WORK-ENERGY-POWER MCQs for JEE PHYSICS — Mixed difficulty, Page 13

361 Medium

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Application +4 | -1

A body dropped from a height

H

reaches the ground with a speed of

1.2 \sqrt{g H}

. Calculate the work done by air-friction.

View answer and explanation

Correct option: D

-0.28 \mathrm{~mgH}

Option D (

-0.28 \mathrm{~mgH}

) is correct because the work done by air friction is the difference between the total work done and the work done by gravity.

Solution approach

Correct Answer + Core Reason: $-0.28 \mathrm{~mgH}$ . The work done by friction is the difference between the kinetic energy gained and the potential energy lost.

Step-by-Step Solution:

Step 1: Identify the given information: Height = H, Final speed = $1.2 \sqrt{{gH}}$ .

Step 2: Calculate the kinetic energy gained: $KE = \frac{{1}}{{2}}mv^2 = \frac{{1}}{{2}}m(1.2\sqrt{{gH}})^2 = 0.72mgH$ .

Step 3: Calculate the work done by gravity: $W_g = mgH$ .

Step 4: Calculate the work done by air friction: $W_{friction} = KE - W_g = 0.72mgH - mgH = -0.28mgH$ .

Final Answer: $-0.28 \mathrm{~mgH}$

Why other options are wrong:
- Option A: Incorrect sign and magnitude of work done by friction.
- Option B: Incorrect magnitude of work done by friction.
- Option C: Incorrect sign of work done by friction.
- Option D: Correct answer.

Key Concepts:
- Work-energy theorem
- Work done by gravity
- Work done by friction

Memory Aids:
- Remember that friction opposes motion, so the work done by friction is negative.

Quick Tips:
- Always consider the signs of work done by different forces.

362 Medium

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Analyzing +4 | -1

An object of mass

m

slides down a hill of arbitrary shape and after travelling a certain horizontal path stops because of friction. The total vertical height descended is

h

. The friction coefficient is different for different segments for the entire path but is independent of the velocity and direction of motion. The work that a tangential force must perform to return the object to its initial position along the same path is

View answer and explanation

Correct option: D

2 m g h

Option D (

2 m g h

) is correct because the work done against friction during the descent is

mgh

, and the work required to return the object to its initial position is also

mgh

, totaling

2mgh

.

Solution approach

Correct Answer + Core Reason: $2 m g h$ . The work done against friction during descent is $mgh$ , and the work to return is also $mgh$ .

Step-by-Step Solution:

Step 1: During the slide down, the object loses potential energy $mgh$ . This potential energy is converted into work done against friction.

Step 2: Work done against friction during descent: $W_{friction} = mgh$ .

Step 3: To return the object to its initial position, we need to overcome friction again and also provide the potential energy it had initially.

Step 4: Work done against friction during ascent: $W'_{friction} = mgh$ (since the path is the same and friction coefficient is independent of velocity).

Step 5: Work done to increase potential energy: $W_{potential} = mgh$ .

Step 6: Total work required: $W_{total} = W'_{friction} + W_{potential} = mgh + mgh = 2mgh$ .

Why other options are wrong:
- Option A: $mgh$ is the work done against friction during descent or the work to raise the object without friction.
- Option B: $-mgh$ represents the potential energy lost during descent.
- Option C: $-2mgh$ is incorrect as work done is always positive in this case.

Key Concepts:
- Work-energy theorem
- Potential energy
- Work done against friction

Memory Aids:
- Friction always opposes motion, so work done against it is always positive.

Quick Tips:
- Remember that the work done against friction is path-dependent.
- Consider both the work done against friction and the change in potential energy.

363 Medium

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Analyzing +4 | -1

Which of the following statements is correct?

View answer and explanation

Correct option: A

Kinetic energy of a system can be changed without changing its momentum.

Option A (Kinetic energy of a system can be changed without changing its momentum) is correct because internal forces can change kinetic energy while conserving momentum.

Solution approach

Correct Answer + Core Reason: Kinetic energy of a system can be changed without changing its momentum. This occurs in situations like explosions where internal forces change KE while conserving momentum.

Step-by-Step Solution:

Step 1: Recall the definitions of kinetic energy and momentum: Kinetic energy (KE) is related to the speed of the object, and momentum (p) is related to the velocity of the object.

Step 2: Consider an explosion: In an explosion, a stationary object breaks into multiple pieces. The initial momentum is zero. After the explosion, the vector sum of the momenta of all the pieces is still zero (conservation of momentum). However, the pieces now have kinetic energy.

Step 3: Analyze the change in KE and momentum: The kinetic energy of the system has increased (from zero to some positive value), while the momentum of the system remains constant (zero).

Step 4: Conclusion: Therefore, kinetic energy can change without a change in momentum.

Why other options are wrong:
- Option B: Kinetic energy of a system cannot be changed without changing its momentum: This is incorrect, as demonstrated by the explosion example.
- Option C: Momentum of a system cannot be changed without changing its kinetic energy: This is also incorrect. A force acting perpendicular to the motion of an object changes its momentum (direction) but not its kinetic energy (speed).
- Option D: A system cannot have energy without having momentum: A system can have potential energy without having momentum.

Key Concepts:
- Kinetic energy
- Momentum
- Conservation of momentum
- Internal forces

Memory Aids:
- Remember the explosion example to relate changes in KE and momentum.

Quick Tips:
- Consider scenarios involving internal forces (explosions, collisions) to understand how KE and momentum can change independently.

364 Medium

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Understanding +4 | -1

Roads are banked on curves so that

View answer and explanation

Correct option: A

The speeding vehicles may not fall outwards

Option A is correct because banking roads helps provide the necessary centripetal force to prevent speeding vehicles from skidding outwards.

Solution approach

Correct Answer + Core Reason: The speeding vehicles may not fall outwards. Banking provides a component of the normal force that contributes to the centripetal force.

Step-by-Step Solution:

Step 1: Understand the concept of centripetal force: A force required to keep an object moving in a circular path.

Step 2: Analyze the forces acting on a vehicle on a banked road: Weight (mg), Normal force (N).

Step 3: Resolve the normal force into horizontal and vertical components: $N\sin\theta$ (horizontal) and $N\cos\theta$ (vertical), where $\theta$ is the banking angle.

Step 4: The horizontal component of the normal force provides the necessary centripetal force: $N\sin\theta = \frac{mv^2}{r}$ , where $m$ is the mass of the vehicle, $v$ is its speed, and $r$ is the radius of the curve.

Step 5: Banking helps to reduce the reliance on friction to provide the centripetal force, especially at higher speeds.

Why other options are wrong:
- Option B: The frictional force between the road and vehicle may be decreased. While banking reduces reliance on friction, it doesn't necessarily decrease the frictional force itself. The primary purpose is to provide centripetal force.
- Option C: The wear and tear of tyres may be avoided. Banking reduces the need for excessive friction, which can contribute to tire wear, but it's not the primary reason.
- Option D: The weight of the vehicle may be decreased. Banking does not change the weight of the vehicle.

Key Concepts:
- Centripetal Force
- Banking of Roads
- Normal Force

Memory Aids:
- Remember that banking provides a component of the normal force towards the center of the curve.

Quick Tips:
- Always consider the forces acting on an object moving in a circular path.
- Banking helps to increase the safe speed on curves.

365 Medium

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Application +4 | -1

A bullet fired from a gun with a velocity of

{10^4}\mathrm{\;m}{\mathrm{s}^{ - 1}}

goes through a bag full of straw. If the bullet loses half of its kinetic energy in the bag, its velocity when it comes out of the bag will be

View answer and explanation

Correct option: A

7071.06\mathrm{\;m}{\mathrm{s}^{ - 1}}

Option A (7071.06 m/s⁻¹) is correct because the velocity is calculated by considering the loss of kinetic energy, resulting in a reduced velocity.

Solution approach

Correct Answer + Core Reason: 7071.06 m/s⁻¹. The velocity is calculated based on the kinetic energy loss.

Step-by-Step Solution:

Step 1: Define initial kinetic energy:
$KE_1 = \frac{1}{2}mu^2$ , where $u = 10^4 \text{{ m/s}}$

Step 2: Define final kinetic energy (half of the initial):
$KE_2 = \frac{1}{2}KE_1 = \frac{1}{2}(\frac{1}{2}mu^2) = \frac{1}{4}mu^2$

Step 3: Define final velocity $v$ :
$KE_2 = \frac{1}{2}mv^2$

Step 4: Equate the two expressions for $KE_2$ :
$\frac{1}{2}mv^2 = \frac{1}{4}mu^2$

Step 5: Solve for $v$ :\nv² = $\frac{1}{2}$ u² $\nv = \frac{u}{\sqrt{2}}$

Step 6: Substitute the value of $u$ :\nv = $\frac{10^4}{\sqrt{2}}$ = $\frac{10000}{1.414}$ ≈ 7071.06 \text{{ m/s}} $\n\n**Why other options are wrong:**\n- **Option B:** 707 m/s⁻¹ is incorrect. This is off by a factor of 10, likely a calculation error.\n- **Option C:** 70.71 m/s⁻¹ is incorrect. This is off by a factor of 100, likely a calculation error.\n- **Option D:** 707.06 m/s⁻¹ is incorrect. This is off by a factor of 10, likely a calculation error.\n\n**Key Concepts:**\n- Kinetic energy:$ KE = $\frac{1}{2}$ mv² $\n- Relationship between kinetic energy and velocity\n\n**Memory Aids:**\n- Kinetic energy is proportional to the square of the velocity.\n\n**Quick Tips:**\n- When kinetic energy is halved, velocity is divided by$ $\sqrt{2}$ $.

366 Medium

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Understanding +4 | -1

Two particles are seen to collide and move jointly together after the collision. During such a collision, for the total system,

View answer and explanation

Correct option: B

linear momentum is conserved but not the mechanical energy.

Option B is correct because in an inelastic collision, linear momentum is always conserved, but kinetic energy is not, as some energy is converted into other forms like heat or sound.

Solution approach

Correct Answer + Core Reason: linear momentum is conserved but not the mechanical energy. This is characteristic of an inelastic collision.

Step-by-Step Solution:

Step 1: Define linear momentum.\nLinear momentum ( $p$ ) is the product of mass ( $m$ ) and velocity ( $v$ ): $p = mv$ .

Step 2: State the law of conservation of linear momentum.\nIn a closed system, the total linear momentum remains constant if no external forces act on the system.

Step 3: Define mechanical energy.\nMechanical energy is the sum of kinetic energy (KE) and potential energy (PE).

Step 4: Consider the collision scenario.\nThe two particles collide and move jointly together. This is an example of a perfectly inelastic collision.

Step 5: Apply the conservation laws.
- Linear momentum is always conserved in collisions, regardless of whether they are elastic or inelastic.
- Mechanical energy (specifically kinetic energy) is not conserved in inelastic collisions. Some of the kinetic energy is converted into other forms of energy, such as heat, sound, or deformation.

Why other options are wrong:
- Option A: Mechanical energy is not conserved in this type of collision.
- Option C: Linear momentum is conserved, so this option is incorrect.
- Option D: Linear momentum is conserved, and mechanical energy is not, so this option is incorrect.

Key Concepts:
- Linear Momentum
- Conservation of Linear Momentum
- Mechanical Energy
- Kinetic Energy
- Inelastic Collision

Memory Aids:
- Momentum is always conserved in collisions (in a closed system).
- "Inelastic" implies kinetic energy is not conserved.

Quick Tips:
- Identify the type of collision (elastic, inelastic, perfectly inelastic) to determine which conservation laws apply.

367 Medium

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Understanding +4 | -1

When work done on a particle is positive
a) K.E remains constant
b) Momentum increases
c) K.E increases
d) K.E decreases

View answer and explanation

Correct option: B

b, c

Option B is correct because positive work done on a particle increases its kinetic energy, and consequently, its momentum also increases.

Solution approach

Correct Answer + Core Reason: b, c. Positive work increases kinetic energy and momentum.

Step-by-Step Solution:

Step 1: Recall the work-energy theorem: The work done on an object is equal to the change in its kinetic energy: $W = \Delta KE = KE_f - KE_i$ .

Step 2: Apply the work-energy theorem: If the work done is positive ( $W > 0$ ), then $KE_f > KE_i$ , meaning the kinetic energy increases.

Step 3: Relate kinetic energy to momentum: Kinetic energy is given by $KE = \frac{1}{2}mv^2$ , and momentum is given by $p = mv$ . Therefore, $KE = \frac{p^2}{2m}$ .

Step 4: Analyze the relationship: If kinetic energy increases, then momentum must also increase (since $KE = \frac{p^2}{2m}$ ).

Why other options are wrong:
- Option A: Kinetic energy does not remain constant when work is done on a particle. It changes according to the work-energy theorem.
- Option C: Kinetic energy increases, not decreases, when positive work is done.
- Option D: Kinetic energy does not remain constant when work is done on a particle. It changes according to the work-energy theorem.

Key Concepts:
- Work-energy theorem
- Kinetic energy
- Momentum

Memory Aids:
- Positive work = increase in KE = increase in momentum.

Quick Tips:
- Remember the work-energy theorem and the relationship between kinetic energy and momentum.

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Analyzing +4 | -1

The slope of kinetic energy and displacement curve for a particle in motion will be

View answer and explanation

Correct option: B

directly proportional to the acceleration of the particle

Option B is correct because the slope of the kinetic energy and displacement curve is directly proportional to the acceleration of the particle.

Solution approach

Correct Answer + Core Reason: directly proportional to the acceleration of the particle. The slope of the kinetic energy vs. displacement curve is directly proportional to the acceleration.

Step-by-Step Solution:
Step 1: Write the expression for kinetic energy. $E = \frac{1}{2}mv^2$ .
Step 2: Differentiate the kinetic energy with respect to displacement. $\frac{dE}{dx} = \frac{d}{dx}(\frac{1}{2}mv^2) = \frac{1}{2}m \cdot 2v \frac{dv}{dx} = mv \frac{dv}{dx}$ .
Step 3: Use the chain rule to express $\frac{dv}{dx}$ in terms of acceleration. $\frac{dv}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx} = a \cdot \frac{1}{v}$ .
Step 4: Substitute this back into the expression for $\frac{dE}{dx}$ . $\frac{dE}{dx} = mv \cdot \frac{a}{v} = ma$ .
Step 5: Analyze the result. The slope of the kinetic energy vs. displacement curve is equal to $ma$ , which is directly proportional to the acceleration $a$ .

Why other options are wrong:
- Option A: Incorrect because the slope is equal to $ma$ , not just $a$ .
- Option C: Incorrect because the slope is directly proportional to $a$ , not inversely proportional.
- Option D: Incorrect because the slope is related to the acceleration.

Key Concepts:
- Kinetic energy
- Differentiation
- Acceleration

Memory Aids:
- Remember the definition of kinetic energy: $E = \frac{1}{2}mv^2$ .

Quick Tips:
- Use the chain rule when differentiating with respect to a different variable.

369 Medium

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Application +4 | -1

A particle moves along x-axis from x=1m to x=3m under the effect of the force

F = 3{x^2} - 2x + 5\mathrm{ }N

work done in process is

View answer and explanation

Correct option: D

28 J

Option D (28 J) is correct because it is the result of integrating the force function over the given displacement range.

Solution approach

Correct Answer + Core Reason: 28 J. The work done is calculated by integrating the force function with respect to displacement.

Step-by-Step Solution:

Step 1: Given: $F = 3{x^2} - 2x + 5 \text{{ N}}$ , $x_1 = 1 \text{{ m}}$ , $x_2 = 3 \text{{ m}}$

Step 2: Work done: $W = \int_{{x_1}}^{{x_2}} {Fdx} = \int_1^3 {\left( {3{x^2} - 2x + 5} \right)dx}$

Step 3: Integrate: $W = \left[ {{x^3} - {x^2} + 5x} \right]_1^3$

Step 4: Evaluate the integral at the limits: $W = \left( {{3^3} - {3^2} + 5(3)} \right) - \left( {{1^3} - {1^2} + 5(1)} \right) = \left( {27 - 9 + 15} \right) - \left( {1 - 1 + 5} \right) = 33 - 5 = 28 \text{{ J}}$

Why other options are wrong:
- Option A: 24 J: Incorrect due to a calculation error during integration or evaluation of limits.
- Option B: 38 J: Incorrect due to a calculation error during integration or evaluation of limits.
- Option C: 18 J: Incorrect due to a calculation error during integration or evaluation of limits.

Key Concepts:
- Work done by a variable force
- Integration

Memory Aids:
- Remember the formula: $W = \int {Fdx}$

Quick Tips:
- Ensure correct integration and evaluation of limits.

370 Medium

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Application +4 | -1

Two spherical bodies of the same mass M are moving with velocities

{v_1}\,and\,{v_2}

.These collide perfectly inelastically , then the loss in kinetic energy is

View answer and explanation

Correct option: C

\frac{1}{4}M{\left( {{v_1} - {v_2}} \right)^2}

Option C (

\frac{1}{4}M{\left( {{v_1} - {v_2}} \right)^2}

) is correct because it represents the kinetic energy lost during a perfectly inelastic collision of two equal masses.

Solution approach

Correct Answer + Core Reason: $\frac{1}{4}M{\left( {{v_1} - {v_2}} \right)^2}$ . This is the correct formula for the loss of kinetic energy in a perfectly inelastic collision of two equal masses.

Step-by-Step Solution:

Step 1: Define the initial kinetic energy:
- $KE_{{initial}} = \frac{1}{2}Mv_1^2 + \frac{1}{2}Mv_2^2$

Step 2: Find the final velocity after the collision using conservation of momentum:
- $Mv_1 + Mv_2 = (M+M)v_f$
- $v_f = \frac{{Mv_1 + Mv_2}}{{2M}} = \frac{{v_1 + v_2}}{2}$

Step 3: Calculate the final kinetic energy:
- $KE_{{final}} = \frac{1}{2}(2M)v_f^2 = M \left( \frac{{v_1 + v_2}}{2} \right)^2 = \frac{M}{4}(v_1 + v_2)^2$

Step 4: Calculate the loss in kinetic energy:
- $\Delta KE = KE_{{initial}} - KE_{{final}} = \frac{1}{2}Mv_1^2 + \frac{1}{2}Mv_2^2 - \frac{M}{4}(v_1 + v_2)^2$
- $\Delta KE = \frac{M}{4}[2v_1^2 + 2v_2^2 - (v_1^2 + 2v_1v_2 + v_2^2)]$

371 Medium

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Application +4 | -1

The potential energy of

1kg

particle free to move along the

x - axis

is given by

U\,(x) = \left[ {\frac{{{x^4}}}{4} - \frac{{{x^2}}}{2}} \right]J

. The total mechanical energy of the particle is

2J

. What is the maximum speed

\left( {in\,\,m{s^{ - 1}}} \right)

of the particle?

View answer and explanation

Correct option: A

\frac{3}{{\sqrt 2 }}m{s^{ - 1}}

Option A (

\frac{3}{{\sqrt 2 }}m{s^{ - 1}}

) is correct because it is calculated by finding the minimum potential energy and using the conservation of energy to determine the maximum kinetic energy and hence the maximum speed.

Solution approach

Correct Answer + Core Reason: $\frac{3}{{\sqrt 2 }}m{s^{ - 1}}$ . The maximum speed is calculated by finding the minimum potential energy and using conservation of energy.

Step-by-Step Solution:

Step 1: Find the potential energy function: $U(x) = \frac{x^4}{4} - \frac{x^2}{2}$

Step 2: Find the minimum potential energy by taking the derivative and setting it to zero:
$\frac{dU}{dx} = x^3 - x = 0$
$x(x^2 - 1) = 0$
$x = 0, x = \pm 1$

Step 3: Determine the minimum potential energy:
$U(0) = 0$
$U(1) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \, \text{J}$
$U(-1) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \, \text{J}$
The minimum potential energy is $U_{min} = -\frac{1}{4} \, \text{J}$

Step 4: Use conservation of energy to find the maximum kinetic energy:
Total energy $E = KE + U$
$KE_{max} = E - U_{min} = 2 - \left(-\frac{1}{4}\right) = 2 + \frac{1}{4} = \frac{9}{4} \, \text{J}$

Step 5: Calculate the maximum speed:
$KE_{max} = \frac{1}{2}mv_{max}^2$
$\frac{9}{4} = \frac{1}{2}(1)v_{max}^2$
$v_{max}^2 = \frac{9}{2}$
$v_{max} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}} \, \text{m/s}$

Step 6: Final Answer: $\frac{3}{{\sqrt 2 }}m{s^{ - 1}}$

Why other options are wrong:
- Option B: $8 \, m{s^{ - 1}}$ : This is incorrect due to a miscalculation in finding the minimum potential energy or applying the conservation of energy.
- Option C: $5 \, m{s^{ - 1}}$ : This is incorrect due to a miscalculation in finding the minimum potential energy or applying the conservation of energy.
- Option D: $10 \, m{s^{ - 1}}$ : This is incorrect due to a miscalculation in finding the minimum potential energy or applying the conservation of energy.

Key Concepts:
- Potential Energy
- Kinetic Energy
- Conservation of Energy
- Finding Minimum of a Function

Memory Aids:
- Remember that total energy is the sum of kinetic and potential energy.

Quick Tips:
- Always find the minimum potential energy to determine the maximum kinetic energy.
- Be careful with the units and calculations.

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Application +4 | -1

A body of mass

{m}

attached to a thread is revolved along a vertical circle of radius

r

. If its velocity at the topmost point of the circle is

v

, the tension of the thread at the instant will be :

View answer and explanation

Correct option: C

({mv} / {r})-{mg}

Option C ((

mv} / {r})-{mg}

) is correct because it represents the net force required to provide the centripetal acceleration at the topmost point.

Solution approach

Correct Answer + Core Reason: $({mv} / {r})-{mg}$ . The tension is calculated by considering the forces at the topmost point of the circle.

Step-by-Step Solution:

Step 1: Draw a free-body diagram of the mass at the topmost point. The forces acting on the mass are tension ( $T$ ) downwards and weight ( $mg$ ) downwards.

Step 2: Apply Newton's second law in the vertical direction: $T + mg = \frac{{mv^2}}{r}$ , where $\frac{{mv^2}}{r}$ is the centripetal force.

Step 3: Solve for the tension $T$ : $T = \frac{{mv^2}}{r} - mg$ .

Why other options are wrong:
- Option A: ${mg}-{mv}^2 {r}$ : This is incorrect because the signs are reversed. The centripetal force must be greater than the weight for the mass to complete the circle.
- Option B: ${mv}^{2 / {r}}$ : This is incorrect because it doesn't account for the weight of the mass.
- Option D: $\left(m v^2 / r\right)+m g$ : This is incorrect because the weight acts in the same direction as the tension, so it should be subtracted from the centripetal force.

Key Concepts:
- Circular motion
- Centripetal force
- Newton's second law

Memory Aids:
- Remember the formula for centripetal force: $F_c = \frac{{mv^2}}{r}$

Quick Tips:
- Always draw a free-body diagram to identify all the forces acting on the object.
- Remember that the net force towards the center of the circle is the centripetal force.

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Application +4 | -1

The position x of a particle of mass 2 kg moving along straight line is given as

x = {\left( {t - 2} \right)^2}

, where t is in second and x is in metre. The work done by the force in first 2s is

View answer and explanation

Correct option: B

-16 J

Option B (-16 J) is correct because the work done is equal to the change in kinetic energy, which is negative in this case.

Solution approach

Correct Answer + Core Reason: -16 J. The work done is calculated as the change in kinetic energy using the work-energy theorem.

Step-by-Step Solution:

Step 1: Find the velocity as a function of time by differentiating the position function: $x = (t - 2)^2$ . Therefore, $v = \frac{dx}{dt} = 2(t - 2) = 2t - 4$ .

Step 2: Calculate the initial velocity at $t = 0$ : $v(0) = 2(0) - 4 = -4 \text{{ m/s}}$ .

Step 3: Calculate the final velocity at $t = 2$ : $v(2) = 2(2) - 4 = 0 \text{{ m/s}}$ .

Step 4: Calculate the initial kinetic energy: $K_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(2)(-4)^2 = 16 \text{{ J}}$ .

Step 5: Calculate the final kinetic energy: $K_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(2)(0)^2 = 0 \text{{ J}}$ .

Step 6: Calculate the work done using the work-energy theorem: $W = \Delta K = K_f - K_i = 0 - 16 = -16 \text{{ J}}$ .

Why other options are wrong:
- Option A (0 J): This would be the answer if the initial and final velocities were the same, which is not the case here.
- Option C (16 J): This is the magnitude of the work done, but the sign is incorrect. The work done is negative because the kinetic energy decreases.
- Option D (32 J): This is incorrect and likely arises from a miscalculation of the kinetic energy or a misunderstanding of the work-energy theorem.

Key Concepts:
- Work-energy theorem
- Kinetic energy
- Differentiation to find velocity

Memory Aids:
- Work-energy theorem: $W = \Delta K$

Quick Tips:
- Always remember to consider the sign of the work done.
- Differentiate the position function to find the velocity function.

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Analyzing +4 | -1

Assertion (A): Collision between two particles is not necessarly associated with physical contact between them
Reason(R): Only in physical contact, momentum transfer takes place

View answer and explanation

Correct option: C

If Assertion is true but Reason is false

Option C (If Assertion is true but Reason is false) is correct because collisions don't require physical contact, but momentum transfer still occurs.

Solution approach

Correct Answer + Core Reason: If Assertion is true but Reason is false. Collisions do not necessarily require physical contact, but momentum transfer does take place.

Step-by-Step Solution:

Step 1: Analyze the Assertion: The assertion states that a collision between two particles does not necessarily require physical contact. This is true. For example, the interaction between charged particles (e.g., alpha particle scattering) can be considered a collision without direct physical contact.

Step 2: Analyze the Reason: The reason states that momentum transfer only takes place in physical contact. This is false. Momentum transfer can occur through forces acting at a distance, such as electromagnetic forces.

Step 3: Final Answer: The assertion is true, but the reason is false.

Why other options are wrong:
- Option A (If both Assertion and Reason are true and Reason is correct explanation of Assertion): This is incorrect because the reason is false.
- Option B (If both Assertion and Reason are true but Reason is not the correct explanation of Assertion): This is incorrect because the reason is false.
- Option D (If both Assertion and Reason are false): This is incorrect because the assertion is true.

Key Concepts:
- Definition of collision
- Momentum transfer
- Forces acting at a distance

Memory Aids:
- Remember that collisions can occur without physical contact.
- Momentum transfer can occur through various types of forces.

Quick Tips:
- Consider examples of collisions without physical contact, such as interactions between charged particles.

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Understanding +4 | -1

If a particle of mass

m_1

approaches another particle of mass

m_2

, initially at rest, with speed

v_v

, the speed of the

\mathrm{cm}

of the system is

v_{\mathrm{cm}}=\frac{m_1 v_1}{m_1+m_2}

before collision takes place. If the collision is elastic, the speed of the

\mathrm{cm}

of the system after the collision is

View answer and explanation

Correct option: B

\frac{m_1 v_1}{m_1+m_2}

Option B (

\frac{{m_1 v_1}}{{m_1+m_2}}

) is correct because in an elastic collision with no external forces, the velocity of the center of mass remains constant.

Solution approach

Correct Answer + Core Reason: $\frac{{m_1 v_1}}{{m_1+m_2}}$ . The velocity of the center of mass remains constant in the absence of external forces.

Step-by-Step Solution:
Step 1: The initial velocity of the center of mass is given as $v_{\mathrm{cm}} = \frac{{m_1 v_1}}{{m_1+m_2}}$ .
Step 2: In an elastic collision, momentum and kinetic energy are conserved.
Step 3: Since there are no external forces acting on the system, the velocity of the center of mass remains constant before, during, and after the collision.
Step 4: Therefore, the speed of the center of mass after the collision is the same as before the collision.

Why other options are wrong:
- Option A: $\frac{{m_2 v_1}}{{m_1+m_2}}$ : This would be the correct answer if the initial velocity was associated with $m_2$ instead of $m_1$ .
- Option C: $\frac{{m_2-m_1}}{{m_1+m_2}}$ : This expression does not represent the velocity of the center of mass.
- Option D: Zero: The center of mass will only be zero if the initial momentum is zero, which is not the case here.

Key Concepts:
- Center of mass velocity
- Conservation of momentum
- Elastic collision

Memory Aids:
- No external forces implies constant center of mass velocity.

Quick Tips:
- Remember that the velocity of the center of mass is conserved in the absence of external forces, regardless of the type of collision.

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Understanding +4 | -1

Which of the following forces does not act on a body moving in uniform circular motion?

View answer and explanation

Correct option: A

Centripetal force

Option A (Centripetal force) is correct because centripetal force is not a fundamental force but rather the net force causing circular motion, provided by other forces.

Solution approach

Correct Answer + Core Reason: Centripetal force. Centripetal force is the net force causing circular motion, not a separate force acting on the body.

Step-by-Step Solution:

Step 1: Define uniform circular motion: Motion with constant speed along a circular path.

Step 2: Identify the forces involved: Weight, normal reaction, and friction can all act on the body.

Step 3: Explain centripetal force: Centripetal force is the net force directed towards the center of the circle, which causes the body to change direction continuously. It is provided by other forces like tension, friction, or gravitational force.

Step 4: Conclusion: The question asks for a force that *does not* act on the body. While weight, normal reaction, and friction can be present, the centripetal force is the *result* of these forces, not an additional force.

Why other options are wrong:
- Option B: The weight of the body acts downwards.
- Option C: The normal reaction acts perpendicular to the surface.
- Option D: The force of friction can act on the body, especially if there is no external driving force.

Key Concepts:
- Uniform circular motion
- Centripetal force
- Free-body diagrams

Memory Aids:
- Centripetal force is not a fundamental force; it's the net force.

Quick Tips:
- Always identify the real forces acting on the body before considering centripetal force.

377 Medium

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Application +4 | -1

A ball is spun with angular acceleration

\alpha = 6{t^2} - 2t

, where

t

is in second and

\alpha

is in

rad{s^{ - 2}}

. At

t = 0

, the ball has angular velocity of

10\,rad{s^{ - 1}}

and angular position of 4 rad. The most appropriate expression for the angular position of the ball is:

View answer and explanation

Correct option: B

\frac{{{t^4}}}{2} - \frac{{{t^3}}}{3} + 10t + 4

Option B (

\frac{{{t^4}}}{2} - \frac{{{t^3}}}{3} + 10t + 4

) is correct because it is obtained by integrating the angular acceleration twice with respect to time, using the given initial conditions for angular velocity and angular position.

Solution approach

Correct Answer + Core Reason: $\frac{{{t^4}}}{2} - \frac{{{t^3}}}{3} + 10t + 4$ . The angular position is found by integrating the angular acceleration twice.

Step-by-Step Solution:

Step 1: Given angular acceleration $\alpha = 6t^2 - 2t$ .

Step 2: Integrate angular acceleration to find angular velocity: $\omega(t) = \int \alpha dt = \int (6t^2 - 2t) dt = 2t^3 - t^2 + C_1$ .

Step 3: Use the initial condition $\omega(0) = 10 \text{{ rad/s}}$ to find $C_1$ : $10 = 2(0)^3 - (0)^2 + C_1 \Rightarrow C_1 = 10$ . So, $\omega(t) = 2t^3 - t^2 + 10$ .

Step 4: Integrate angular velocity to find angular position: $\theta(t) = \int \omega(t) dt = \int (2t^3 - t^2 + 10) dt = \frac{t^4}{2} - \frac{t^3}{3} + 10t + C_2$ .

Step 5: Use the initial condition $\theta(0) = 4 \text{{ rad}}$ to find $C_2$ : $4 = \frac{(0)^4}{2} - \frac{(0)^3}{3} + 10(0) + C_2 \Rightarrow C_2 = 4$ . So, $\theta(t) = \frac{t^4}{2} - \frac{t^3}{3} + 10t + 4$ .

Why other options are wrong:
- **Option A ( $\frac{3}{4}{t^4} - {t^2} + 10t$ ): This is incorrect because it does not include the initial angular position and the coefficients of the terms are wrong.
- Option C ( $\frac{{2{t^4}}}{3} - \frac{{{t^3}}}{6} + 10t + 12$ ): This is incorrect because the coefficients of the terms and the constant term are wrong.
- Option D ( $2{t^4} - \frac{{{t^3}}}{2} + 5t + 4$ ): This is incorrect because the coefficients of the terms are wrong.

Key Concepts:
- Angular acceleration, angular velocity, and angular position
- Integration
- Initial conditions

Memory Aids:
- Remember that angular velocity is the integral of angular acceleration, and angular position is the integral of angular velocity.

Quick Tips:**
- Always use the initial conditions to find the constants of integration.

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Application +4 | -1

A solid sphere of mass

2 \mathrm{~kg}

is rolling on a frictionless horizontal surface with velocity 6

\mathrm{m} / \mathrm{s}

. It collides on the free end of an ideal spring whose other end is fixed. The maximum compression produced in the spring will be (Force constant of the spring

=36 \mathrm{~N} / \mathrm{m}

).

View answer and explanation

Correct option: B

\sqrt{2.8} \mathrm{~m}

Option B (

\sqrt{2.8} \mathrm{~m}

) is correct because it's the maximum compression calculated by equating the kinetic energy of the rolling sphere to the potential energy stored in the spring.

Solution approach

Correct Answer + Core Reason: $\sqrt{2.8} \mathrm{~m}$ . The maximum compression is found by equating the kinetic energy of the rolling sphere to the potential energy of the spring.

Step-by-Step Solution:

Step 1: Identify the given information: mass $m = 2 \mathrm{~kg}$ , velocity $v = 6 \mathrm{~m} / \mathrm{s}$ , spring constant $k = 36 \mathrm{~N} / \mathrm{m}$ .

Step 2: Calculate the kinetic energy of the rolling sphere: $KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ , where $I = \frac{2}{5}mr^2$ and $v = r\omega$ .

Step 3: Substitute $I$ and $\omega$ into the KE equation: $KE = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$ .

Step 4: Calculate the potential energy stored in the spring: $PE = \frac{1}{2}kx^2$ , where $x$ is the maximum compression.

Step 5: Equate the kinetic energy of the sphere to the potential energy of the spring: $\frac{7}{10}mv^2 = \frac{1}{2}kx^2$ .

Step 6: Solve for $x$ : $x^2 = \frac{7}{5}\frac{mv^2}{k} = \frac{14}{10}\frac{mv^2}{k}$ .

Step 7: Substitute the given values: $x^2 = \frac{14}{10} \times \frac{2 \times 6^2}{36} = \frac{14}{10} \times \frac{2 \times 36}{36} = \frac{14}{5} = 2.8$ .

Step 8: Calculate $x$ : $x = \sqrt{2.8} \mathrm{~m}$ .

Why other options are wrong:
- **Option A ( $\sqrt{14} \mathrm{~m}$ ):** This is incorrect because it likely misses the factor of $\frac{7}{10}$ in the kinetic energy calculation or makes a calculation error.
- **Option C ( $\sqrt{1.4} \mathrm{~m}$ ): This is incorrect because it likely only considers translational kinetic energy and neglects rotational kinetic energy.
- Option D ( $\sqrt{0.7} \mathrm{~m}$ ): This is incorrect because it likely has a significant error in the calculation or formula used.

Key Concepts:**
- Kinetic Energy of Rolling Motion
- Potential Energy of a Spring
- Conservation of Energy
- Moment of Inertia of a Sphere: $I = \frac{2}{5}mr^2$

Memory Aids:
- Remember to include both translational and rotational kinetic energy for rolling objects.

Quick Tips:
- Always consider the type of rolling motion (solid sphere, hollow sphere, cylinder, etc.) when calculating kinetic energy.
- Ensure consistent units throughout the calculation.

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Analyzing +4 | -1

A body is acted upon by a constant force directed towards a fixed point. The magnitude of the force varies inversely as the square of the distance from the fixed point. What is the nature of the path?

View answer and explanation

Correct option: C

Circle

Option C is correct because a force directed towards a fixed point, with magnitude inversely proportional to the square of the distance, results in circular motion.

Solution approach

Correct Answer + Core Reason: Circle. A force directed towards a fixed point, inversely proportional to the square of the distance, leads to circular motion.

Step-by-Step Solution:

Step 1: The problem describes a central force, i.e., a force always directed towards a fixed point.

Step 2: The magnitude of the force is given as inversely proportional to the square of the distance from the fixed point: $F \propto \frac{{1}}{{r^2}}$ .

Step 3: This is the form of the gravitational force or electrostatic force. If the initial velocity is perpendicular to the force, the path will be a circle.

Step 4: If the initial velocity is not perpendicular, the path will be an ellipse. However, the question does not specify the initial conditions, and circular motion is a possible solution.

Why other options are wrong:
- Option A: Straight line: This would only occur if the initial velocity is zero or directly towards/away from the fixed point.
- Option B: Parabola: Parabolas are associated with constant acceleration, not a force inversely proportional to the square of the distance.
- Option D: Hyperbola: Hyperbolas also result from inverse square forces, but with higher initial kinetic energy than needed for a closed orbit.

Key Concepts:
- Central forces
- Inverse square law
- Circular motion
- Kepler's Laws

Memory Aids:
- Inverse square forces often lead to conic sections (circles, ellipses, parabolas, hyperbolas).

Quick Tips:
- Consider the initial conditions when determining the path of an object under a central force.

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Understanding +4 | -1

Assertion(A): If work done by conservative force is negative, then potential energy associated with that force should increase.

Reason(R): This is according to the relation,

\Delta U=-W

. Here,

\Delta U

is change in potential energy and

W

is work done by conservative force.

View answer and explanation

Correct option: A

If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

The correct answer is A (If both Assertion and Reason are correct and Reason is the correct explanation of Assertion) because the change in potential energy is the negative of the work done by a conservative force.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct and Reason is the correct explanation of Assertion. The change in potential energy is the negative of the work done by a conservative force.

Step-by-Step Solution:

Step 1: Recall the relationship between work done by a conservative force and the change in potential energy: $\Delta U = -W$ .

Step 2: If the work done by a conservative force is negative (W < 0), then the change in potential energy is positive ( $\Delta U > 0$ ), meaning the potential energy increases.

Step 3: The reason correctly explains this relationship.

Why other options are wrong:
- Option A: If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
- Option B: The reason is the correct explanation of the assertion.
- Option C: The reason is correct.
- Option D: The assertion is correct.

Key Concepts:
- Conservative Forces
- Potential Energy
- Work-Energy Theorem

Memory Aids:
- $\Delta U = -W$ (for conservative forces).

Quick Tips:
- Remember that potential energy is only defined for conservative forces.

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Understanding +4 | -1

A ball hits a floor and rebounds after an inelastic collision. In this case,
a) Total energy of the ball and the earth remains the same
b) Total momentum of the ball and the earth is conserved
c) Momentum of the ball just after the collision is same as that just before the collision
d) Mechanical energy of the ball remains the same during the collision

View answer and explanation

Correct option: C

a, b

Option C is correct because in an inelastic collision, the total energy of the ball and earth remains the same, and the total momentum of the ball and earth is conserved.

Solution approach

Correct Answer + Core Reason: a, b. Total energy of the ball and the earth remains the same, and total momentum of the ball and the earth is conserved.

Step-by-Step Solution:

Step 1: Consider the system as the ball and the Earth.

Step 2: In any collision (elastic or inelastic), the total momentum of the system is conserved if there are no external forces acting on the system. In this case, we assume no external forces.

Step 3: In an inelastic collision, kinetic energy is not conserved. Some of the kinetic energy is converted into other forms of energy, such as heat and sound.

Step 4: However, the total energy of the system (ball + Earth) *is* conserved, according to the law of conservation of energy. The kinetic energy lost by the ball is converted into other forms of energy within the system.

Step 5: The momentum of the ball alone is *not* conserved because there is an impulse acting on it during the collision.

Step 6: The mechanical energy of the ball is *not* conserved because the collision is inelastic.

Why other options are wrong:
- Option A: Mechanical energy of the ball is not conserved in an inelastic collision.
- Option B: Momentum of the ball alone is not conserved.
- Option D: Mechanical energy of the ball is not conserved, and momentum of the ball alone is not conserved.

Key Concepts:
- Conservation of Momentum
- Conservation of Energy
- Inelastic Collision
- Impulse

Memory Aids:
- Total momentum is always conserved in the absence of external forces.
- Kinetic energy is not conserved in inelastic collisions.

Quick Tips:
- Always consider the system when applying conservation laws.

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Application +4 | -1

A projectile is fired from the top of a

40 \mathrm{~m}

high cliff with an initial speed of

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

at an unknown angle. Find its speed when it hits the ground.

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the final speed is calculated using the conservation of energy principle.

Solution approach

Correct Answer + Core Reason: $58 \mathrm{~m} / \mathrm{s}$ . The final speed is calculated using conservation of energy.

Step-by-Step Solution:

Step 1: Given $h = 40 m$ and $u = 50 m/s$ .

Step 2: Using conservation of energy, $mgh + \frac{1}{2}mu^2 = \frac{1}{2}mv^2$ .

Step 3: Substituting the values, $10 \times 40 + \frac{1}{2} \times 2500 = \frac{1}{2}v^2$ .

Step 4: Simplifying, $400 + 1250 = \frac{1}{2}v^2$ , so $v^2 = 3300$ .

Step 5: Therefore, $v = \sqrt{3300} \approx 57.4 m/s \approx 58 m/s$ .

Why other options are wrong:
- **Option B ( $85 \mathrm{~m} / \mathrm{s}$ ): This is incorrect because it results from an incorrect calculation or misunderstanding of the conservation of energy.
- Option C ( $29 \mathrm{~m} / \mathrm{s}$ ): This is incorrect because it results from an incorrect calculation or misunderstanding of the conservation of energy.
- Option D ( $37 \mathrm{~m} / \mathrm{s}$ ): This is incorrect because it results from an incorrect calculation or misunderstanding of the conservation of energy.

Key Concepts:
- Conservation of energy
- Projectile motion

Memory Aids:
- Remember the conservation of energy principle.

Quick Tips:**
- Use the conservation of energy principle directly.
- Ensure consistent units before substituting values.

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Understanding +4 | -1

For a body in uniform circular motion, the

View answer and explanation

Correct option: C

Angular velocity is constant

Option C (Angular velocity is constant) is correct because uniform circular motion implies constant angular speed.

Solution approach

Correct Answer + Core Reason: Angular velocity is constant. In uniform circular motion, the angular speed remains constant.

Step-by-Step Solution:

Step 1: Define uniform circular motion: Motion in a circle with constant speed.

Step 2: Relate linear velocity and angular velocity: $v = r\omega$ , where $v$ is linear speed, $r$ is the radius, and $\omega$ is angular speed.

Step 3: Since the speed is constant and the radius is constant, the angular speed must also be constant. Therefore, the angular velocity is constant.

Why other options are wrong:
- Option A (Velocity is constant): Velocity is not constant because the direction is constantly changing.
- Option B (Acceleration is constant): Acceleration is not constant because the direction is constantly changing (centripetal acceleration).
- Option D (Displacement is constant): Displacement is not constant because the object is continuously moving around the circle.

Key Concepts:
- Uniform circular motion
- Angular velocity
- Centripetal acceleration

Memory Aids:
- Uniform circular motion: constant speed, changing velocity.

Quick Tips:
- Remember that velocity is a vector, so changing direction means changing velocity.

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Understanding +4 | -1

Statement I In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision. Statement II Inan elastic collision, the linear momentum of the system is conserved.

View answer and explanation

Correct option: B

Statement I is true, Statement II is true; statement II is not correct explanation for statement I

Option B (Statement I is true, Statement II is true; statement II is not correct explanation for statement I) is correct because both statements are true about elastic collisions, but the conservation of momentum doesn't directly explain the relative speed relationship.

Solution approach

Correct Answer + Core Reason: Statement I is true, Statement II is true; statement II is not correct explanation for statement I. Both statements are true for elastic collisions, but one doesn't explain the other.

Step-by-Step Solution:

Step 1: Analyze Statement I: In an elastic collision between two bodies, the relative speed of the bodies after the collision is equal to the relative speed before the collision. This is true. The coefficient of restitution (e) is 1 for elastic collisions, meaning relative speed of approach = relative speed of separation.

Step 2: Analyze Statement II: In an elastic collision, the linear momentum of the system is conserved. This is also true. Linear momentum is always conserved in collisions when no external forces are acting.

Step 3: Determine if Statement II explains Statement I. While both are true, the conservation of linear momentum doesn't directly explain why the relative speeds are equal. The equality of relative speeds is a consequence of the conservation of kinetic energy in an elastic collision.

Why other options are wrong:
- Option A (Statement I is true ,statement II is true; statement II is a correct explanation for statement I): Statement II does not directly explain Statement I.
- Option C (Statement I is true, Statement II is false): Statement II is true.
- Option D (Statement I is false, Statement II is True): Statement I is true.

Key Concepts:
- Elastic collision
- Coefficient of restitution
- Conservation of linear momentum
- Conservation of kinetic energy

Memory Aids:
- Elastic collision: e = 1, KE conserved.

Quick Tips:
- Remember that elastic collisions conserve both momentum and kinetic energy.

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Understanding +4 | -1

A boy carrying a box on his head is walking on a level road from one place to another on a straight road is doing no work. The statement is

View answer and explanation

Correct option: B

correct

Option B (correct) is correct because the work done is zero when the force (weight of the box) is perpendicular to the displacement (horizontal movement).

Solution approach

Correct Answer + Core Reason: correct. Work done is zero because the force is perpendicular to the displacement.

Step-by-Step Solution:

Step 1: Define Work Done: Work done ( $W$ ) is defined as $W = Fd\cos{\theta}$ , where $F$ is the force, $d$ is the displacement, and $\theta$ is the angle between the force and the displacement.

Step 2: Analyze the Situation: The force acting on the box is the weight, which acts vertically downwards. The displacement is horizontal.

Step 3: Angle Between Force and Displacement: The angle between the force (weight) and the displacement is 90 degrees.

Step 4: Calculate Work Done: $W = Fd\cos{90°} = Fd \times 0 = 0$ .

Step 5: Conclusion: Therefore, the boy is doing no work.

Why other options are wrong:
- Option A: The statement is completely correct, not partly correct.
- Option C: The statement is correct, not incorrect.
- Option D: The data is sufficient to determine that the work done is zero.

Key Concepts:
- Work Done
- Force and Displacement
- Angle Between Force and Displacement

Memory Aids:
- Work = Force x Distance x Cos(theta)

Quick Tips:
- Remember that work done is zero when the force is perpendicular to the displacement.

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Understanding +4 | -1

Which of the following statement(s) is/are correct?

I. Spring force is a conservative force.

II. Potential energy is defined only for conservative forces.

View answer and explanation

Correct option: C

Both I and II

Option C (Both I and II) is correct because spring force is a conservative force, and potential energy is defined only for conservative forces.

Solution approach

Correct Answer + Core Reason: Both I and II. Spring force is conservative, and potential energy is defined only for conservative forces.

Step-by-Step Solution:

Step 1: Define conservative force: A force is conservative if the work done by the force is independent of the path taken.

Step 2: Spring force is a conservative force because the work done by the spring force depends only on the initial and final positions of the spring.

Step 3: Potential energy is defined only for conservative forces because the work done by a conservative force can be expressed as the change in potential energy.

Step 4: Therefore, both statements are correct.

Why other options are wrong:
- Option A: Statement II is also correct.
- Option B: Statement I is also correct.
- Option D: Both statements are correct.

Key Concepts:
- Conservative Force
- Potential Energy
- Spring Force

Memory Aids:
- Remember that potential energy is only defined for conservative forces.

Quick Tips:
- Examples of conservative forces include gravitational force, spring force, and electrostatic force.

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Application +4 | -1

A simple pendulum is oscillating with an angular amplitude

{60^0}

. If mass of bob is 50 gm, then the tension in the string at mean position (in N) is

\left( {\mathrm{g = 10}\,\,\mathrm{m/}{\mathrm{s}^\mathrm{2}}} \right)

View answer and explanation

Correct option: A

1

Option A (1) is correct because the tension in the string at the mean position is calculated considering both the weight and the centripetal force.

Solution approach

Correct Answer + Core Reason: 1. The tension in the string at the mean position is calculated using T = mg + (mv²)/l, where v is the velocity at the mean position.

Step-by-Step Solution:

Step 1: Given: Angular amplitude $\theta = 60^\circ$ , mass of bob $m = 50 \text{{ gm}} = 0.05 \text{{ kg}}$ , $g = 10 \text{{ m/s}}^2$ .

Step 2: The velocity at the mean position is given by $v = \sqrt{2gl(1 - \cos\theta)}$ .

Step 3: The tension in the string at the mean position is $T = mg + \frac{{mv^2}}{l} = mg + \frac{m}{l}2gl(1 - \cos\theta) = mg + 2mg(1 - \cos\theta)$ .

Step 4: Substituting the given values: $T = 0.05 \times 10 + 2 \times 0.05 \times 10 (1 - \cos 60^\circ) = 0.5 + 1(1 - 0.5) = 0.5 + 0.5 = 1 \text{{ N}}$ .

Why other options are wrong:
- Option B: This option is incorrect because it does not match the calculated tension using the correct formula and given values.
- Option C: This option is incorrect because it does not match the calculated tension using the correct formula and given values.
- Option D: This option is incorrect because it does not match the calculated tension using the correct formula and given values.

Key Concepts:
- Tension in a string of a simple pendulum.
- Velocity at the mean position.

Memory Aids:
- Remember the formula for tension at the mean position: $T = mg + \frac{{mv^2}}{l}$ .

Quick Tips:
- Convert all units to SI units before performing calculations.

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Application +4 | -1

Find W.D in stretching a storing of spring constant

\mathrm{K}

from extension

x_0

to extension

3 x_0

?

View answer and explanation

Correct option: D

4kx_0^2

Option D (

4kx_0^2

) is correct because it is the work done in stretching the spring from

x_0

to

3x_0

, calculated as the change in potential energy.

Solution approach

Correct Answer + Core Reason: $4kx_0^2$ . The work done is the change in potential energy of the spring.

Step-by-Step Solution:

Step 1: Identify the given information: Spring constant = K, initial extension = $x_0$ , final extension = $3x_0$ .

Step 2: Recall the formula for the potential energy stored in a spring: $U = \frac{1}{2}kx^2$ .

Step 3: Calculate the initial potential energy: $U_i = \frac{1}{2}k(x_0)^2$ .

Step 4: Calculate the final potential energy: $U_f = \frac{1}{2}k(3x_0)^2 = \frac{1}{2}k(9x_0^2)$ .

Step 5: Calculate the work done: $W = U_f - U_i = \frac{1}{2}k(9x_0^2) - \frac{1}{2}k(x_0)^2 = \frac{1}{2}k(8x_0^2) = 4kx_0^2$ .

Why other options are wrong:
- Option A: Incorrect calculation of the change in potential energy.
- Option B: Incorrect calculation of the change in potential energy.
- Option C: Incorrect calculation of the change in potential energy.

Key Concepts:
- Work Done by a Spring
- Potential Energy of a Spring

Memory Aids:
- Remember the formula: $U = \frac{1}{2}kx^2$ .

Quick Tips:
- Work done is equal to the change in potential energy.

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Understanding +4 | -1

A body of mass m kg is lifted by a man to a height of one metre in 30 sec. Another man lifts the same mass to the same height in 60 sec. The work done by them are in the ratio

View answer and explanation

Correct option: B

1:1

Option B (

1:1

) is correct because the work done only depends on the force (weight) and the displacement (height), not the time taken.

Solution approach

Correct Answer + Core Reason: $1:1$ . Work done is independent of time.

Step-by-Step Solution:

Step 1: Define work done: $W = F \times d \times \cos{\theta}$ , where $F$ is the force, $d$ is the displacement, and $\theta$ is the angle between them.

Step 2: In this case, the force is the weight of the object ( $mg$ ), the displacement is the height ( $h$ ), and the angle is 0 degrees (since the force and displacement are in the same direction).

Step 3: Calculate the work done by the first man: $W_1 = mgh$ .

Step 4: Calculate the work done by the second man: $W_2 = mgh$ .

Step 5: Find the ratio of the work done: $\frac{{W_1}}{{W_2}} = \frac{{mgh}}{{mgh}} = 1$ .

Step 6: Express the ratio as $1:1$ .

Why other options are wrong:
- Option A: $1:2$ is incorrect because it assumes work is inversely proportional to time, which is related to power, not work itself.
- Option C: $2:1$ is incorrect because it assumes work is directly proportional to time, which is related to power, not work itself.
- Option D: $4:1$ is incorrect as work done is independent of time.

Key Concepts:
- Work Done: $W = Fd\cos{\theta}$
- Gravitational Potential Energy: $U = mgh$

Memory Aids:
- Work is force times distance. Time is irrelevant for calculating work done.

Quick Tips:
- Distinguish between work and power. Power is the rate of doing work (work/time).

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Application +4 | -1

A particle of mass 100 gm moves in a potential well given by

U=8 x^2-4 x+400

Joule. Find its acceleration at a distance of

25 \mathrm{~cm}

from equilibrium in the positive direction

View answer and explanation

Correct option: C

-40 \mathrm{~ms}^{-1}

Option C (

-40 \mathrm{~ms}^{-1}

) is correct because it is the acceleration calculated from the force derived from the potential well, evaluated at the given displacement from equilibrium.

Solution approach

Correct Answer + Core Reason: $-40 \mathrm{~ms}^{-1}$ . The acceleration is calculated from the force derived from the potential energy function.

Step-by-Step Solution:

Step 1: Find the force $F$ from the potential energy $U$ : $F = -\frac{dU}{dx}$ .

Step 2: Calculate the derivative of the potential energy: $\frac{dU}{dx} = \frac{d}{dx}(8x^2 - 4x + 400) = 16x - 4$ .

Step 3: Calculate the force: $F = -(16x - 4) = 4 - 16x$ .

Step 4: Find the equilibrium position $x_0$ by setting $F = 0$ : $4 - 16x_0 = 0 \Rightarrow x_0 = \frac{4}{16} = \frac{1}{4} \mathrm{~m} = 0.25 \mathrm{~m} = 25 \mathrm{~cm}$ .

Step 5: Calculate the position $x$ at a distance of $25 \mathrm{~cm}$ from equilibrium in the positive direction: $x = x_0 + 25 \mathrm{~cm} = 25 \mathrm{~cm} + 25 \mathrm{~cm} = 50 \mathrm{~cm} = 0.5 \mathrm{~m}$ .

Step 6: Calculate the force at $x = 0.5 \mathrm{~m}$ : $F = 4 - 16(0.5) = 4 - 8 = -4 \mathrm{~N}$ .

Step 7: Calculate the acceleration $a$ using Newton's second law: $a = \frac{F}{m} = \frac{-4 \mathrm{~N}}{0.1 \mathrm{~kg}} = -40 \mathrm{~ms}^{-2}$ .

Why other options are wrong:
- Option A: This answer is incorrect. It might be obtained by not considering the negative sign of the force or by making a calculation error.
- Option B: This answer is incorrect. It might be obtained by not considering the negative sign of the force or by making a calculation error.
- Option D: This answer is incorrect. It might be obtained by not considering the negative sign of the force or by making a calculation error.

Key Concepts:
- Potential energy
- Force as the negative gradient of potential energy
- Equilibrium
- Newton's second law

Memory Aids:
- $F = -\frac{dU}{dx}$
- $F = ma$

Quick Tips:
- Remember to find the equilibrium position before calculating the force at a given displacement.
- Pay attention to the units and convert them to the SI system if necessary.

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