

Option B (zero) is correct because when t=4, x=0, and therefore dx/dt at t=4 is zero.

Solution approach

Correct Answer + Core Reason: zero. When t=4, x=0, so dx/dt is zero.

Step-by-Step Solution:

Step 1: Express x in terms of t: Given $t = \sqrt{{x}} + 4$ , we can rearrange to solve for x: $\sqrt{{x}} = t - 4$ , so $x = (t - 4)^2$ .

Step 2: Differentiate x with respect to t: $\frac{{dx}}{{dt}} = 2(t - 4)$ .

Step 3: Evaluate at t = 4: $\left(\frac{{dx}}{{dt}}\right)_{{t=4}} = 2(4 - 4) = 2(0) = 0$ .

Step 4: Conclusion: Therefore, $\left(\frac{{dx}}{{dt}}\right)_{{t=4}} = 0$ .

Why other options are wrong:
- Option A: This is incorrect. The correct derivative evaluated at t=4 is 0.
- Option C: This is incorrect. The correct derivative evaluated at t=4 is 0.
- Option D: This is incorrect. The correct derivative evaluated at t=4 is 0.

Key Concepts:
- Differentiation
- Implicit Differentiation

Memory Aids:
- Remember the power rule for differentiation: $\frac{{d}}{{dx}}(x^n) = nx^{{n-1}}$ .

Quick Tips:
- Always solve for the dependent variable before differentiating if possible. Check your algebra carefully.

14 Medium

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Analyzing +4 | -1

Assertion: If two particles are moving with same speed but their velocity vectors are opposite, then the distance between the particles must be changing.
Reason: If the velocity of one particle w.r.t. other is zero, then separation between the particle must be constant.

View answer and explanation

Correct option: B

If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Option B (If both Assertion and Reason are true but Reason is not the correct explanation of Assertion) is correct because both statements are true, but the reason does not explain the assertion.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. Both statements are true, but the reason does not explain the assertion.

Step-by-Step Solution:

Step 1: Analyze the assertion: If two particles are moving with same speed but their velocity vectors are opposite, then the distance between the particles must be changing. This is true. If they are moving towards each other, the distance decreases. If they are moving away from each other, the distance increases.

Step 2: Analyze the reason: If the velocity of one particle w.r.t. other is zero, then separation between the particle must be constant. This is true. If the relative velocity is zero, the distance between them is constant.

Step 3: Determine the relationship between the assertion and the reason. Both are true, but the reason does not explain the assertion. The assertion is about particles moving with opposite velocities, while the reason is about particles with zero relative velocity.

Why other options are wrong:
- Option A: The reason is not the correct explanation for the assertion.
- Option C: The reason is true.
- Option D: The assertion is true.

Key Concepts:
- Relative velocity
- Motion in one dimension

Memory Aids:
- Relative velocity is the velocity of one object as observed from another.

Quick Tips:
- Consider different scenarios to understand relative motion.

15 Medium

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Application +4 | -1

An object accelerates from rest to a velocity

27.5\;\mathrm{m}{\mathrm{s}^{ - 1}}

in 10 s, then the distance covered after next 10 s is

View answer and explanation

Correct option: C

412.5 m

Option C (412.5 m) is correct because it is calculated by first finding the acceleration and then using the kinematic equation for distance with constant velocity.

Solution approach

Correct Answer + Core Reason: 412.5 m. The distance is calculated using kinematics equations, first finding acceleration and then distance covered during constant velocity motion.

Step-by-Step Solution:

Step 1: Find the acceleration during the first 10 seconds.
$a = \frac{{\Delta v}}{{\Delta t}} = \frac{{27.5 \;\mathrm{m}{\mathrm{s}^{ - 1}} - 0}}{{10 \;\mathrm{s}}} = 2.75 \;\mathrm{m}{\mathrm{s}^{ - 2}}$

Step 2: Find the distance covered during the first 10 seconds.
$s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2.75)(10)^2 = 137.5 \;\mathrm{m}$

Step 3: After 10 seconds, the object moves with a constant velocity of $27.5 \;\mathrm{m}{\mathrm{s}^{ - 1}}$ for another 10 seconds.
$s_2 = vt = 27.5 \times 15 = 275 \;\mathrm{m}$

Step 4: Total distance covered after the next 10 seconds is the distance covered with constant velocity.
$s_2 = 27.5 \times 15 = 412.5 \;\mathrm{m}$

Why other options are wrong:
- Option A: Incorrect calculation or misunderstanding of the problem.
- Option B: This is the distance covered in the first 10 seconds, not the next 10 seconds.
- Option D: Incorrect calculation or misunderstanding of the problem.

Key Concepts:
- Kinematics equations of motion
- Constant acceleration
- Constant velocity

Memory Aids:
- SUVAT equations: $v = u + at$ , $s = ut + \frac{1}{2}at^2$ , $v^2 = u^2 + 2as$

Quick Tips:
- Break the problem into different stages based on the type of motion (constant acceleration, constant velocity).

16 Easy

16 of 192Question permalink

Application +4 | -1

A boy standing at the top of a tower of 20 m height drops a stone. Assuming

g = 10\,m/{s^2},

the velocity with which it hits the ground is

View answer and explanation

Correct option: C

20.0 m/s

Option C (20.0 m/s) is correct because it accurately calculates the final velocity using the kinematic equation

v^2 = u^2 + 2as

.

Solution approach

Correct Answer + Core Reason: 20.0 m/s. The final velocity is calculated using the kinematic equation.

Step-by-Step Solution:

Step 1: Initial velocity, $u = 0 \,m/s$

Step 2: Acceleration, $a = g = 10 \,m/s^2$

Step 3: Distance, $s = 20 \,m$

Step 4: Using the equation $v^2 = u^2 + 2as$

Step 5: $v^2 = 0^2 + 2 \times 10 \times 20 = 400$

Step 6: $v = \sqrt{400} = 20 \,m/s$

Why other options are wrong:
- Option A: Incorrect calculation.
- Option B: Incorrect calculation.
- Option C: Correct calculation of final velocity.
- Option D: Incorrect calculation.

Key Concepts:
- Kinematic equations of motion

Memory Aids:
- Remember the kinematic equations.

Quick Tips:
- Choose the appropriate kinematic equation based on given values.

17 Hard

17 of 192Question permalink

Application +4 | -1

A particle moves according to the equation

\frac{d v}{d t}=\alpha-\beta v

, where

\alpha

and

\beta

are constants. Find velocity as a function of time. Assume, body starts from rest

View answer and explanation

Correct option: B

v=(\alpha / \beta)\left(1-e^{-\beta t}\right)

Option B (

v=(\alpha / \beta)(1-e^{-\beta t})

) is correct because it is the solution to the given differential equation with the initial condition that the body starts from rest.

Solution approach

Correct Answer + Core Reason: $v=(\alpha / \beta)(1-e^{-\beta t})$ . This is the solution to the differential equation with the given initial condition.

Step-by-Step Solution:

Step 1: Given the equation $\frac{dv}{dt} = \alpha - \beta v$ .

Step 2: Separate the variables: $\frac{dv}{\alpha - \beta v} = dt$ .

Step 3: Integrate both sides: $\int \frac{dv}{\alpha - \beta v} = \int dt$ .

Step 4: The integral on the left side is $-\frac{1}{\beta} \ln |\alpha - \beta v|$ , and the integral on the right side is $t + C$ , where C is the constant of integration.\nSo, $-\frac{1}{\beta} \ln |\alpha - \beta v| = t + C$ .

Step 5: Multiply both sides by $-\beta$ : $\ln |\alpha - \beta v| = -\beta t - \beta C$ .

Step 6: Exponentiate both sides: $|\alpha - \beta v| = e^{-\beta t - \beta C} = e^{-\beta C} e^{-\beta t}$ .\nLet $A = e^{-\beta C}$ , so $|\alpha - \beta v| = A e^{-\beta t}$ .

Step 7: Since the body starts from rest, $v(0) = 0$ . Therefore, $|\alpha - \beta (0)| = A e^{-\beta (0)}$ , which simplifies to $\alpha = A$ .\nSo, $\alpha - \beta v = \alpha e^{-\beta t}$ .

Step 8: Rearrange to solve for $v$ : $\beta v = \alpha - \alpha e^{-\beta t} = \alpha (1 - e^{-\beta t})$ .

Step 9: Divide by $\beta$ : $v = \frac{\alpha}{\beta} (1 - e^{-\beta t})$ .

Why other options are wrong:
- Option A: This option is incorrect because the ratio of $\alpha$ and $\beta$ is inverted.
- Option C: This option is incorrect because it does not have the correct form of the solution, lacking the (1 - ...) term.
- Option D: This option is incorrect because it does not have the correct form of the solution, lacking the (1 - ...) term and has the ratio of $\alpha$ and $\beta$ inverted.

Key Concepts:
- Differential equations
- Integration
- Kinematics

Memory Aids:
- Remember the initial condition to solve for the constant of integration.

Quick Tips:
- Separate the variables and integrate to solve first-order differential equations.

18 Medium

18 of 192Question permalink

Understanding +4 | -1

The ratio of the numerical values of the average velocity and average speed of a body is always

View answer and explanation

Correct option: B

Unity or less

Option B (Unity or less) is correct because the magnitude of average velocity is always less than or equal to the average speed.

Solution approach

Correct Answer + Core Reason: Unity or less. The magnitude of average velocity is always less than or equal to the average speed.

Step-by-Step Solution:

Step 1: Define average velocity: Average velocity is the total displacement divided by the total time.

Step 2: Define average speed: Average speed is the total distance divided by the total time.

Step 3: The magnitude of displacement is always less than or equal to the distance.

Step 4: Therefore, the magnitude of average velocity is always less than or equal to the average speed.

Step 5: The ratio of the numerical values of the average velocity and average speed is always unity or less.

Why other options are wrong:
- Option A: Unity - The ratio is unity only when the object moves in a straight line without changing direction.
- Option C: Unity or more - The average speed is always greater than or equal to the magnitude of the average velocity.
- Option D: Less than unity - The ratio can be unity if the object moves in a straight line without changing direction.

Key Concepts:
- Average velocity
- Average speed
- Displacement
- Distance

Memory Aids:
- Displacement ≤ Distance

Quick Tips:
- Remember that average speed and average velocity are different concepts.

19 Medium

19 of 192Question permalink

Understanding +4 | -1

The distance travelled by a body is directly proportional to the time taken. Its speed

View answer and explanation

Correct option: D

Remains constant

Option D (Remains constant) is correct because if distance is directly proportional to time, the speed is constant.

Solution approach

Correct Answer + Core Reason: Remains constant. Distance directly proportional to time implies constant speed.

Step-by-Step Solution:

Step 1: Define speed. Speed is defined as the distance traveled divided by the time taken: $v = \frac{d}{t}$ , where $v$ is the speed, $d$ is the distance, and $t$ is the time.

Step 2: Analyze the given information. The distance traveled by a body is directly proportional to the time taken. This means that $d = kt$ , where $k$ is a constant of proportionality.

Step 3: Calculate the speed. Substituting $d = kt$ into the speed equation, we get $v = \frac{kt}{t} = k$ . Since $k$ is a constant, the speed remains constant.

Why other options are wrong:
- Option A: Increases. This is incorrect because the speed remains constant.
- Option B: Decreases. This is incorrect because the speed remains constant.
- Option C: Becomes zero. This is incorrect because the speed remains constant.

Key Concepts:
- Speed
- Distance
- Time
- Direct proportionality

Memory Aids:
- Speed = Distance / Time

Quick Tips:
- If distance is directly proportional to time, the speed is constant.

20 Medium

20 of 192Question permalink

Analyzing +4 | -1

A particle moves along a straight line such that its position x at any time t is

x = 6{t^2} - {t^3}.

. Where x in metre and t is in second, then

View answer and explanation

Correct option: D

Both (1) and (2) are correct

Option D (Both (1) and (2) are correct) is correct because at t=0 the acceleration is indeed 12 m/s², and the x-t curve has a maximum at t=4s.

Solution approach

Correct Answer + Core Reason: Both (1) and (2) are correct. The acceleration at t=0 is 12 m/s², and the x-t curve has a maximum at t=4s.

Step-by-Step Solution:

Step 1: Given the position function $x = 6t^2 - t^3$ .

Step 2: Find the velocity function by taking the first derivative of the position function with respect to time: $v = \frac{dx}{dt} = 12t - 3t^2$ .

Step 3: Find the acceleration function by taking the first derivative of the velocity function with respect to time: $a = \frac{dv}{dt} = 12 - 6t$ .

Step 4: Evaluate the acceleration at t=0: $a(0) = 12 - 6(0) = 12 \;\mathrm{m}{\mathrm{s}^{ - 2}}$ . Thus, statement (1) is correct.

Step 5: To find the maximum of the x-t curve, set the velocity to zero: $12t - 3t^2 = 0$ . This gives $3t(4 - t) = 0$ , so $t = 0$ or $t = 4$ .

Step 6: Check the second derivative of the position function (which is the acceleration) at t=4: $a(4) = 12 - 6(4) = -12 \;\mathrm{m}{\mathrm{s}^{ - 2}}$ . Since the acceleration is negative, the x-t curve has a maximum at t=4s. Thus, statement (2) is correct.

Step 7: Therefore, both statements (1) and (2) are correct.

Why other options are wrong:
- Option A: This option is incorrect because it only considers statement (1) to be correct, while statement (2) is also correct.
- Option B: This option is incorrect because it only considers statement (2) to be correct, while statement (1) is also correct.
- Option C: This option is incorrect because both statements (1) and (2) are correct.

Key Concepts:
- Kinematics
- Differentiation
- Finding maxima and minima

Memory Aids:
- Velocity is the rate of change of position.
- Acceleration is the rate of change of velocity.
- To find maxima/minima, set the first derivative to zero.

Quick Tips:
- Always differentiate the position function to find velocity and acceleration.
- Remember to check the second derivative to confirm maxima or minima.

21 Medium

21 of 192Question permalink

Analyzing +4 | -1

The velocity of a particle depends upon the time

\mathrm{t}

according to the equation:

\mathrm{v}=\sqrt{\mathrm{a b}}+\mathrm{b t}+\frac{\mathrm{c}}{\mathrm{d+t}}

. The physical quantities which are represented by

\mathrm{a, b, c}

and

\mathrm{d}

, are in the following order:

View answer and explanation

Correct option: B

Distance, acceleration, distance, time

Option B (Distance, acceleration, distance, time) is correct because, based on dimensional analysis, 'a' must represent distance, 'b' must represent acceleration, 'c' must represent distance, and 'd' must represent time to maintain consistent units for velocity.

Solution approach

Correct Answer + Core Reason: Distance, acceleration, distance, time. The physical quantities are determined by dimensional analysis.

Step-by-Step Solution:

Step 1: Analyze the given equation: $v = \sqrt{ab} + bt + \frac{c}{d+t}$ .

Step 2: Consider the dimensions of each term. The dimensions of velocity, $v$ , are $[LT^{-1}]$ .

Step 3: Analyze the first term: $\sqrt{ab}$ must have dimensions of $[LT^{-1}]$ . Therefore, $[ab] = [L^2T^{-2}]$ .

Step 4: Analyze the second term: $bt$ must have dimensions of $[LT^{-1}]$ . Since $t$ has dimensions of $[T]$ , $b$ must have dimensions of $[LT^{-2}]$ , which is acceleration.

Step 5: Analyze the third term: $\frac{c}{d+t}$ must have dimensions of $[LT^{-1}]$ . Since $t$ has dimensions of $[T]$ , $d$ must also have dimensions of $[T]$ (time). Therefore, $\frac{c}{[T]} = [LT^{-1}]$ , which means $c$ must have dimensions of $[L]$ (distance).

Step 6: From step 3 and 4, since b is acceleration $[LT^{-2}]$ , then a must be distance $[L]$ to satisfy $[ab] = [L^2T^{-2}]$ .

Step 7: Therefore, a is distance, b is acceleration, c is distance, and d is time.

Why other options are wrong:
- Option A: This option is incorrect because it assigns distance to both a and b, which is inconsistent with the dimensions of the equation.
- Option C: This option is incorrect because it assigns acceleration to a and distance to b, which is inconsistent with the dimensions of the equation.
- Option D: This option is incorrect because one of the other options is correct.

Key Concepts:
- Dimensional analysis
- Velocity
- Acceleration

Memory Aids:
- Remember the dimensions of basic physical quantities: length [L], time [T], mass [M]

22 Easy

22 of 192Question permalink

Analyzing +4 | -1

Statement-1: Two balls of different masses are thrown vertically upwards with same speed. They will pass through their point of projection in the downward direction with the same speed in absence of air resistance.
Statement-2: In absence of air resistance, the mechanical energy of a projectile is conserved.

View answer and explanation

Correct option: A

Statement-1 and Statement-2 are true, Statement-2 is the correct explanation for Statement-1

Option A (Statement-1 and Statement-2 are true, Statement-2 is the correct explanation for Statement-1) is correct because both statements are true and the conservation of mechanical energy explains why the balls return with the same speed.

Solution approach

Correct Answer + Core Reason: Statement-1 and Statement-2 are true, Statement-2 is the correct explanation for Statement-1. The conservation of mechanical energy explains why the balls return with the same speed.

Step-by-Step Solution:

Step 1: Analyze Statement-1: Two balls of different masses are thrown vertically upwards with the same speed. They will pass through their point of projection in the downward direction with the same speed in the absence of air resistance.

Step 2: Consider the motion of the balls. As they move upwards, their kinetic energy is converted into potential energy. At the highest point, all the kinetic energy is converted into potential energy. As they fall back down, the potential energy is converted back into kinetic energy. In the absence of air resistance, the total mechanical energy (kinetic + potential) is conserved.

Step 3: Since the initial speed is the same, the initial kinetic energy is the same ( $\frac{1}{2}mv^2$ ). When they return to the point of projection, all the potential energy has been converted back into kinetic energy. Therefore, their speed will be the same as their initial speed, regardless of their mass.

Step 4: Analyze Statement-2: In the absence of air resistance, the mechanical energy of a projectile is conserved.

Step 5: This statement is true. Mechanical energy is the sum of kinetic and potential energy. In the absence of non-conservative forces like air resistance, the total mechanical energy remains constant.

Step 6: Determine if Statement-2 is the correct explanation for Statement-1. The conservation of mechanical energy is the reason why the balls return with the same speed. The initial kinetic energy is converted to potential energy, and then back to kinetic energy. Since the mechanical energy is conserved, the final kinetic energy (and thus the final speed) is the same as the initial kinetic energy (and thus the initial speed).

Step 7: Conclusion: Both statements are true, and Statement-2 is the correct explanation for Statement-1.

Why other options are wrong:
- Option B (Statement-1 and Statement-2 are true, Statement-2 is not the correct explanation for Statement-1): Statement-2 is the correct explanation for Statement-1.
- Option C (Statement-1 is true, Statement-2 is false): Statement-2 is true.
- Option D (Statement-1 is false, Statement-2 is true): Statement-1 is true.

Key Concepts:
- Conservation of mechanical energy
- Projectile motion
- Kinetic energy
- Potential energy

Memory Aids:
- In the absence of air resistance, mechanical energy is conserved.

Quick Tips:
- Remember that mass does not affect the final velocity in this scenario due to energy conservation.

23 Easy

23 of 192Question permalink

Application +4 | -1

Two stones are falling at a place from heights in the ratio 2 : 3. Their velocities on reaching the ground are in the ratio

View answer and explanation

Correct option: C

\sqrt 2 :\sqrt 3

Option C (

\sqrt 2 :\sqrt 3

) is correct because the final velocity is proportional to the square root of the height from which the stone falls, as derived from the kinematic equation

v^2 = 2gh

.

Solution approach

Correct Answer + Core Reason: $\sqrt 2 :\sqrt 3$ . The velocities are proportional to the square root of the heights.

Step-by-Step Solution:

Step 1: Identify the relevant formula: We can use the kinematic equation $v^2 = u^2 + 2as$ , where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, and $s$ is the displacement.

Step 2: Apply the formula to the falling stones: Since the stones are falling from rest, $u = 0$ . The acceleration is due to gravity, so $a = g$ . The displacement is the height, $h$ . Therefore, $v^2 = 2gh$ , and $v = \sqrt{2gh}$ .

Step 3: Find the ratio of the velocities: Let $h_1$ and $h_2$ be the heights of the two stones, and $v_1$ and $v_2$ be their respective velocities. We are given that $h_1 : h_2 = 2 : 3$ . Then, $v_1 = \sqrt{2gh_1}$ and $v_2 = \sqrt{2gh_2}$ . The ratio of the velocities is $\frac{v_1}{v_2} = \frac{\sqrt{2gh_1}}{\sqrt{2gh_2}} = \sqrt{\frac{h_1}{h_2}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$ .

Step 4: Conclusion: The ratio of the velocities is $\sqrt 2 : \sqrt 3$ .

Why other options are wrong:
- Option A: 0.085416667: This is a numerical value that doesn't represent the correct ratio.
- Option B: 0.126388889: This is a numerical value that doesn't represent the correct ratio.
- Option D: $\sqrt 3 :\sqrt 2$ : This is the inverse of the correct ratio.

Key Concepts:
- Kinematics
- Free fall
- Constant acceleration

Memory Aids:
- Remember the kinematic equations for constant acceleration.

Quick Tips:
- When dealing with ratios, simplify the expressions as much as possible before substituting values.

24 Medium

24 of 192Question permalink

Application +4 | -1

An athlete completes one round of a circular track or radius R in 40 sec. What will be his displacement at the end of 2 min. 20 sec

View answer and explanation

Correct option: B

2R

Option B (2R) is correct because after 2 minutes and 20 seconds, the athlete completes 3.5 rounds, ending up diametrically opposite the starting point, hence the displacement is 2R.

Solution approach

Correct Answer + Core Reason: 2R. The displacement is the shortest distance between the initial and final positions.

Step-by-Step Solution:

Step 1: Calculate the total time in seconds: $2 \mathrm{~min}. 20 \mathrm{~sec} = (2 \times 60) + 20 = 140 \mathrm{~sec}$

Step 2: Calculate the number of rounds completed: $\frac{140 \mathrm{~sec}}{40 \mathrm{~sec/round}} = 3.5 \mathrm{~rounds}$

Step 3: After 3.5 rounds, the athlete is at the opposite end of the circle from the starting point.

Step 4: The displacement is the diameter of the circle, which is $2R$ .

Why other options are wrong:
- Option A: Zero displacement would occur after an integer number of complete rounds.
- Option C: $2\pi R$ is the circumference of the circle, which represents the distance traveled in one complete round, not the displacement.
- Option D: $7\pi R$ is not a meaningful quantity in this context.

Key Concepts:
- Displacement
- Circular motion

Memory Aids:
- Displacement is the shortest distance between the initial and final points, while distance is the total path length.

Quick Tips:
- Visualize the motion to determine the final position relative to the starting point.

25 Hard

25 of 192Question permalink

Analyzing +4 | -1

A stone dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after

{t_1}

and

{t_2}

seconds respectively, then

View answer and explanation

Correct option: C

t = \sqrt {{t_1}{t_2}}

Option C (

t = \sqrt {{t_1}{t_2}}

) is correct. By analyzing the equations of motion for each scenario and eliminating the initial velocity, we find that the time for the dropped stone is the geometric mean of the times for the stones thrown upwards and downwards.

Solution approach

Correct Answer + Core Reason: $t = \sqrt {{t_1}{t_2}}$ . This relationship is derived from the kinematic equations for the three scenarios: dropped, thrown upwards, and thrown downwards.

Step-by-Step Solution:

Step 1: Define the variables: h = height of the building, u = initial velocity, t = time for the dropped stone, t1 = time for the stone thrown upwards, t2 = time for the stone thrown downwards.

Step 2: Write the equation for the dropped stone: $h = \frac{1}{2}gt^2$ (since initial velocity is 0)

Step 3: Write the equation for the stone thrown upwards: $-h = ut_1 - \frac{1}{2}gt_1^2$ (negative h because displacement is downwards)

Step 4: Write the equation for the stone thrown downwards: $-h = -ut_2 - \frac{1}{2}gt_2^2$ (negative h and u because both are downwards)

Step 5: From the dropped stone equation, $t = \sqrt{\frac{2h}{g}}$

Step 6: From the upwards thrown stone equation: $h = -ut_1 + \frac{1}{2}gt_1^2$

Step 7: From the downwards thrown stone equation: $h = ut_2 + \frac{1}{2}gt_2^2$

Step 8: Substitute $h = \frac{1}{2}gt^2$ into the equations from steps 6 and 7:
$\frac{1}{2}gt^2 = -ut_1 + \frac{1}{2}gt_1^2$ and $\frac{1}{2}gt^2 = ut_2 + \frac{1}{2}gt_2^2$

Step 9: Multiply the two equations from step 8:
$\frac{1}{4}g^2t^4 = (\frac{1}{2}gt_1^2 - ut_1)(\frac{1}{2}gt_2^2 + ut_2)$

Step 10: Rearrange the equations from step 8 to isolate 'u':
$u = \frac{g}{2}(\frac{t_1^2 - t^2}{t_1})$ and $u = \frac{g}{2}(\frac{t^2 - t_2^2}{t_2})$

Step 11: Equate the two expressions for 'u':
$\frac{g}{2}(\frac{t_1^2 - t^2}{t_1}) = \frac{g}{2}(\frac{t^2 - t_2^2}{t_2})$

Step 12: Simplify and solve for $t^2$ :
$t_1^2t_2 - t^2t_2 = t^2t_1 - t_2^2t_1$
$t^2(t_1 + t_2) = t_1t_2(t_1 + t_2)$
$t^2 = t_1t_2$
$t = \sqrt{t_1t_2}$

Why other options are wrong:
- Option A: $t = {t_1} - {t_2}$ : This is incorrect because it doesn't account for the initial velocity and the relationship between the times.
- Option B: $t = \frac{{{t_1} + {t_2}}}{2}$ : This is incorrect because it represents the arithmetic mean, not the geometric mean, and doesn't properly account for the initial velocity.
- Option D: $t = t_1^2t_2^2$ : This is incorrect because it is dimensionally inconsistent and doesn't follow from the kinematic equations.

Key Concepts:
- Kinematic equations
- Projectile motion

Memory Aids:
- Remember the equations of motion: $s = ut + \frac{1}{2}at^2$

Quick Tips:
- Break down the problem into different scenarios and apply the appropriate kinematic equations.

26 Medium

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Understanding +4 | -1

Two balls of same size but the density of one is greater than that of the other are dropped from the same height, then which ball will reach the earth first (air resistance is negligible)?

View answer and explanation

Correct option: C

Both simultaneously

Option C (Both simultaneously) is correct because in the absence of air resistance, the acceleration due to gravity is the same for all objects, regardless of their mass or density.

Solution approach

Correct Answer + Core Reason: Both simultaneously. The acceleration due to gravity is the same for all objects.

Step-by-Step Solution:

Step 1: The problem states that air resistance is negligible.

Step 2: In the absence of air resistance, the only force acting on the balls is gravity.

Step 3: The acceleration due to gravity (g) is constant for all objects, regardless of their mass or density.

Step 4: The equation for the distance fallen is $h = ut + \frac{1}{2}g{t^2}$ , where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time.

Step 5: Since both balls are dropped from the same height (h) with the same initial velocity (u = 0) and experience the same acceleration due to gravity (g), the time taken to reach the earth will be the same for both balls.

Why other options are wrong:
- Option A (Heavy ball): This is incorrect because the time taken is independent of mass in the absence of air resistance.
- Option B (Light ball): This is incorrect because the time taken is independent of mass in the absence of air resistance.
- Option D (Will depend upon the density of the balls): This is incorrect because the time taken is independent of density in the absence of air resistance.

Key Concepts:
- Free fall
- Acceleration due to gravity

Memory Aids:
- Remember that in a vacuum, all objects fall with the same acceleration due to gravity.

Quick Tips:
- Always consider the effect of air resistance if it is not negligible.

27 Medium

27 of 192Question permalink

Application +4 | -1

A bomb is dropped from an aeroplane flying horizontally with a velocity of 720 kmph at an altitude of 980 m. Time taken by the bomb to hit the ground is

View answer and explanation

Correct option: C

14.14 S

Option C (14.14 S) is correct because it calculates the time it takes for the bomb to fall from a height of 980m under the influence of gravity, independent of its horizontal velocity.

Solution approach

Correct Answer + Core Reason: 14.14 S. The time taken depends only on the vertical distance and gravity.

Step-by-Step Solution:

Step 1: The horizontal velocity of the aeroplane is 720 kmph, but this does not affect the time taken for the bomb to hit the ground.

Step 2: The bomb is dropped from a height of 980 m. The time taken to fall can be calculated using the equation $h = \frac{1}{2} g t^2$ , where $h = 980 \mathrm{~m}$ and $g = 9.8 \mathrm{~m/s^2}$ .

Step 3: Substitute the values: $980 = \frac{1}{2} (9.8) t^2$ .

Step 4: Solve for $t^2$ : $t^2 = \frac{2 \times 980}{9.8} = 200$ .

Step 5: Take the square root: $t = \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2} \approx 14.14 \mathrm{~s}$

Why other options are wrong:
- Option A: This option is incorrect. It is too small and does not account for the height from which the bomb is dropped.
- Option B: This option is incorrect. It is not the correct time calculated using the given height and acceleration due to gravity.
- Option D: This option is incorrect. It is too small and does not account for the height from which the bomb is dropped.

Key Concepts:
- Projectile motion
- Free fall

Memory Aids:
- Remember that the horizontal and vertical motions of a projectile are independent of each other.

Quick Tips:
- When dealing with projectile motion problems, separate the horizontal and vertical components of motion.

28 Easy

28 of 192Question permalink

Understanding +4 | -1

The velocity of a body at a given instant of time during motion is:

View answer and explanation

Correct option: C

Instantaneous velocity

=\lim \limits_{{\Delta t \rightarrow 0}} \frac{\Delta \vec{s}}{\Delta t}

Option C (Instantaneous velocity

=\lim \limits_{{\Delta t \rightarrow 0}} \frac{\Delta \vec{s}}{\Delta t}

) is correct because it represents the definition of instantaneous velocity as the limit of the average velocity as the time interval approaches zero.

Solution approach

Correct Answer + Core Reason: Instantaneous velocity $=\lim \limits_{{\Delta t \rightarrow 0}} \frac{\Delta \vec{s}}{\Delta t}$ . This is the definition of instantaneous velocity.

Step-by-Step Solution:

Step 1: Define instantaneous velocity. Instantaneous velocity is the velocity of an object at a specific instant in time. It is defined as the limit of the average velocity as the time interval approaches zero.

Step 2: Express the mathematical definition. Mathematically, instantaneous velocity is given by: $\vec{v} = \lim_{\Delta t \rightarrow 0} \frac{\Delta \vec{s}}{\Delta t}$ , where $\Delta \vec{s}$ is the displacement and $\Delta t$ is the time interval.

Why other options are wrong:
- Option A: Instantaneous velocity $=\frac{\Delta \vec{s}}{\Delta t}$ . This represents average velocity, not instantaneous velocity.
- Option B: Average velocity $=\lim \limits_{{\Delta t \rightarrow 0}} \frac{\Delta \vec{s}}{\Delta t}$ . This is the definition of instantaneous velocity, not average velocity.
- Option D: None of these. Option C is the correct definition of instantaneous velocity.

Key Concepts:
- Instantaneous velocity
- Average velocity
- Limits

Memory Aids:
- Instantaneous velocity is the velocity at a specific instant.
- Average velocity is the velocity over a time interval.

Quick Tips:
- Remember the difference between average and instantaneous velocity. Instantaneous velocity involves a limit as the time interval approaches zero.

29 Medium

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Application +4 | -1

The displacement of the particle varies with time according to the relation

x = \frac{k}{b}\left[ {1 - {e^{ - bt}}} \right].

Then the velocity of the particle is

View answer and explanation

Correct option: A

k\left( {{e^{ - bt}}} \right)

Option A (

k\left( {{e^{ - bt}}} \right)

) is correct because it represents the derivative of the given displacement function with respect to time, which gives the velocity.

Solution approach

Correct Answer + Core Reason: $k\left( {{e^{ - bt}}} \right)$ . The velocity is the derivative of the displacement function.

Step-by-Step Solution:

Step 1: Write down the displacement function: $x = \frac{k}{b}[1 - e^{-bt}]$

Step 2: Find the velocity function by taking the derivative of the displacement function with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(\frac{k}{b}[1 - e^{-bt}]$

30 Easy

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Application +4 | -1

A body moves from one corner of an equilateral triangle of side 10 cm to the same corner along the sides. Then the distance and displacement are respectively

View answer and explanation

Correct option: B

30 cm & 0 cm

Option B (30 cm & 0 cm) is correct because the distance is the total length traveled along the sides (10 cm + 10 cm + 10 cm = 30 cm), and the displacement is zero since the body returns to its starting point.

Solution approach

Correct Answer + Core Reason: 30 cm & 0 cm. Distance is the total path length, and displacement is zero since it returns to the starting point.

Step-by-Step Solution:

Step 1: Define Distance: The body moves along the sides of the equilateral triangle, covering a distance of 10 cm + 10 cm + 10 cm.

Step 2: Calculate Distance: Distance = 10 cm + 10 cm + 10 cm = 30 cm.

Step 3: Define Displacement: Since the body returns to the same corner, the initial and final positions are the same.

Step 4: Calculate Displacement: Displacement = 0 cm.

Step 5: Conclusion: The distance is 30 cm, and the displacement is 0 cm.

Why other options are wrong:
- Option A: The displacement is zero, not 10 cm.
- Option C: The distance is 30 cm, not 0 cm.
- Option D: The displacement is zero, not 30 cm.

Key Concepts:
- Distance
- Displacement

Memory Aids:
- Returning to the start = zero displacement.

Quick Tips:
- Always distinguish between distance (total path length) and displacement (shortest distance between initial and final points).

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