CBSE PHYSICS, Class-XI

Sample Question PaperFor 2020 Examination

SOLUTIONS

SECTION ‘A’

1. (d) Explanation: because for SHM, ax ∝ (– x) 2. (d) 3. (a, c) 4. (c) Explanation: Just because road does not move at all so the

work done by the cycle on the road must be zero.

OR (d) Explanation: In the whole process, the man exerts a variable

force (F) on the ground to set his body in motion. This force is in addition to the force required to support his weight (mg). Once the man is in standing position, F become zero.

5. (c) Explanation : Tension = Force = [MLT–2]

Surface Tension =

Forcelength

ML T= −[ ]0 2

6. (c)OR

(c) 7. (a) 8. (a) Lift is coming downwards, so it’s retarding.

Thus, a is acting downwards, so a > 0 and the value of x becomes less here negative, i.e., x < 0, velocity is downwards (i.e. negative) so v < 0.

9. (b) for the value of displacement, two timings are there. Therefore for one time, the average velocity is positive and for other time is equal but negative. Due to this average velocity can vanish.

10. (b) Explanation

µ =

ML m

kg kg/m= =2 5

200 125

..

v = T

m/sµ= 40

t = L

sy

= 0 5.

OR(c) Explanation Here, m = 150 g = 0.15 kg

3 4 m/su i j∧ ∧ = +

3 4 m/sv i j∧ ∧ = + −

Initial momentum, Pi = mu

( )∧ ∧

∧ ∧

= + = +

0.15 3 4 m/s

0.45 0.6 kg m/s

ip kg i j

i j

Final momentum, =fp ( )0.15 kg 3 4 m/si j∧ ∧

− −

0.45 0.6 kg m/si j∧ ∧ = − −

Change in momentum,

∧ ∧

∧ ∧

∧ ∧

∆ = = +

− − −

= +

−

−

0.45 0.6 kg m/s

0.45 0.6 kg m/s

0.9 1.2 kg m/s

ifp p i jp

i j

i j

11. The time period becomes 2 times the

original value since T ∝ l . 12. No, the equation of kinematics are applicable

only so long as the acceleration is uniform. The acceleration due to gravity is uniform only near the surface of the earth.

2 | Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XI

13. The winning team is performing work over the losing team. Work done by winning team is positive while that of losing team is negative.

OR No, the mechanical energy is not conserved,

Because resistive force of air also acts on the body which is a non-conservative force, So the gain in KE would be smaller than the loss in PE.

14. Surface tension of water decreases when soap is dissolved in it so spraying becomes easier because less work is required to spray it.

OR Let the volume of iceberg be Vi and volume of

water displaced by iceberg be Vw in floating condition. Weight of iceberg = Weight of water displaced by submerged ice.

ρwVwg = ρiVig [∴ W = mg = Vρg]

ρ= = =ρ

w i

i w

V 0.9170.917.

V 1

15. Motion of a point cylinder capable of rotating without friction about its own axis is the required example.

16. The groups of stars resembling the shapes of animals, common daily use objects, figures etc. are called constellations, e.g., hunter, great bear, small dipper etc.

17. The r.m.s. speed of the molecules of a gas is defined as the square root of the mean of the squared velocities of the molecules of a gas. No, r.m.s. speed is different from the average speed.

½

c = vrms = 2 2 21 2 3

3v v v+ +

and vavg = 1 2 3

3v v v+ +

OR O2 has 5 degrees of freedom. Therfore, energy

per mole = 52

RT

∴ For 2 moles of O2, energy = 5RT Neon has 3 degrees of freedom

∴ Energy per mole =32

RT

∴ For 4 mole of neon, energy = × =34 6

2RT RT

∴ Total energy = 11RT. 18. No, it is a circular and periodic motion but

not to and fro about a mean position which is essential for S.H.M, ∴ it cannot be taken S.H.M.

19. Air expands on sudden bursting of tyre which is an adiabatic process. Air does work against atmosphere at the cost of its own internal energy.

20. The substance which can be stretched to large values of strain are called elastomers, e.g., elastic tissue of aorta, the largest artery carrying blood from the heart.

SECTION ‘B’

2

21.

t

x°

O

x

(i)

t-x°

O

x

(ii)

t

-x°

O

x

(iii)

tx°O

x

(iv)

22. The two vectors A→

and B→

are perpendicular to each other. 1

Explanation—Let θ = angle between the two

vectors A→

and B→

.

Therefore component of vector A→

along the

direction of vector B→

is obtained resolving

vector A→

i.e. A cosθ.

A

A cos B

Now according to statement A cos θ = 0 cos θ = 0 = cos 90° ∴ θ = 90°

i.e., A→

^ B→

OR

| | (| | | |)a b a b→ → → →

− − +2 2

⇒ | | | | | || |a b a b→ → → →

+ −2 2 2 cos | | | | | || |θ − − −→ → → →a b a b2 2 2

cos | | | | | || |θ − − −→ → → →a b a b2 2 2 1

SOLUTIONS | 3

⇒

− 4

22| || |cosa b

θ

= a negative quantity

Hence, | | | | | |a b a b→ → → →

− < + 1

Commonly made error Generally students are not able to distinguish between | |a b− and ( a b )| | | |−

ANSWERING TIP...

Practice is required for dot and cross products of two vectors.

23. Let initial momenta of the particle A and B be

p→

A and p→

B respectively. The particles collide

and separate out with final momentum p→

A '

and p→

B ' respectively. Using Newton’s second

law of motion, the force of on particle A due to B is given by,

F AB→

= p p→ →

−A A't

where, t is the time interval.

Similarly, F BA→

=

p p→ →

−B B't

As per Newton’s third law of motion,

F AB→

= −→F BA

i.e., p p→ →

−A A't

= –

p p→ →

−B B't

i.e., p p→ →

A B'+ ' = p p→ →

+B A ∴ Linear momentum is conserved. 2 24. We are given that Tmax = maximum tension in the string so that

it does not break = 2 kg wt = 2 × 10 N = 20 N Let T1 be the tension in the string when the

stone is in its lowest position of its circular path. We know that

T1 =

mvl

12

+ mg. ½

T1 would have its minimum value when v1

equal to its minimum value = 5gl , needed

by the stone, to complete its vertical circular path. ½

Hence (T1)min = mvlmin2

+ mg = 6mg

= 6 × 0.4 × 10 = 24 N We thus see that (T1)min is more than the

breaking strength of the string. Hence the particle cannot describe the vertical circle.

25. In normal conditions, the water rises in the capillary tube due to surface tension upto a height, becomes equal to the weight of water column raised in the tube. In the case of weightlessness the effective weight of water column raised becomes zero. Hence, the water will rise upto the other end of the capillary tube, howsoever long the capillary is. The water meniscus will adjust its curvature at the upper end of capillary tube so that there is no over flowing. 2

26. In an elastic collision, (a) Total momentum is conserved, i.e., total

final momentum is equal to the total initial momentum.

(b) Total mechanical energy is conserved, i.e., total final energy is equal to the total initial energy.

(c) Total kinetic energy is conserved, i.e., total final kinetic energy is equal to the total initial kinetic energy.

(d) All the forces are of conservative nature, i.e., work done does not depend upon the actual paths. (½ mark each)

OR

When body (B2) is at rest, we get

v1 = 1 2

1 2

–m mm m+

u1

and v2 = 1

1 2

2mm m

u1

Putting m1 = m2 = m, we get v1 = 0 and v2 = u1, i.e., body B1 comes to rest whereas body B2

moves with the velocity of body B1

27. The resonance will take place in air column if the compression or rarefaction produced by the vibration of the tuning fork travels from open end of the air column to lower closed end (water level in the resonance apparatus) and back to the upper open end in the same time in which the prong goes from one extreme to other. 2

4 | Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XI

SECTION ‘C’

28. A mind set moulded in a particular set of thinking and is known as scientific temperament. It is not only based on logic and facts but also on reliable observations. 1

The ultimate test of truth in science is the experimental verification. A scientific way of doing things involve the following steps : ½

(i) Identifying the problem or aim. ½ (ii) Collecting all relevant information or data

related to the problem. ½ (iii) Hypothesising or proposing a possible

theory. ½ 29. (i) We know, area covered by velocity-time

graph represents distance travelled ∴ Distance travelled in t = 0 to t = 5 sec

=

12

× 2 × 20 + 3 × 20 = 80 m

1½

(ii) Distance travelled in t = 5 to t = 10 sec

= 12

× 5 × 20 = 50 m

∴ Total distance = 130 m. 1½ 30. Let the body of mass m be taken at height h

above the surface of earth. At any instant of time t it reaches at a distance x from the centre of earth. The work done in raising through dx is.

dW = GM

2m

x.dx

= ∆(P.E.) 1 Hence the work done in taking the body from

surface of the earth (x = R) to a height h (x = R + h) is

P.E. = W

= GM

R

R m

xdx

h

2

+∫

= GMm12xdx

h

R

R+∫

= – GMm1x

h

R

R+

= – GMm1 1

R R+h−

= −+

−

GMmR h

R

1

11

= − +

−

−GMmRRh

1 11

1

Since, R > > h

∴ P.E. = − − −

GMmRRh

1 1

= GM

R

mh2

= GM

R2

mh as g=

GM

R2

= mgh P.E. above the surface of earth 1 = mgh

ANSWERING TIP...

More practice is required for finding the expression for gravitational field by integral method.

OR At the surface of earth, weight w = mg = GMm/R2 1 At height h = R, weight w’ = mg’

=

( )2GM

R+

m

h =

( )2GMR+R

m

∴ 'ww

= ( )

2

2R2R

=

14

or w’ = 4w

It means the weight would reduce to one-fourth of the weight on the surface of earth.

31. (i) Conduction is the process of flow of heat from one point to another through a substance without the actual transfer of particle. 1

Rate of flow of heat,

(ii)

ddQT

=

KA T T1 2−( )d

where K is coefficient of conductivity. A is the area of cross-section of the face, T T1 2−

d is temperature gradient. 1

(iii) Coefficient of thermal conductivity (K) is defined as the amount of heat flowing in a second between two opposite faces of a cube of side 1 m kept at a temperature difference of 1 K. S.I. unit of K is Js–1m–1K–1. 1

32. From the relation

h = 1 2

1−

TT

SOLUTIONS | 5

or

TT

2

1 =

1

30100

− =

710

or T1 = 10

72T

=

10 2007

×

= 285.71K 1 new efficiency is now 50%

h’ = 1 2

1−

TT

TT

2

1 = 1 – h’ ½

TT

2

1 =

1

50100

12

− or

1

or 2T2 = T1

T1 = 2 × 200K T1 = 400K Now increase in temperature of source = 400K – 285.71K = 114.3K ½ 33. Boyle’s law : Pressure exerted by one mole of gas

P = 13

2ρvrms ,

P = 13

2MV

=MV

vrms [but, ]ρ

PV = 13

2Mvrms 1

Here M is fixed and v2 ∝T If temperature T is fixed then PV = constant

or P ∝ 1V

½

If temperature of a given mass of a gas is kept constant its pressure is inversely proportional to its volume.

Charles’s law Pressure exerted by one mole of gas

P = 13

2ρvrms ½

=

13

2MVvrms ,

or V = 13

2MPvrms

v2rms ∝ T and M is fixed

V ∝ T ½ If pressure of a given mass of a gas is kept

constant its volume is directly proportional to the temperature of the gas. ½

OR Terminal velocity : It is maximum constant

velocity acquired by the body while falling freely in a viscous medium.

When a small spherical body falls freely through a viscous medium, three forces act on it.

(i) Weight of the body acting vertically downwards.

(ii) Upward thrust due to buoyancy equal to weight of liquid displaced.

W

FT FV

Motion

(iii) Viscous drag acting in the direction opposite to the motion of body. According to Stoke’s law, F ∝ v, i.e., the opposing viscous drag goes on increasing with the increasing velocity of the body. 1

As the body falls through a medium, its velocity goes on increasing due to gravity. Therefore, the opposing viscous drag which acts upwards also goes on increasing. A stage reaches when the true weight of the body is just equal to the sum of the upward thrust due to buoyancy and the upward viscous drag. At this stage, there is no net force to accelerate the body. Hence it starts falling with a constant velocity, which is called terminal velocity.

Let ρ be the density of the material of the spherical body of radius r and ρ0 be the density of the medium.

∴ True weight of the body, W = Volume × Density × g

= 43

πr3ρg

Upward thrust due to buoyancy, FT = Weight of the medium displaced ∴FT = Volume of the medium displaced × Density × g

= 43

πr3ρ0g

If v is the terminal velocity of the body, then according to Stoke’s law, upward viscous drag,

FV = 6πhrv When body attains terminal velocity, then

6 | Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XI

FT + FV = W

∴ 43πr3ρ0g + 6πηrv =

43πr3ρg

or 6πηrv = 43πr3(ρ – ρ0) g

or v = 2

9

20r gρ ρ

η−( )

1

It depends upon the terminal velocity as it varies directly as the square of the radius of the body and inversely as the coefficient of viscosity of the medium. It also depends upon densities of the body and the medium. 1

34. Lengths of the segments should be in the inverse ratio of fundamental frequencies, i.e.,

l1 : l2 : l3 = 1 1 1

1 2 3ν ν ν: :

X Y1 2 3

or l1 : l2 : l3 = 1 1 1

: :1 2 3

= 6 : 3 : 1 1

l1 =

66 3 1

100+ +( ) ×

= 60 cm ½

l2 =

36 3 1

100+ +( ) ×

= 30 cm.

and l3 =

16 3 1

100+ +( ) ×

= 10 cm

½

Thus, first bridge should be placed at a distance

of 60 cm from end X and the second bridge should be placed at a distance of 90 cm from end X. 1

SECTION ‘D’

35. A body rolling on an inclined plane of height h, is shown in the following figure :

θ

h

R

m = Mass of the body R = Radius of the body K = Radius of gyration of the body v = Translational velocity of the body h = Height of the inclined plane

g = Acceleration due to gravity Total energy at the top of the plane, El = mgh Total energy at the bottom of the plane, Eb = KErot + KEtrans

=

12

12

2 2Iω + mv

But I = mK2 and w = vR

∴ Eb = 12

12

22

22( )m

vmvK

R

+

=

12

12

2 2mv mvK

R

2

2+

=

12

12mv +

K

R

2

2

From the law of conservation of energy, we have:

ET =Eb

mgh = 12

12mv +

K

R

2

2

∴ v = 2

1

gh

( )+ K /R2 2

Hence, the given result is proved. 5

Commonly made error Students consider only kinetic energy formula for translatory motion for rolling object through inclined plane.

ANSWERING TIP...

Before to apply the law of conservation of energy, students should write all types of energies separately.

OR For smooth inclined plane Let s = length of inclined plane or mg sin q = ma or a = g sin q Applying 2nd kinematic equation of motion

s =

ut at+

12

2

or s = 0 × t +

12

(g sin q) t2

( t = T)

SOLUTIONS | 7

or s =

12

(g sin θT2)

or s = 1

2 22gT

... (i) (∴ θ = 45°)

For rough inclined plane.

N

mg sin �mg cos �

mg

m f

�

�

f = µN = µmg cos θ mg cos θ – f = ma1

or a1 = (sin θ – µ cos θ) g

=

12

(1 – µ) g

Using, s = ut at+

12

2

or s = 0 + (pT) +

12

12

1 2 2−( )

×µ g p T

or s =

12 2

(1 – µ) gp2T2 ...(ii)

∴ by (i) & (ii), we get

12 2

2gT =

12 2

(1 – µ) gp2T2

or 1 = (1 – µ) p2

or µ = 112

−

p 5

36. Suppose L = length of each of the two wires A and B made of steel and aluminium respectively.

Also suppose A1, A2 be the areas of cross-section of A and B respectively.

∴ A1 = 1 mm2 = (10–3)2 = 10–6 m2

A2 = 2 mm2 = 2 × 10–6 m2 Y1 for Steel = 2 × 1011 Nm–2 Y2 for Aluminium= 7 × 1010 Nm–2 Suppose F1 and F2 be the forces acting at their

lower ends.

(a) ∴ Stresses on A and B are

FA

1

1 and

FA2

2

respectively.

If these stresses are equal, then

FA

1

1 =

FA2

2

or FF2

1

=

AA2

1 ...(1)

Suppose F1 and F2 be produced by a weight mg suspended from the rods at a distance x and y from the two rods. Then we have

F1x = F2y

or FF2

1

=

yx

...(2)

∴ From equations (1) and (2),

yx

=

AA2

1

or x = AA1

2 y ...(3)

Now x + y = 1.05 m (length of the rod) ∴ y = 1.05 – x ...(4) ∴ From equations (3) and (4),

x = AA1

2 (1.05 – x)

or A1x = 1.05 A2 – xA2

or x (A1 + A2) = 1.05 A2

or x = 1 05 2

1 2

. AA A+

=

1 05 2 10

1 2 10

6

6.

( )

× ×+ ×

−

−

=

1 05 23

. × = 0.70 m

or y = 1.05 – 0.70 = 0.35 m So, the mass m must be suspended at a

distance of 0.70 m from A (steel wire) or 0.35 m from B. 2½

(b) From the relation,

Y = StressStrain

,

or Strain =

StressY

for wire A, Strain = F /A

Y1 1

1

and for wire B, Strain = F /A

Y2 2

2 As the strains are equal.

Hence F

A Y1

1 1 =

FA Y

2

2 2

8 | Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XI

or FF

2

1 =

A YA Y

2 2

1 1 ...(5)

Also from equation (2), we get

FF

2

1 =

xy

...(6)

∴ From equation (5) and (6),

xy

=

A YA Y

2 2

1 1

or x = A YA Y

2 2

1 1(1.05 – x)

[∴ y=1.05 – x] by using eqn. (4), or A1Y1x = A2Y2 × 1.05 – A2Y2x or x(A1Y1 + A2Y2) = 1.05 A2Y2

∴ x = 1.05A Y

A Y +A Y2 2

1 1 2 2 ...(7)

Putting the values of Y1, Y2 and A1, A2 from above in equation (7),

x = 1 05 2 10 7 10

10 2 10 2 10 7 10

6 10

6 11 6 10. × × × ×

× × + × × ×

−

− −

x = 1 05 14 10

2 10 1 4 10

4

5 5.

.

× ×× + ×

m

x =

1 05 14 10

3 4 10

4

5.

.

× ××

m

x = 1 05 14

34. ×

m = 0.43 m

y = 1.05 – x = 1.05 – 0.43 = 0.62 m. 2½

OR Let us consider the diagram below, Given : α = 1.2 × 10–5°C–1, L = 10m, ΔT = 20°C. By pythagoras theorem,

x2 = L LL

+ ∆( )

−

12 2

2 2

=

14

[L2 + DL2 + 2LDL] – L2

4

x = 12

2∆ ∆L L L2 +

DL2 << L, ∴ neglecting DL2

x = 12

2L L∆

L

x

1/2 (L L)+ ∆

90°90°

But ΔL = LΔ t

∴ x =

12

2L L× α∆t

= 12

2L α∆t

=

102

2 1 2 10 205× × ×−.

= 5 4 1 2 10 4× × × −.

= 5 × 2 × 1.1 × 10–2

= 0.11 m x = 11 cm 5 37.

(i) Maximum height : It is the maximum vertical height attained by the object above the point of projection during its flight denoted by h.

y = u sin θ, ay = – g, y0= 0 y = h, t = T/2 = u sin θ/g Using relation,

y = y0 + uyt + 12

uy t2,

We have

h = 0 + u sin θ×

ug

gug

sin( )

sin+ −

12

2θ

or h = ug

ug

22

221

2sin sinθ θ−

h = u

g

2 2

2sin θ

(ii) Time of flight : The total time for which

the projectile is in flight, taking vertical downward motion of object from O to C.

Let time taken to complete the trajectory = T

As the projectile is reaching the same level of projection vertical displacement, y = 0

SOLUTIONS | 9

We have, S = ut at+12

2

0 = u gtsin .θ T −12

2

T = 2u

gsinθ

(iii) Horizontal range : It is the horizontal distance travelled by projectile during its flight i.e. horizontal distance covered by object while going from O to C.

Distance = velocity × time = u cos θ × T

= u u

gcos sinθ θ×2

=

ug

2 22

sin cosθ θ

=

ug

2 2sin θ

OR It states that if two vectors can be represented

completely (i.e., both in magnitude and direction) by the two adjacent sides of a parallelogram drawn from a point then their resultant is represented completely by its diagonal drawn from the same point. ½

Proof : Let P→

and Q→

be the two vectors

represented completely by the adjacent sides OA and OB of the parallelogram OACB such that,

OA→

= P OB Q→ → →

=,

or OA = P, OB = Q

B C

O D

R→

P→ A

90°θθ β

Q→

½ θ = angle between them ÐAOB If R be their resultant, then it will be represented

completely by the diagonal OC through point

O such that OC→

= R→

Magnitude of R→

: Draw CD ^ to OA produced ∴ ÐDAC = ÐAOB = θ Now in right angled triangle ODC OC2 = OD2 + DC2

= (OA + AD)2 + DC2

= OA2 + AD2 + 2OA.AD + DC2

= OA2 + (AD2 + DC2) + 2OA.AD ...(i)

Also in right angled triangle ADC, AC2 = AD2 + DC2 ...(ii)

Also, ADAC

= cos θ

or AD = AC cos θ ...(iii)

and

DCAC

= sin θ

or DC = AC sin θ ...(iv) ∴ From eqn. (i), (ii), (iii), we get OC2 = OA2 + AC2 + 2OA.AC cos θ

or OC = OA AC OA AC2 2 2+ + . cosθ ...(v)

As OC = R, OA = P, AC = OB = Q ...(vi) ∴ From equation (v) and (vi), we get

R = P Q PQ2 2 2+ + cosθ ...(vii)

eq. (vii) gives the magnitude of R→

. 1

Direction of R→

: Let β the angle made by R

with P→

∴ In right angled triangle ODC,

tan β = DCOD

=

DCOA+ AD

=AC

OA ACsin

cosθ

θ+

[by using (iii) and (iv)]

tan β = Q

P Q

sin

cosθθ+

...(viii)

Special cases : (a) When two vectors are acting in same direction :

Then θ = 0°

∴ R = ( )P Q + 2

= P + Q

and tan β = Q

P Q.0

0 +

=

or β = 0° Thus, the magnitude of the resultant vector is

equal to the sum of the magnitudes of the two vectors acting in the same direction and their resultant acts in the direction of P and Q. 1

(b) When two vectors act in the opposite directions :

Then θ = 180° ∴ cos θ = – 1 and sin θ = 0

∴ R = P Q PQ2 2 2 1+ + −( )

10 | Oswaal CBSE Chapterwise & Topicwise Question Bank, PHYSICS, Class – XI

= P Q PQ2 2 2+ −

= ( )P Q− 2 or ( )Q P− 2

= (P – Q) or (Q – P)

and tan β = Q

P Q.( )0

10

+ −=

= tan 0° ∴ β = 0° Thus, the magnitude of the resultant of

two vectors acting in the opposite direction to the difference of the magnitude of two vectors and it acts in the direction of bigger vector. 1

(c) If two vectors act perpendicular to each other :

θ = 90°, i.e., if P Q→ →⊥ , then cos 90° = 0

and sin 90° = 1

∴ R = P Q2 2+

and tan β =QP

·

Commonly made error Students generally commit error while using the different angles associated with the diagram and their corresponding trigonometric functions.

ANSWERING TIP...

Students should do the calculations step by step for determining the magnitude and then the direction of resultant vector using the diagram of a parallelogram.

qqq