Question 1

   Question: A balloon and its content having mass M is moving up with an acceleration $a$. The mass that must be released from the content so that the balloon starts moving up with an acceleration $3a$ will be (Take $g$ as acceleration due to gravity). (Note: The source text contains some formatting artifacts, but the physics context and solution indicate these standard variables).

   Options: 

    A. $\frac{3Ma}{2a+g}$

    B. $\frac{2Ma}{3a+g}$

    C. $\frac{3Ma}{2a-g}$

    D. $\frac{2Ma}{3a-g}$

   Correct Answer: B

   Year: JEE Main 2025 (Online) 28th January Evening Shift

   Solution: 

    $F - mg = ma$

    $F = ma + mg$

    $F - (m - x)g = (m - x)3a$

    Put $F$:

    $ma + mg - mg + xg = 3ma - 3xa$

    $x = \frac{2ma}{g + 3a}$ 

   Step Solution:

    1.  Equation 1: For the initial state moving up with acceleration $a$: $F - Mg = Ma$, which gives the upward buoyancy force $F = M(g + a)$.

    2.  Equation 2: Let $x$ be the mass released. The new mass is $(M - x)$, and it moves up with acceleration $3a$: $F - (M - x)g = (M - x)3a$.

    3.  Substitution: Substitute $F$ from Step 1 into Equation 2: $M(g + a) - (M - x)g = (M - x)3a$.

    4.  Expansion: $Mg + Ma - Mg + xg = 3Ma - 3xa$.

    5.  Solving for x: Rearrange terms to get $xg + 3xa = 2Ma \rightarrow x(g + 3a) = 2Ma \rightarrow x = \frac{2Ma}{g + 3a}$.

   Difficulty Level: Medium

   The Concept Name: Newton’s Second Law ($F = ma$)

   Short cut solution: Use the general formula for mass change in a balloon: $m_{released} = \frac{M(a_{final} - a_{initial})}{g + a_{final}}$. Here, $\frac{M(3a - a)}{g + 3a} = \frac{2Ma}{g + 3a}$.

 Question 7

   Question: A body of mass 2 kg moving with velocity of $\vec{v}_{in} = 3\hat{i} + 4\hat{j} \text{ ms}^{-1}$ enters into a constant force field of 6N directed along positive Z-axis. If the body remains in the field for a period of $\frac{5}{3}$ seconds, then velocity of the body when it emerges from force field is.

   Options: 

    A. $3\hat{i} + 4\hat{j} + \sqrt{5}\hat{k}$

    B. $4\hat{i} + 3\hat{j} + 5\hat{k}$

    C. $3\hat{i} + 4\hat{j} - 5\hat{k}$

    D. $3\hat{i} + 4\hat{j} + 5\hat{k}$

   Correct Answer: D

   Year: JEE Main 2025 (Online) 8th April Evening Shift

   Solution: 

    $\vec{a} = \frac{\vec{F}}{m} = \frac{6\hat{k}}{2} = 3\hat{k} \text{ ms}^{-2}$

    $\vec{u} = 3\hat{i} + 4\hat{j} \text{ ms}^{-1}$

    $t = \frac{5}{3} \text{ seconds}$

    $\vec{v} = \vec{u} + \vec{a}t = (3\hat{i} + 4\hat{j}) + 3\hat{k}(\frac{5}{3}) = 3\hat{i} + 4\hat{j} + 5\hat{k}$

   Step Solution:

    1.  Identify Force Vector: The force is 6N along the positive Z-axis, so $\vec{F} = 6\hat{k}$.

    2.  Calculate Acceleration: Use $\vec{a} = \frac{\vec{F}}{m} = \frac{6\hat{k}}{2} = 3\hat{k} \text{ ms}^{-2}$.

    3.  Use Kinematic Equation: Apply $\vec{v} = \vec{u} + \vec{a}t$ for vector addition.

    4.  Substitute Values: $\vec{v} = (3\hat{i} + 4\hat{j}) + (3\hat{k} \times \frac{5}{3})$.

    5.  Simplify: The 3 in the acceleration and the 3 in the denominator of the time cancel out, leaving $5\hat{k}$. Result: $3\hat{i} + 4\hat{j} + 5\hat{k}$.

   Difficulty Level: Easy

   The Concept Name: Force and Kinematics in Vector Form

   Short cut solution: Since the force is only in the Z-direction, the X and Y components of velocity remain unchanged ($3\hat{i} + 4\hat{j}$). Simply calculate the change in the Z component: $\Delta v_z = a_z t = (\frac{6}{2}) \times (\frac{5}{3}) = 5$. Add this to the initial velocity to get $3\hat{i} + 4\hat{j} + 5\hat{k}$.

 Question 22

   Question: A body of mass 4 kg experiences two forces. (Note: Total force is given in the solution as $8\hat{i} + 4\hat{j} + 4\hat{k}$). The acceleration acting on the body is.

   Options: 

    A.$-2\hat{i} + \hat{j} + \hat{k}$

    B. $4\hat{i} + 2\hat{j} + 2\hat{k}$

    C. $2\hat{i} + \hat{j} + \hat{k}$

    D. $2\hat{i} + 3\hat{j} + 3\hat{k}$

   Correct Answer: C

   Year: 1-Feb-2024 Shift 2

   Solution: 

    Net force $= 8\hat{i} + 4\hat{j} + 4\hat{k}$

    $\vec{a} = \frac{\vec{F}}{m} = \frac{8\hat{i} + 4\hat{j} + 4\hat{k}}{4} = 2\hat{i} + \hat{j} + \hat{k}$

   Step Solution:

    1.  Determine Net Force: Sum the individual forces acting on the body (given as $8\hat{i} + 4\hat{j} + 4\hat{k}$).

    2.  Newton's Second Law: Use the vector form of the law: $\vec{a} = \frac{\vec{F}_{net}}{m}$.

    3.  Substitute Values: Divide the net force by the mass (4 kg): $\vec{a} = \frac{8\hat{i} + 4\hat{j} + 4\hat{k}}{4}$.

    4.  Vector Division: Divide each component individually: $a_x = \frac{8}{4}, a_y = \frac{4}{4}, a_z = \frac{4}{4}$.

    5.  Final Vector: Combine the components to get $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$.

   Difficulty Level: Easy

   The Concept Name: Newton’s Second Law (Vector Addition)

   Short cut solution: Divide the magnitude of the net force components directly by the mass: $8/4=2, 4/4=1, 4/4=1$. The acceleration is $2\hat{i} + 1\hat{j} + 1\hat{k}$.

 Question 28

   Question: Force acts for 20 s on a body of mass 20 kg, starting from rest, after which the force ceases and then body describes 50m in the next 10s. The value of force will be.

   Options: 

    A. 40N

    B. 5N

    C. 20N

    D. 10N

   Correct Answer: B

   Year: 29-Jan-2023 Shift 2

   Solution: 

    $50 = V \times 10 \Rightarrow V = 5 m/s$

    $V = 0 + a \times 20$

    $5 = a \times 20 \Rightarrow a = \frac{1}{4} m/s^2$

    $F = ma = 20 \times \frac{1}{4} = 5 N$

   Step Solution:

    1.  Find Final Velocity: After the force ceases, the body moves with constant velocity. $v = \frac{\text{distance}}{\text{time}} = \frac{50}{10} = 5 \text{ m/s}$.

    2.  Determine Acceleration: Use the first kinematic equation for the period the force was active: $v = u + at \rightarrow 5 = 0 + a(20)$.

    3.  Solve for $a$: $a = \frac{5}{20} = 0.25 \text{ m/s}^2$.

    4.  Newton's Second Law: Apply $F = m \times a$.

    5.  Final Calculation: $F = 20 \text{ kg} \times 0.25 \text{ m/s}^2 = 5 \text{ N}$.

   Difficulty Level: Medium

   The Concept Name: Newton’s Second Law and Kinematics

   Short cut solution: The impulse $F \cdot \Delta t$ equals the change in momentum $m \cdot \Delta v$. Since it starts from rest, $F \times 20 = 20 \times 5$, which directly gives $F = 5 \text{ N}$.

 Question 33

   Question: As shown in figure, a 70 kg garden roller is pushed with a force of $\vec{F} = 200 \text{ N}$ at an angle of $30^\circ$ with horizontal. The normal reaction on the roller is.

   Options: 

    A. 800√2N

    B. 600N

    C. 800N

    D. 200√3N

   Correct Answer: C

   Year: 2023 (Session Context)

   Solution: 

    $N = mg + F \sin 30^\circ$

    $= 700 + 200 \times \frac{1}{2} = 800 \text{ newton}$

   Step Solution:

    1.  Identify Vertical Forces: The forces acting vertically are Weight ($mg$) down, the vertical component of push ($F \sin \theta$) down, and Normal reaction ($N$) up.

    2.  Equilibrium Equation: In the vertical direction, $N = mg + F \sin 30^\circ$.

    3.  Calculate Weight: $mg = 70 \text{ kg} \times 10 \text{ m/s}^2 = 700 \text{ N}$.

    4.  Calculate Push Component: $F \sin 30^\circ = 200 \text{ N} \times 0.5 = 100 \text{ N}$.

    5.  Calculate Total $N$: $N = 700 + 100 = 800 \text{ N}$.

   Difficulty Level: Easy

   The Concept Name: Resolution of Forces and Statics

   Short cut solution: Pushing increases the normal reaction by the vertical component of the force. $N = \text{Weight} + F \sin \theta = 700 + 200(0.5) = 800 \text{ N}$.

 Question 45

   Question: The position vector of a particle related to time $t$ is given by $\vec{r} = (10t \hat{i} + 15t^2 \hat{j} + 7 \hat{k}) \text{ m}$. The direction of net force experienced by the particle is.

   Options: 

    A. Positive z-axis

    B. In x-y plane

    C. Positive y-axis

    D. Positive x-axis

   Correct Answer: C

   Year: 15-Apr-2023 Shift 1

   Solution: 

    $\vec{v} = \frac{d\vec{r}}{dt} = 10 \hat{i} + 30t \hat{j}$

    $\vec{a} = \frac{d\vec{v}}{dt} = 30 \hat{j}$

    $\vec{F}$ along (+) y-axis

   Step Solution:

    1.  Differentiate Position: Find the velocity vector $\vec{v}$ by differentiating $\vec{r}$ with respect to $t$: $\vec{v} = 10 \hat{i} + 30t \hat{j} + 0 \hat{k}$.

    2.  Differentiate Velocity: Find the acceleration vector $\vec{a}$ by differentiating $\vec{v}$ with respect to $t$: $\vec{a} = 0 \hat{i} + 30 \hat{j}$.

    3.  Apply Newton's Law: Force direction is the same as acceleration direction ($\vec{F} = m\vec{a}$).

    4.  Identify Direction: The acceleration is only in the $+ \hat{j}$ direction.

    5.  Final Result: The direction is the Positive y-axis.

   Difficulty Level: Easy

   The Concept Name: Vector Differentiation and Newton’s Second Law

   Short cut solution: Look for the term with the highest power of $t$ that is at least $t^2$. Only $15t^2 \hat{j}$ will produce a non-zero acceleration upon double differentiation. Since the coefficient is positive and attached to $\hat{j}$, the force is along the positive y-axis.

Question 47

   Question: A force on an object of mass 100g is $(10 \hat{i} + 5 \hat{j}) \text{ N}$. The position of that object at $t = 2\text{ s}$ is $(a\hat{i} + b\hat{j}) \text{ m}$ after starting from rest. The value of $a/b$ will be.

   Options: (None provided in the source for this specific numerical-type question).

   Correct Answer: 2.

   Year: 25-Jun-2022-Shift-1.

   Solution: 

    $\vec{F} = m\vec{a}$

    $\Rightarrow \vec{a} = 100\hat{i} + 50\hat{j}$

    So, $\vec{S} = \frac{1}{2}\vec{a}t^2 = \frac{1}{2}(100\hat{i} + 50\hat{j})2^2 = 200\hat{i} + 100\hat{j} \text{ m}$

    so $a = 200 \text{ m}$ and $b = 100 \text{ m}$ so $a/b = 2$.

   Step Solution:

    1.  Convert Mass: Convert the mass to kilograms: $m = 100 \text{ g} = 0.1 \text{ kg}$.

    2.  Find Acceleration: Use Newton's Second Law: $\vec{a} = \frac{\vec{F}}{m} = \frac{10\hat{i} + 5\hat{j}}{0.1} = 100\hat{i} + 50\hat{j} \text{ m/s}^2$.

    3.  Apply Displacement Formula: Since the object starts from rest, use $\vec{s} = \frac{1}{2}\vec{a}t^2$.

    4.  Calculate Position: Substitute $t = 2$: $\vec{s} = \frac{1}{2}(100\hat{i} + 50\hat{j})(2)^2 = 2(100\hat{i} + 50\hat{j}) = 200\hat{i} + 100\hat{j} \text{ m}$.

    5.  Calculate Ratio: Identify $a = 200$ and $b = 100$, then $\frac{a}{b} = \frac{200}{100} = 2$.

   Difficulty Level: Easy

   The Concept Name: Newton’s Second Law and Kinematics (Vector Form)

   Short cut solution: Since the object starts from rest and both components of force are constant, the ratio of displacements $a/b$ is equal to the ratio of the force components $F_x/F_y = 10/5 = 2$.

 Question 59

   Question: A wire of length $L$ is hanging from a fixed support. The length changes to $L_1$ and $L_2$ when masses 1 kg and 2 kg are suspended respectively from its free end. Then the value of $L$ is equal to.

   Options: 

    A. $\sqrt{L_1L_2}$

    B. $L_1 + L_2$

    C. $2L_1 - L_2$

    D. $3L_1 - 2L_2$.

   Correct Answer: C.

   Year: 29-Jun-2022-Shift-1.

   Solution: 

    By Hooke's Law so $F \propto \Delta L$

    $\frac{F_1}{F_2} = \frac{\Delta L_1}{\Delta L_2}$

    $\frac{10}{20} = \frac{(L_1 - L)}{(L_2 - L)}$

    $L = 2L_1 - L_2$.

   Step Solution:

    1.  State Proportionality: According to Hooke's Law, force is proportional to elongation: $F \propto (L_{final} - L_{original})$.

    2.  Set up Ratio: Create a ratio for the two cases: $\frac{F_1}{F_2} = \frac{L_1 - L}{L_2 - L}$.

    3.  Substitute Forces: Since $F = mg$, the ratio of forces is the ratio of masses: $\frac{1}{2} = \frac{L_1 - L}{L_2 - L}$.

    4.  Cross-Multiply: $L_2 - L = 2(L_1 - L) \rightarrow L_2 - L = 2L_1 - 2L$.

    5.  Rearrange for L: $2L - L = 2L_1 - L_2 \rightarrow L = 2L_1 - L_2$.

   Difficulty Level: Easy

   The Concept Name: Hooke’s Law (Elasticity)

   Short cut solution: The elongation for a 1 kg mass is $(L_1 - L)$. Doubling the mass to 2 kg must double the elongation: $2(L_1 - L) = (L_2 - L)$. Solving this linear equation immediately gives $L = 2L_1 - L_2$.

 Question 60

   Question: A block of metal weighing 2 kg is resting on a frictionless plane. It is struck by a jet releasing water at a rate of $1 \text{ kg s}^{-1}$ and at a speed of $10 \text{ m s}^{-1}$. Then, the initial acceleration of the block, in $\text{m s}^{-2}$, will be.

   Options: 

    A. 3

    B. 6

    C. 5

    D. 4.

   Correct Answer: C.

   Year: 29-Jun-2022-Shift-1.

   Solution: 

    $F = \rho v^2 a$ (Note: This is given as the impact force formula in the source)

    $\Rightarrow 10 \times 1 = 2 \times \text{acceleration}$

    $\Rightarrow \text{Acc.} = 5 \text{ m/s}^2$.

   Step Solution:

    1.  Identify Flow Rate: The rate at which water hits the block is $\frac{dm}{dt} = 1 \text{ kg/s}$.

    2.  Calculate Impact Force: The force exerted by the water jet is $F = v \frac{dm}{dt}$ (rate of change of momentum).

    3.  Substitute Values: $F = 10 \text{ m/s} \times 1 \text{ kg/s} = 10 \text{ N}$.

    4.  Apply Second Law: The acceleration of the block is $a = \frac{F}{M_{block}}$.

    5.  Final Calculation: $a = \frac{10 \text{ N}}{2 \text{ kg}} = 5 \text{ m/s}^2$.

   Difficulty Level: Easy

   The Concept Name: Newton’s Second Law (Thrust/Variable Mass)

   Short cut solution: Acceleration $a = \frac{\text{Jet Velocity} \times \text{Flow Rate}}{\text{Block Mass}} = \frac{10 \times 1}{2} = 5 \text{ m/s}^2$.

 Question 72

   Question: A balloon has mass of 10g in air. The air escapes from the balloon at a uniform rate with velocity 4.5 cm/s. If the balloon shrinks in 5s completely. Then, the average force acting on that balloon will be (in dyne).

   Options: 

    A. 3

    B. 9

    C. 12

    D. 18

   Correct Answer: B

   Year: 28-Jul-2022-Shift-1

   Solution: $F = \frac{dm}{dt} v$; $\Gamma = \frac{10g}{5s} (4.5 \frac{cm}{s}) = 9 \frac{g \cdot cm}{s^2} = 9$ dyne.

   Step Solution:

    1.  Identify Given Values: Initial mass of air $m = 10 \text{ g}$, time to escape $t = 5 \text{ s}$, and velocity $v = 4.5 \text{ cm/s}$.

    2.  Calculate Mass Flow Rate: Determine the rate of mass change ($\frac{dm}{dt}$) by dividing total mass by time: $10 \text{ g} / 5 \text{ s} = 2 \text{ g/s}$.

    3.  Newton's Second Law for Variable Mass: Use the thrust force formula: $F = v \times (\frac{dm}{dt})$.

    4.  Substitute and Calculate: $F = 4.5 \text{ cm/s} \times 2 \text{ g/s}$.

    5.  Final Result: $F = 9 \text{ dyne}$.

   Difficulty Level: Easy

   The Concept Name: Newton’s Second Law (Variable Mass/Thrust Force)

   Short cut solution: Average Force = $\frac{\Delta P}{\Delta t} = \frac{m \cdot v}{t} = \frac{10 \times 4.5}{5} = 9 \text{ dyne}$.

 Question 77

   Question: A boy pushes a box of mass 2kg with a force $\vec{F} = (20\hat{i} + 10\hat{j}) \text{ N}$ on a frictionless surface. If the box was initially at rest, then ........ m is displacement along the X-axis after 10s.

   Options: (Numerical type question; no options provided in the source).

   Correct Answer: 500

   Year: 26 Feb 2021 Shift 1

   Solution: $m = 2 \text{ kg}$; $F = 20\hat{i} + 10\hat{j} \text{ N}$; $u = 0$; $t = 10 \text{ s}$. $a = F/m = (20\hat{i} + 10\hat{j})/2 = 10\hat{i} + 5\hat{j}$. $s_x = u_x t + \frac{1}{2} a_x t^2 = 0 + \frac{1}{2} \times 10 \times (10)^2 = 500 \text{ m}$.

   Step Solution:

    1.  Find Acceleration Vector: Use $\vec{a} = \vec{F}/m = \frac{20\hat{i} + 10\hat{j}}{2} = 10\hat{i} + 5\hat{j} \text{ m/s}^2$.

    2.  Extract X-component: The acceleration along the X-axis ($a_x$) is $10 \text{ m/s}^2$.

    3.  Initial Conditions: The box starts from rest, so $u_x = 0$.

    4.  Kinematic Equation: Apply $s_x = u_x t + \frac{1}{2} a_x t^2$ for the X-axis.

    5.  Final Calculation: $s_x = 0 + \frac{1}{2}(10)(10)^2 = 5 \times 100 = 500 \text{ m}$.

   Difficulty Level: Easy

   The Concept Name: Newton’s Second Law and Kinematics (Component Method)

   Short cut solution: Displacement $s_x = \frac{1}{2} (\frac{F_x}{m}) t^2 = \frac{1}{2} (\frac{20}{2}) \times 100 = 5 \times 100 = 500 \text{ m}$.

 Question 83

   Question: A body of mass 2kg moves under a force of $(2\hat{i} + 3\hat{j} + 5\hat{k}) \text{ N}$. It starts from rest and was at the origin initially. After 4s, its new coordinates are (8, b, 20). The value of b is.

   Options: (Numerical type question; answer rounded to nearest integer).

   Correct Answer: 12

   Year: 16 Mar 2021 Shift 2

   Solution: $\vec{a} = \frac{\vec{F}}{m} = \frac{2\hat{i} + 3\hat{j} + 5\hat{k}}{2} = 1\hat{i} + 1.5\hat{j} + 2.5\hat{k}$. $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(1\hat{i} + 1.5\hat{j} + 2.5\hat{k})(4)^2 = 8\hat{i} + 12\hat{j} + 20\hat{k}$. Comparing coordinates: $b = 12$.

   Step Solution:

    1.  Calculate Acceleration: $\vec{a} = \frac{\vec{F}}{m} = \frac{2\hat{i} + 3\hat{j} + 5\hat{k}}{2} = 1\hat{i} + 1.5\hat{j} + 2.5\hat{k} \text{ m/s}^2$.

    2.  Focus on Y-component: To find '$b$', we only need the displacement along the Y-axis: $a_y = 1.5 \text{ m/s}^2$.

    3.  Kinematic Equation for Y: Since $u_y = 0$, $s_y = \frac{1}{2} a_y t^2$.

    4.  Substitute Values: $s_y = \frac{1}{2} \times 1.5 \times (4)^2 = 0.75 \times 16$.

    5.  Identify coordinate: $s_y = 12 \text{ m}$, therefore the coordinate $b = 12$.

   Difficulty Level: Easy

   The Concept Name: Newton’s Second Law and Kinematic Equations in Vector Form

   Short cut solution: $b = \frac{1}{2} (\frac{F_y}{m}) t^2 = \frac{1}{2} (\frac{3}{2}) \times 16 = \frac{3}{4} \times 16 = 12$.

 Question 96

   Question: A force $\vec{F} = (40\hat{i} + 10\hat{j}) \text{ N}$ acts on a body of mass 5 kg. If the body starts from rest, its position vector $\vec{r}$ at time $t = 10\text{ s}$, will be :

   Options:

    A. $(100\hat{i} + 400\hat{j})\text{ m}$

    B. $(100\hat{i} + 100\hat{j})\text{ m}$

    C. $(400\hat{i} + 100\hat{j})\text{ m}$

    D. $(400\hat{i} + 400\hat{j})\text{ m}$

   Correct Answer: C

   Year: 25 Jul 2021 Shift 2

   Solution: $\frac{d\vec{v}}{dt} = \vec{a} = \frac{\vec{F}}{m} = (8\hat{i} + 2\hat{j}) \text{ m/s}^2$. $\vec{r} = (8\hat{i} + 2\hat{j})\frac{t^2}{2}\text{ m}$. At $t = 10 \text{ sec}$, $\vec{r} = [(8\hat{i} + 2\hat{j}) 50]\text{ m} \Rightarrow \vec{r} = (400\hat{i} + 100\hat{j})\text{ m}$.

   Step Solution:

    1.  Find Acceleration: Use Newton's Second Law in vector form: $\vec{a} = \frac{\vec{F}}{m} = \frac{40\hat{i} + 10\hat{j}}{5} = 8\hat{i} + 2\hat{j} \text{ m/s}^2$.

    2.  State Kinematic Equation: Since the body starts from rest ($\vec{u} = 0$), the displacement is $\vec{r} = \frac{1}{2}\vec{a}t^2$.

    3.  Substitute Values: $\vec{r} = \frac{1}{2}(8\hat{i} + 2\hat{j})(10)^2$.

    4.  Simplify Squared Term: $\vec{r} = \frac{1}{2}(8\hat{i} + 2\hat{j}) \times 100$.

    5.  Calculate Final Vector: $\vec{r} = (4\hat{i} + 1\hat{j}) \times 100 = 400\hat{i} + 100\hat{j} \text{ m}$.

   Difficulty Level: Easy

   The Concept Name: Newton’s Second Law and Kinematics (Vector Form)

   Short cut solution: The ratio of the components of the position vector ($x/y$) must be the same as the ratio of the components of the force ($40/10 = 4$). Only option C ($400/100 = 4$) satisfies this.

 Question 100

   Question: A steel block of 10 kg rests on a horizontal floor as shown. When three iron cylinders are placed on it as shown, the block and cylinders go down with an acceleration $0.2 \text{ m/s}^2$. The normal reaction $R'$ by the floor if mass of the iron cylinders are equal and of 20 kg each, is N. [Take $g = 10 \text{ m/s}^2$ and $\mu_s = 0.2$]

   Options:

    A. 716

    B. 686

    C. 714

    D. 684

   Correct Answer: B

   Year: 20 Jul 2021 Shift 1

   Solution: $a = 0.2 \text{ m/s}^2$. Writing force equation in vertical direction: $Mg - N = Ma$. $\Rightarrow 70g - N = 70 \times 0.2$. $\Rightarrow N = 70[g - 0.2] = 70 \times 9.8$. $N = 686$ Newton.

   Step Solution:

    1.  Calculate Total Mass: Total mass $M = 10\text{ kg (steel block)} + 3 \times 20\text{ kg (iron cylinders)} = 70\text{ kg}$.

    2.  Identify Downward Force: The weight acting downwards is $Mg = 70 \times 10 = 700\text{ N}$.

    3.  Apply Newton’s Law: For a system accelerating downwards, the equation is $Mg - N = Ma$.

    4.  Rearrange for Normal Force: $N = M(g - a) = 70(10 - 0.2)$.

    5.  Calculate Final Value: $N = 70 \times 9.8 = 686\text{ N}$.

   Difficulty Level: Easy

   The Concept Name: Newton’s Second Law (Vertical Motion)

   Short cut solution: Apparent weight $R'$ in a downward accelerating frame is $M(g - a)$. $70 \times (10 - 0.2) = 70 \times 9.8 = 686\text{ N}$.

 Question 111

   Question: An object of mass m is being moved with a constant velocity under the action of an applied force of 2N along a frictionless surface with following surface profile. The correct applied force versus distance graph will be.

   Options: 

A.

B.

C.

D.

   Correct Answer: B

   Year: 1 Sep 2021 Shift 2

   Solution: During the upward direction $F = 2N = (-ve)$ constant. During the upward motion, the force is positive constant and during the downward motion the force is negative constant.;;

   Step Solution:

    1.  Constant Velocity Condition: Constant velocity means net force is zero ($F_{net} = 0$).

    2.  Analyze Upward Motion: To move up at constant velocity, the applied force must balance the component of gravity down the slope: $F_{applied} = mg \sin \theta$.

    3.  Analyze Downward Motion: To move down at constant velocity, gravity's component acts in the direction of motion, so the applied force must act in reverse to prevent acceleration: $F_{applied} = -mg \sin \theta$.

    4.  Identify Force Sign: This results in the applied force being a positive constant for upward slopes and a negative constant for downward slopes.

    5.  Select Graph: The graph must show discrete horizontal segments switching between positive and negative values based on the surface profile.

   Difficulty Level: Medium

   The Concept Name: Newton's First Law and Resolution of Forces

   Short cut solution: Moving at constant velocity on a frictionless slope requires an applied force exactly equal and opposite to the gravitational component ($mg \sin \theta$). Since the slope angle is constant for each section, the force magnitude is constant, but its sign flips when the direction of the slope changes.

 Question 116

   Question: A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate $\frac{dM(t)}{dt} = bv^2(t)$, where $v(t)$ is its instantaneous velocity. The instantaneous acceleration of the satellite is :

   Options:

       A. $-bv^3(t)$

       B. $- \frac{bv^3}{M(t)}$

       Note: Options C and D involve different coefficients, but the correct derivation leads to B.

   Correct Answer: B

   Year: Sep. 05, 2020 (II)

   Solution: From Newton's second law, $F = \frac{dp}{dt} = \frac{d(mv)}{dt} = v\left(\frac{dm}{dt}\right)$. We are given $\frac{dM(t)}{dt} = bv^2(t)$. Thrust on the satellite $F = -v\left(\frac{dm}{dt}\right) = -v(bv^2) = -bv^3$. Using $F = M(t)a$, we get $a = \frac{-bv^3}{M(t)}$.

   Step Solution:

    1.  Rate of Mass Change: Identify the given mass increase rate: $\frac{dm}{dt} = bv^2$.

    2.  Thrust Force: Apply the thrust formula for a system accumulating stationary mass: $F = -v_{rel} \frac{dm}{dt}$. Here, $v_{rel} = v$.

    3.  Substitution: Substitute the mass rate into the force formula: $F = -v(bv^2) = -bv^3$.

    4.  Newton's Law: Relate the force to instantaneous acceleration: $M(t)a = F$.

    5.  Calculate Acceleration: $a = \frac{-bv^3}{M(t)}$.

   Difficulty Level: Medium

   The Concept Name: Newton’s Second Law (Variable Mass Systems)

   Short cut solution: Acceleration $a = \frac{Force}{Mass}$. The retarding force due to accumulating stationary dust is $v(\frac{dm}{dt})$. Thus, $a = -\frac{v(bv^2)}{M} = -\frac{bv^3}{M}$.

 Question 138

   Question: A body of mass 5kg under the action of constant force $\vec{F} = F_x \hat{i} + F_y \hat{j}$ has velocity at $t = 0\text{ s}$ as $\vec{v} = (6\hat{i} - 2\hat{j}) \text{ m/s}$ and at $t = 10\text{ s}$ as $\vec{v} = + 6\hat{j} \text{ m/s}$. The force $\vec{F}$ is:

   Options:

       A. $(-3\hat{i} + 4\hat{j}) \text{ N}$

       B. $\left(- \frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}\right) \text{ N}$

       C. $(3\hat{j} - 4\hat{j}) \text{ N}$

       D. $\left(\frac{3}{5}\hat{i} - \frac{4}{5}\hat{j}\right) \text{ N}$

   Correct Answer: A

   Year: Online April 11, 2014

   Solution: $u = (6\hat{i} - 2\hat{j}) \text{ m/s}$; $v = 6\hat{j} \text{ m/s}$. Acceleration $\vec{a} = \frac{\vec{v} - \vec{u}}{t} = \frac{6\hat{j} - (6\hat{i} - 2\hat{j})}{10} = \frac{-6\hat{i} + 8\hat{j}}{10} = \frac{-3\hat{i} + 4\hat{j}}{5} \text{ m/s}^2$. Force $\vec{F} = m\vec{a} = 5 \times \frac{(-3\hat{i} + 4\hat{j})}{5} = (-3\hat{i} + 4\hat{j}) \text{ N}$.

   Step Solution:

    1.  Initial & Final Vectors: Define $\vec{u} = 6\hat{i} - 2\hat{j}$ and $\vec{v} = 6\hat{j}$.

    2.  Velocity Change: Calculate $\Delta \vec{v} = \vec{v} - \vec{u} = 6\hat{j} - 6\hat{i} + 2\hat{j} = -6\hat{i} + 8\hat{j}$.

    3.  Acceleration: Divide by time ($t=10$): $\vec{a} = \frac{-6\hat{i} + 8\hat{j}}{10} = -0.6\hat{i} + 0.8\hat{j}$.

    4.  Newton's Second Law: Multiply by mass ($m=5$): $\vec{F} = 5 \times (-0.6\hat{i} + 0.8\hat{j})$.

    5.  Result: $\vec{F} = -3\hat{i} + 4\hat{j} \text{ N}$.

   Difficulty Level: Easy

   The Concept Name: Newton’s Second Law and Kinematics (Vector Form)

   Short cut solution: Use Impulse-Momentum: $\vec{F} \Delta t = m(\vec{v} - \vec{u})$. $\vec{F}(10) = 5(-6\hat{i} + 8\hat{j})$. Divide both sides by 10 to get $\vec{F} = \frac{1}{2}(-6\hat{i} + 8\hat{j}) = -3\hat{i} + 4\hat{j}$.

 Question 158

   Question: Two fixed frictionless inclined planes making an angle $30^\circ$ and $60^\circ$ with the vertical are shown in the figure. Two blocks A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B ?

   Options:

       A. $4.9\text{ ms}^{-2}$ in horizontal direction

       B. $9.8\text{ ms}^{-2}$ in vertical direction

       C. Zero

       D. $4.9\text{ ms}^{-2}$ in vertical direction

   Correct Answer: D

   Year: 2010

   Solution: $mg \sin \theta = ma \rightarrow a = g \sin \theta$. Vertical component of acceleration $= a \sin \theta = g \sin^2 \theta$. Relative vertical acceleration of A with respect to B is $g(\sin^2 60^\circ - \sin^2 30^\circ) = g\left(\frac{3}{4} - \frac{1}{4}\right) = \frac{g}{2} = 4.9\text{ m/s}^2$ in vertical direction.

   Step Solution:

    1.  Acceleration along Planes: For an angle $\theta$ with horizontal, $a = g \sin \theta$.

    2.  Determine Horizontal Angles: Since given angles are with the vertical, the horizontal angles are $(90-30)=60^\circ$ and $(90-60)=30^\circ$.

    3.  Vertical Acceleration Component: $a_y = a \sin \theta = g \sin^2 \theta$.

    4.  Individual Values: $a_{yA} = g \sin^2 60^\circ = 9.8 \times \frac{3}{4} = 7.35$; $a_{yB} = g \sin^2 30^\circ = 9.8 \times \frac{1}{4} = 2.45$.

    5.  Relative Calculation: $a_{rel} = a_{yA} - a_{yB} = 7.35 - 2.45 = 4.9\text{ ms}^{-2}$.

   Difficulty Level: Medium

   The Concept Name: Resolution of Acceleration on Inclined Planes

   Short cut solution: Relative vertical acceleration $= g(\sin^2 \theta_1 - \sin^2 \theta_2)$. For horizontal angles $60^\circ$ and $30^\circ$, this is $g(0.75 - 0.25) = 0.5g = 4.9\text{ ms}^{-2}$.

Question 162

   Question: A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes up to 2 m height further, find the magnitude of the force. (Consider $g = 10 \text{ m/s}^2$).

   Options: 

    A. 4 N

    B. 16 N

    C. 20 N

    D. 22 N

   Correct Answer: D

   Year: 2006

   Solution: For the motion of ball, just after the throwing $v = 0$, $s = 2 \text{ m}$, $a = -g = -10 \text{ ms}^{-2}$. $v^2 - u^2 = 2as$ for upward journey $\Rightarrow -u^2 = 2(-10) \times 2 \Rightarrow u^2 = 40$. When the ball is in the hands of the thrower $u = 0$, $v = \sqrt{40} \text{ ms}^{-1}$, $s = 0.2 \text{ m}$. $v^2 - u^2 = 2as \Rightarrow 40 - 0 = 2(a)0.2 \Rightarrow a = 100 \text{ m/s}^2$. $\therefore F = ma = 0.2 \times 100 = 20 \text{ N}$. $\Rightarrow N - mg = 20 \Rightarrow N = 20 + 2 = 22 \text{ N}$.

   Step Solution:

    1.  Velocity at Release: Calculate the velocity ($u$) of the ball when it leaves the hand using the height it reaches afterward: $v^2 - u^2 = 2gh \rightarrow 0^2 - u^2 = 2(-10)(2) \rightarrow u^2 = 40$.

    2.  Acceleration in Hand: Use the distance the hand moves ($0.2 \text{ m}$) to find the acceleration ($a$) applied by the hand: $v^2 - u^2 = 2as \rightarrow 40 - 0 = 2(a)(0.2)$.

    3.  Solve for $a$: $a = 40 / 0.4 = 100 \text{ m/s}^2$.

    4.  Equation of Motion: Apply Newton’s Second Law for the period the hand is pushing: $F_{net} = ma \rightarrow F - mg = ma$.

    5.  Calculate Total Force: $F = m(g + a) = 0.2(10 + 100) = 0.2(110) = 22 \text{ N}$.

   Difficulty Level: Medium

   The Concept Name: Newton’s Second Law and Kinematics

   Short cut solution: Use the Work-Energy Theorem: $W_{hand} + W_{gravity} = \Delta K$. Since it starts and ends at rest, $\Delta K = 0$. Therefore, $F \times d_{hand} - mg \times (d_{hand} + h_{flight}) = 0 \Rightarrow F(0.2) = 0.2(10)(0.2 + 2) \Rightarrow F = 22 \text{ N}$.

 Question 180

   Question: A solid sphere, a hollow sphere and a ring are released from top of an inclined plane (frictionless) so that they slide down the plane. Then maximum acceleration down the plane is for (no rolling).

   Options: 

    A. solid sphere

    B. hollow sphere

    C. ring

    D. all same

   Correct Answer: D

   Year: 2002

   Solution: This is a case of sliding (if plane is frictionless) and therefore the acceleration of all the bodies is same.

   Step Solution:

    1.  Analyze Conditions: The plane is frictionless and the Question specifies "no rolling," which means all objects are sliding.

    2.  Identify Force: The only force acting along the inclined plane is the component of gravity: $F = mg \sin \theta$.

    3.  Newton's Second Law: Set the force equal to mass times acceleration: $ma = mg \sin \theta$.

    4.  Solve for $a$: Dividing by mass ($m$) shows that $a = g \sin \theta$.

    5.  Conclusion: Because $a$ depends only on gravity and the angle of the plane, it is independent of the object's mass or shape. Thus, all are the same.

   Difficulty Level: Easy

   The Concept Name: Newton’s Second Law on an Inclined Plane

   Short cut solution: For any object sliding down a frictionless incline without rolling, the acceleration is always $a = g \sin \theta$, regardless of geometry.