Physics — Class 11 practice questions

Worked examples for Physics — Class 11

1. Question 1

   In still air, a source of sound of frequency $n$ and a listener move along the same straight line with the same speed $u$, in the same direction, the listener ahead of the source.

   The speed of sound in air is $v$, with $u < v$.

   The frequency heard by the listener is

   1. A. $n\dfrac{v + u}{v - u}$

      You treated both motions as approach, adding $u$ on top and subtracting it below. Here the listener is moving away from the source while the source chases it, so both terms carry the same sign and cancel.

   2. B. $n\dfrac{v - u}{v}$

      You counted only the listener receding from the source and left the source at rest. The source is advancing at the same speed, which supplies the matching $v - u$ in the denominator.

   3. C. $n\dfrac{v}{v - u}$

      You kept the source's motion towards the listener and forgot that the listener is moving too. That form holds only for a stationary listener.

   4. D. $n$Correct answer

      The distance between the source and the listener never changes, so wavefronts arrive at exactly the rate they are emitted: $n' = n\frac{v - u}{v - u} = n$.

   Explanation

   The Doppler effect in sound depends on the motion of the source and of the listener relative to the medium, and the two enter the formula $n' = n\frac{v \pm v_{o}}{v \mp v_{s}}$ in different places — the listener's speed in the numerator, the source's in the denominator. When both travel in the same direction with the same speed, the listener's recession and the source's pursuit contribute identical factors that cancel, so the apparent frequency equals the emitted one. Physically, the separation between them is constant, so the wavefronts are neither crowded nor stretched.

2. Question 2

   A sound note of fixed frequency travels through a given mass of air.

   The speed of sound in that air is increased by

   1. A. an increase in the frequency of the source

      From $v = f\lambda$ you took the speed to follow the frequency. It is the other way round: the medium fixes $v$, and raising $f$ simply shortens $\lambda$ in the same proportion.

   2. B. a rise in the temperature of the airCorrect answer

      $v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma R T}{M}}$, so for a given gas the speed rises with the absolute temperature — the only one of these changes that alters the medium's $\frac{P}{\rho}$.

   3. C. a rise in the pressure of the air at constant temperature

      You read $v = \sqrt{\frac{\gamma P}{\rho}}$ as though only $P$ were in it. At constant temperature a given mass of air that is compressed raises its density in the same proportion as its pressure, so $\frac{P}{\rho}$, and hence $v$, is unchanged.

   4. D. an increase in the loudness of the source

      Loudness is set by the amplitude of the wave, which carries more energy but does not change how fast a compression is handed on from layer to layer. A loud note and a faint one of the same frequency reach you together.

   Explanation

   The speed of a sound wave is a property of the medium, not of the source. Laplace's corrected formula $v = \sqrt{\frac{\gamma P}{\rho}}$ can be written $v = \sqrt{\frac{\gamma R T}{M}}$, which shows that for a given gas only the absolute temperature matters: a pressure change at constant temperature alters $P$ and $\rho$ together and leaves the ratio alone. Frequency and amplitude belong to the source and merely fix the wavelength and the loudness of the note that travels at that speed.

3. Question 3

   A ball strikes a wall normally and rebounds along the same line.

   $m = 0.2\ \text{kg}$ (mass of the ball)

   $u = 10\ \text{m/s}$ (speed before it strikes the wall)

   $v = 6\ \text{m/s}$ (speed after it rebounds)

   $t = 0.02\ \text{s}$ (duration of the contact)

   The average force exerted by the wall on the ball is

   1. A. $40\ \text{N}$

      You subtracted the two speeds, $m(u - v)$, as if the ball kept moving in the same direction. A rebound reverses the velocity, so the two momenta have opposite signs and their magnitudes add.

   2. B. $100\ \text{N}$

      You used only the incoming momentum, $\frac{mu}{t}$, treating the ball as if it stopped dead at the wall. The wall also has to push it back out at $6\ \text{m/s}$, and that part of the momentum change counts too.

   3. C. $160\ \text{N}$Correct answer

      The velocity reverses, so $\lvert \Delta p \rvert = m(u + v) = 0.2 \times 16 = 3.2\ \text{kg m/s}$, and $F = \frac{\Delta p}{t} = \frac{3.2}{0.02} = 160\ \text{N}$.

   4. D. $3.2\ \text{N s}$

      You stopped at the momentum change itself. That quantity is the impulse; dividing it by the contact time is what turns it into an average force.

   Explanation

   By Newton's second law in its momentum form, the average force over a collision is the change in momentum divided by the contact time. The direction of the velocity is the whole difficulty here: a rebound along the same line reverses the sign, so the initial and final momenta subtract to $m(u + v)$ and not $m(u - v)$. The momentum change alone is an impulse, which becomes a force only after division by the time.

4. Question 4

   A point charge $q$ is kept at the centre of a hollow cube, and there is no other charge anywhere.

   The edge of the cube is now doubled, the charge staying at the centre.

   The electric flux through one face of the cube

   1. A. Becomes one half

      This takes the flux through a face as falling in proportion to the edge length. No such dependence exists: Gauss' law fixes the total flux by the enclosed charge, and symmetry divides it equally among the faces.

   2. B. Becomes one fourth

      You used the inverse square law, as though doubling the distance to a face quartered the flux. That is how the FIELD at a point falls, but flux is the field summed over the whole face, and the face has grown in exactly the compensating proportion.

   3. C. Becomes four times

      You multiplied by the growth in area alone, since each face is now four times as large, and forgot that the field over that face is correspondingly weaker.

   4. D. Remains unchangedCorrect answer

      By Gauss' law the flux through the whole closed surface is $\frac{q}{\varepsilon_{0}}$, which depends only on the enclosed charge. By symmetry each of the six faces carries $\frac{q}{6\varepsilon_{0}}$, whatever the size of the cube.

   Explanation

   Gauss' law says the net flux through any closed surface is the charge enclosed divided by $\varepsilon_{0}$, and nothing else about the surface enters it, neither its size nor its shape. With the charge at the centre, the six faces are equivalent, so each carries one sixth of that total. Enlarging the cube changes neither the enclosed charge nor the symmetry, so the flux per face is untouched.

5. Question 5

   A metal block at $100,^\circ\text{C}$ is dropped into water at $20,^\circ\text{C}$ in a calorimeter of negligible heat capacity, and the mixture settles at $40,^\circ\text{C}$.

   $m_{M} = 2\ \text{kg}$ (metal)

   $m_{W} = 0.4\ \text{kg}$ (water)

   $c_{W} = 4200\ \text{J/(kg K)}$

   The specific heat capacity of the metal is

   1. A. $168\ \text{J/(kg K)}$

      You divided by the block's starting temperature instead of by its fall in temperature. The block cools from $100,^\circ\text{C}$ to $40,^\circ\text{C}$, a drop of $60\ \text{K}$, not $100\ \text{K}$.

   2. B. $2520\ \text{J/(kg K)}$

      You gave the water's rise of $60\ \text{K}$ and the metal's drop of $20\ \text{K}$. Each body carries its own change of temperature: the water rises by $20\ \text{K}$ and the metal falls by $60\ \text{K}$.

   3. C. $560\ \text{J/K}$

      You stopped at the heat capacity of the whole block, $\frac{Q}{\Delta T}$ in joule per kelvin. A specific heat is heat per kelvin per kilogram, so one more division, by the $2\ \text{kg}$ of metal, is still owed.

   4. D. $280\ \text{J/(kg K)}$Correct answer

      Heat gained by the water is $0.4 \times 4200 \times 20 = 33600\ \text{J}$; the metal falls through $60\ \text{K}$, so $c = \frac{33600}{2 \times 60} = 280\ \text{J/(kg K)}$.

   Explanation

   In a calorimeter of negligible heat capacity the heat lost by the hot body equals the heat gained by the cold one, each written as $mc,\Delta T$ with its own change of temperature. Equating the two gives one equation in the unknown specific heat; dividing by the mass of the metal is what turns a heat per degree into a specific heat.

6. Question 6

   Two thin lenses are kept in contact on a common principal axis.

   $f_{1} = 10\ \text{cm}$

   $f_{2} = -15\ \text{cm}$

   An object stands erect on the axis in front of the combination.

   $h = 2\ \text{cm}$ (height of the object)

   $u = -45\ \text{cm}$ (position of the object)

   The height of the image is

   1. A. $4\ \text{cm}$

      You took $m = -\frac{v}{u}$, which belongs to a mirror. For a thin lens $m = \frac{v}{u}$, so the sign is lost and a real, inverted image is reported as erect.

   2. B. $-2$

      You stopped at the magnification. It is a pure number; the question asks for a height, which needs the last step $h' = mh$.

   3. C. $0.2\ \text{cm}$

      You added the focal lengths, $10 + (-15) = -5\ \text{cm}$. Lenses in contact add their POWERS, $\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$, which here gives a converging combination of $F = 30\ \text{cm}$, not a diverging one.

   4. D. $-4\ \text{cm}$Correct answer

      $\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{1}{10} - \frac{1}{15}$, so $F = 30\ \text{cm}$. Then $\frac{1}{v} = \frac{1}{F} + \frac{1}{u} = \frac{1}{30} - \frac{1}{45}$ gives $v = 90\ \text{cm}$, $m = \frac{v}{u} = -2$, and $h' = mh = -4\ \text{cm}$ — inverted and twice as tall.

   Explanation

   Thin lenses in contact combine through their powers, so the reciprocals of the focal lengths add and the pair behaves as one thin lens of focal length $F$. That focal length feeds the thin lens formula, which locates the image, and the image height then follows from $h' = mh$ with $m = \frac{v}{u}$. The sign carries the physics: a negative height means the image is inverted with respect to the object.

7. Question 7

   Two adjacent sides of a parallelogram are represented by the displacement vectors $\vec{A}$ and $\vec{B}$, inclined to each other at an angle $\theta$, where

   $\lvert\vec{A}\rvert = 8\ \text{m}$

   $\lvert\vec{B}\rvert = 5\ \text{m}$

   $\theta = 30^\circ$

   The area of the parallelogram is

   1. A. $20\ \text{m}$

      The number is right, but you have quoted a length where an area was asked for: multiplying two displacements in metres gives square metres.

   2. B. $34.6\ \text{m}^{2}$

      You used $AB\cos\theta$. The cosine belongs to the scalar product, which measures how much of one vector lies along the other; the area of the parallelogram is the vector product, and it carries the sine.

   3. C. $10\ \text{m}^{2}$

      You computed $\frac{1}{2}AB\sin\theta$, which is the area of the triangle on these two sides. The parallelogram is made of two such triangles, so it is twice as large.

   4. D. $20\ \text{m}^{2}$Correct answer

      The area is $\lvert\vec{A}\times\vec{B}\rvert = AB\sin\theta = 8 \times 5 \times \frac{1}{2} = 20\ \text{m}^{2}$.

   Explanation

   The magnitude of the vector product of two vectors is exactly the area of the parallelogram they span, $AB\sin\theta$ — that is the geometric meaning of the cross product. The scalar product $AB\cos\theta$ measures the projection of one vector along the other instead, and so can never be an area of this parallelogram. Since two lengths are multiplied, the result must be expressed in square metres.

8. Question 8

   The dimensional formula of surface tension is

   1. A. $[L^{-1}M^{1}T^{-2}]$

      You divided the force by an area instead of by a length; force per unit area is pressure, whose unit is $\text{N/m}^{2}$.

   2. B. $[L^{1}M^{1}T^{-2}]$

      You stopped at force. Surface tension is the force acting per unit length of the free surface, so one power of length must still be divided out.

   3. C. $[L^{0}M^{1}T^{-2}]$Correct answer

      Surface tension is force per unit length, so its dimensions are $\frac{[M^{1}L^{1}T^{-2}]}{[L^{1}]} = [M^{1}T^{-2}]$, i.e. the length exponent is zero. Its SI unit N/m carries exactly these dimensions.

   4. D. $[L^{2}M^{1}T^{-2}]$

      These are the dimensions of work or energy. Surface tension may be read as surface energy per unit area, so the energy dimensions still have to be divided by $[L^{2}]$.

   Explanation

   A dimensional formula follows from the defining relation of the quantity. Surface tension is defined as the force per unit length acting along a line in the liquid surface, so dividing the dimensions of force by one power of length leaves mass and time only. Reading it instead as surface energy per unit area gives the same result, since energy per area also reduces to $[M^{1}T^{-2}]$.