Class Xii Physics Dpp Set (05) - Prev Chaps - Electrostatics.pdf

TARGET : JEE (Main + Advanced) 2016 O

Course : VIJETA (JP)

Date : 27-04-2015

DPP No. : 14 (JEE-ADVANCED) Total Marks : 45 Single choice Objective ('–1' negative marking) Q.1 to Q.2 Multiple choice objective ('–1' negative marking) Q.3 to Q.5 Subjective Questions ('–1' negative marking) Q.6 to Q.7

Max. Time : 47 min. (3 marks 3 min.) [6, 6] (4 marks 4 min.) [12, 12] (4 marks 5 min.) [8, 10]

Comprehension ('–1' negative marking) Q.8 to Q.10 Match the Following (no negative marking)

(3 marks 3 min.) (8 marks 10 min.)

[9, 9] [8, 10]

ANSWER KEY OF DPP NO. : 14 (JEE-ADVANCED) 1. 6.

(C) 2. = /3

(B) 7.

3. 6

(B) (C)(D) 8. (C)

4. 9.

(A) (B) (C) (D) (A) 10. (A)

5. (A) (C)(D)

1.

An infinitely long rectangular strip is placed on principal axis of a concave mirror as shown in figure. One end of the strip coincides with centre of curvature as shown. The height of rectangular strip is very small in comparison to focal length of the mirror. Then the shape of image of strip formed by concave mirror is

      

F

(A) Rectangle (A)  Sol.

C

(B) Trapezium (B) 

(C*) Triangle (C*) 

(D) Square (D) 

Draw an incident ray along the top side of rectangular strip,which happens to be parallel to the principal axis. After reflection this ray passes through focus. Hence image of all points (for e.g. O1, O2, O3, .......) on top side of the strip lie on this reflected ray (at I1, I2, I3, .......) in between focus and (Moderate)

centre of curvature. Thus the image of this strip is a triangle as shown in figure O1 O2 O3 F C

I I3 I 2

I1

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PAGE NO.-1

2.

Sol.

A block of mass 10 kg is released on a fixed wedge inside a cart which is moved with constant velocity 10 m/s towards right. Take initial velocity of block with respect to cart zero. Then work done by normal reaction (with respect to ground )on block in two seconds will be: (g = 10 m/s2).  10 m/s      10 kg                    2     : (g = 10 m/s2).

(A) zero (B*) 960 J (C) 1200 J (D) none of these  Because the acceleration of wedge is zero, the normal reaction exerted by wedge on block is N = mg cos37° . The acceleration of the block is g sin 37° along the incline and initial velocity of the block is v = 10 m/s horizontally towards right as shown in figure.

The component of velocity of the block normal to the incline is v sin 37°. Hence the displacement of the block normal to the incline in t = 2 second is 3 S = v sin 37° × 2 = 10 × × 2 = 12 m. 5 The work done by normal reaction 4 W = mg cos 37° S = 100 × × 12 = 960 J 5 3.

Consider the P-T diagram for an ideal monatomic gas. P-T 

(A) The process is isochoric. (B*) Volume of the gas is increasing (C*) Internal energy of the gas is increasing (D*) Workdone by the gas is positive (A)  (B*)  (C*)  (D*) 

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PAGE NO.-2

V1=  V2=  P 1

Sol.

2

T PV = nRT

Here  V2 > V1

4.

The specific heat of a diatomic ideal gas can be (R is universal gas constant) ( R   (A*) 7R/2 (B*) 5R/2 (C*) 0 (D*) Infinite

5.

Which of the following is true about moment of inertia ( ) : ( )  : (A*) If about an axis is minimum, then it must pass through centre of mass (B) All axis passing through centre of mass have same (C*) Perpendicular axis theorem can't be applied for 3 dimensional body (D*) Parallel axis theorem can be applied for 3 dimensional body (A*)    (B)    (C*)  (D*)  In all the values of moment of inertias about different axis, moment of inertia about an axis passing through centre of mass is minimum.

Sol.

             6.

Two particles perform SHM with the same amplitude and same frequency about the same mean position and along the same line. If the maximum distance between them during the motion is A (A is the amplitude of either) then the phase difference between their SHM is _ _ _ _ _ _ _.

          A  (A  )    _ _ _ _ _ _ _     [Ans. = /3 ] Y1 = A sin t Y2 = A sin( t + ) = 2 /3 As, Y1/2 = A sin ( t + ) here A = 2A sin /2 Since A = A = 60° Now, solve and use the condition of maxima. Alternate : by symmetry both should be on opposite side of mean position at equal distance from mean (for max. seperation) Sol.

As projection of SHM on circular path

The phase difference from figure is

6

6

3

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PAGE NO.-3

7.

AB wire is vibrating in its fundamental mode. Wire AB is in resonance with resonance tube in which air column is also vibrating with its fundamental mode. Sound speed is 400 m/sec and linear mass density of AB wire is 10–4 kg/m and g = 10 m/sec2, value of mass m = [ (10–1)] kg, then find value of . Neglect the masses of wires in comparison to block's mass 'm'.  AB   AB,        400 m/sec  AB  10–4 kg/m  g = 10 m/sec2  m = [ (10–1)] kg   'm'  

Ans.

6

Sol.

T1 = 2T0 =

2m( 2m) g m 2m

8m 80m g = 3 3 In resonance,  fwire = ftube

T1 =

(1)V1 2 1

................(i)

(1)V2 4 2

T1 ( 400 ) x 4 2 T1 = (16 × 104) 2( x )

=

From (i),  80 m = 10–4 (16 × 104) 3 m = 0.6 kg.

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PAGE NO.-4

COMPREHENSION A glass prism with a refracting angle of 600 has a refractive index 1.52 for red and 1.6 for violet light. A parallel beam of white light is incident on one face at an angle of incidence, which gives minimum deviation for red light. Find : [Use: sin (50º) = 0.760; sin (31.6º) = 0.520 ; sin (28.4º) = 0.475; sin (56º) = 0.832 ; = 22/7] 600      1.52    1.6 

     [Use: sin (50º) = 0.760; sin (31.6º) = 0.520 ; sin (28.4º) = 0.475; sin (56º) = 0.832 ; 8.

The angle of incidence at the prism is : 



 (A) 30º

(B) 40º

(C*) 50º

= 22/7]

(D) 60º

Sol.

µR = 1.52 µv = 1.6 Minimum deviation condition for red is r = 30° i = 50º, = (50º) 2 – 60° R 9.

(1) sin i = (1.52) sin30° = 40°

The angular width of the spectrum is :

 (A*) 6º (B) 4.8º Sol. For violet light (1) sin 50° = (1.6) sin r r = 28.4° r = 31.6° ( r+r =A) (1) sin e = (1.6) sin 31.6° e = 56°, = i + e – A = 50º + 56° – 60° = 46° v angular width = v – R = 6°

(C) 9.6º

(D) 12º

10.

The length of the spectrum if it is focussed on a screen by a lens of focal length 100 cm is : 100 cm  10 10 5 (A*) cm (B) m (C) cm (D) m 3 3 3 3 10 Sol. if = 100 × 6 × cm = m 180 3

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PAGE NO.-5

11.

An object O (real) is placed at focus of an equi-biconvex lens as shown in figure 1. The refractive index of lens is = 1.5 and the radius of curvature of either surface of lens is R. The lens is surrounded by air. In each statement of column-I some changes are made to situation given above and information regarding final image formed as a result is given in column-II. The distance between lens and object is unchanged in all statements of column-I. Match the statements in column-I with resulting image in column-II. O () -1 (equi-biconvex)  = 1.5    R    -I 

          -II   -I       -I -II 

Column-I (A) If the refractive index of the lens is doubled (that is, made 2 ) then (B) If the radius of curvature is doubled (that is, made 2R) then (C) If a glass slab of refractive index = 1.5 is introduced between the object and lens as shown, then

R

Column-II (p) final image is real (q) final image is virtual (r) final image becomes smaller in size in comparison to size of image before the change was made

R

O slab (D) If the left side of lens is filled with a medium of refractive index = 1.5 as shown, then

R

(s) final image is of same size of object.

R air

O (t) final image is inverted

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PAGE NO.-6

-I (A)   ( 2 )  (B)  ( 2R )  (C)  = 1.5    R



-II (p)  (q)  (r)  

  

R

O

 (D)   = 1.5

 R

(s) 

 

R air

O

Ans. Sol.

(t)   (A) p,r,t (B) q,r (C) q,r, (D) q,r (Tough) Initially the image is formed at infinity. (A) As m is increased the focal length decreases. Hence the object is at a distance larger than focal length. Therefore final image is real. Also final image becomes smaller is size in comparison to size of image before the change was made. (B) If the radius of curvature is doubled, the focal length decreases. Hence the object is at a distance lesser than focal length. Therefore final image is virtual. Also final image becomes smaller is size in comparison to size of image before the change was made. (C) Due to insertion of slab the effective object for lens shifts right wards. Hence final image is virtual. Also final image becomes smaller is size in comparison to size of image before the change was made. (D) The object comes to centre of curvature of right spherical surface as a result. Hence the final image is virtual. Also final image becomes smaller is size in comparison to size of image before the change was made.

 

(A)      

                    (B)               (C)                 (D)       

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website: www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029

PAGE NO.-7

TARGET : JEE (Main + Advanced) 2016 O

Course : VIJETA (JP)

Date : 27-04-2015

DPP No. : 15 (JEE-MAIN) Total Marks : 60 Single choice Objective ('–1' negative marking) Q.1 to Q.20

Max. Time : 60 min. (3 marks 3 min.) [60, 60]

ANSWER KEY OF DPP No. : 15 1. 8. 15.

(B) (B) (D)

1.

A symmetrical uniform solid cube of side 5 m is placed on a horizontal surface beside a vertical wall, one side of cube is making an angle 45° with the floor as shown. If coefficient of friction is the same for both wall and floor, the minimum value of for cube not to slip?             5           45° 



2. 9. 16.

(C) (D) (D)

3. 10. 17.

(D) (C) (A)

4. 11. 18.

(C) (D) (A)

5. 12. 19.

(A) (A) (A)

6. 13. 20.

(C) (D) (B)

7. 14.

(C) (B)

 

5m 45°

(A) (C)

Sol.

=1 = 1/3

(B*) = 0 (D) impossible to balance for any value of . (D) 

At angles 45º N and Mg balance each other and they both pass through COM, so no need of friction to balance it 45°   Mg  N          

 N

Mg

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PAGE NO.-8

2.

The angular momentum of a particle relative to a certain point O varies as L are constant vectors with a perpendicular to b . The torque particle when angle between

a bt 2 where a and b

relative to the point O acting on the

and L is 45º is :

 O   L a bt 2  a  b   a  b O   L 45º   (A)

2a

a b

(B)

2b

b a

(C*)

2b

a b

.

a b

(D) zero 

Sol. dL dt

tan = 3.

Ans.

2bt = 2bt a bt

2

= tan45º

t=

a b

2b

Two points of a rod move with velocities 3 v & v perpendicular to the rod and in the same direction, separated by a distance ' r'. Then the angular velocity of the rod is:            3 v  v           ' r'   3v 4v 5v 2v (A) (B) (C) (D*) r r r r v rel` ; vrel. being the velocity of one point w.r.t. other. rod = point = r vrel.  =

3v

v r

and ‘r’ being the distance between them. ‘r’ 

2v = r

4.

The angular velocity of a rigid body about any point of that body is same:

 (A) only in magnitude (B) only in direction (C*) both in magnitude and direction necessarily (D) both in magnitude and direction about some points, but not about all points. (A)  (B)  (C*)   (D)  

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PAGE NO.-9

Sol.

Suppose a rod is having angular velovity w about point C .

v + r1

B r1 C

v

r2 r2

v

A

Choose two points A and B as shown in the fig. velocity of B w.r.t A = (v + r1) – (v – r2)

VB ,A = (r1 + r2)

5.

v B,A

(r r ) = 1 2 = = Ans (C) AB r1 r2 A particle of mass m is performing simple harmonic motion as shown in figure (a) and (b). In figure (b), the charge q of particle is such that qE = mg. If their velocities are same at mean position and let A1 and A2 be their amplitudes and T1 and T2 be their time periods then.

Angular velocity of B w.r.t A =

(a) (b) m (b)  q  qE = mg      A1  A2    T1  T2 :

Sol.

(A*) A1 = A2 , T1 = T2 (B) A1 > A2 , T 1 = T2 (C) A1 < A2 , T1 = T2 (D) A1 = A2 , T 1 > T2 Let velocity at mean position be .   1 1 1 m 2= KA12 = KA22 2 2 2 A1 = A2 and T = 2

6.

m . K

T1 = T2

Which of the following represents a simple harmonic motion ?

 (A) e 7.

t

(B) tan t

(C*) sin t + cos t

(D) sin t + sin 2 t

A force of 6.4 N stretches a vertical spring by 0.1 m. The mass that must be suspended from the spring so that it oscillates with a period of /4 second, is 6.4 N  0.1 m   

: (A) /4 kg

(B) / kg

(C*) 1 kg

(D) 10 kg

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PAGE NO.-10

4

8.

In a cyclic process ABCA for an ideal gas. In AB, BC and CA process 50 J, 20 J and 5 J heat is supplied to an ideal gas. In process AB internal energy of gas increases by 60 J and in process BC work done by gas is 30 J. The increase in internal energy of gas in process CA is :    ABCA  AB, BC  CA    50 J, 20 J  5 J  AB   60 J     BC   30 J  CA : (A) 50 J (B*) – 50 J (C) 75 J (D) 55 J

Sol.

9.

Sol.

One mole of an ideal gas is taken from state A to state B by three different processes, (a) ACB (b) ADB (c) AEB as shown in the P V diagram. The heat absorbed by the gas is: (      A   B     (a) ACB (b) ADB (c) AEB  P V )

(A) greater in process (b) then in (a) ( (b)  (a) ) (B) the least in process (b) ((b) ) (C) the same in (a) and (c) ((a)  (c)  ) (D*) less in (c) than in (b) ((c) (b)  ) Heat absorbed by gas in three processes is given by

 QACB = U + WACB QADB = U QAEB = U + WAEB The change in internal energy in all the three cases is same. And WACB is +ve, VAEB is –ve. Hence QACB > QADB > QAEB WACB  WAEB  QACB > QADB > QAEB 10.

Heat energy absorbed by a system in going through a cyclic process is shown in the figure [ V in litres and p in kPa ] is:        [ V  p , kPa  ] :

V(Litre)

P(k Pa )

(A) 107

J

(B) 104

J

(C*) 102

J

(D) 10

7

J

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PAGE NO.-11

11.

The correct curve between V/T and

1 V

for an ideal gas at constant pressure is : 1 V

V/T 

(A)

Sol.

(B)

PV = nRT V nR = = constant. T P



(C)

(D*)



 12.

A wire of density 9 × 103 kg/m3 is stretched between two clamps 1 m apart and is subjected to an extension of 4.9 × 10–4 m. What will be the lowest frequency of transverse vibrations in the wire ? (Y = 9 × 1010 N/m2)  9 × 103 kg/m3  1 m      4.9 × 10–4 m  (Y = 9 × 1010 N/m2) (A*) 35 Hz (B) 43 Hz (C) 40 Hz (D) 50 Hz

13.

The length of the wire shown in figure between the pulleys is 1.5 m and its mass is 12.0 g. The frequency of vibration with which the wire vibrates in three loops forming antinode at the mid point of the wire is (Take g = 9.8 m/s2)    1.5  12.0    (g = 9.8 m/s2)



9kg

(A) 210 Hz

9kg

(B) 140 Hz

(C) 70 Hz

(D*) none of these   

 Sol.

=1m 9 9 . 8 1 .5 v= = 105 m/s 12 10 3 105 1 f= = 105 Hz 1

14.

Amplitude of a travelling wave on a string is 1mm. If linear mass density of string is 10–4 kg m–1 , tension in the string is 1N and frequency of vibration is 10Hz, then average power needed to maintain such waves in string is : ( 2 = 10)   1mm  10–4 kg m–1    1N 10Hz  ( 2 = 10) (A) 3 × 10–5 W (B*) 2 × 10–5 W (C) 4 × 10–5 W (D) 10–5 W

Sol.

Pav. =

2

A 2v w 10 = 2

4

4

2

10 2 10 2

6

10 4

=2

2

× 10–6 = 2 × 10–5 W

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PAGE NO.-12

15.

The moment of inertia of a uniform thin rod of mass m and length



through centre of rod C and in the plane of the rod are PQ and RS respectively. Then PQ + RS is equal to m  L      C       PQ  RS  PQ  RS  PQ + RS  

(A)

m 2 3

(B)

m 2 2

(C)

about two axis PQ and RS passing

m 2 4

(D*)

m 2 12

Sol.



The M of rod about axis PQ figure(a) and M of rod about axis P Q figure (b) are same by symmetry. PQ (a) P Q (b) 

m 2 PQ 12 by perpendicular axis theorem. ()

+ PQ

16.

= RS

+

= RS

A point charge + Q is placed at the centroid of an equilateral triangle. When a second charge + Q is placed at a vertex of the triangle, the magnitude of the electrostatic force on the central charge is 8 N. The magnitude of the net force on the central charge when a third charge + Q is placed at another vertex of the triangle is:  + Q    + Q      8 N + Q  

 (A) zero

(B) 4 N



(C) 4 2 N

(D*) 8 N

Sol.

R=

42

42

2.44 cos 120

= 4N

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PAGE NO.-13

17.

The ratio of the intensities of the mechanical waves propagating in the same medium Y1=10 sin( t – kx) and Y2= 5 [sin ( t – kx) + 3 cos (kx – t)] is Y1=10 sin( t – kx)  Y2= 5 [sin ( t – kx) + 3 cos (kx – t)]    

 (A*) 1:1 [ Ans. 1:1 ] 18.

Sol.

(B) 1:2

(C) 2:1

(D) 4:1

The deviation caused for red, yellow and violet colours for crown glass prism are 2.840, 3.280 and 3.720 respectively. The dispersive power of prism material is : 2.840, 3.280  3.720  : 11 92 117 22 (A*) (B) (C) (D) 41 250 250 57 Dispersive power is given by v

r y

3.72 2. 84 3 .28

19.

=

11 41

The average density of Earth’s crust 10 km beneath the surface is 2.7 gm/cm3. The speed of longitudnal seismic waves at that depth is 5.4 km/s. The bulk modulus of Earth’s crust considering its behaviour as fluid at that depth, is :  10   2.7 3     5.4   (B)

 (A*) 7.9 × 1010 Pa (A*) 7.9 × 1010 

(B) 5.6 × 1010 Pa (B) 5.6 × 1010 

(C) 7.9 × 107 Pa (C) 7.9 × 107 

(D) 1.46 × 107 Pa (D) 1.46 × 107 

B B = V2 p = (5.40 × 103 m/s)2 (2.7 × 103) = 7.9 × 1010 Pa.

Sol.

(A) V =

20.

Select the Incorrect alternative :   (A) The charge gained by the uncharged body from a charged body due to conduction is of the same nature of charge initially present on charged body.

        (B*) The magnitude of charge increases with the increase in velocity of charge

 (C) Charge can not exist without matter although matter can exist without charge

 (D) Between two non–magnetic substances repulsion is the true test of electrification (electrification means body has net charge)  () NCERT Questions to be discussed Q. No. 9.11 to 9.14, 9.24 to 9.28,

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website: www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029

PAGE NO.-14

Board Level Question 1.

How can a convex lens behave like a diverging lens.

 Ans.

When a convex lens is placed inside a liquid of refractive index greater than that of the material of the lens, it behaves like a diverging lens.

        2.

Why are danger signal red in colour ? Give reasons .

 Ans.







The scattering (Rayleigh scattering) of light is inversely proportional to the fourth power of the wavelength of light. So, the scattering of red light is much less than the yellow light and accordingly, the red light signals can be seen upto a longer distance without much loss in their intensity. For this reason, the danger signals are red in colour.  (   )      

             3.

 4.

Draw a labelled ray diagram of an compound microscope. Define its magnifying power. Why should both the objective and the eyepiece have small focal lengths in microscope?

  Draw a graph to show the variation of the angle of deviation with that of the angle of incidence i for a monochromatic ray of light passing through a glass prism of refracting angle A. Hence, deduce the relation,    A    i  

 =

sin(A m ) / 2 sin A / 2

 5.

Does dispersive power of the material of a prism depend on the shape, size and angle of the prism ? Explain?





Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website: www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029

PAGE NO.-15

TARGET : JEE (Main + Advanced) 2016 O

Course : VIJETA (JP)

Date : 27-04-2015

DATE : 17.05.2015

PART TEST-1 (ADVANCED)

Syllabus : Geometrical Optics complete with optical instruments, Electrostatics upto Electric field due spherical shell & solid sphere, Simple Harmonic motion, String Waves, Sound waves, KTG & Heat & Thermodynamics, Calorimetry & Thermal Expansion complete

This DPP is to be discussed in the week (27-04-2015 to 02-05-2015)

DPP No. : 16 (JEE-ADVANCED) Total Marks : 38

Max. Time : 41 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.4 Multiple choice objective ('–1' negative marking) Q.5 to Q.6 Subjective Questions ('–1' negative marking) Q.7 Comprehension ('–1' negative marking) Q.9 to Q.10 Match the Following (no negative marking) Q.11

(3 marks 3 min.) (4 marks 4 min.) (4 marks 5 min.) (3 marks 3 min.) (8 marks 10 min.)

[12, 12] [08, 08] [04, 05] [06, 06] [08, 10]

ANSWER KEY OF DPP No. : 16 1. 7.

1.

Sol.

(C)

2. 6q

4

0

(B)

3.

(B)

4.

(A)

8.

(C)

9.

(C)

10. (A)

5.

(A,C,D)

6.

(A,C)

(p, s),(C)

(r, s),(D)

2

a2

(p, r, s), (B)

(q, s)

1 mole of an ideal gas undergoes an isothermal expansion as energy is added to it as heat Q. Graph shows the volume V versus Q. The gas temperature is nearly equal to : (use R = 8.31 J/K.mole) QV Q   (R = 8.31 J/K.mole)

(A) 208.4 K (B) 268.2 K For isothermal process v Q = nRT n 2 v1

(C*) 312.6 K

(D) 353.8 K

1800 = 1 × 8.3 T n z get

T = 312.6 K Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website: www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029

PAGE NO.-16

2.

A disc is hinged in a vertical plane about a point on its radius. What will be the distance of the hinge from the disc centre so that the period of its small oscillations under gravity is minimum?

          (A) R Sol.

R

(B*)

(C)

2

R 2

(D)

R 4

MR 2 R2 K2 = + Mx2 K2 = +x2 2 2 K2 As is max. (For T to be min.) x hinged CM line

R

1 R2 ( x 2

d R2 ( dx 2 x

2

x ) is max.

R2 1 +1=0 2 x2 R2 2

x2 = and also 3.

d2 dx

2

R2 2x

x=

R2 2x 2 R 2

x

x)=0

vertical plane

=1

.

x = + ve at x =

R 2

.so T will be minimum

The dispersive powers of two materials are 0.30 & 0.28. They are used to construct two lenses which are kept in contact to eliminate chromatic aberration (that means the fv = fr, the focal length of combination is same for red and violet) If the focal length) (for av. color) of the lens made of the material of dispersive power 0.30 is 10 cm, then the focal length (for av. color) of the lens of other material is :    0.30 & 0.28          (Chromotic aberration) ( fv = fr ,

 ) 0.3    10 cm  

(A) 28/3 cm

4.

A uniform rod of mass M and length L leans against a frictionless wall, with quarter of its length hanging over a corner as shown. Friction at corner is sufficient to keep the rod at rest. Then the ratio of magnitude of normal reaction on rod by wall and the magnitude of normal reaction on rod by corner is

(B*)

0.28/3 m

(C) 0.75/7 m

(D) none of these

     M   L                                    (A*)

1 2 sin

(B)

2 sin

(C)

1 2 cos

(D)

2 cos

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website: www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029

PAGE NO.-17

Sol. The FBD of rod is FBD  Taking moment of force about centre of mass

 L L – N1 sin × =0 4 2 N1 1 N2 2 sin

N2 ×

5.



Sol.

6.

An electron is placed just in the middle between two long fixed line charges of charge density + each. The wires are in the xy plane (Do not consider gravity) +       xy    (e)  ()

(A*) The equilibrium of the electron will be unstable along x-direction (B) The equilibrium of the electron will be stable along y-direction (C*) The equilibrium of the electron will be neutral along y-direction (D*) The equilibrium of the electron will be stable along z-direction (A*) x- (B) y- (C*)  y- (D*)  z- If we displace the electron slightly toward x direction, it will thrown away toward right. So eql. is unstable along x direction. If we displace the electron slightly towards y direction, No extra force will act. So eql. is neutral along y axis If we displace the electron toward z direction, it will be attracted and try to come to eql. positron. So eql. is stable along z direction. Two free point charges +q and +4q are placed a distance x apart. A third charge is so placed that all the three charges are in equilibrium. Then   +q  +4q, x       

 (A*) unknown charge is -4q/9  -4q/9  (B) unknown charge is -9q/4  -9q/4  (C*) It should be at (x/3) from smaller charge between them (x/3)  (D) It should be placed at (2x/3) from smaller charge between them. (2x/3) 

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PAGE NO.-18

Sol. For equilibrium of all the changes net force on each charge should be zero. kqQ 4kqQ – = 0 (Net force on Q) d = x/3 2 d ( x d) 2 &

kqQ d

2

Q =

+

4kq 2

= 0 (Net force on A )

x2

4 x q & d = 9 3

 

7.

Calculate the magnitude of electrostatic force on a charge placed at a vertex of a triangular pyramid (4 vertices, 4 faces), if 4 equal point charges are placed at all four vertices of pyramid of side ‘a’.  (4 , 4 )      'a' 

Sol.

cos

Fnet = 3Fcos = 3 [ Ans.

6 q2 4

0

a2

Kq2 a

2

2 3

=

2 3

6 q2 4

0

a2

]

COMPREHENSION An extended object of size 2 cm is placed at a distance of d (cm) in medium (refractive index n = 3) from pole, on the principal axis of a spherical curved surface.The medium on the other side of refracting surface is air (refractive index n = 1). 2    d (cm) (  n = 3)  n = 1  n=3

2cm

n=1

ROC =20cm

Sol.

n=1

d

d

8.

n=3

2cm

=20cm

For d = 20 cm, the distance of the image from the pole is d = 20 cm  (A) 2 cm (B) 3 cm (C*) 4 cm

(D) 5 cm

From formula for refraction at curved surface

 n 2 n1 n 2 n1 v = – 4 cm v u R image is formed in denser medium at a distance 4 cm from pole. 4 cm 

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PAGE NO.-19

9.

Sol.

For d = 20 cm, The size of image is d = 20 cm  1 2 6 3 (A) cm (B) cm (C*) cm (D) cm 6 15 5 2 4 3 6 v n1 Size of image = × size of object = 2= cm. u n2 20 1 5

 = 10.

v u

n1 n2

=

4 20

3 6 2 = cm. 1 5

In each situation of column-I a mass distribution is given and information regarding x and y-coordinate of centre of mass is given in column-II. Match the figures in column-I with corresponding information of centre of mass in column-II. -I   -II  x  y- -II  Column-I

Column-II

(A) An equilateral triangular wire frame is made using three thin uniform rods of mass per unit lengths , 2 and 3 as shown

(p)

xcm > 0

(B) A square frame is made using per unit length lengths , 2 , 3 and 4 as shown

(q)

ycm > 0

(C) A circular wire frame is made of two uniform semicircular wires of same radius and of mass per unit length and 2 as shown

(r)

xcm < 0

(D) A circular wire frame is made wires of same radius and mass per unit length , 2 , 3 and 4 as shown

(s)

ycm < 0

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PAGE NO.-20

-I

-II

(A)  , 2

 3        

(p)

xcm > 0

(B)  , 2 , 3  4  

(q)

ycm > 0

 2                

(r)

xcm < 0

 , 2 , 3  4                       

(s)

ycm < 0

          (C) 

(D)

Ans. Sol.



(A) q,r (B) p,s (C) p,s (D) p,s (A) Centre of mass lies in second quadrant. (B), (C) and (D) Centre of mass lies on y-axis and below x-axis. (A)  (B), (C)  (D)  y-x-

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website: www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029

PAGE NO.-21

TARGET : JEE (Main + Advanced) 2016 O

Course : VIJETA (JP)

Date : 27-04-2015

This DPP is to be discussed in the week (27-04-2015 to 02-05-2015)

DPP No. : 17 (JEE-MAIN) Total Marks : 60 Single choice Objective ('–1' negative marking) Q.1 to Q.20

Max. Time : 60 min. (3 marks 3 min.) [60, 60]

ANSWER KEY OF DPP NO. : 17 (JEE-MAIN) 1. 8. 15.

(D) (C) (A)

1.

When a point charge of

Sol.

2. 9. 16.

(B) (D) (A)

3. 10. 17.

(B) (C) (B)

4. 11. 18.

(A) (D) (C)

5. 12. 19.

(C) (C) (B)

6. 13. 20.

(B) (B) (D)

7. 14.

(C) (C)

1 2 C is placed along the axis of a thin disc of total charge C (uniform 3 3 distribution) and radius 3.95 cm such that distance between point charge and centre of disc is 1 m, then force experienced by disc is approximately : 2      3.95 cm      C        3 1 C 1m  3  : (A) 4mN (B) 6mN (C) 3mN (D*) 2mN Behaviour of disc is like a point charge because distance between the disc and the point charge is very large in comparison to radius of disc.

 9 10 9

F= 2.

2 1 C C 3 3 = 2mN 12

A wire having a linear density 0.1kg/m is kept under a tension of 490 N. It is observed that it resonates at a frequency of 400Hz and the next higher frequency 450Hz. Find the length of the wire 0.1kg/m 490 N  400Hz  450Hz  (A) 0.4m (B*) 0.7m (C) 0.6m (D) 0.49m

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PAGE NO.-22

3.

Consider the arrangement shown in the figure. Assuming frictionless contacts, then the magnitude of external horizontal force P applied at the lower end for equilibrium of the rod will be : (The rod is uniform and its mass is ' m ')      P  (' m ' )

mg mg (B*) cot 2 2 The F.B.D. of rod is as shown For rod to be in translational equilibrium N1 = P ....(1) N2 = W = mg ....(2)

(A) Sol.

(C)

mg tan 2

(D)

mg sec 2

For rod to be in rotational equilibrium, net torque on rod about any axis is zero. Net torque on rod about B is zero i.e., mg

cos – N2 cos + P sin 2 from equation (2) and (3) solving we get mg P= cot 2 4.

= 0 .......(3)

A uniform rod of length 4L and mass M is suspended from a horizontal roof by two light strings of length L and 2L as shown. Then the tension in the left string of length L is 4L   M    L  2L       L 

(A*)

Mg 2

(B)

Mg 3

(C)

3 Mg 5

(D)

Mg 4

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PAGE NO.-23

Sol.

The free body daigram of rod is Net torque about centre of mass is zero



  L L cos = T2 × cos 2 2 Mg T1 = T2 = 2

T1 ×

5.

A uniform disc of mass m and radius r and a point mass m are arranged as shown in the figure. The acceleration of point mass is: (Assume there is no slipping between pulley and thread and the disc can rotate smoothly about a fixed horizontal axis passing through its centre and perpendicular to its plane)

 m  r   m   ( )

g g (B) 2 3 2g (C*) (D) none of these  3 Let a & be linear and angular acceleration of disc respectively r a=r .........(i) T Xm Torque about centre of disc = 1 mgr = mr 2 mr 2 2

(A)

Sol.

3 mgr = mr 2 2 From eqn. (i) & (ii) 3 a mgr = mr 2 2 r

a=

m

.........(ii)

mg

2g 3

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PAGE NO.-24

6.

Two identical discs of mass m and radius r are arranged as shown in the figure. If is the angular acceleration of the lower disc and acm is acceleration of centre of mass of the lower disc, then relation between acm, & r is : m r        acm    ,  ,acm,  r   :

r 2 (C) acm = r

(A) acm =

(B*) acm = 2

r

(D) none of these 

Sol.

Tr =

mr 2 2

Tr =

..........(1)

1

mr 2 2

.......... (2)

= .......... (3) 1 From (1) & (2) accn . of point b = accn of point a r 1 = acm – r .......... (4) Hence 2 r = acm Ans. (B) 7.

A force of 6.4N stretches a vertical spring by 0.1m. The mass that must be suspended from the spring so that it oscillates with a time period of /4 second. 6.4 N  0.1 m  /4  

 (A) Sol.

4

kg

(B)

4

kg

Spring constant ( ) K = Now () T = 2

m k

or

4

2

(C*) 1 kg

(D) 10 kg

6 .4 = 64 N/m. 0. 1 m 64

m = 1 kg

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PAGE NO.-25

8.

Sol.

9.

PV = constant for a given amount of a gas is true for : T PV  =   T (A) isothermal change only. (B) adiabatic change only. (C*) both isothermal & adiabatic changes. (D) neither isothermal nor adiabatic change. (A)   (B)  (C*)   (D)  As PV = nRT PV For n = constant : = constant T for all changes. Hence (C)

The gas law

If a tuning fork of frequency (f0) 340 Hz and tolerance 1% is used in resonance column method [v = 2f0 ( 2 – 1)], the first and the second resonance are measured at 1 = 24.0 cm and 2 = 74.0 cm. Find max. permissible error in speed of sound.  [v = 2f0 ( 2 – 1)]     ( 0) 340 Hz

 (tolerance ) 1%  1 = 24.0 cm   (permissible) Ans. Sol.

(A) 1.5% (D) v = 2f0 ( 2 – 1), v v

10.

(B) 1.3%

= max

f0

(C) 1.2%

1

f0

2

2

=

1

2

= 74.0 cm 

(D*) 1.4%

1 0.1 0. 1 + = 1.4%. 100 74 24

Two vibrating strings of same length, same cross section area and stretched to same tension is made of materials with densities & 2 . Each string is fixed at both ends. If v1 represents the fundamental mode of vibration of the one made with density and v2 for another, then v1/v2 is:

                2    v 1   v2 v1/v2  (A) Sol. 11.

V1 V2

1 2

(B) 2

(C*)

(D)

2

1 2

2

=

=

2

=

Ans.

2

1

What is the percentage change in the tension necessary in a sonometer of fixed length to produce a note one octave lower (half of original frequency) than before

          Sol.

(A) - 25% (B) - 50% In Sonometer () V

(C) - 67%

(D*) - 75%

T

V1 =2= V2

T1 T2

T2 =

T1 4

%   T1

T2 T1

T1 4

T1

× 100 =

T1

100 =

75%

Ans.

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PAGE NO.-26

12.

13.



The tension, length, diameter and density of a string B are double that of another string A. Which of the following overtones of B is same as the fundamental frequency of A. (They are fixed at both the ends)  B  A    B  A

  



(A) 1st

(D) 4th

(B) 2nd

(C*) 3rd

A string 1m long fixed at one end is made to oscillate by a 300Hz vibrator attached to its other end. The string vibrates in 3 loops. The speed of transverse waves in the string is equal to     1m  300Hz       

 (A) 100 m/s 14.

(B*) 200 m/s

(C) 300 m/s

(D) 400 m/s

A person calculates electric field due to a point charge q at a distance r from the point charge. If value of q is known with 2% error and r is measured with 1% error, then the percentage error in calculation of 1 electric field is (assume that there is no error in k = ): 4 0

  q   r  q   2%  r 1% ,  (k =

1 4

): 0

(A) 3% Sol.

E=

(B) 0

kq r2

dE E

(C*) 4%

(D) 1%

dq dr –2 q r

dE dq dr +2 E q r Two point charges 2q and q are placed at some distance as shown in the figure. If the charge q is moved towards right then electric field at a point which on line AB and equidistant from both point charges at any instant:    2q  q   q  AB

For percentage error 15.



(A*) decrease (A*)   Sol.

(B) increase (B)  

(C) remains same (C) 

(D) may increase or decrease (D) 

Electric field at mid point

=

kq 2

=

4kq 2

a a 2 On increasing a E will decrease

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PAGE NO.-27

16.

A parallel beam of light is incident on a lens of focal length 10 cm. A parallel slab of refractive index 1.5 and thickness 3 cm is placed on the other side of the lens. Find the distance of the final image from the lens.  10 cm    1.5  3 cm   

(A*) 11 cm

(B) 9 cm

(C) 4 cm

(D) 15 cm

Sol.

As rays are parallel to the principal axis, image is created by lens at the focus. By placing of glass-slab, 1 1 Shift = 1 .t = 1 3 = 1 cm. 1. 5 Irrespective of separation, Image is shifted to the right by 1 cm. Total distance from lens 10 + 1 = 11 cm Ans.

17.

A body is located on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens. The lens is placed at a distance d ahead of second wall, then the required focal length will be: d d (A) Only (B*) Only 4 2 d d (C) More than but less than (D) Less than 4 4   d   d  d d d d d (A)  (B*)   (C)   (D)  4 2 4 2 4 The lens formula can be written as  1 1 1 = .... (i) f v u Given , v=d For equal size image  |v|=|u|=d By sign convention , u = – d 1 1 1 d = or  f= f d d 2

Sol.



Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website: www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029

PAGE NO.-28

18.

An equiconvex lens is cut into two halves along (i) XOX’ and (ii) YOY’ as shown in the figure. Let f, f ’,f ’’ be the focal lengths of the complete lens, of each half in case (i), and of each half in case (ii), respectively.

Choose the correct statement from the following : (A) f ’ = f, f ’’ = f (B) f ’ = 2f, f ’’ = 2f (C*) f ’ = f, f ’’ = 2f (D) f ’ = 2f, f ’’ = f (i) XOX’ (ii) YOY’f, f ’,f ’’  (i) (ii)  

 Sol.

   



(A) f ’ = f, f ’’ = f (B) f ’ = 2f, f ’’ = 2f (C*) f ’ = f, f ’’ = 2f (D) f ’ = 2f, f ’’ = f Initially, the focal length of equiconvex lens is  1 1 = ( – 1) R1 f

1 R2

1 1 1 2(u 1) = ( – 1) = R1 R 2 f R Case I :  When lens is cut along XOX’ then each half is again equiconvex with  I XOX’  R1 = + R, R2 = – R 1 1 = ( – 1) f R

Thus,

= ( – 1)

1 R

1 ( R)

1 2 1 = ( – 1) = R R f'

f’=f Case II : When lens is cut along YOY’, then each half becomes plano-convex with  I YOY’  R1 = R, R2 = 1 1 = ( – 1) R1 f'

Thus, = ( – 1) =

(

Hence 

1) R

1 R

1 R2

1

1 2f f ’ = f, f ’’ = 2f

=

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website: www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029

PAGE NO.-29

19.

Two springs of same spring constants are arranged as shown in figure. A block of mass m strikes one of the spring with velocity v. Find the period of oscillation of the block. (The block does not stick to the spring)      m  v 

 v

(A) 2 20.



m k

m k

(B*) 2

V

m

2 V

(C) 2

m 2k

2 V

(D) 2

2m k

2 V

A ring of mass m and radius R rolls on a horizontal rough surface without slipping due to an applied force ‘F’. The friction force acting on ring is : –  m  R  F     



(A) Sol.

(D)

F 3

(B) F + f = ma

Also ; FR – fR = F – f = ma From (1) & (2) f = 0.

2F 3

(C)

F 4

(D*) Zero   

.... (1) a R

.... (2)

[ = mR 2 ]

NCERT Questions to be discussed Q. No. 9.6, 9.21, 9.23, 9.29, 9.30, 9.31, 9.32, 9.33, 9.34, 9.35, 9.36, 9.38

BOARD LEVEL QUESTIONS 1.

A person looking at a mesh of crossed wires. He is able to see the vert ical wires more distinctly than the horizontal wires. What is the reason of this defect, By what name this defect is known. How is such a defect of vision can be corrected?

     Ans.

The defect (called astigmatism) arises because the curvature of the cornea plus eyelens refracting system is not the same in different planes. [The eye-lens is usually spherical i.e., has the same curvature on different planes but in some cases the curv ature in the v ertical plane is enough, so sharp images of vertical wires can be formed on the retina, but the curvature is insufficient in the horizontal planes, so horizontal wires appear blurred. This defect can be corrected by using a cylindrical lens with its axis along the vertical. Clearly, parallel rays in the vertical plane will suffer no extra refraction, but those in the horizontal plane can get the required extra convergence due to refraction by the curv ed surface of the cylindrical lens if the curvature o the cylindrical surface is chosen appropriately.

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website: www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029

PAGE NO.-30

                                                        2.

Draw a labelled ray diagram of a reflecting (Cassegrain) telescope and explain its working. (Cassegrain) 

3.

List some advantages of a reflecting telescope, especially for high resolution astronomy.

 Ans.

No chromatic aberration due to the objective because only reflection is involved; spherical aberration reduced by using a mirror of the shape of paraboloid ; brighter image than in a refracting telescope of equivalent size because in the latter intensity of light is partially lost due to reflection and absorption by the objectiv e lens glass; mirror entails grinding and polishing of only one side; high resolution (as well as brightness of a point object) achieved by using a mirror of large aperture which is easier to support (its back side being av ailable) than a lens o the same aperture.

                        (brighter image)                     (    )           (support)    4.

Why is rainbow formed in the sky ?

 Ans.

Rainbow is formed due to the dispersion of sun rays, when they fall on the suspended tiny droplets of water (which acts as prisms of small angles). The rainbow will be visible to an observer having sun at his back.

            

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