Class XI

Midterm Examination 2017-18

Physics marking key

SET-A1/A2

Time 3hrs MM70

General instructions:

i) All questions are compulsory.

ii) Marks for each question are indicated against it.

iii) Q No 1-5 is 1mark each; Q No 6-10 are 2 marks each; Q No 11-22 is 3 mark each; Q No23 isvalue basedand of 4 mark; Q No 24-26 is 5 marks each .Internal choice is provided for one question each of 2. 3 marks and all three Qs of 5 marks

iv)Use log tables, if necessary. Calculators are not allowed.

Q1.Name the strongest and weakest fundamental force in nature.? 1

A1. Strong nuclear force and Gravitational force (1/2 each)

Q2. Give one example each of the situations for the motion of a ball having i) negative acceleration andpositive velocity? ii) velocity and the acceleration are perpendicular to each other? (1)

A2.i) ball is thrown up.During upward motion ii)During projectile motion of ball, at topmost point (1/2) each

Q3.The free body diagram for the various forces acting on a body is given.

Find the direction of its acceleration. 1

A3.tan θ =4/3

Q4. What is the time period of the rotation of the earth about its axis if the unit of time adopted is β second? 1

A4.Time period =24 X60X60/β s =86400/β s

Q5. State work-energy theorem for a variable force.Express it mathematically also.1

A5.The change in kinetic energy of a particle is equal to the work done on it by the net force.

1

Q6. .According to Galileos law of odd numbers The distances traversed ,during equal intervals of time,by a body falling from rest,stand to one another in the same ratio as the odd numbers(1:3:5:7… Prove it.2 A6.S1=g/2 S1+S2=2g S2=3g/2 S1+S2+S3=9g/2 S3=5g/2 S1:S2+:S3=1:3:5 Q7. A physical quantity P is related to four observables a, b, c and d as follows

The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%,respectively. What is the percentage error in the quantity P ? 2

A7.13%

Q8.The motion of a particle of mass m = 0.1 kg is described by the following equation y(t) = - 5.0 + 10.0 t - 4.9 t2 (m).What is the force acting on the body? 2

A8. a =-9.8m/s2

Force =m a =0.1 x9.8=0.98N

Q 9. Find the potential energy of a system of four particles each of mass m placed at the vertices of a square of side l. Also obtain the potential at the centre of the square. 2

OR

Obtain an expression for the acceleration due to gravity at a depth ‘d’ below the surface of the earth.

A9.U=4Gm2/l +(2)1/2Gm2/l=(4+ √2) Gm2/l

or

Density is uniform, therefore, ρ=33 )(3/43/4 dR

mR

MVolume

Mass−′

==ππ

3

3)(R

dRMm −=′

Force on mass m

323

3

2

)/1()()(

)( RmRdGMR

dRRmdRGM

dRmmGgm −

=−−

=−′

=′ )/1( Rdgg −=′

Q-10. Give expressions for the normal reaction on a lawn mover

1) When it is pushed with a force F

2) When it is pulled with a force F 2

A – 10. i)N = w + F Sinθ

ii) N = w - F Sinθ

Q -11. A partical P, of mass m is moving with uniform velocity v along a straight path ,as shown

a) Write an expression for the angular momentum about the origin

b) Show that the areal velocity of the particle is half the angular momentum per unit mass

A-11.a)L=r x p

b) Area=( r xvt)/2

Areal velocity= Area /t=(r xp)/2m= L/2m 3

Q-12. State theorem of perpendicular axis.

What is the moment of inertia of a rod of mass M, length l about an an axis perpendiular to it through one end?

A12. It states the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axis concurrent with perpendicular axis .

I cm = 12

2ml

Using parallel axis theorem:

I= Icm + md2 = 3412

222 mlmlml=+

Q13. Free body diagram of a block of wood of mass 3 kg resting on the surface ,inclined at an angle θ =30O is shown in fig . Name the forces1,2and 3 (b If coefficient of static friction is 0.2, calculate the value of all the three forces. 3 3 2 1

A13.Forces 1 is weight = 30 N,

Forces 2 is friction=mgSinθ=30x½=15 N

andForces 3 is normal reaction=mgCos θ =30x√3/2=15√3

Q14. The frequency ‘f ’of vibration of a stretched string depends upon(1)its lengthl,(2)its mass per unit length’m’ and(3)the tension T in the string.Obtain dimensionally an expression for frequency f. 3

A14. f = K la mb TC

a=-1 b=1/2 c=-1/2

Q15. A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10ms1? . 3

A15.

0.082 J in each half ; 1

WG+WR = ½ mv2 1

WR =– 0.163 J 1

Q16. For the motion of any satellite in circular orbit of the earth, derive the expression for the distance from its centre (:RE is radius of earth and h is the height of the satellite above its surface

: 3

A16.Derivation mv2/RE+h =GMEM/(RE+h)2 1

On putting v=(RE+h). 2π/T 1

1

Q17. A projectile is thrown, from the origin (0, 0), at an angle θ to the horizontal level and with speed V under the action of constant gravity g. Obtain expressions for maximum height and the range of the projectile.Show that the projection angle θ0 for a projectile launched from the origin

is given by θ0 = tan -1

RH4

A17. : For a projectile fired with speed V0 at angle θ0

R = g

v 02 2sin θ

and H =g

v2

)sin( 20θ ⇒ tan θ0 =

RH4

OR

(a) Derive an expression for centripetal acceleration of an object in uniform circular motion in a plane. What will be the direction of the velocity and acceleration at any instant?

(b) Find vector product of two vectors

a = 3 i− 4 j + 5 k and b = − 2 i + j− 3 k

b)Vector product =7i-j-5k

Q18.Derive an expression for the i) maximum ii) recommended speed (i.e without depending on friction) at which a car or truck can go on a banked circular turn of radius R. The coefficient of friction between the wheels and the road being µ and the banking angle is θ. 3

A 18.

recommended speed =(Rg tanθ)1/2

Q19. The angular speed of the motor wheel increased from 1200rpm to 3120rpm in 16 seconds. (1)What is it’s angular acceleration, assuming the acceleration to be uniform?

(2)How many revolutions does the engine make during this time ? A19.

Angular displacement =1152 π rad

No. of revolution=1152 π /2π=576

Q20. A bob of mass m is suspended by a light string of length L

It is imparted a horizontal velocity vo at the lowest point A such

that it completes a semi -circular trajectory in the vertical plane

with the string becoming slack only on reaching the top most

point, C. Obtain an expression for (i) vo; (ii) the speeds at

points B and C;

A20.

Q21. The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets. (3)

A21.Total energy on the surface of the earth:

Ei =

Total energy in space = ½ mv2

=½ mv2

∴ ½ mve2 =

RGMm

=½ mv2

8 ve2 =v2

v= 2 (2)1/2ve= 2x(2)1/2x11.2=31.7k m/s

Q22. A balloon is ascending at the rate of 14 ms-1 at a height of 98 m above the ground, when a packet is dropped from it. After how much time and with what velocity does it reach the ground? A22. u=14 ms-1 x2-x1=ut+1/2 at2

-98 = 14t -1/2x10xt2

5t2-14 t-98=0 t=6s

V=u+at=14-10x6=-46m/s

Q23.

Suraj went to Big Bazaar to purchase certain goods .There he has noticed an old lady

struggling with her shopping. Immediately he showed her the lift but she was

reluctant to go by lift as she did not know how to operate it.Suraj helped her in

carrying the load up to the lift and showed her how to operate it.

That old lady was very happy.

A23.

(a)What values does Suraj possess?

(b)An elevator can carry a maximum load of 1800 kg is moving up with a constant

speed of 2 m/s , The frictional force opposing the motion is 4000 N. Determine the

minimum power delivered by the motor to the elevator in watts .

Ans: (a) Suraj is sympathetic and also has the attitude of helping others. He has

patience

(b)The downward force on the elevator is F= mg + Ff = (1800 X 10) + 4000 =

22000N

The motor must supply enough power to balance this force.

Hence P = F.V = 22000 X 2 = 44000W

Q-24. A block of 0.3 kg mass is taken up an inclined plane making angle 300 with the horizontal,

and then allowed to slide down to the bottom again. The coefficient of friction between the block

and the slope is 0.15. Calculate the

(1)work done by the gravitational force over the round trip

(2)work done by the applied force over the round trip

(3)work done by the frictional force over the round trip

(4)Kinetic energy of the body at the end of the trip?

A24.(1).W = -mgSin θ.l –mgSin θ =0

(2) W = mg(Sin θ + µCos θ)l

=18.5J

(3) W =-2µmgCos θ l =-7.6J

(4)W =mg(Sin θ -µCos θ)l

=10.9J

=K.E.

Or

Explain the terms (1)elastic collision(2)Inelastic collision(3)Completely inelastic

collision.Obtain the expression of loss in kinetic energy in completely inelastic collision in

one dimension. 5

A24. For ‘Elastic collisions’ KE is conserved whereas for ‘inelastic case it is not conserved.

Consider a completely inelastic collision in one dimension

Q-25.The variation of velocity of a particle along a straight line, with time is as shown:

• For how much time does it accelerate and what is the acceleration?

• What are the average speed and average velocity of the body for entire motion?

• For how much distance is it having uniform motion?

Or

Derive the equation S = ut + ½ 2at using graphical method.

A hundred meter sprinter increases her speed from rest uniformly at the rate of 1ms-2 upto three

quarters of the total run and covers the last quarter with uniform speed.How much time does she

take to cover the first half and the second half of the run? 5

Time(s)

47 10 12

8m/s

A25.

a)T =0-4s a =2m/s2

And t =7-12s a = -8/3m/s2

b)velocity corresponding to lowermost point =5.3 m/s

average speed=16+24+12+5.3/12=57.3/12=4.7m/s

and average velocity=16+24+12-5.3/12=3.8m/s

c)24m for T =4s -7s

OR

.

Time does she take to cover the first half = 10 s

and the second half of the run = 2.2+2.04=4.24s

Q26. To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass m moving with a speed v on a rough road(coefficient of friction μ)and colliding with a horizontally mounted spring of spring constant k.

1) What are the forces acting on the car?(represent their direction and magnitude in the diagram).

2) Obtain the expression of maximum compression of the spring? 3

OR

a) Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary, given that the co-efficient of static friction between the box and the train’s floor is 0.15 2

A body of mass 60 kg suspended by means of three strings P, Q and R as shown in the figure is in equilibrium. The tension in the string P is

a)

b)Let us denote the tensions in the three strings by P, Q and R. The concurrent forces P, Q and R keep the common meeting point of the strings in equilibrium so that they can be represented by the three sides of a triangle taken in order as shown in the adjoining figure. From this right angled triangle, Q/P = tan 30º where Q = weight of the mass M = Mg =60g newton.

Substituting for Q, we have 60g/P = 1/√3, from which P = 60g×√3 = 103.9g N.