cycle 2 quize 1 solutions.pdf

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[ CYCLE - 02 ]

[ QUIZE - 01]

SOLUTION

[PHYSCIS]

1. (a)l

P R Q

Let a be the acceleration and l be the distance between P and Q, then using 2 2 2 ,v u as= + we get,2 2

2 2 17 717 7 22

al la

= + =

Therefore, velocity at R is

2 22R PlV V a= +

2PV al= +

2 22 17 77

2

= +

2 27 17 132+

= =

Therefore, average velocity between P and R is

1 2P RV VV +< >= [ ] is const.a

7 132+

=

10 / .m s=Therefore, average velocity between R and Q is

213 17

2 2R QV VV+ +

< >= =

15 / .m s=

LLLLLOOOOOTSTSTSTSTS


2. (a)u2

u11

2g

Y

X

u2

u 21

12

u1

Y

X

Both the projectiles in the ground frame Projectile 2 in the frame of projectile 1.Relative acceleration = 0.

21 = speed of 2 wrt 1.

Therefore, equation of the path of 2 wrt 1 isy mx=

2 2 1 1

1 2 1 1

sin sincos cos

y u umx u u

= =

3. (b) Speed is always nonzero. While falling speed must increase at a constant rate. While rising speed must decrease at a constant rate. After each collision speed must decrease by a factor of 2.

4. (a) Using conservation ofmechanical energy, we have,los in P.E. = gain in K.E.

212

mgy mv =

Ry

N

v

v /R2

mg21sin =

2mgR mv

2 2 sinv gR = ... (i)

Centripetal force, 2

CmvF

R=

Again, using Newtons second law along the radially inward direction, we have,2

sin mvN mgR

=

2

sin= mvN mgR

2

2

2

/ 1/ sinsin / 1C

mv Rx F N gRmg mv Rv

= = =+ +

1 21 312

= =

+


5. (a) 7xUFx

= =

24yUFy

= =

2 2 25x yF F F= + =

6. See Example 18 on page 39 of Locus Study Material for work Energy Power

7. (a) We have, dvdt

= rate of change of speed

ta= = a

v at =Again, 2 /na v r=

2 22 a tbt

r =

2 /r a b =8. (d) When speed becomes maximum, all the power developed by the external agent is disssipitated by the friction

and hence there is no further increase in speed, as the power of external agent is constant. Therefore, if maxV bethe maximum speed, the

maxmgV p =

maxpVmg

=

9. (a) We have

212

mv p t=

where v is speed at some time t and p is constant power.

2 2pt pv tm m

= =

2ds p tdt m

=

0 0

2s tpds t dtm

= 3

22 ps tm

=

322

2

p tms tv p t

m

= =


10. (d) 1

1N

11N

N2

N1

N12 2N rough

External impulsiveforces : , N .N2 2 2

Impulsive forces acting on the ball and the wedge are shown in figure. Internal forces cancel each other whenthe ball and the wedge are considered as a single system. But external impulsive force 2N (normal contact

force from the horizontal surface) and 2 2N ( frictional force from the horizontal surface) change the momentumof the system along the vertical and horizontal directions, respectively.

11. (b) 5atx atx conU U W= =

5

x

xx

F dx=

= xF x= [ ] is const.xF

( 6)( 5)x= ( ) 0 6 30U x x = ( ) 6 30U x x =

12. (a) When we consider ball + earth as a single system, there is no external force is acting on the system during thecollision. Hence, momentum of the system is conserved.

13. (b) Using conservation of linear momentum,

fin inP P=" "

1 2 3 0P P P + + ="" " "

3 0P =""

1 2 0P P + = "" "

K.E. released 1 2 3K K K= + +

1 2K K= + 3[ 0]K =

2 21 12 22 2

m v m v= + +

22mv=


14. (a) thrust reldmF Vdt

=

( )( )60 / 2050 /kg s m s=12,3000 N=At 15 sec.,t =

thrustFa gm

=

( )12,3000 10

5000 60 15=

+

210.8 /m s=

15. (a) As the external horizontal force on the system is zero and initially the system is at rest,

0cmX =Assuming direction horizontallyaway from the pulley

X 1 2 3 0block man wedgem x m x m x + + =

( )2 2 0M l M l M l + + + = [ ]2 3Assuming x x l = = +24

MlM

=

0.5 m=

16. (d) At F acceleration is +ve. At D velocity is zero and acceleration is negative. Particle moves away from origin, comes back towards origin by some distance and again moves awayfrom origin.Hence, only (d) is true

17. (a) We have 2

.2xy x= Comparing it with

2

2 2tan 24 cosgxy x=

We get,

tan 1 45 = =

and 2

2 2

12 cos 2 2

g ugu

= =

1using cos2

= 2u g =

Now, time of flight

12 22 sin 2

gu

g g

= =

2g

=


18. (a) 0Bl =( 4) ( 4 5) 0BV + =

13 / .BV m s =

4 /m sA

VB

-4-5

5 /m s4 /m s

B

VB-4

19. (d)

N2N1

mg sinmg cos

SMOOTH

h

We have, 2 sin sinN mg = ... (i)and 1 2 cos cosN N mg + =

2 1cos cosN mg N = ... (ii)When h is minimum, the sphere is about to leave the inclined surface. Therefore, in that case in equation (ii) weshould put 1 0N = . Hence, from (ii),

2 cos cosN mg = ... (iii)(i) / (iii) gives,

sin sin =cos cos =

cosR hR

=

( )1 cosh R =


20. (d) Let x be the elongation in the spring and distance AC be y.As the rod is in equilibrium,

1 2F k x k x= + ...(1)

and ( )1 2k x y k x l y = ...(2)Solving (1) and (2), we get,

2

1 2

lkyk k

=

+

l

k x1 k x2 C

y

F

21. (a) As 0,extr =

in finl l=

in in fin finI I = 2 2

1 1 1 2 2 2M r m r = 2

1 12 2

2

rr

= [ ]1 2m m=

2 14 = 1

22r r

=

22. (b) When A reaches maximum height, its speed wrt to the wedge B is zero, i.e, wrt the wedge B it is at rest. Let atthat moment the combined system (rim + wedge) moves horizontally with speed .uAs there is no external force is acting on the system along the horizontal direction, linear momentum of thesystem is conserved along the horizontal direction.Therefore,

( ) fmv m m v= +

2fvv =

If h be the maximum height attained by the rim, using conservation of mechanical energy, we have,loss in K.E. = gain in P.E.

( )2

21 1 22 2 2

vmv m mgh = 2 21 1

2 4mv mv mgh =

214

mv mgh =

2

4vh g =


23. (b) We have ,( ) 22T M d=

222M dP

= d

2MT

2d

2

2

8 MdP

=

24. (b) When the rod makes minimum possible (while in equilibrium) with the surface, the frictional force acting on itmust be limiting. As the rod is in rotational equilibrium, net torque on it about end A must be zero.Therefore,

N1 B

hmg

uN2

N2

C

A

N1 B

hmg

uN2

N2

C

A

N1 B

hmg

uN2

N2

C

A

N1 B

hmg

uN2

N2

C

A

1 cosec cos2lN h mg = [ l is length of the rod]

1cos sin2

mglNh

= ...(1)

As the rod is in translational equilibrium along the vertical direction, we have, 2 1 cosN N mg+ =

2 1 cosN mg N = ...(2)As the rod is in translational equilibrium along the horizontal direction, we have,

1 2sinN N =

1

2

sinNN

=

2

2

cos sin2cos sin

2

mglh

mglmgh

=

[Using (1) and (2) ]

2

2

cos sin2 cos sin

lh l

=


25. (c) While rising during the jump let the centre of mass accelerate with average acceleration a and t, be the timetaken to make the jump. If v be the upward speed immediately after making the jump then, we have

.v a t= ... (i)

and 21 .2

s a t= ... (ii)

(ii)/(i) gives, 2s tv

=

2stv

= ... (iii)

Again, we have,

212

P t mv =

2 2P tvm

=

3 4Psvm

= ... (iv) [ using (iii)]

If h be the rise in height after the jumps then,2/ 32 1 4

2 2v Pshg g m

= =

26. (a)

B

= MgF

A

x

D

y

E

( )2 2 22 1 cosAD x r = = ( )2 2 22 1 cos 2 2cos / 2yAE y r

x= = =

/ 2, FAD FAE = = (by geometry)


( )( )

22

2

1 cos / 2 cos21 cos / 2cos / 22

x g t y nxx g t

=

=

=

22cos 2cos / 2 cos 2cos 2 / 2

cos / 2n n

= =

2 cos 1 2cos 2 / 2 1 cosn = =

( )2 1 cos 1n =27. (a) Let the particle leaves the surface at some anlge . At this moment normal contact force becomes zero. Therefore,

vmg

m v0y

v /R2

vmg

m v0y

v /R2

vmg

m v0y

v /R2

vmg

m v0y

v /R2

2

cos mvmgR

=

2 cosv gR = ... (i)Applying conservation of mechanical energy, we have,loss in P.E = gain in K.E

2 20

1 12 2

mg y mv mv =

( ) 201 11 cos cos2 2mg R mgR mv = [using (i)]20

3 1cos2 2

mgR mgR mv = +

203cos

2 3vgR

= +

21 02cos

3 3vgR

= +


28. (b) We have,2

2vhg

=

M

um

M+m

stops

hv

M

um

M+m

stops

hv

M

um

M+m

stops

hv

M

um

M+m

stops

hv

2

2vhg

=

2

0.752 10

v =

15 / .v m s =Applying conservation of linear momentum, we get,

( )mu m M v= +0.015 2 15

0.015m Mu v

m+ +

= =

2000 515 /15

m s# #

29. See Example 11 on page 21 of Locus Study Material for Work Energy Power

30. (c) When the man moves on the trolley, let the trolley moves in the opposite direction with velocity v with respectot the horizontal rails.Applying conservation of linear momentum, we get,

,sin ,sys sym inP P=1 m/sman

v trolley+ve

( )80 1 320. 0v v =80 1 /400 5

v m s = =

Therefore velocity of the man wrt grout is1 41 1 /5 5

v m s = =

distance travelled in 4 seconds is44 3.25

= m


[ CHEMISTRY ]

1 (c) Is wrong representation of resonance structure of p-nitrophenoride ion, since total charge on the compoundis 1 unit but in (c) the total charge is zero. Resonating structures show the delocalisation of charge. The totalcharge does not change.

2 (d) CH CH CH O2 CH CH CH O2

CH2 CH CH O+

+

3 (c)5

4 2

67 9

10

138 5, 6-Diethyl -3-methylendec -4-ene

4. (d) (I) and (II) are mirror images which are non-superimposable, therefore are enantiomass.

5 (d)C

CCl

Br

I

FC

CCl

Br

I

FC

CCl

BrI

F

C

CCl

BrI

F

C

C

Cl Br

IF

C

C

Cl Br

I F

6 (a) Let the ratio of A to B be x : y. we have20 20 ( ) 10x y x y = + 10 30x y=

31

xy=

3 1to

7 (a) ClBr Na/Ether Cl Mg/Ether ClMg MgCl

(I) (II)

D O2

DD

(A)

Cl

8 (d) C C CH Cl2 CH3CH3

CH3 CH3

CH3(B)

since all other can form more than one product after chlorination


9 (b) Anti-elimination through E2 mechanism, because Br is attached to the secondary carbon. The product formedin this case is trans.

C C

CH3

H CH3

H

If the -Br atom was attached with a primary carbon, a cis product would have been form because of just 2nS

attack by I to replace Br and then anti-elimination by 2E mechanism.

10 (c) 2 2/Cl H O gives Cl+ which attachs on the double bond

R CH CH2Cl+ R CH CH2

Cl+

11 (b) O /z3 n O +O

+ O = CH2H O2O

12 (b) CH3 CH CH 2BD3 CH3 CH CH 2+

D 2BD

CH3 CH CH 2

D BD2

Similarly we get;

(CH3 CHD CH ) B 2 3 H O2 2 (CH3 CH CH OH + B(OH) 2 3

D

13 (b) Because intermediate & final product after dehydration will be more stable in this order due to conjugatedsystem

III

> >

II I

Stability order

14 (a)

III

Mg Br + CH3 C OH + MgBr

OC(CH )3 3

CH3

CH3

Answer was wrong for this question


15 (d) Negative charge of oxygen atom will be stabilised more by NO2 group at &Q P position, therefore (4) will bemore acidic than (3).(1) will be more acidic than (2) because CH3 is an activating group which increases electron density. (3) willbe more acidic than (1) because NO2 is deactivating. Therefore order will be 4 > 3 > 1 > 2.

16 (b) C H O2 5 +

OH

C H OH2 5 +

O

Now C2H5O and PhO both are nucleophiles but C2H5O

is a stronger nucleophile because in case ofPhO negative charge is delocalised in the ring.

C2H5O attacks C2H5I

Product is C2H5OC2H5

17 (a) Carbocation is the intermediate and therefore rearrangement takes place to stabilise the carbocation.

18 (a)

19 (d) 6 electrons, delocalised

20 (b) Meta position to both COOH group

21 (b) Carbocation intermediate and rearrengement takes place

CH CH3 2 CH CH2 2 CH CH3 2 CH CH3+

+

22. (c) Benzene ring gets deactivated in the order.Bromination is electrophillic substitution

23 (b)

24 (c)

25 (c) + charge on carbon gets stabilised in the order.

26 (b)

27 (b)

28 (d)

29 (c) MeO group at para position stabilises the positive charge generated because it activates the ring

30 (a) Cannizaro reactionOH attacks carbon atom of for maldehyde.


[ MATHEMATICS ]

1 (a) Given 3( ) ( ) ,f x f y x y x y $Let x y t= + , then

3( ) ( ) ,f y t f y t t y R+

2( ) ( )f y t f y tt

+

2

0 0

( ) ( )lim limt t

f y t f y tt

+

'( ) 0 '( ) 0f y f y =

'( ) 0 ( )f y f y = = constantHence, f(x) is periodic with no fundamental period

2 (a) y = mx is continuous on all real x. Now, if y = mx satisfies the equation 20

( ) 0x

y f t dt+ =then 2 2

0

( ) ( ) 0 ( ) 0x

m x f t d t h x+ = = [where 2 20

( ) ( )x

h x m x f t dt= + ]So, its clear for 0,x =

0

( )x

f t dt and 2 2 0m x = , Hence ( ) 0h x =

Thus y mx= meets curve 2

0

( ) 0x

y f t dt+ = at origin real values of m.

3 (b) 10 1 1( ) ...n n

nf x a x a x a x

= + +

(0) ( ) 0f f= =Then there is a 0,( ) such that '( )f must be equal to zeroSo 1 20 1 1( 1) ... 0

n nnna x n a x a

+ + + = has a root such that 0 < <


4 (c) If ( ) 0f x = (f(x) is continuous) has two roots & , then there exists a < < , such that '( ) 0f y =For limiting value if then . So If ( ) 0f x = has repeated roots at then '( ) 0f =The graph will be as follows

for eg. 2( ) ( 2)f x x=

Here 2

( ) 1 ...11 21

nx x xf xn

= + + + is continoues

let is repeated root. So

( ) 0 1 ... 0! ! !

nf

n

2 = + + + =

1 2...(i)

1'( ) 0 1 ... 0

! ! ( 1)!

nf

n

2 = + + + =

1 2 ...(ii)

from (i) & (ii), 0!

n

n

= = 0

Thus if ( ) 0f x = has a repeated root it has to be at 0x =but 0x = cannot be a root. So ( ) 0f x = doesnt have repeated roots

5 (a) ( )f x is even, continuous function on [ , ]a a which is non-differentiable at 0x = . and ( )g x is continousdifferentiable function in [0, ]bSo, ( ( ))g f x is continuous and even function on [ , ]a a

Now,

For [ )0, , ( ) [0, ]x a f x b which is domain of ( )g x . Thus ( ( ))g f x is differentiable for ( ]0,x a . (verifyfrom figure given in question) ( ( ))g f x being an even function is also differentiable in [ ),0x a

To check differentiability of ( ( ))g f x at 0x = .

We have for 0 , ( )h f h b+ , from graph of ( )g x we can see when parameter of ( )g x [here ( )f h ]

tends to b , Then derivative of ( )g x goes towards 0. As ( ( ))g f x is even, same is case for 0h , as it

also implies ( )f h b . Thus ( ( ))g f x is differentiable at 0x = and its derivative is 0.[Infact, derivative

being equal to 0 is only possibility when an even function can be differentiable at 0] But if '( ) 0.g b Then( ( ))g f x would not have been differentiable at 0x = .

An example of above case is

( )f x x= and 2( ) sing x x=


6 (a) For 0y > , 2x y y y x+ = =For 0y < , 2 , / 3x y y y x = =

So ( )y f x= is continuous at 0x =

7 (d) ( )h x can be defined using ( )f x and ( )g x as {2 , when is rational2 , when is irrational( ) x xx xh x =2 & 2x x are one-one functions, and rationals and irrationals numbers can never coincide. So ( )h x is one-one. Now between any two rational nos there are infinite irrational numbers and vice-versa. Hence graphcannot be drawn.

8 (a) 0 1,a< < So 2sin !lim1

a

n

n nn +

2

1in !lim a an

s nh n

=

+

1 00

aa >

So for 1 2cos 1 x & 1sin x to be identical 1sin 0x < . So the function will be identical for [ 1, 0]


18 (a)12 ( 1) 1f x f xx

= ...(i)

replace x by 1/x

1 12 1 ( 1)f f xx x

= ...(ii)

Solving (i) & (ii) we get

1 2( 1)3

f x xx

= +

1 2 1( ) ( 1) ( )3 ( 1) 3

f x x h xx

= + + = +

for 2( ) 1

1h x x

x= + +

+, we have

12 2

2 2

( )( ) , 2 2 2 2,h x

Thus 2 2 2 2( ) , ,

3 3f x

2 2 2 2( ) ,3 3

f x

$

19 (b) 2 1 2 1( ) ( ) { ( )}n nf x y f x f y+ ++ = +

put 2 10, 0 (0) (0) { (0)} nx y f f f += = = +(0) 0f =

put 0, 1x y= = 2 1(1) (0) (1) nf f f += +2(1)(1 (1) ) 0nf f =

(1) 0 (1) 1f or f = =If (1) 0f =

2 1( 1) ( ) { (1)} nf x f x f ++ = +( 1) ( )f x f x + =(1) (2)f f = ...= 0

(but ( )f x is not identically zero)So (1) 1f =


( 1) ( ) 1 '( 1) '( )f x f x f x f x + = + + =

also 0 0( 1) ( )'(1) lim lim1 1( 1) ( )x x

f x f xfx x +

= = =

+

So (10) '(9) .... '(1) 1f f f= = = =

20 (c) sin cos'( ) n 2(2 cos 2 sin )x xf x l x x=

For '( ) 0f x >sin cos2 cos 2 sin 0x xx x >

(i) sin cos2 tanx x x >3when cos 0, 0, , 22

x x > 2 (ii) sin cos2 tanx x x < 3when cos 0, ,

2 2x x <

So 50, , 24

x 4 (verify from graph given below)

For '( ) 0f x 3when cos 0, ,

2 2x x <

So, 5,

4 4x (verify from graph given below)

Thus ( )f x decreases in 5,

4 4

Use graph

O 4

2

34

54

32

74

2

tan x

2sinx cosn


21 (c) 2 2 24 3 0 4 3 0r x x y x + = + + = ...(i)(i) is a circle with centre (2, 0) and radius

2,0A B

( )3,0(1 ),0O

pr

r is distance of point in circle from originOA r OB

[1,3]r

22 (c) 21xy

a x+

=

+2 1ya yx x + = +

2 ( 1) 0yx x ay + = ...(i)y contains interval [0, 1]for ,x R D of (i) 0

1 4 ( 1) 0y ay 24 4 1 0ay y

Thus 24 4 1 0ay y < or 24 4 1 0ay y =

D of 24 4 1ay y is 16 16 16( 1)a a+ = +

(i) when 21, 0, 4 4 1 0,a D ay y y R So y contains (0, 1)

(ii) for 1 0a following will be graph of 24 4 1,ay y two possibilities as shown in fig.

( )0 1 ( )0 1

For this graph Conditions for y to contain (0, 1) are

1 and (1) 02

b fa

> + >

2 [ 2] [ ] 2y x x x= + = + = + also [ 2] [ ] 2 [ ] 2y x x x= + = + = +

(using [ 2] [ ] 2x x+ = + )

Thus, a0 = 2, for 2, 2 and [ 2]x y x y x> = + = + have same graph

24 (b)2 4

2 2 21

(1 )(1 )....(1 )lim{(1 )(1 )....(1 )]

n

nx

n n nn n n+

+

42

nnC=

tem independent of x is also 4 2n

nCso a = 1

25 (c) This is possible if functions are continous then there sum may remove the cusp from the non differentiable point

eg. ( ) 1f x x= +

( ) 1g x x= +

26 (b)//

df df dxd d dx

=

for L.H.D. / (2 )/ 3

df df dx xd d dx

= =

(using definitions for 1x < )

for R.H.D. 2/ 2/ 3

df df dxd d dx x

= =

(using definitions for 1x > )

At 1,x = L.H.D = R.H.D = 2/3

27 (a) 10 0ydx xdyxydt dt

= + =

given, 51 / ,

2dx dy y dxunit sdt dt x dt

= = = at (2, 5)


28 (a) 1 40

1, put1

dxI xzx

= =

+4 2

1 24 2 20 01 1

z dz z dzI Iz z z

= = =

+ + 2 2

1 2 4 20 0

11121 1 2

x xI I I dx dxx

xx

+ ++ = = =

+ +

22 2 2

I I = =

29 (d) 2 11 ...2 2

+ + +

2 3ln(1 )

2 3x xx x = + + +

Inlegrating2 3 4

(1 ) ln(1 ) (1 )2 6 12x x xx x x = + +

2 3....

2 6 12x x xx

= + + + put

2 12

x =

30 (c) ( ) ( )a t a

t t

f x dx f t z dz x t z+

= + = +

0

( )a

zf z d=

0

( )a

f x dx= 0

0

( ) ( ) ( ) ( )a f t a t

a a t

I f x dx f x dx f x dx f x dx+ +

= = + +

0 0 0

( ) ( ) ( )a t a

f x dx f x dx f x dx= + +

0

( )t

f x dx= Thus, its independent of x

cycle 2 quize 1 solutions.pdf

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