Q.A stone dropped from the top of a tower of height $300\ \text{m}$ splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is $340\ \text{m s}^{-1}$? ($g = 9.8\ \text{m s}^{-2}$)

Showing the 12 most recent of 20 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If a source of sound initially at rest is moving away from a stationery observer with an acceleration of $11\,\mathrm{ms}^{-2}$, then the time taken for the frequency of sound heard by the observer to become $10\%$ less than the frequency of source is (Speed of sound in air $= 330\,\mathrm{ms}^{-1}$) (A) $4.4\,\mathrm{s}$ (B) $1.1\,\mathrm{s}$ (C) $2.2\,\mathrm{s}$ (D) $3.3\,\mathrm{s}$

   ›Reveal solutionSolution

   The problem uses the Doppler effect for a source moving away from a stationary observer. The observed frequency drops to 90% of the source frequency when the source reaches a certain speed; using the relation between speed, acceleration, and time gives the answer as 3.3 s, option (D).

   Concept & Intuition

   When a source of sound moves away from a stationary observer, the observed frequency is lower than the source frequency. The Doppler formula tells us exactly how much lower based on the source’s speed. Here, the source starts from rest and accelerates uniformly, so its speed increases linearly with time. We need the time when the observed frequency is 10% less — meaning it is 90% of the source frequency. That gives a specific source speed, and from acceleration we find the time.

   Step-by-step solution

   1. Write the Doppler effect formula for a source moving away from a stationary observer. For a stationary observer and a moving source, the observed frequency $f'$ is:

   $f' = \frac{v}{v + v_s} \, f$

   where $v = 330\,\mathrm{m/s}$ is the speed of sound, $v_s$ is the speed of the source (positive when moving away), and $f$ is the source frequency.

   2. Set the condition for a 10% decrease. A 10% decrease means the observed frequency is 90% of the source frequency:

   $f' = 0.9 f$

   Substitute into the Doppler formula:

   $0.9 f = \frac{330}{330 + v_s} \, f$

   Cancel $f$ (assuming non-zero):

   $0.9 = \frac{330}{330 + v_s}$

   3. Solve for the source speed $v_s$. Rearranging:

   $0.9 (330 + v_s) = 330$

   $297 + 0.9 v_s = 330$

   $0.9 v_s = 33$

   $v_s = \frac{33}{0.9} = \frac{330}{9} = 36.\overline{6} \,\mathrm{m/s}$

   So $v_s = \frac{110}{3} \,\mathrm{m/s}$ exactly.

   4. Relate speed to time under constant acceleration. The source starts from rest ($u = 0$) and accelerates at $a = 11\,\mathrm{m/s}^2$ away from the observer. Using $v = u + at$:

   $v_s = a t$

   $\frac{110}{3} = 11 t$

   $t = \frac{110}{3 \times 11} = \frac{10}{3} = 3.\overline{3} \,\mathrm{s}$

   5. Match with the options. $3.\overline{3}\,\mathrm{s}$ is exactly 3.3 s, which corresponds to option (D).

   Tip

   A common mistake is to think a 10% decrease means $f' = 0.1 f$ instead of $0.9 f$. Always read “10% less than” as $100\% - 10\% = 90\%$ of the original.

   Watch out

   Another pitfall: forgetting that the source is moving away, so the denominator is $v + v_s$, not $v - v_s$. Using the wrong sign gives a different (incorrect) time.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If $t_1$ is time taken for a body to cool from temperature of $80\,^\circ\mathrm{C}$ to $75\,^\circ\mathrm{C}$ and $t_2$ is time taken to cool from $75\,^\circ\mathrm{C}$ to $70\,^\circ\mathrm{C}$, then $t_1 : t_2 =$ (Temperature of surroundings $= 30\,^\circ\mathrm{C}$) (A) $19 : 21$ (B) $17 : 19$ (C) $7 : 9$ (D) $9 : 11$

   ›Reveal solutionSolution

   By Newton's law of cooling each interval drops the same $5\,^\circ$C, so the time is inversely proportional to the mean excess temperature: $t_1 : t_2 = 17 : 19$.

   Setup. For a small temperature drop, Newton's law of cooling gives

   $\frac{\Delta\theta}{t} = k\left(\theta_{\text{mean}} - \theta_s\right),\qquad \theta_s = 30\,^\circ\mathrm{C}.$

   Both intervals have the same drop $\Delta\theta = 5\,^\circ$C, so $t \propto \dfrac{1}{\theta_{\text{mean}} - \theta_s}$.

   Interval 1 ($80 \to 75$): mean $= 77.5\,^\circ$C, excess $= 77.5 - 30 = 47.5$.

   $t_1 \propto \frac{5}{47.5}.$

   Interval 2 ($75 \to 70$): mean $= 72.5\,^\circ$C, excess $= 72.5 - 30 = 42.5$.

   $t_2 \propto \frac{5}{42.5}.$

   Ratio.

   $\frac{t_1}{t_2} = \frac{1/47.5}{1/42.5} = \frac{42.5}{47.5} = \frac{85}{95} = \frac{17}{19}.$

   ✓Final answer

   $t_1 : t_2 = 17 : 19$ — option (B).

3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.A body P of mass 3 kg at rest is dropped from a height of 250 m from the ground. At the same moment another body Q of mass 2 kg is thrown vertically upwards from the ground with a velocity of $50\,\mathrm{ms}^{-1}$. Both the bodies travel along the same straight line in opposite directions. The velocity of body Q when the centre of mass of the system of the bodies P and Q reaches the maximum height is (Acceleration due to gravity $= 10\,\mathrm{ms}^{-2}$) (A) $25\,\mathrm{ms}^{-1}$ (B) $30\,\mathrm{ms}^{-1}$ (C) $40\,\mathrm{ms}^{-1}$ (D) $20\,\mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   The centre of mass starts upward at $20\,\mathrm{m/s}$ and decelerates at $g$; it peaks when $v_{\text{cm}}=0$ at $t=2\,\mathrm{s}$, when Q's velocity is $30\,\mathrm{m/s}$ — option (B).

   Take upward as positive. Only gravity acts, so both bodies have acceleration $-g=-10\,\mathrm{m/s^2}$.

   Velocities as functions of time.

   $v_P = -10t \quad(\text{P dropped from rest}),\qquad v_Q = 50 - 10t.$

   Velocity of the centre of mass ($m_P=3$, $m_Q=2$):

   $v_{\text{cm}} = \frac{3(-10t)+2(50-10t)}{5} = \frac{100-50t}{5} = 20-10t.$

   COM reaches maximum height when $v_{\text{cm}}=0$:

   $20-10t = 0 \;\Rightarrow\; t = 2\,\mathrm{s}.$

   Velocity of Q at that instant.

   $v_Q = 50 - 10(2) = 30\,\mathrm{m/s}\ \text{(upward)}.$

   ✓Final answer

   Q's velocity when the COM is at its maximum height is $30\,\mathrm{m/s}$ — option (B).

4. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.A solid sphere is rolling down without slipping on an inclined plane of length 21 m with an acceleration of $5\ \mathrm{ms}^{-2}$. The time taken by a circular disc to roll down without slipping to reach the bottom from the top of the same inclined plane is (A) $5\ \mathrm{s}$ (B) $9\ \mathrm{s}$ (C) $3\ \mathrm{s}$ (D) $6\ \mathrm{s}$

   ›Reveal solutionSolution

   The key idea is that the acceleration of a rolling body down an incline depends on its moment of inertia. Using the sphere’s acceleration to find the incline’s slope, then applying the disc’s acceleration to find its time, gives 3 seconds.

   When a rigid body rolls without slipping down an inclined plane, its linear acceleration is not simply $g \sin \theta$ (as for a sliding block). Part of the gravitational potential energy goes into rotational kinetic energy, so the acceleration is reduced by a factor that depends on the body’s moment of inertia.

   For any rolling object of radius $R$ and moment of inertia $I = k m R^2$ (where $k$ is a dimensionless constant), the acceleration down an incline of angle $\theta$ is:

   $a = \frac{g \sin \theta}{1 + k}$

   This is a standard result derived from combining Newton’s second law for translation and rotation, with the no-slip condition $a = \alpha R$.

   For a solid sphere, $k = \frac{2}{5}$, so $a_{\text{sphere}} = \frac{g \sin \theta}{1 + \frac{2}{5}} = \frac{5}{7} g \sin \theta$.

   For a circular disc (or solid cylinder), $k = \frac{1}{2}$, so $a_{\text{disc}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{2}{3} g \sin \theta$.

   The problem gives the sphere’s acceleration as $5\ \mathrm{m/s}^2$. This lets us find $g \sin \theta$, which is the same for both bodies on the same incline. Then we can find the disc’s acceleration and, using the constant-acceleration equation, the time to cover the 21 m length.

   1. Find $g \sin \theta$ from the sphere’s motion.

      For the sphere: $a_s = \frac{5}{7} g \sin \theta = 5$.

      So $g \sin \theta = 5 \times \frac{7}{5} = 7\ \mathrm{m/s}^2$.

   2. Find the disc’s acceleration.

      For the disc: $a_d = \frac{2}{3} g \sin \theta = \frac{2}{3} \times 7 = \frac{14}{3}\ \mathrm{m/s}^2$.

   3. Find the time taken by the disc.

      Both start from rest, so using $s = \frac{1}{2} a t^2$ with $s = 21$ m:

   $21 = \frac{1}{2} \times \frac{14}{3} \times t^2$

   $21 = \frac{7}{3} t^2$

   $t^2 = 21 \times \frac{3}{7} = 9$

   $t = 3\ \mathrm{s}$

   Watch out

   A common mistake is to forget that the acceleration depends on the moment of inertia and use $g \sin \theta$ directly. Here, $g \sin \theta = 7$, not 5 — the sphere’s acceleration is only $5/7$ of that value. Using $a = 7$ for the disc would give a wrong, smaller time.

   Tip

   Notice that the ratio of accelerations is $\frac{a_{\text{disc}}}{a_{\text{sphere}}} = \frac{2/3}{5/7} = \frac{14}{15}$, so the disc is slightly slower. The time ratio is the inverse square root: $t_{\text{disc}} = t_{\text{sphere}} \times \sqrt{\frac{15}{14}}$, but here we didn’t need the sphere’s time directly.

   ✓Final answer

   The time taken by the disc is $\boxed{3\ \mathrm{s}}$, which corresponds to option (C).

5. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.One second after projection, the horizontal and vertical velocities of a projectile are found to be equal and after one more second, the motion of the projectile is along the horizontal. The horizontal range of the projectile is (Acceleration due to gravity $= 10\, \mathrm{ms}^{-2}$) (A) 10 m (B) 20 m (C) 30 m (D) 40 m

   ›Reveal solutionSolution

   The key is to use the given velocity conditions to find the initial velocity components. The horizontal range is $40\,\mathrm{m}$, so the correct option is (D).

   The problem gives you two snapshots of a projectile’s motion. At $t = 1\,\mathrm{s}$, the horizontal and vertical velocity components are equal. One second later — at $t = 2\,\mathrm{s}$ — the velocity is purely horizontal, meaning the vertical component has become zero. That second condition tells you exactly when the projectile reaches its highest point: at $t = 2\,\mathrm{s}$. From there, you can work backwards to find the initial vertical velocity, and then use the first condition to find the horizontal velocity. The range follows directly.

   Let’s go step by step.

   1. Interpret the “motion along the horizontal” condition. At $t = 2\,\mathrm{s}$, the projectile’s velocity is horizontal. That means $v_y = 0$ at $t = 2\,\mathrm{s}$. For a projectile under constant gravity $g = 10\,\mathrm{m/s^2}$ (taking upward as positive), the vertical velocity obeys

   $v_y = u_y - g t,$

   where $u_y$ is the initial vertical velocity. Setting $v_y = 0$ at $t = 2\,\mathrm{s}$:

   $0 = u_y - (10)(2) \quad\Rightarrow\quad u_y = 20\,\mathrm{m/s}.$

   2. Use the “equal velocities” condition at $t = 1\,\mathrm{s}$. At $t = 1\,\mathrm{s}$, the horizontal and vertical speeds are equal. The horizontal velocity is constant: $v_x = u_x$ (no acceleration horizontally). The vertical velocity at $t = 1\,\mathrm{s}$ is

   $v_y(1) = u_y - g(1) = 20 - 10 = 10\,\mathrm{m/s}.$

   The condition $|v_x| = |v_y|$ at this instant gives

   $u_x = 10\,\mathrm{m/s}.$

   (Both are positive, so no sign confusion.)

   3. Find the time of flight. The total time of flight $T$ is twice the time to reach the highest point (since the motion is symmetric if launch and landing are at the same height). The highest point occurs at $t = 2\,\mathrm{s}$, so

   $T = 2 \times 2 = 4\,\mathrm{s}.$

   4. Compute the horizontal range. Range $R = u_x \times T = 10 \times 4 = 40\,\mathrm{m}$.

   Watch out

   A common mistake is to think “equal velocities at $t=1\,\mathrm{s}$” means the initial velocities are equal — that would give $u_x = u_y$, which is false here. Always plug the given time into the vertical velocity equation first.

   Tip

   The phrase “motion along the horizontal” is a clean way of saying the projectile is at the top of its trajectory. That single fact gives you $u_y$ directly.

   ✓Final answer

   The horizontal range of the projectile is $\boxed{40\,\mathrm{m}}$, which corresponds to option (D).

6. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.The frequency of sound heard by an observer moving towards a stationary source with certain speed is $n_1$ and if the observer moves away from the same source with same speed, the frequency of sound heard by the observer is $n_2$. If the speed of sound in air is $340\,\mathrm{ms}^{-1}$ and $n_1 : n_2 = 71 : 65$, then speed of observer is (A) 36 kmph (B) 27 kmph (C) 15 kmph (D) 54 kmph

   ›Reveal solutionSolution

   The moving-observer Doppler ratio $\dfrac{n_1}{n_2}=\dfrac{v+v_o}{v-v_o}=\dfrac{71}{65}$ gives $v_o=15\ \mathrm{m/s}=54\ \mathrm{kmph}$.

   Concept

   With a stationary source and an observer moving at speed $v_o$, the frequency heard is raised when approaching and lowered when receding:

   $n_1 = n\,\frac{v+v_o}{v}, \qquad n_2 = n\,\frac{v-v_o}{v}$

   Their ratio depends only on $v_o$ and the speed of sound $v$:

   $\frac{n_1}{n_2} = \frac{v+v_o}{v-v_o} = \frac{71}{65}$

   Solving for the observer's speed

   $65(340 + v_o) = 71(340 - v_o)$

   $22100 + 65v_o = 24140 - 71v_o$

   $136v_o = 2040 \;\Rightarrow\; v_o = 15\ \mathrm{m/s}$

   Converting to km/h: $v_o = 15 \times 3.6 = 54\ \mathrm{kmph}$.

   ✓Final answer

   The speed of the observer is $54\ \mathrm{kmph}$ — option (D).

7. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.The frequency of sound heard by an observer moving towards a stationary source with certain speed is $n_1$ and if the observer moves away from the same source with same speed, the frequency of sound heard by the observer is $n_2$. If the speed of sound in air is $340\,\mathrm{ms}^{-1}$ and $n_1 : n_2 = 71 : 65$, then speed of observer is (A) 27 kmph (B) 15 kmph (C) 54 kmph (D) 36 kmph

   ›Reveal solutionSolution

   The Doppler effect for a moving observer gives frequencies $n_1 = n_0 \frac{v+v_o}{v}$ and $n_2 = n_0 \frac{v-v_o}{v}$. Their ratio $71/65$ yields $v_o = 15\,\text{m/s} = 54\,\text{km/h}$, so the correct option is (C).

   The key idea is the Doppler effect for a moving observer and a stationary source. When the observer moves, the effective speed of sound relative to the observer changes, altering the perceived frequency. The ratio of frequencies when moving toward vs. away directly gives the observer’s speed without needing the source frequency.

   1. Set up the Doppler formulas

      For a stationary source emitting frequency $n_0$ and speed of sound $v = 340\,\text{m/s}$, the frequency heard by an observer moving with speed $v_o$ is:

      • Toward the source: $n_1 = n_0 \frac{v + v_o}{v}$
      • Away from the source: $n_2 = n_0 \frac{v - v_o}{v}$ (The plus/minus is because the observer’s motion changes the relative speed of sound.)

   2. Use the given ratio

      We are told $n_1 : n_2 = 71 : 65$, so:

   $\frac{n_1}{n_2} = \frac{71}{65}$

   Substitute the expressions:

   $\frac{n_0 \frac{v+v_o}{v}}{n_0 \frac{v-v_o}{v}} = \frac{v+v_o}{v-v_o} = \frac{71}{65}$

   3. Solve for $v_o$ Cross-multiply:

   $65(v + v_o) = 71(v - v_o)$

   Expand:

   $65v + 65v_o = 71v - 71v_o$

   Bring terms together:

   $65v_o + 71v_o = 71v - 65v$

   $136 v_o = 6v$

   $v_o = \frac{6}{136} v = \frac{3}{68} \times 340$

   Simplify:

   $v_o = \frac{3 \times 340}{68} = \frac{1020}{68} = 15\,\text{m/s}$

   4. Convert to km/h Since $1\,\text{m/s} = 3.6\,\text{km/h}$:

   $v_o = 15 \times 3.6 = 54\,\text{km/h}$

   Watch out

   A common mistake is to use the formula for a moving source instead of a moving observer. Here the source is stationary, so the frequency shift depends linearly on $v_o$, not on $v/(v \pm v_s)$.

   Tip

   Notice that the source frequency $n_0$ cancels out immediately — you never need its value. The ratio alone is enough.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

8. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.A steel pendulum clock manufactured at $32^\circ \mathrm{C}$ and working at $47^\circ \mathrm{C}$ is nearly (Coefficient of linear expansion of steel $= 12 \times 10^{-6} / ^\circ \mathrm{C}$) (A) 7.8 s slow per day (B) 7.8 s fast per day (C) 15.6 s slow per day (D) 15.6 s fast per day

   ›Reveal solutionSolution

   A pendulum clock runs slower at higher temperatures because the pendulum rod expands, increasing its length and thus its period. The fractional change in time is half the fractional change in length. For a 15°C rise, the clock loses about 7.8 seconds per day, so the correct option is (A).

   The key concept here is thermal expansion of the pendulum rod. A pendulum’s period depends on its length: $T = 2\pi \sqrt{L/g}$. When the temperature rises, the rod expands, making $L$ larger, so the period increases. A longer period means the clock ticks less frequently — it runs slow. The question asks how many seconds per day it loses.

   We need the fractional change in period. For small changes, we can use calculus or a simple approximation.

   ---

   1. Find the change in length.

      The rod’s length changes by $\Delta L = L_0 \, \alpha \, \Delta T$, where $\alpha = 12 \times 10^{-6} /^\circ\mathrm{C}$ and $\Delta T = 47 - 32 = 15^\circ\mathrm{C}$.

      So $\frac{\Delta L}{L_0} = \alpha \Delta T = 12 \times 10^{-6} \times 15 = 180 \times 10^{-6} = 1.8 \times 10^{-4}$.

   2. Relate period change to length change.

      From $T = 2\pi \sqrt{L/g}$, take the natural log: $\ln T = \ln(2\pi) + \frac{1}{2}\ln L - \frac{1}{2}\ln g$.

      Differentiate: $\frac{dT}{T} = \frac{1}{2} \frac{dL}{L}$.

      For small changes, $\frac{\Delta T}{T} \approx \frac{1}{2} \frac{\Delta L}{L}$.

      So $\frac{\Delta T}{T} = \frac{1}{2} \times 1.8 \times 10^{-4} = 9.0 \times 10^{-5}$.

   3. Interpret the sign.

      Since $\Delta L > 0$, $\Delta T > 0$ — the period increases. The clock ticks less often, so it loses time. It will be slow.

   4. Calculate the time lost per day.

      One day has $24 \times 3600 = 86400$ seconds.

      The clock’s period is longer by a fraction $9.0 \times 10^{-5}$, so in one real day, the clock completes fewer oscillations. The time it shows is less than real time by:

   $\text{loss per day} = 86400 \times \frac{\Delta T}{T} = 86400 \times 9.0 \times 10^{-5}$

   Compute: $86400 \times 9 = 777600$, then $777600 \times 10^{-5} = 7.776$ seconds.

   Rounding gives 7.8 seconds slow per day.

   Watch out

   A common mistake is to forget the factor of 1/2. Students sometimes think $\Delta T/T = \Delta L/L$, which would give 15.6 s — that’s option (C) or (D). But the period depends on the square root of length, so the fractional change in period is half the fractional change in length.

   Tip

   You can also think: if length increases by 0.018%, the period increases by 0.009%. Multiply 86400 s by 0.00009 to get ~7.8 s. No calculator needed if you notice $864 \times 9 = 7776$ and shift decimal.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

9. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.A car moving towards a cliff emits sound of frequency ‘n’. If the difference in frequencies of the horn and its echo heard by the driver of the car is 10% of ‘n’, then the speed of the car is nearly (Speed of sound in air is $336\,\mathrm{ms}^{-1}$) (A) $16\,\mathrm{ms}^{-1}$ (B) $18\,\mathrm{ms}^{-1}$ (C) $30\,\mathrm{ms}^{-1}$ (D) $33\,\mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   The problem involves the Doppler effect for sound: the driver hears both the direct horn frequency and the echo reflected from the cliff. The difference between these two frequencies is given as 10% of the original frequency. Solving the Doppler equations yields the car’s speed as approximately $16\,\mathrm{m/s}$, which corresponds to option (A).

   Concept and Intuition

   The driver emits a sound of frequency $n$ while moving toward a stationary cliff. The cliff acts as a “listener” that receives a higher frequency (because the source is approaching). The cliff then reflects that sound, becoming a stationary source emitting that higher frequency. The driver, still moving toward the cliff, now hears this reflected sound at an even higher frequency. The echo frequency is thus shifted twice: once on the way to the cliff, and once on the way back. The problem states that the difference between the echo frequency and the original horn frequency is $10\%$ of $n$, i.e., $0.1n$. We set up the Doppler equations and solve for the car’s speed.

   Step-by-step solution

   1. Frequency heard by the cliff (first Doppler shift) The car (source) moves toward the stationary cliff (observer) with speed $v_c$. The frequency received at the cliff is:

   $n_1 = n \cdot \frac{v}{v - v_c}$

   where $v = 336\,\mathrm{m/s}$ is the speed of sound. The cliff is a stationary observer, so the formula uses the source moving toward observer.

   2. Frequency heard by the driver from the echo (second Doppler shift) The cliff now acts as a stationary source emitting frequency $n_1$. The driver (observer) moves toward this source with speed $v_c$. The frequency heard by the driver is:

   $n_2 = n_1 \cdot \frac{v + v_c}{v}$

   Here the observer moves toward a stationary source, so we add the observer’s speed.

   3. Combine the two shifts Substitute $n_1$ into the expression for $n_2$:

   $n_2 = n \cdot \frac{v}{v - v_c} \cdot \frac{v + v_c}{v} = n \cdot \frac{v + v_c}{v - v_c}$

   So the echo frequency is simply $n_2 = n \cdot \frac{v + v_c}{v - v_c}$.

   4. Use the given frequency difference The difference between the echo frequency and the original horn frequency is:

   $n_2 - n = 0.1\,n$

   Substitute $n_2$:

   $n \cdot \frac{v + v_c}{v - v_c} - n = 0.1\,n$

   Divide through by $n$ (nonzero):

   $\frac{v + v_c}{v - v_c} - 1 = 0.1$

   5. Solve for $v_c$ Simplify the left side:

   $\frac{v + v_c}{v - v_c} - 1 = \frac{v + v_c - (v - v_c)}{v - v_c} = \frac{2v_c}{v - v_c}$

   So we have:

   $\frac{2v_c}{v - v_c} = 0.1$

   Multiply both sides by $v - v_c$:

   $2v_c = 0.1(v - v_c) = 0.1v - 0.1v_c$

   Bring terms together:

   $2v_c + 0.1v_c = 0.1v \quad \Rightarrow \quad 2.1v_c = 0.1v$

   Thus:

   $v_c = \frac{0.1}{2.1} v = \frac{1}{21} \times 336$

   Calculate:

   $v_c = \frac{336}{21} = 16\,\mathrm{m/s}$

   Watch out

   A common mistake is to forget that the echo undergoes two Doppler shifts — one when the sound reaches the cliff and another when it returns to the moving car. Using only a single shift gives a different (incorrect) answer.

   Tip

   Notice the neat result: the combined shift formula $n_2 = n \cdot \frac{v+v_c}{v-v_c}$ is the same as if the car were approaching its own sound reflected from a stationary wall. This is a handy shortcut for such problems.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

10. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.A car moving towards a cliff emits sound of frequency 'n'. If the difference in frequencies of the horn and its echo heard by the driver of the car is 10% of 'n', then the speed of the car is nearly (Speed of sound in air is $336\,\mathrm{ms}^{-1}$) (A) $30\,\mathrm{ms}^{-1}$ (B) $18\,\mathrm{ms}^{-1}$ (C) $16\,\mathrm{ms}^{-1}$ (D) $33\,\mathrm{ms}^{-1}$

    ›Reveal solutionSolution

    The driver hears both the direct horn frequency and the echo from the cliff (which is Doppler-shifted twice). The difference is 10% of the original frequency, leading to a car speed of about $16\,\mathrm{m/s}$.

    The key here is to track the frequency shifts carefully. The horn emits frequency $n$ from the moving car. The driver hears two sounds: the direct sound from the horn (which is Doppler-shifted because the car is moving toward the driver — but wait, the driver is in the car, so the source and observer move together). The echo comes from the sound reflecting off the cliff, which acts like a stationary observer then a stationary source.

    Let’s break it down.

    1. Direct sound heard by the driver

       The car is the source of frequency $n$, and the driver is the observer — both move together at speed $v_c$. Since source and observer have zero relative velocity, the direct sound is heard at the original frequency $n$. No Doppler shift here.

    2. Echo heard by the driver

       The sound travels to the cliff and back. First, the cliff (stationary) receives sound from the approaching car. The frequency heard by the cliff is:

    $n_1 = n \frac{v}{v - v_c}$

    where $v = 336\,\mathrm{m/s}$ is the speed of sound and $v_c$ is the car’s speed. The cliff then acts as a stationary source re-emitting $n_1$ toward the approaching car. The driver (observer) moves toward this source, so the frequency heard by the driver is:

    $n_2 = n_1 \frac{v + v_c}{v} = n \frac{v}{v - v_c} \cdot \frac{v + v_c}{v} = n \frac{v + v_c}{v - v_c}$

    3. Difference in frequencies The driver hears the direct sound at $n$ and the echo at $n_2$. The difference is:

    $n_2 - n = n \left( \frac{v + v_c}{v - v_c} - 1 \right) = n \left( \frac{2v_c}{v - v_c} \right)$

    This difference is given as 10% of $n$, i.e., $0.1 n$. So:

    $\frac{2v_c}{v - v_c} = 0.1$

    4. Solve for $v_c$ Multiply both sides by $v - v_c$:

    $2v_c = 0.1 (v - v_c)$

    $2v_c = 0.1v - 0.1v_c$

    $2.1 v_c = 0.1 v$

    $v_c = \frac{0.1}{2.1} v = \frac{1}{21} \times 336 = 16\,\mathrm{m/s}$

    Watch out

    A common mistake is to forget that the echo involves two Doppler shifts — one for the wave reaching the cliff and another for the reflected wave reaching the car. Treating it as a single shift gives a wrong answer.

    Tip

    Notice that the direct sound has no shift because source and observer move together. The echo formula $n_2 = n \frac{v+v_c}{v-v_c}$ is a handy shortcut for a moving source and moving observer approaching a stationary reflector.

    ✓Final answer

    The speed of the car is nearly $16\,\mathrm{m/s}$, which corresponds to option (C).

11. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
    Q.A body is allowed to fall freely under gravity from a height of 15 m from the ground. At a point in its path, if the kinetic energy of the body is 200% more than its potential energy, then the velocity of the body at that point is (Acceleration due to gravity $= 10~\mathrm{ms}^{-2}$) (A) $6~\mathrm{ms}^{-1}$ (B) $20~\mathrm{ms}^{-1}$ (C) $10~\mathrm{ms}^{-1}$ (D) $15~\mathrm{ms}^{-1}$

    ›Reveal solutionSolution

    When a body falls freely, its total mechanical energy (kinetic + potential) remains constant. By using the given condition that kinetic energy is 200% more than potential energy at a certain point, and applying conservation of energy from the initial height, we find the velocity at that point to be $\boxed{15~\mathrm{ms}^{-1}}$.

    When a body falls freely under gravity, assuming no air resistance, its total mechanical energy remains constant. This is a fundamental principle known as the conservation of mechanical energy. Mechanical energy is the sum of kinetic energy and potential energy.

    • Potential Energy ($PE$): This is the energy stored in an object due to its position or state. For an object at a height $h$ above a reference level (usually the ground), its gravitational potential energy is given by $PE = mgh$, where $m$ is the mass and $g$ is the acceleration due to gravity.
    • Kinetic Energy ($KE$): This is the energy an object possesses due to its motion. For an object with mass $m$ moving with velocity $v$, its kinetic energy is given by $KE = \frac{1}{2}mv^2$.

    As the body falls, its height decreases, so its potential energy decreases. Since total mechanical energy is conserved, this decrease in potential energy must be compensated by an increase in kinetic energy, meaning the body speeds up.

    The problem states a specific condition: at a certain point, the kinetic energy is 200% more than its potential energy. This means if the potential energy is $PE'$, the kinetic energy $KE'$ is $PE' + 200\%$ of $PE'$, which simplifies to $KE' = PE' + 2PE' = 3PE'$. We can use this relationship along with the conservation of mechanical energy to find the velocity.

    Here's how to solve the problem step-by-step:

    1. Calculate the initial total mechanical energy:

       The body starts falling freely from a height of $15~\mathrm{m}$. "Freely falling" implies its initial velocity is $0~\mathrm{ms}^{-1}$.

       • Initial height, $H = 15~\mathrm{m}$.
       • Initial velocity, $u = 0~\mathrm{ms}^{-1}$.
       • Acceleration due to gravity, $g = 10~\mathrm{ms}^{-2}$.

       The initial potential energy ($PE_{initial}$) is:

    $PE_{initial} = mgH = m \times 10~\mathrm{ms}^{-2} \times 15~\mathrm{m} = 150m~\mathrm{J}$

    The initial kinetic energy ($KE_{initial}$) is:
    

    $KE_{initial} = \frac{1}{2}mu^2 = \frac{1}{2}m(0)^2 = 0~\mathrm{J}$

    The total initial mechanical energy ($E_{total}$) is:
    

    $E_{total} = PE_{initial} + KE_{initial} = 150m + 0 = 150m~\mathrm{J}$

    2. Express energies at the point in question: Let the body be at a height $h'$ from the ground at the point where the given condition applies, and let its velocity at this point be $v'$.
       • Potential energy at this point ($PE'$) is:

    $PE' = mgh' = m \times 10~\mathrm{ms}^{-2} \times h' = 10mh'~\mathrm{J}$

    *   Kinetic energy at this point ($KE'$) is:
    

    $KE' = \frac{1}{2}mv'^2~\mathrm{J}$

    3. Apply the given condition: The problem states that the kinetic energy is 200% more than its potential energy.

    $KE' = PE' + (200\% \text{ of } PE')$

    $KE' = PE' + 2PE'$

    $KE' = 3PE'$

    Substitute the expressions for $KE'$ and $PE'$:
    

    $\frac{1}{2}mv'^2 = 3(10mh')$

    Divide both sides by $m$:
    

    $\frac{1}{2}v'^2 = 30h'$

    This gives us a relationship between $v'^2$ and $h'$:
    

    $v'^2 = 60h' \quad \text{(Equation 1)}$

    4. Apply the principle of conservation of mechanical energy: The total mechanical energy at the initial point must be equal to the total mechanical energy at the point in question.

    $E_{total} = PE' + KE'$

    $150m = 10mh' + \frac{1}{2}mv'^2$

    Divide both sides by $m$:
    

    $150 = 10h' + \frac{1}{2}v'^2 \quad \text{(Equation 2)}$

    5. Solve for the velocity ($v'$): We have two equations (Equation 1 and Equation 2) and two unknowns ($h'$ and $v'$). We can substitute $h'$ from Equation 1 into Equation 2. From Equation 1, $h' = \frac{v'^2}{60}$. Substitute this into Equation 2:

    $150 = 10\left(\frac{v'^2}{60}\right) + \frac{1}{2}v'^2$

    $150 = \frac{v'^2}{6} + \frac{v'^2}{2}$

    To combine the terms on the right, find a common denominator (which is 6):
    

    $150 = \frac{v'^2}{6} + \frac{3v'^2}{6}$

    $150 = \frac{v'^2 + 3v'^2}{6}$

    $150 = \frac{4v'^2}{6}$

    $150 = \frac{2v'^2}{3}$

    Now, solve for $v'^2$:
    

    $2v'^2 = 150 \times 3$

    $2v'^2 = 450$

    $v'^2 = \frac{450}{2}$

    $v'^2 = 225$

    Take the square root to find $v'$:
    

    $v' = \sqrt{225}$

    $v' = 15~\mathrm{ms}^{-1}$

    Watch out

    Be careful with the phrasing "200% more than". It means $PE + 200\% \text{ of } PE$, not just $200\% \text{ of } PE$. If it were "kinetic energy is 200% of its potential energy", then $KE = 2PE$. But "200% more than" means $KE = PE + 2PE = 3PE$.

    ✓Final answer

    The velocity of the body at that point is $\boxed{15~\mathrm{ms}^{-1}}$.

12. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
    Q.When a long hollow steel pipe is struck with a hammer at one end, two sounds are heard at the other end. If the time interval between the two sounds heard is 0.53 s, then the length of the pipe is (Speed of sound in steel is $5100~\mathrm{ms}^{-1}$ and speed of sound in air is $330~\mathrm{ms}^{-1}$) (A) 153 m (B) 174 m (C) 187 m (D) 270 m

    ›Reveal solutionSolution

    The two sounds travel through steel and air at different speeds; the time difference gives the pipe length. The length is 187 m.

    The key idea is that when you strike one end of a long steel pipe, the sound reaches the other end through two different paths: one through the steel itself (the pipe material) and one through the air inside the pipe. Since sound travels much faster in steel than in air, the sound through steel arrives first, and the sound through air arrives later. The time interval you hear is the difference between these two travel times.

    This is a classic problem of two media with different speeds. The distance travelled is the same — the length of the pipe — so we can write the time for each path as distance divided by speed. The difference between these times is given, and we solve for the length.

    1. Let the length of the pipe be $L$ metres.

       Time taken by sound through steel: $t_s = \dfrac{L}{v_s}$, where $v_s = 5100~\mathrm{m/s}$.

       Time taken by sound through air: $t_a = \dfrac{L}{v_a}$, where $v_a = 330~\mathrm{m/s}$.

    2. The sound through air arrives later, so the time interval is:

    $t_a - t_s = 0.53~\mathrm{s}$

    3. Substitute the expressions:

    $\frac{L}{330} - \frac{L}{5100} = 0.53$

    4. Take $L$ common:

    $L \left( \frac{1}{330} - \frac{1}{5100} \right) = 0.53$

    5. Compute the bracket. Find a common denominator (LCM of 330 and 5100). $330 = 33 \times 10$, $5100 = 51 \times 100$. It's easier to compute directly:

    $\frac{1}{330} = \frac{5100}{330 \times 5100} \quad \text{and} \quad \frac{1}{5100} = \frac{330}{330 \times 5100}$

    So:

    $\frac{1}{330} - \frac{1}{5100} = \frac{5100 - 330}{330 \times 5100} = \frac{4770}{330 \times 5100}$

    6. Simplify the fraction. Divide numerator and denominator by 30:

    $\frac{4770 \div 30}{330 \times 5100 \div 30} = \frac{159}{330 \times 170}$

    (Check: $330 \times 5100 \div 30 = 330 \times 170$).

    Now simplify further: $330 \times 170 = 56100$. So:

    $\frac{1}{330} - \frac{1}{5100} = \frac{159}{56100}$

    7. Reduce $\frac{159}{56100}$. Both are divisible by 3: $159 \div 3 = 53$, $56100 \div 3 = 18700$. So:

    $\frac{1}{330} - \frac{1}{5100} = \frac{53}{18700}$

    Tip

    A faster way: compute the bracket as $\frac{5100 - 330}{330 \times 5100} = \frac{4770}{1,683,000}$. Simplify by dividing numerator and denominator by 30: $\frac{159}{56,100}$, then by 3: $\frac{53}{18,700}$. Same result.

    8. Now the equation becomes:

    $L \times \frac{53}{18700} = 0.53$

    9. Solve for $L$:

    $L = 0.53 \times \frac{18700}{53}$

    10. Notice $0.53 = \frac{53}{100}$. So:

    $L = \frac{53}{100} \times \frac{18700}{53} = \frac{18700}{100} = 187$

    Thus the length of the pipe is 187 metres.

    Watch out

    A common mistake is to subtract the speeds directly or to use the wrong order in the time difference. Always remember: the faster medium gives the shorter time, so the slower medium's time minus the faster medium's time gives a positive interval.

    ✓Final answer

    The length of the pipe is $\boxed{187~\mathrm{m}}$, which corresponds to option (C).