Students can access the CBSE Sample Papers for Class 11 Physics with Solutions and marking scheme Set 5 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 11 Physics Set 5 with Solutions
Time Allowed : 3 hours
Maximum Marks : 70
General Instructions:
- There are 33 questions in all. All questions are compulsory.
- This question paper has five sections: Section A, Section B, Section C, Section D and Section E.
- All the sections are compulsory.
- Section A contains sixteen questions, twelve MCQ and four Assertion Reasoning based of 1 mark each, Section B contains five questions of two marks each, Section C contains seven questions of three marks each, Section D contains two case study based questions of four marks each and Section E contains three long answer questions of five marks each.
- There is no overall choice. However, an internal choice has been provided in one question in Section B, one question in Section C, one question in each CBQ in Section D and all three questions in Section E. You have to attempt only one of the choices in such questions
- Use of calculators is not allowed.
- You may use the following values of physical constants where ever necessary
- c = 3 × 108 m/s
- me = 9.1 × 10-31 kg
- µ0 = 4π × 10-7 TmA-1
- ε0 = 8.854 × 10-12 × C2 N-1 m-2
- Avogadro’s number = 6.023 × 1023 per gram mole
Section – A
Question 1.
The order of the magnitude of speed of light in SI unit is
(A) 10
(B) 8
(C) 3
(D) 12
Answer:
(B) 8
Explanation:
SI unit of Speed of light is 3 × 108 ms-1. So, the order of magnitude is 8.
Question 2.
The ratio of the numerical values of the average velocity and average speed of a body is
(A) unity
(B) less than unity
(C) unity or less
(D) unity or more
Answer:
(C) unity or less
Explanation:
Velocity ≤ speed
So, \(\frac{\text { Velocity }}{\text { Speed }}\) ≤ 1
Question 3.
Two particles of equal masses are revolving in circular paths of radii and r2 respectively with the same speed. The ratio of their centripetal forces is
(A) \(\frac{r_2}{r_1}\)
(B) \(\frac{r_1}{r_2}\)
(C) \(\sqrt{\frac{r_1}{r_2}}\)
(D) \(\sqrt{\frac{r_2}{r_1}}\)
Answer:
(A) \(\frac{r_2}{r_1}\)
Explanation:
Centripetal force = F = \(\frac{m v^2}{r}\)
Mass and speed being same for the two particles,
F ∝ \(\frac{1}{r}\)
∴ \(\frac{F_1}{F_2}=\frac{r_1}{r_2}\)
Question 4.
A body of mass m collides against a wall with velocity v and rebounds with same speed. Its change of momentum is
(A) 2mv
(B) mv
(C) – mv
(D) zero
Answer:
(A) 2mv
Explanation:
Initial velocity = v
Final velocity = – v,
if mass = m, then
Initial momentum = Pi = mv
Final momentum = Pf = – mv
So |∆P| = |Pf – Pi| = 2mv
Question 5.
A particle is moving with a constant velocity along a straight line. Force is not required to
(A) Change its direction
(B) Decrease its momentum
(C) Increase its speed
(D) Keep it moving with uniform velocity
Answer:
(D) Keep it moving with uniform velocity
Question 6.
A person is holding a bucket by applying a force of 10 Newton. He moves a horizontal distance of 5 m and then climbs up a vertical height of 10 m. The total work done by him is
(A) 100 J
(B) 200 J
(C) 150 J
(D) 60 J
Answer:
(A) 100 J
Explanation:
For horizontal motion,
F = 10 N, s = 5 m and θ = 90°.
So work done = 0
For vertical motion,
F = 10 N, s = 10 m and θ = 0°.
So work done = 10 × 10 = 100 J
Total work done = 100 J
Question 7.
The ratio of lengths of two wires of same material is 1 : 2. The ratio of their diameters is 2 : 1. They are stretched by the same force. The ratio of their increase in lengths is
(A) 1 : 8
(B) 8 : 1
(C) 1 : 4
(D) 4 : 1
Answer:
(A) 1 : 8
Explanation:
Increase in length = ∆l = \(\frac{F L}{A γ}\)
Or, ∆l = \(\frac{F L}{\pi \frac{\mathrm{d}^2}{4} Y}\)
F and Y being constants,
∆l ∝ \(\frac{L}{d^2}\)
Or, \(\frac{\Delta l_A}{\Delta l_B}=\frac{L_A}{L_B} \times \frac{d_B^2}{d_a^2}\)
∴ \(\frac{\Delta l_A}{\Delta l_B}=\frac{1}{2} \times \frac{1}{4}\)
= 1 : 8
Question 8.
Two water pipes A and B of diameters 2 cm and 4 cm respectively are joined with the main supply line. The ratio of velocities flow of water is
(A) 4 : 1
(B) 1 : 4
(C) 1 : 2
(D) 2 : 1
Answer:
(A) 4 : 1
Explanation:
From equation of continuity,
av = constant.
∴ aAvA = aBvB
Or, \(\frac{v_A}{v_B}=\frac{a_B}{a_A}\)
Or, \(\frac{v_A}{v_B}=\frac{d_B^2}{d_A^2}=\frac{4^2}{2^2}\)
= 4 : 1
Question 9.
Which of the following rods of same material will conduct maximum heat when their ends are maintained at the same steady temperature?
(A) Length 1 m, radius 1 cm
(B) Length 1 m, radius 2 cm
(C) Length 2m, radius 1 cm
(D) Length 2m, radius 2 cm
Answer:
(B) Length 1 m, radius 2 cm
Explanation:
\(\frac{Q}{t}=\frac{K A \Delta \theta}{L}\)
K and ∆θ remaining constant
\(\frac{Q}{t} \propto \frac{A}{L}\)
Or, \(\frac{Q}{t} \propto \frac{r^2}{L}\)
\(\frac{r^2}{L}\) is maximum for rod of length 1 m and radius 2 cm.
Question 10.
The equation of motion of a particle \(\frac{d^2 y}{d x^2}\) + ky = 0, where k is a positive constant. The time period of the motion
(A) 2π√k
(B) \(\frac{2 \pi}{\sqrt{k}}\)
(C) 2πk
(D) \(\frac{2 \pi}{k}\)
Answer:
(B) \(\frac{2 \pi}{\sqrt{k}}\)
Explanation:
Comparing the given equation with \(\frac{d^2 y}{d t^2}\) + ω2y = 0
ω2 = k
Or, \(\left(\frac{2 \pi}{T}\right)^2\) = k
Or, \(\frac{2 \pi}{T}\) = √k
∴ T = \(\frac{2 \pi}{\sqrt{k}}\)
Question 11.
Speed of sound wave in air
(A) is independent of temperature.
(B) increases with pressure.
(C) increases with increase in humidity
(D) decreases with increase in humidity.
Answer:
(C) increases with increase in humidity
Question 12.
The relation between acceleration and displacement of four particles are given below
(A) ax = + 2x
(B) ax = + 2x2
(C) ax = – 2x
(D) ax = – 2x2
Answer:
(C) ax = – 2x
Explanation:
For S.H. M, ax ∝ (- x)
For Questions 13 to 16 two statements are given – one labelled Assertion (A) and other labelled Reason (R).
Select the correct answer to these questions from the options as given below.
(A) If both Assertion and Reason are true and Reason is correct explanation of Assertion.
(B) If both Assertion and Reason are true and Reason is not the correct explanation of Assertion.
(C) If Assertion is true but Reason is false.
(D) If both Assertion and Reason are false.
Question 13.
Assertion : Parallax method cannot be used for measuring distances of stars more than 100 light years away.
Reason : Because parallax angle reduces so much that it cannot be measured accurately.
Answer:
(A) If both Assertion and Reason are true and Reason is correct explanation of Assertion.
Explanation:
As the distance of the star increases, the parallax angle decreases. It is a practical limitation of measurement of parallax angle accurately if the distance of the star is more than loo light years. Hence, the assertion and reason both are true and the reason explains the assertion.
Question 14.
Assertion : Two particles of different mass, projected with same velocity at same angles. The maximum height attained by both the particle will be same.
Reason : The maximum height of projectile is independent of mass of the particle.
Answer:
(A) If both Assertion and Reason are true and Reason is correct explanation of Assertion.
Explanation:
The maximum height attained by a projectile is
Hmax = \(\frac{u^2 \sin ^2 \theta}{2 g}\)
It is independent of mass.
Hence, the assertion and reason both are true and the reason explains the assertion.
Question 15.
Assertion : The speed of sound in solids is maximum though their density is large.
Reason : The coefficient of elasticity of solid is large.
Answer:
(A) If both Assertion and Reason are true and Reason is correct explanation of Assertion.
Explanation:
The velocity of sound through solid
v = \(\sqrt{\frac{E}{\rho}}\)
Though the density (p) of solids is large by their coefficient of elasticity (E) is much larger compared to that of gases and liquids. For this reason, The speed of sound in solids is maximum though their density is large.
So, the assertion and reason both are true and the reason explains the assertion.
Question 16.
Assertion : On a rainy day, sound travel slower than on a dry day. .
Reason : When moisture is present in air the density of air increases.
Answer:
(C) If Assertion is true but Reason is false.
Explanation:
Presence of moisture reduces the density of air. It is because the density of water vapour is less than that of dry air. The velocity of sound is inversely proportional to the square root of density. So, speed of sound is more in humid air than that in the dry air. Therefore, on a rainy day sound travels faster than on a dry day.
So, assertion and reason both are false.
Section – B
Question 17.
A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the sun and the earth in terms of the new unit if light takes 8 minutes and 20 second to cover the distance?
Answer:
Distance between sun and earth = Speed of light × times taken by light the travel the distance
= 1 unit × 8 min 20 s
= 1 × 500 s
= 500 unit
Question 18.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms-1. How long does the body take to stop?
Answer:
F = ma
Or, – 50 = 20 × a
∴ a = \(\frac{-50}{20}\) = – 2.5 m/s2
V = u + at
Or, 0 = 15 – 2.5 × t
t = \(\frac{15}{2.5}\) = 6s
Question 19.
In HCl molecule, the separation between the nuclei of hydrogen and Chlorine atom is 1.27 Å. Find the location of C.M. of the molecule. Chlorine atom is 35.5 times massive compared to Hydrogen atom.
Answer:
Mass of H atom = m
Mass of Cl atom = 35.5 m
Distance of C.M. from Chlorine atom = x Å
So, distance of C.M. from H atom = (1.27 – x) Å
If C.M. is located at the origin, then
\(\frac{(1.27-x) m+35.5 m x}{m+35.5 m}\) = 0
∴ x = – 0.037 Å
Hence the C.M. is located at a distance 0.037 Å left of the Chlorine atom.
OR
Show that the area of the triangle contained between the vectors a and b is half of the magnitude of a x b.
Answer:
In triangle POQ,
sin θ = \(\frac{P N}{O P}=\frac{P N}{|\vec{b}|}\)
∴ PN = \(|\vec{b}|\) sin θ
Now, \(|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}|\) = sin θ
= OQ × PN
= 2 × — × OQ × PN
= 2 × Area of ∆POQ
∴ Area of ∆POQ = – \(|\vec{a} \times \vec{b}|\)
Question 20.
A steel wire has length 12 m and mass 2.1 kg. What should be the tension in the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at 20°C? (speed of sound in dry air at 20°C = 343 m/s)
Answer:
Mass per unit length = µ = mass/length
∴ µ = 0.175 kg/m
If T is the tension of the wire, then
Velocity = ν = \(\sqrt{\frac{T}{\mu}}\)
Or, (343)2 = T/0.175
∴ T = (343)2 × 0.175 = 20588.575 N
Question 21.
A tuning fork A, marked 512 Hz, produces 5 beats per second, when sounded with another unmarked tuning fork B. If B is loaded with wax the number of beats is again 5 per second. What is the frequency of the tuning fork B when not loaded?
Answer:
Frequency of tuning fork A,
υA = 512 Hz
Now, frequency of tuning fork B may be
υB = υA ± 5
= 312 ± 5
= 517 Hz or 507 Hz
As on loading B, frequency decreases.
If υB = 517 Hz, it might reduced to 507 Hz, to give again a beat of 5 Hz.
If υB = 507 Hz, reduction will increase the frequency of beat.
But that has not happened.
So, υB = 517 Hz.
Section – C
Question 22.
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Find the magnitude and direction of the acceleration produced in the body.
Answer:
Resultant of the forces applied = Fnet
= \(\sqrt{8^2+6^2}\)
= 10 N
Acceleration = \(\frac{F_{n e t}}{m}=\frac{10}{5}\) = 2 m/s2
tan θ = \(\frac{8}{6}=\frac{4}{3}\)
∴ θ = tan-1 \(\left(\frac{4}{3}\right)\)
= 53° from 6 N
Acceleration will be in the direction of Fnet.
Question 23.
A rocket is fired from earth towards the sun. At what distance 8 N and 6 N. Find the magnitude and direction of from earth’s centre the gravitational force on the rocket be zero?
[Mass of sun = 2 × 1030 kg
Mass of earth = 6 × 1024 kg
Orbital radius of earth = 1.5 × 1011 m
Neglect the effect of other heavenly bodies.]
Answer:
At a distance x from the centre of the earth the gravitational force on the earth will be zero. At that point, the gravitational force due to the earth will be equal to the gravitational force due to the sun.
∴ x = 0.00259 × 1011
= 2.59 × 108 m
Question 24.
(i) Following figure shows the stress-strain graph of a given material.
Calculate its
(a) Young’s Modulus and
(b) yield strength.
Answer:
(a) From graph, strain = 0.002 when stress = 150 × 106 Nm-2
So, Young’s modulus = Y = stress / strain
= \(\frac{150 \times 10^6}{0.002}\)
= 7.5 × 1010 Nm-2
(b) Yield strength = 300 × 106 Nm-2
(ii) From the following stress strain graph identify which material is brittle and which one is ductile.
Answer:
Material A is brittle and material B is ductile.
Question 25.
(i) Glycerin flows steadily through a horizontal tube of length 1.5 m and radius 1 cm. If the amount of glycerin collected per second at one end is 4 × 10-3 kg. What is the pressure difference between the two ends of the tube?
[Density of glycerin = 1.3 × 103 kgm-3
Viscosity of glycerin = 0.83 pascal second]
Answer:
Volume of glycerin flowing per second = V
= \(\frac{\text { Mass of glycerin }}{\text { Density of glycerin }}\)
∴ V = \(\frac{4 \times 10^{-2}}{1.3 \times 10^3}\)
= 3.08 × 106 m3
From Poiseuille’s law
V = \(\frac{\pi \Delta p r^4}{8 \eta l}\)
Or, ∆p = \(\frac{8 V \eta l}{\pi r^4}\)
Or, ∆p = \(\frac{8 \times 3.08 \times 10^6 \times 0.83 \times 1.5}{\pi \times\left(\frac{1}{100}\right)^4}\)
∴ ∆p = 980 pa
(ii) Can Bernoulli’s equation be used to describe the flow of through a rapid in a river?
Answer:
No, Bernoulli’s equation cannot be used to describe the flow of water through a rapid in a river because of the turbulent flow of water. This equation can only be applied to streamline flow.
Question 26.
(i) A copper block of mass 2.5 kg is heated in a furnace to a temperature 500° C and then placed on a large ice block. Find the maximum amount of ice which it can melt.
[Specific heat of copper = 0.39 J g-1 K-1
Latent heat of fusion of ice = 335 J g-1]
Answer:
Maximum heat lost by copper = Q = MC∆θ
= 2500 × 0.39 × 500
=487500 J
Heat required to melt in gram of ice = Q = mL
Or, 487500 = m × 335
∴ m = \(\frac{487500}{335}\) = 1455.22 g
This is the maximum amount of ice that can melt.
(ii) The coefficient of volume expansion of glycerin is 49 × 10-5 K-1 . What is the fractional change of its density for 300° C rise of temperature?
Answer:
Fractional change in density = \(\frac{\rho_t-\rho_o}{\rho_o}\) = – y∆θ
= – 49 × 10-5 × 30
= – 1470 × 10-5
(Negative sign indicates decrease of density with temperature rise.)
Question 27.
What amount of heat is to be supplied to 2 × 10-2 kg of Nitrogen to raise its temperature by 45°C at constant pressure?
[Molecular mass of Nitrogen = 28
R = 8.3 J mol-1 K-1]
Answer:
Number of moles in 2 × 10-2 kg Nitrogen = n = m/M
= 2 × 10-2 × 103 / 28
= 0.714
Molar specific heat at constant pressure of Nitrogen = C\(\hat{P}\)
= \(\frac{7 R}{2}\)
= 7 × \(\frac{7 \times 8.3}{2}\)
= 29.05 J mol-1K-1
Total amount of heat to be supplied = ∆Q = nC\(\hat{P}\) ∆T
= 0.714 × 29.05 × 45
= 933.38 J
OR
Volume of an air bubble is 1 cm3 when at the bottom of a lake 40 m deep at temperature 12°C. What will be its volume when it reaches the surface of the lake where the temperature is 35°C?
Answer:
Pressure on the surface of the lake = PS = 1 atm
= 1.013 × 105 pa
Pressure on the bottom of the lake = P\(\hat{B}\)
= P\(\hat{S}\)
= hρg
= 1.013 × 105 + 40 × 103 × 9.8
= 493300 pa
Now applying \(\frac{PV}{T}\) = constant
\(\frac{P_S V_S}{T_S}=\frac{P_B V_B}{T_B}\)
Or, \(\frac{1.013 \times 10^5 \times V_S}{(37+273)}=\frac{493300 \times 1 \times 10^{-6}}{(12+273)}\)
Or, VS = \(\frac{493300 \times 1 \times 10^{-6}}{(12+273)} \times \frac{(35+273)}{1.013 \times 10^5}\)
Or, VS = 526268.68 × 10-11 m3
∴ VS = 526268.68 × 10-11 × 106 cm3
= 5.26 cm3
Question 28.
(i) A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound?
Speed of sound in air is 340 m s-1
Speed of sound and in water 1486 ms-1
Answer:
Wavelength of reflected sound in air = vair / v
= \(\frac{340}{10^6}\)
= 3.4 × 10-4 m
Wavelength of refracted sound in water = vwater / v
= 1486/106
= 1.46 × 10-3 m
(ii) Two sitar strings A and B playing a tone “Ga” are slightly out of tune and produce beats of frequency 6 Hz. The tension of the string A is slightly reduced and the beat frequency is found to be 3 Hz. If the initial frequency of A was 324 Hz, what was the frequency of B?
Answer:
Beat frequency = n = |fA ± fA|
Or, 6 = 324 ± fB
∴ fB = 330 Hz or 318 Hz
Frequency of A decreases with decrease of tension.
Since the beat frequency reduces, so fB cannot be 330 Hz. So fB = 318 Hz initially.
Section – D
Question 29.
Read the following text and answer the following questions on the basis of the same:
Since earth is approximately an ellipsoid (flattened at the poles and bulged at the equator), the equatorial radius is greater than polar radius by 21 km.
Value of acceleration due to gravity (g) on the surface of the earth is given by
g = GM / r2
i.e. g ∝ 1 / r2
So, the value of g at equator < value of g at pole.
Acceleration due to gravity at the height ‘h’ from the surface of the earth is given by,
gh = g (1 – \(\frac{2 h}{r}\))
Acceleration due to gravity at depth ‘d’ is given by, gd = g (1 – \(\frac{d}{R}\))
(i) A piece of stone is weighed at pole and at equator using spring balance. Which of the following statement is true?
(A) Weight at pole is more than that at equator
(B) Weight at pole is less than that at equator
(C) Weights are same at both the places
(D) It cannot be predicted
Answer:
(A) Weight at pole is more than that at equator
Explanation:
Value of acceleration due to gravity is more at pole than that at equator. Hence, weight of any substance at pole is more compared to that at equator.
(ii) Which of the following statement is true?
(A) value of acceleration due to gravity decreases with increase in the height above the surface of the earth.
(B) value of acceleration due to gravity increases with increase in the height above the surface of the earth.
(C) value of acceleration due to gravity increases with increase in depth below the surface of the earth.
(D) value of acceleration due to gravity is constant everywhere.
Answer:
(A) value of acceleration due to gravity decreases with increase in the height above the surface of the earth.
Explanation:
Acceleration due to gravity at the height ‘h’ from the surface of the earth is given by,
g\(\widehat{H}\) = g (1 – \(\frac{2 h}{R}\))
So, as height from the earth’s surface increases, acceleration due to gravity decreases.
(iii) At a height R/4, the value of acceleration due to gravity is gh and at a depth R/4, the value of acceleration due to
gravity is gd. Which of the following statement is true?
(A) gh > gd
(B) gd > gh
(C) gd = gh
(D) None of these
Answer:
(B) gd > gh
Explanation:
Acceleration due to gravity at the height ‘h’ from the surface of the earth is given by,
gh = g (1 – \(\frac{2 h}{R}\))
when h = R/4
gh = g (1 – (2 × R/4)/R)
∴ gh = g/2
Acceleration due to gravity at depth ‘d’ is given by,
gd= g (1 – \(\frac{d}{R}\))
when d = R/4
gd = g (1 – (R/4)/R) = 3 g/4
∴ gd > gh
OR
Acceleration due to gravity at the surface of the earth is
(A) g = \(\frac{G M}{r^2}\)
(B) g = \(\frac{G M}{r}\)
(C) g = \(\frac{G M}{r^3}\)
(D) g = GMr
Answer:
(A) g = \(\frac{G M}{r^2}\)
(iv) Which of the following graphs properly represent the variation of acceleration due to gravity inside and outside the earth?
(A)
(B)
(C)
(D)
Answer:
(A)
Question 30.
Read the following text and answer the following questions on the basis of the same:
Tabu lives at A. He was supposed to go to his uncle’s house at B. A and B is connected by a straight road 5 km long. But that day the road was under repair. So, all the buses were following a diversion via C. A to B via C is 7 , km. Moreover this route is congested. There is a traffic signal at C also.
Tabu got a seat just behind the driver. Once he noticed that the reading of the speedometer was 15 km/h. But ultimately the bus took 1 hour to reach B. He could not understand the fallacy.
(i) What is the distance and displacement of Tabu?
(A) 7 km, 5 km
(B) 5 km, 7 km
(C) 5 km, 5 km
(D) 7 km, 7 km
Answer:
(A) 7 km, 5 km
Explanation:
Distance is the actual path covered i.e., 3 + 4 = 7 km.
Displacement_is the straight line distance from A to B i.e., \(\sqrt{3^2+4^2}\) = 5 km
(ii) Why the speedometer reading was 15 km/h, but actual time required to cover 7 km was 1 hour?
(A) Speedometer was erratic.
(B) Actual distance was more than 7 km.
(C) Halt timing at the traffic signal, slow speed at the congested areas and halt-timing at the bus stops are also to be taken into account.
(D) Both (A) and (B)
Answer:
(C) Halt timing at the traffic signal, slow speed at the congested areas and halt-timing at the bus stops are also to be taken into account.
Explanation:
The bus might have run at minimum speed 15 km/h for sometime. But halt timings at the traffic signal, slow speed at the congested areas and halt-timing at the bus stops increase the actual time taken.
(iii) Speedometer measures
(A) Average speed
(B) Instantaneous speed
(C) Distance traversed
(D) None of these
Answer:
(B) Instantaneous speed
Explanation:
A speedometer is a gauge that measures and displays the instantaneous speed of a vehicle.
(iv) Which instrument is used the distance covered by a vehicle?
(A) Odometer
(B) Speedometer
(C) Synchrometer
(D) Barometer
Answer:
(A) Odometer
OR
Which of the following graphs represents the motion of the bus if it covers AC distance at a speed of 15 km/h, CB distance at a speed 20 km/h and total distance is covered in 1 hour including halt at traffic signal?
(A)
(B)
(C)
(D)
Answer:
(D)
Explanation:
It is a time-speed graph.
For AC :
speed = 15 km/h, km/h,
AC = 3 km
So, time taken = \(\frac{3}{15}\)
= \(\frac{1}{5}\) h
For CB, speed = 20 km/h, CB = 4 km,
So,time taken = \(\frac{4}{20}\)
= \(\frac{1}{5}\) h
Total time taken = 1 h
So, halt time at traffic signal = 1 – \(\left(\frac{1}{5}+\frac{1}{5}\right)\)
= \(\frac{3}{5}\) h
Section – E
Question 31.
Deduce the expressions for the maximum height, time of flight, range for a projectile. Also, find the equation of trajectory.
Answer:
A projectile is launched with initial velocity u, making an angle θ with the horizontal.
X component of the velocity is u cos θ and Y component of the velocity is u sin θ.
Acceleration due to gravity is acting downwards only. It has no component along x-axis.
Determination of maximum height :
The vertical component of velocity at maximum height is 0.
If maximum height is represented by H, then applying the equation
v2 = u2 + 2as,
0 = u2 sin2 θ – 2gH
∴ H = \(\frac{u^2 \sin ^2 \theta}{2 g}\)
Determination o time of flight :
T be the time required to reach the maximum height, then applying the equation
y = u + at
0 = u sin θ – gT’
u sin O
∴ T’ = \(\frac{u \sin \theta}{g}\)
Since it is considered that there is no air resistance, the time of fall is also T’.
So, time of flight = T = T’ + T’
= \(\frac{2 u \sin \theta}{g}\)
Determination of range:
The horizontal component of velocity = u cos θ
So, range is the horizontal distance covered in T time.
R = u cos θ × T
Or, R = u cos θ × \(\frac{2 u \sin \theta}{g}\)
∴ R = \(\frac{u^2 \cos 2 \theta}{g}\)
Equation of trajectory:
Let after f time of launching the projectile, P be a point on the trajectory whose co-ordinate is (x, y).
Since there is no acceleration along x-axis, then
x = u cos θ × T
∴ t = \(\frac{x}{u \cos \theta}\)
and y = u sin θ × t – gt2
putting t = \(\frac{x}{u \cos \theta}\)
y = \(\frac{x u \sin \theta}{u \cos \theta}\) – g \(\frac{x^2}{u^2 \cos ^2 \theta}\)
∴ y = x tan – \(\frac{g}{2 u^2 \cos ^2 \theta}\) × 2
OR
Show that the maximum possible speed of a car on a banked road is greater than that on a flat road.
Answer:
Motion of a car on a flat circular road:
Three forces act on the car, of mass in, when it moves on a level circular road, of radius R, with velocity V:
(i) The weight of the car, mg
(ii) Normal reaction, N
(iii) Frictional force, fS
As there is no acceleration in the vertical direction
N – mg = 0
∴ N = mg
Static friction that provides the centripetal acceleration.
fC = \(\frac{m v^2}{r}\) ≤ µS N
v2 ≤ \(\frac{\mu_S R N}{m}\) = µSRg
It is independent of mass of the car.
Maximum_possible speed is
vmax = \(\sqrt{\mu_S R g}\) …………….(i)
Motion of a car on a banked circular road:
There is no acceleration along the vertical direction.
So, the net force along this direction must be zero.
Hence, N cos θ = mg + fS sin θ …………….(ii)
The centripetal force is provided by the horizontal components of N and f.
N sin θ + fS cos θ = \(\frac{m v^2}{R}\) ……………..(iii)
But fS ≤ µSN
To obtain vmax, we put
fS = µS N
So, equation (ii) becomes,
N cos θ = mg + µS N sin θ ……………..(iv)
∴ N = \(\frac{m g}{\cos \theta-\mu_S \sin \theta}\) ……………(v)
Equation (iii) becomes,
N sin θ + µS N cos θ = \(\frac{m v^2}{R}\)
Substituting N from equation (v),
Comparing this with Eq. (i) we find that maximum possible speed of a car on a banked road is greater than that on a flat road.
Question 32.
Derive an expression of acceleration due to gravity on the surface of the earth, at an altitude h and in a depth d.
Answer:
Let us consider that the earth is a sphere of radius R and mass M.
An object of mass in is at a distance r from the centre of the earth.
F = GMm/r2
From Newton’s second law of motion, due to this force an acceleration is produced in the object which is directed towards the centre of the earth
and the magnitude is a = \(\frac{F}{m}\)
This acceleration is known as the acceleration due to gravity and is represented by g.
So, g = \(\frac{F}{m}\)
Or, g = \(\frac{G M m}{m r^2}\)
∴ g = \(\frac{G M}{r^2}\)
On the surface on the earth, r = R
So, acceleration due to gravity on the surface of the earth = \(\frac{G M}{r^2}\)
Variation of g at an altitude :
L.et us consider,
g = acceleration due to gravity on the surface of the earth
g’ = acceleration due to gravity at an altitude h from the surface of earth.
So, g = \(\frac{G M}{r^2}\)
g’ = \(\frac{\mathrm{GM}}{(R+h)^2}\)
variation of g at a depth:
Let us consider,
g = acceleration due to gravity on the surface of the earth
g” = acceleration due to gravity at a depth d from the surface of earth.
So, g = \(\frac{G M}{R^2}\)
The acceleration due to gravity at this depth is for a sphere of radius (R – d).
Volume of this sphere = 4/3π (R – d)3
Density of earth = ρ = \(\frac{M}{-\pi R}\)
So, mass of the sphere of radius (R – d) is
M’= Volume × density
Or, M’ = – π (R – d)3 × \(\frac{\mathrm{M}}{\frac{4}{3} \pi R^3}\)
∴ M’ = \(\frac{M R d^3}{R^3}\)
Gravitational force on a object of mass m placed at the depth d is
F = \(\frac{G M^{\prime} m}{(R-d)^2}\)
Substituting M’
OR
A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
Answer:
R1 and R2 are the reaction forces exerted by the ground on the front and back wheel respectively.
R1 + R2 = 1800 × 9.8 = 17640 N ……………… (i)
Distance of C.G. from front axle = 1.05 m
Distance of C.G. from back axle = 1.8 – 1.05 = 0.75 m
Taking moment about the center of gravity,
R1 × 1.05 = R2 × 0.75
∴ R1 = \(\frac{5 R_2}{7}\) …………….(ii)
Substituting R1 in equation (1),
\(\frac{5 R_2}{7}\) + R2 = 17640
∴ R2 = 10290 N
And R1 = \(\frac{5 R_2}{7}\)
= 5 × 10290/7
= 7350 N
Force exerted by the level ground on each front wheel = – × 10290
= 5145 N
Force exerted by the level ground on each back wheel = – × 7350
= 3675 N
Question 33.
(a) State and explain principle of continuity.
Answer:
Principle of continuity:
When an incompressible and non-viscous fluid flows in streamlines motion through a tube of non uniform cross-section, then the product of the cross- section and the velocity of flow of liquid remains constant at every point of the tube.
Explanation:
An incompressible and non-viscous fluid flows in streamlined motion through a non-uniform tube XY.
At X, the area of cross-section is A1 and velocity v1.
So, the volume of fluid entering in 1 second is A1v1.
Mass of fluid entering in 1 second is A1v1ρ.
At Y, the area of cross-section is A2 and velocity v2. So, the volume of fluid emerging in 1 second is A2v2. Mass of fluid entering in 1 second is A2v2ρ.
The mass of fluid which enters at one end should leave at the other end i.e.
A1v1ρ = A2v2ρ
Or, A1v1 = A2v2
∴ Av = constant
(b) State and prove Bernoulli’s theorem.
Answer:
Bernoulli’s theorem:
During the steam-lined motion of an incompressible and non-viscous fluid the total energy per unit volume comprising of gravitational potential energy due to elevation, the energy due to fluid pressure, and the kinetic energy of the fluid due to motion remains constant.
Proof:
An incompressible and non-viscous fluid flows in streamlined motion through a non-uniform tube XY.
A1 and A2 are the areas of cross-section at points X any respectively.
h1 and h2 are the heights of points X any respectively.
P1 and P2 are the pressures at points X any respectively.
v1 and v2 are the velocities of fluid at points X any respectively.
The force acting on the fluid at X = P1A1
The fluid which enters at point X travels v1 distance in 1 second.
Work done per second on the liquid = P1A1v1
Similarly work done per second on the fluid emerging from Y = P2A2v2
∴ Net work done = P1A1v1 – P2A2v2 …………(i)
From the principle of continuity,
A1v1 = A2v2 = m/ρ
m is the mass of fluid entering at X and emerging at Y in 1 second.
P is the density of the fluid.
Substituting in equation (1),
Net work done = (P1 – P2) m/ρ
Kinetic energy of the fluid entering at X in 1 second = – mv12
Kinetic energy of the fluid emerging from Y in 1 second = – mv22
∴ Increase in kinetic energy = m(v22 – v12)
Potential energy of the fluid at X = mgh1
Potential energy of the fluid at Y = mgh2
∴ Decrease in potential energy = mg (h1 – h2)
So, net increase in energy of the fluid = – m(v22 – v12) – mg (h1 – h2)
Net increase in energy = Net work done
∴ – m(v22 – v12) – mg (h1 – h2) = (P1 – P2) m/ρ
Or, P1 + – ρv12 + ρgh1 = P2 + – ρv22 + ρgh2
∴ P1 + – ρv2 + ρgh = constant
OR
Determine the equivalent thermal conductivity
(a) when two conducting blocks having different conductivities and lengths but of same area of cross-section are joined in series and
(b) When two conducting blocks having different thermal conductivities and areas of cross-section but of same length are joined in parallel.
Answer:
(a) Equivalent thermal conductivity when two conducting blocks are joined in series:
Two conducting slabs are joined in series.
Area of cross-section of both the slabs = A
Iengths of the slabs are L1 and L2 respectively.
Thermal conductivities are K1 and K2 respectively.
Heat starts flowing.
When the steady state is reached, θ1 and θ2 are the temperatures at the two ends.
θ is the temperature at the interface.
In steady state, the rate of flow of heat through the 1st slab = K1A \(\frac{\theta_1-\theta}{L_1}\) …………….(i)
the rate of flow of heat through the 2nd slab = K1A \(\frac{\theta-\theta_2}{L_1}\) …………….(ii)
Substituting θ in equation (i),
Rate of flow of heat = \(\frac{A(\theta_1-\theta_2)}{\frac{L}{K}+\frac{L}{K}}\) ………….(iii)
If the two blocks are replaced by a single block of thermal conductivity Keq length L1 + L2 and area of cross-section A, then
Rate of flow of heat = Keq A \(\frac{(\theta_1-\theta_2)}{L_1+L_2}\) ………………(iv)
Comparing equations (iii) and (iv),
∴ Keq = \(\frac{L+L}{\frac{L}{K} \frac{L}{K}}\)
(b) Equivalent thermal conductivity when two conducting blocks are joined in parallel:
Two conducting slabs are joined in parallel.
Areas of cross-section of slabs are A1 and A2 respectively.
Length of both the slabs is L.
Thermal conductivities are K1 and K2 respectively.
Heat starts flowing. When the steady state is reached, θ1 and θ2, are the temperatures at the two ends.
In steady state, the rate of flow of heat through the 1st slab = K1A1 \(\frac{\theta}{L}\) ………………(i)
the rate of flow of heat through the 2nd slab = K2A2 \(\frac{\theta}{L}\) ………….(ii)
Combined rate of flow of heat = K1A1 \(\frac{\theta}{L}\) + K2A2 \(\frac{\theta}{L}\)
= \(\frac{K A+K A}{L}\) (θ – θ) ………………(iii)
If the two blocks are replaced by a single block of thermal conductivity Keq, length L and area of cross-section A1 + A2, then
Rate of flow of heat = Keq (A1 + A2) \(\frac{\theta-\theta}{L}\) …………..(iv)
Comparing equations (iii) and (iv),
\(\frac{K A+K A}{L}\) (θ – θ) = Keq (A1 + A2) \(\frac{\theta-\theta}{L}\)
Or, K1A1 + K2A2 = Keq (A1 + A2)
Keq = \(\frac{K A+K A}{L}\)