Q.Reaction time: When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think and act. For example, if a person is driving and suddenly a boy appears on the road, then the time elapsed before he slams the brakes of the car is the reaction time. Reaction time depends on complexity of the situation and on an individual. You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between your thumb and forefinger. After you catch it, find the distance $d$ travelled by the ruler. In a particular case, $d$ was found to be $21.0\ \text{cm}$. Estimate reaction time.

Showing the 12 most recent of 21 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.For a body thrown vertically upwards from the ground, if the time interval between the two instances when the body is at a height of $60\ \mathrm{m}$ is $4\ \mathrm{s}$, then the maximum height reached by the body is (Acceleration due to gravity $= 10\ \mathrm{ms}^{-2}$) (A) $150\ \mathrm{m}$ (B) $90\ \mathrm{m}$ (C) $120\ \mathrm{m}$ (D) $80\ \mathrm{m}$

   ›Reveal solutionSolution

   The problem uses the symmetry of vertical motion to find the velocity at a given height, which then allows us to calculate the initial projection velocity and finally the maximum height. The maximum height reached by the body is $\boxed{80\ \mathrm{m}}$.

   The key to solving this problem efficiently lies in understanding the symmetry of motion under gravity. When a body is thrown vertically upwards, it follows a symmetrical path. This means:

   1. The time taken to reach a certain height while going up is equal to the time taken to fall back to that same height while coming down.
   2. The speed of the body at any given height while going up is equal in magnitude to its speed at the same height while coming down.
   3. The time taken to travel from a certain height to the maximum height is equal to the time taken to fall from the maximum height back to that same height.

   In this problem, the body is at a height of $60\ \mathrm{m}$ at two instances, separated by $4\ \mathrm{s}$. This $4\ \mathrm{s}$ interval represents the time it takes for the body to travel from $60\ \mathrm{m}$ (while going up) to its maximum height, and then fall back down to $60\ \mathrm{m}$ (while coming down). Due to symmetry, the time taken to go from $60\ \mathrm{m}$ to the maximum height is exactly half of this interval.

   Let's use this insight to solve the problem step-by-step. We will take the upward direction as positive and the acceleration due to gravity $g = 10\ \mathrm{ms}^{-2}$ as acting downwards, so $a = -g = -10\ \mathrm{ms}^{-2}$.

   1. Determine the velocity of the body at the height of $60\ \mathrm{m}$: The time interval between the two instances when the body is at $60\ \mathrm{m}$ is $4\ \mathrm{s}$. This means the time taken for the body to travel from $60\ \mathrm{m}$ upwards to the maximum height and then fall back to $60\ \mathrm{m}$ is $4\ \mathrm{s}$. By symmetry, the time taken to travel from the height of $60\ \mathrm{m}$ to the maximum height is half of this interval:

   $t_{\text{up from } 60\mathrm{m}} = \frac{4\ \mathrm{s}}{2} = 2\ \mathrm{s}$

   At the maximum height, the final velocity of the body is $0\ \mathrm{ms}^{-1}$. Let $v_{60}$ be the velocity of the body when it is at $60\ \mathrm{m}$ and moving upwards. We can use the first equation of motion for the journey from $60\ \mathrm{m}$ to the maximum height:
   

   $v = v_{60} + at$

   Here, $v = 0\ \mathrm{ms}^{-1}$ (at max height), $a = -g = -10\ \mathrm{ms}^{-2}$, and $t = 2\ \mathrm{s}$.
   

   $0 = v_{60} + (-10)(2)$

   $0 = v_{60} - 20$

   $v_{60} = 20\ \mathrm{ms}^{-1}$

   So, the speed of the body when it is at a height of $60\ \mathrm{m}$ (while going up) is $20\ \mathrm{ms}^{-1}$.
   

   2. Determine the initial velocity of projection from the ground: …

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2. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.A body is projected from a point which is at a height of $5\,\text{m}$ from the ground with a velocity of $28\,\text{ms}^{-1}$ at an angle of $30^\circ$ above the horizontal. The horizontal distance travelled by the body when it reaches a point which is at a height of $2\,\text{m}$ from the ground is (Acceleration due to gravity $= 10\,\text{ms}^{-2}$) (A) $126\sqrt{3}\ \text{m}$ (B) $21\sqrt{3}\ \text{m}$ (C) $42\sqrt{3}\ \text{m}$ (D) $28\sqrt{3}\ \text{m}$

   ›Reveal solutionSolution

   The key is to treat the motion as a projectile launched from a height, solve the vertical quadratic for the time when height = 2 m, then multiply the horizontal velocity by that time. The horizontal distance is $42\sqrt{3}$ m, so the correct option is (C).

   We start by recalling that projectile motion separates into independent horizontal and vertical components. The horizontal velocity stays constant (no air resistance), while the vertical motion is uniformly accelerated by gravity. The problem gives a launch from 5 m above ground, and we want the horizontal distance when the projectile is at 2 m above ground — that is, a vertical displacement of –3 m from the launch point.

   1. Resolve initial velocity into components

      Launch speed $u = 28\ \text{m/s}$, angle $\theta = 30^\circ$.

      Horizontal: $u_x = u \cos 30^\circ = 28 \cdot \frac{\sqrt{3}}{2} = 14\sqrt{3}\ \text{m/s}$.

      Vertical: $u_y = u \sin 30^\circ = 28 \cdot \frac{1}{2} = 14\ \text{m/s}$.

   2. Set up the vertical displacement equation

      Take upward as positive. Launch height $y_0 = 5$ m, final height $y = 2$ m, so vertical displacement $s_y = y - y_0 = -3$ m.

      Using $s_y = u_y t + \frac{1}{2} a t^2$ with $a = -g = -10\ \text{m/s}^2$:

   $-3 = 14t - 5t^2.$

   3. Solve the quadratic for time Rearranging: $5t^2 - 14t - 3 = 0$. Multiply by –1 if preferred: $5t^2 - 14t - 3 = 0$. Discriminant: $\Delta = (-14)^2 - 4 \cdot 5 \cdot (-3) = 196 + 60 = 256$. $\sqrt{\Delta} = 16$. …

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3. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.A body is projected from certain height with an initial velocity 'u' making an angle 'θ' above the horizontal. The time taken for its vertical and horizontal displacements to become equal is (A) $\dfrac{2u(\sin\theta + \cos\theta)}{g}$ (B) $\dfrac{2u\sin\theta}{g}$ (C) $\dfrac{u(\sin\theta - \cos\theta)}{g}$ (D) $\dfrac{u\sin\theta}{g}$

   ›Reveal solutionSolution

   This problem asks for the time when the horizontal and vertical displacements of a projectile become equal. While "equal displacements" strictly means $x=y$, in multiple-choice questions, it can sometimes imply "equal magnitudes of displacement" ($|x|=|y|$). Assuming the latter, and considering the case where the body has fallen below its projection height, the time taken is $\boxed{\dfrac{2u(\sin\theta + \cos\theta)}{g}}$.

   When a body is projected, its motion can be analyzed by separating it into horizontal and vertical components. The key idea is that these two components of motion are independent, except that they share the same time duration.

   • Horizontal Motion: There is no acceleration in the horizontal direction (assuming negligible air resistance). Thus, the horizontal velocity remains constant.
   • Vertical Motion: The body experiences constant acceleration due to gravity, acting downwards.

   Let's set up the equations for displacement. We'll take the point of projection as the origin $(0,0)$, the horizontal direction as the positive $x$-axis, and the upward vertical direction as the positive $y$-axis.

   1. Resolve the initial velocity:

      The initial velocity 'u' is given at an angle '$\theta$' above the horizontal. We resolve it into its horizontal and vertical components:

      • Horizontal component of initial velocity: $u_x = u \cos\theta$
      • Vertical component of initial velocity: $u_y = u \sin\theta$

   2. Write expressions for horizontal and vertical displacement:

      Using the equations of motion:

      • For horizontal displacement ($x$): Since there is no horizontal acceleration, $x = u_x t$.

   $x = (u \cos\theta) t \quad (1)$

   *   For vertical displacement ($y$): The acceleration is $a_y = -g$ (downwards). Using $s = ut + \frac{1}{2}at^2$:
   

   $y = (u \sin\theta) t - \frac{1}{2}gt^2 \quad (2)$

   3. Interpret "displacements to become equal":

      The phrase "vertical and horizontal displacements to become equal" can be interpreted in two ways:

      • Strict interpretation: The actual displacement values are equal, i.e., $x = y$.
      • Common interpretation in competitive exams (sometimes): The magnitudes of the displacements are equal, i.e., $|x| = |y|$.

      Let's analyze both:

      • Case 1: Strict interpretation ($x = y$) Set equation (1) equal to equation (2):

   $(u \cos\theta) t = (u \sin\theta) t - \frac{1}{2}gt^2$

       Rearrange the terms to form a quadratic equation in $t$:
   

   $\frac{1}{2}gt^2 + (u \cos\theta - u \sin\theta) t = 0$

       Factor out $t$:
   

   $t \left( \frac{1}{2}gt + u (\cos\theta - \sin\theta) \right) = 0$

       This gives two possible solutions for $t$:
       1.  $t = 0$: This is the initial moment when both displacements are zero, so they are equal. This is a trivial solution.
       2.  $\frac{1}{2}gt + u (\cos\theta - \sin\theta) = 0$
           $\frac{1}{2}gt = -u (\cos\theta - \sin\theta)$
           $\frac{1}{2}gt = u (\sin\theta - \cos\theta)$
   

   $t = \frac{2u (\sin\theta - \cos\theta)}{g}$

           For this time $t$ to be positive (a physically meaningful time *after* projection), we must have $\sin\theta - \cos\theta > 0$, which means $\sin\theta > \cos\theta$, or $\tan\theta > 1$. This implies $\theta > 45^\circ$. If $\theta \le 45^\circ$, then $x=y$ only at $t=0$.
           This result, $\frac{2u (\sin\theta - \cos\theta)}{g}$, is not directly available in the given options. Option (C) is $\frac{u (\sin\theta - \cos\theta)}{g}$, which is half of this value. This suggests a potential error in the options or the question's intended meaning.
   
   *   **Case 2: Magnitudes of displacements are equal ($|x| = |y|$)**
       Since the body is projected with an angle $\theta$ above the horizontal (implying $0 < \theta < 90^\circ$), the horizontal displacement $x = (u \cos\theta) t$ will always be positive for $t > 0$. So, $|x| = x$.
       Thus, we need $x = |y|$. This leads to two sub-cases:
       *   **Sub-case 2a: $x = y$** (This is the same as Case 1 above).
           This gives $t = \frac{2u (\sin\theta - \cos\theta)}{g}$, valid for $\theta > 45^\circ$.
       *   **Sub-case 2b: $x = -y$** (This means the vertical displacement is negative, i.e., the body has fallen below its initial projection height).
           Set equation (1) equal to the negative of equation (2):
           $$(u \cos\theta) t = - \left( (u \sin\theta) t - \frac{1}{2}gt^2 \right)$$ …
   

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4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.A helicopter flying horizontally with a velocity of 288 kmph drops a bomb. If the line joining the point of dropping the bomb, and the point where bomb hits the ground makes an angle $45^\circ$ with the horizontal, then the height at which the bomb was dropped is (Acceleration due to gravity = $10~\mathrm{ms}^{-2}$) (A) 1320 m (B) 1280 m (C) 320 m (D) 640 m

   ›Reveal solutionSolution

   The bomb keeps the helicopter's horizontal speed $80\ \mathrm{m/s}$ while falling freely. A $45^\circ$ release-to-impact line means the vertical drop equals the horizontal range, giving $h = 1280\ \mathrm{m}$ — option (B).

   ---

   Concept

   Dropped from the helicopter, the bomb has constant horizontal velocity and free-fall vertical motion. The straight line from release to impact makes $45^\circ$ with the horizontal, so its vertical drop $h$ equals its horizontal range $R$.

   Solution

   Convert the speed: $288\ \mathrm{km/h} = 288 \times \dfrac{1000}{3600} = 80\ \mathrm{m/s}$.

   The $45^\circ$ condition gives …

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5. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The initial and final velocities of a body projected vertically from the ground are $20\ \mathrm{ms}^{-1}$ and $18\ \mathrm{ms}^{-1}$ respectively. The maximum height reached by the body is (Acceleration due to gravity $= 10\ \mathrm{ms}^{-2}$) (A) 20 m (B) 16.2 m (C) 19 m (D) 18.1 m

   ›Reveal solutionSolution

   With air resistance, the upward retardation exceeds the downward one, so the return speed ($18\ \mathrm{ms}^{-1}$) is less than the launch speed ($20\ \mathrm{ms}^{-1}$). Combining both legs gives $H = 18.1\ \mathrm{m}$ — option (D).

   ---

   Concept

   The launch speed is $20\ \mathrm{ms}^{-1}$ but the body returns at only $18\ \mathrm{ms}^{-1}$, so a constant resistive force $f$ acts opposite to motion. Going up the deceleration is $g+f/m$; coming down the acceleration is $g-f/m$. The same maximum height $H$ appears in both legs.

   Solution

   Upward leg (start $20$, top $0$):

   $20^2 = 2\left(g + \tfrac{f}{m}\right)H.$

   Downward leg (top $0$, ground $18$): …

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6. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.A body projected vertically up with an initial speed of $10\,\mathrm{ms}^{-1}$ reaches the point of projection after sometime with a speed of $8\,\mathrm{ms}^{-1}$. The maximum height reached by the body is (Acceleration due to gravity $= 10\,\mathrm{ms}^{-2}$) (A) $5\,\mathrm{m}$ (B) $3.2\,\mathrm{m}$ (C) $4.1\,\mathrm{m}$ (D) $4.5\,\mathrm{m}$

   ›Reveal solutionSolution

   The body loses mechanical energy due to air resistance, so we use the work–energy theorem between launch and the top of the path. The maximum height is found to be $4.1\,\mathrm{m}$, which corresponds to option (C).

   The key idea is that the body does not return to the starting point with the same speed — it comes back at $8\,\mathrm{m/s}$ instead of $10\,\mathrm{m/s}$. This tells us that some non‑conservative force (air resistance) has done negative work. We cannot simply use conservation of mechanical energy. Instead, we apply the work–energy theorem to relate the loss in kinetic energy to the work done against air resistance, and then use that to find the maximum height.

   ---

   1. Identify the energy loss When the body is projected upward with speed $u = 10\,\mathrm{m/s}$, its initial kinetic energy is

   $K_i = \frac12 m u^2 = \frac12 m (10)^2 = 50m.$

   When it returns to the same point, its speed is $v = 8\,\mathrm{m/s}$, so its final kinetic energy is

   $K_f = \frac12 m v^2 = \frac12 m (8)^2 = 32m.$

   The loss in kinetic energy is

   $\Delta K = K_f - K_i = 32m - 50m = -18m.$

   This energy is dissipated by air resistance over the whole round trip.

   2. Work done by air resistance on the upward journey Air resistance always opposes motion. On the way up, it does negative work; on the way down, it also does negative work. Because the path up and down are symmetric in distance (same height $h$), and the resistive force depends only on speed (not direction), the work done by air resistance on the upward leg equals the work done on the downward leg. Let $W_{\text{air}}$ be the work done by air resistance on the upward journey. Then the total work done over the round trip is $2W_{\text{air}}$. By the work–energy theorem, the total work done by all forces equals the change in kinetic energy. Gravity does zero net work over the round trip (since the body returns to the same height), so

   $2W_{\text{air}} = \Delta K = -18m.$

   Hence

   $W_{\text{air}} = -9m.$

   3. Apply work–energy theorem to the upward journey only On the way up, the body starts at ground level with speed $10\,\mathrm{m/s}$ and reaches maximum height $h$ where its speed is $0$. The forces acting are gravity (doing negative work) and air resistance (doing work $W_{\text{air}} = -9m$). …

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7. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.A body X is allowed to fall freely from a height of 125 m from the ground and another body Y is thrown at the same instant vertically upwards with a velocity of $50\,\mathrm{ms}^{-1}$ from the ground. The relative velocity of the body Y with respect to the body X when they cross each other is (A) $125\,\mathrm{ms}^{-1}$ (B) $100\,\mathrm{ms}^{-1}$ (C) $50\,\mathrm{ms}^{-1}$ (D) $25\,\mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   Both bodies have the same acceleration (gravity), so their relative velocity is constant and equals the initial relative velocity. The answer is $50\,\mathrm{ms}^{-1}$.

   The key insight here is that when two objects move under the same uniform acceleration — in this case, gravity — their relative acceleration is zero. That means their relative velocity does not change with time. So instead of finding the moment they cross and calculating velocities then, we can simply use the initial relative velocity.

   Let’s see why.

   1. Set up the motion equations.

      Take upward as positive. For body X (dropped from height $125$ m):

      Initial velocity $u_X = 0$, acceleration $a_X = -g$ (taking $g = 10\,\mathrm{ms}^{-2}$ for convenience, though the exact value cancels).

      For body Y (thrown upward from ground):

      Initial velocity $u_Y = +50\,\mathrm{ms}^{-1}$, acceleration $a_Y = -g$.

   2. Relative acceleration.

      $a_{YX} = a_Y - a_X = (-g) - (-g) = 0$.

      Since relative acceleration is zero, the relative velocity $v_{YX}$ is constant throughout the motion.

   3. Initial relative velocity.

      At $t = 0$,

      $v_{YX}(0) = u_Y - u_X = 50 - 0 = 50\,\mathrm{ms}^{-1}$.

      Because relative acceleration is zero, this remains the relative velocity at every instant — including the moment they cross.

   Watch out

   A common mistake is to first find the time of crossing, then compute individual velocities at that time, and subtract. That works but is unnecessary work. The constancy of relative velocity under equal acceleration is the shortcut.

   4. Check the crossing time (optional, for verification). …

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8. KCET 2024Set D-21 markMCQ
   Q.A body of mass $1\,\mathrm{kg}$ is suspended by a weightless string which passes over a frictionless pulley of mass $2\,\mathrm{kg}$ as shown in the figure. The mass is released from a height of $1.6\,\mathrm{m}$ from the ground. With what velocity does it strike the ground? (A) $16\,\mathrm{ms}^{-1}$ (B) $8\,\mathrm{ms}^{-1}$ (C) $4\sqrt{2}\,\mathrm{ms}^{-1}$ (D) $4\,\mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   Conserve mechanical energy, but remember the pulley is massive — part of the lost potential energy goes into its rotational kinetic energy, not just the block's translational KE.

   1. Model the pulley

   A pulley of mass $M = 2\ \mathrm{kg}$ and radius $R$ is treated as a uniform disc:

   $I = \frac{1}{2}MR^{2}$

   Since the string does not slip, the rim speed equals the block's speed:

   $v = \omega R \quad\Longrightarrow\quad \omega = \frac{v}{R}$

   2. Energy conservation

   The string is weightless and the pulley frictionless, so no energy is dissipated. As the $1\ \mathrm{kg}$ block falls a height $h = 1.6\ \mathrm{m}$ from rest:

   $\underbrace{mgh}_{\text{PE lost}} = \underbrace{\frac{1}{2}mv^{2}}_{\text{block's KE}} + \underbrace{\frac{1}{2}I\omega^{2}}_{\text{pulley's rotational KE}}$

   3. Eliminate $I$ and $\omega$

   $\frac{1}{2}I\omega^{2} = \frac{1}{2}\left(\frac{1}{2}MR^{2}\right)\left(\frac{v}{R}\right)^{2} = \frac{1}{4}Mv^{2}$

   Notice $R$ cancels — the radius never matters. So:

   $mgh = \frac{1}{2}mv^{2} + \frac{1}{4}Mv^{2} = \frac{1}{2}\left(m + \frac{M}{2}\right)v^{2}$

   4. Solve for $v$

   $v = \sqrt{\frac{2mgh}{m + \dfrac{M}{2}}}$

   Substituting $m = 1\ \mathrm{kg}$, $M = 2\ \mathrm{kg}$, $h = 1.6\ \mathrm{m}$, $g = 10\ \mathrm{m\,s^{-2}}$: …

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9. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.Starting from rest a car accelerates with uniform acceleration of $4\ \mathrm{ms}^{-2}$ for some time after which it comes to rest with uniform deceleration of $6\ \mathrm{ms}^{-2}$. If the total time of travel is $5\ \mathrm{s}$, then the total distance travelled by the car is (A) $25\ \mathrm{m}$ (B) $30\ \mathrm{m}$ (C) $60\ \mathrm{m}$ (D) $125\ \mathrm{m}$

   ›Reveal solutionSolution

   The car speeds up and then slows down, both uniformly. The key is that the maximum speed reached is the same for both phases. Using the total time and the two accelerations, we find the distance is 30 m.

   The problem gives you two uniform acceleration phases — first speeding up, then slowing down — with no constant-speed coasting in between. The car starts from rest and ends at rest. The total time is fixed at 5 seconds. The accelerations are $4\ \mathrm{ms}^{-2}$ (forward) and $-6\ \mathrm{ms}^{-2}$ (deceleration, which we treat as a magnitude of $6\ \mathrm{ms}^{-2}$ in the opposite direction).

   The central idea: the car reaches some maximum speed $v_{\text{max}}$ at the end of the acceleration phase, then immediately begins decelerating from that same $v_{\text{max}}$ down to zero. Because both phases are uniform, the distance in each is just the average speed times the time for that phase. And the average speed in each phase is simply $v_{\text{max}}/2$.

   Let’s work it through.

   1. Let the acceleration time be $t_1$ and the deceleration time be $t_2$.

      Total time: $t_1 + t_2 = 5\ \text{s}$.

   2. Relate $v_{\text{max}}$ to each phase.

      For the acceleration phase: $v_{\text{max}} = 4 t_1$.

      For the deceleration phase: $0 = v_{\text{max}} - 6 t_2$, so $v_{\text{max}} = 6 t_2$.

      Equating the two expressions for $v_{\text{max}}$:

   $4 t_1 = 6 t_2 \quad\Rightarrow\quad t_1 = \frac{3}{2} t_2.$

   3. Solve for the times. Substitute into $t_1 + t_2 = 5$:

   $\frac{3}{2} t_2 + t_2 = 5 \quad\Rightarrow\quad \frac{5}{2} t_2 = 5 \quad\Rightarrow\quad t_2 = 2\ \text{s}.$

   Then $t_1 = 3\ \text{s}$.

   4. Find the maximum speed.

      $v_{\text{max}} = 4 \times 3 = 12\ \text{m/s}$ (or $6 \times 2 = 12\ \text{m/s}$ — consistent).

   5. Compute the distance for each phase using average speed. …

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10. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
    Q.The half-life of a radioactive substance is 12 minutes. The time gap between 28% decay and 82% decay of the radioactive substance is (A) 6 minutes (B) 18 minutes (C) 12 minutes (D) 24 minutes

    ›Reveal solutionSolution

    The key idea is that radioactive decay follows an exponential law, so the time between two percentage decays depends only on the ratio of the remaining fractions, not on the initial amount. The time gap is 24 minutes, corresponding to option (D).

    The problem asks for the time interval between two specific decay milestones: when 28% has decayed (so 72% remains) and when 82% has decayed (so 18% remains). Since decay is exponential, the time to go from one fraction to another is determined by the half-life and the ratio of those fractions.

    Why this works:

    Radioactive decay is a first-order process: the number of atoms decreases by a constant fraction per unit time. The half-life ($t_{1/2}$) is the time for half the atoms to decay. The fraction remaining after time $t$ is $N/N_0 = (1/2)^{t/t_{1/2}}$. To find the time between two different remaining fractions, we take the ratio of the two exponential expressions, which cancels the initial amount and gives a simple logarithmic relation.

    Step-by-step solution:

    1. Set up the decay law. Let $N_0$ be the initial number of atoms. After time $t$, the number remaining is

    $N(t) = N_0 \left(\frac{1}{2}\right)^{t / 12},$

    where 12 minutes is the half-life.

    2. Find the time for 28% decay. If 28% has decayed, then 72% remains:

    $\frac{N(t_1)}{N_0} = 0.72 = \left(\frac{1}{2}\right)^{t_1 / 12}.$

    Taking natural logs:

    $\ln(0.72) = \frac{t_1}{12} \ln\left(\frac{1}{2}\right) = -\frac{t_1}{12} \ln 2.$

    So

    $t_1 = -12 \cdot \frac{\ln(0.72)}{\ln 2}.$

    3. Find the time for 82% decay. If 82% has decayed, then 18% remains:

    $\frac{N(t_2)}{N_0} = 0.18 = \left(\frac{1}{2}\right)^{t_2 / 12}.$

    Similarly,

    $t_2 = -12 \cdot \frac{\ln(0.18)}{\ln 2}.$

    4. Compute the time gap $\Delta t = t_2 - t_1$. $\Delta t = -12 \cdot \frac{\ln(0.18)}{\ln 2} - \left(-12 \cdot \frac{\ln(0.72)}{\ln 2}\right) = -12 \cdot \frac{\ln(0.18) - \ln(0.72)}{\ln 2}.$ …

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11. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
    Q.If a substance decays from 32 g to 1 g in 25 days, then its half life is (A) 3 days (B) 4 days (C) 5 days (D) 6 days

    ›Reveal solutionSolution

    The half-life is found by counting how many half-lives it takes to go from 32 g to 1 g, then dividing the total time by that number. The answer is 5 days.

    The key idea here is that radioactive decay follows a simple pattern: after each half-life, the mass is cut in half. So instead of plugging into a complicated exponential formula right away, you can just ask: how many times do you need to halve 32 to get 1? That’s a much cleaner way to think about it.

    1. Start with 32 g. After one half-life, you have 16 g. After two, 8 g. After three, 4 g. After four, 2 g. After five, 1 g. So it takes 5 half-lives to go from 32 g to 1 g.

    2. The problem says this whole process takes 25 days. If 5 half-lives take 25 days, then one half-life is simply $25 \div 5 = 5$ days. …

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12. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.Velocities (V) and accelerations

    (a) in two systems of units 1 and 2 are related as $V_2 = \frac{n}{m^2} V_1$ and $a_2 = \frac{a_1}{mn}$ respectively. Here m and n are constants. Dimensionally relations between distances ($S_1$ and $S_2$) and times ($t_1$ and $t_2$) in the two systems are respectively (A) $S_2 = \left(\frac{n}{m}\right)^3 S_1 \ \&\ t_2 = \frac{n^2}{m} t_1$ (B) $S_2 = \left(\frac{n}{m}\right) S_1 \ \&\ t_2 = \frac{m}{n^2} t_1$ (C) $S_2 = \frac{m}{n^2} S_1 \ \&\ t_2 = \frac{m^2}{n^4} t_1$ (D) $S_2 = \frac{n^2}{m} S_1 \ \&\ t_2 = \frac{m^2}{n^4} t_1$

    ›Reveal solutionSolution

    Let length scale $\lambda_L$ and time scale $\lambda_T$; use the velocity and acceleration ratios to solve for both, giving $S_2=(n/m)^3 S_1$ and $t_2=(n^2/m)t_1$ — option (A).

    Let distances scale as $S_2 = \lambda_L S_1$ and times as $t_2 = \lambda_T t_1$. Then velocity ($\propto L/T$) and acceleration ($\propto L/T^2$) scale as:

    $\frac{V_2}{V_1} = \frac{\lambda_L}{\lambda_T} = \frac{n}{m^2}, \qquad \frac{a_2}{a_1} = \frac{\lambda_L}{\lambda_T^{\,2}} = \frac{1}{mn}.$

    Find $\lambda_T$ by dividing the velocity ratio by the acceleration ratio:

    $\lambda_T = \frac{\lambda_L/\lambda_T}{\lambda_L/\lambda_T^{2}} = \frac{n/m^2}{1/(mn)} = \frac{n}{m^2}\cdot mn = \frac{n^2}{m}.$

    So $t_2 = \dfrac{n^2}{m}\, t_1$. …

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