monksays studmonk the app facebook · pdf file4 monksays studmonk – the app facebook...

1 Monksays StudMonk.com StudMonk – The App Facebook

https://monksays.wordpress.com/

http://studmonk.com/

https://play.google.com/store/apps/details?id=com.infojini.studmonk&hl=en

https://www.facebook.com/StudMonk-MHT-CET-1719764334926380/


1 : (A)

P V

V V

C CR0.4 0.4

C C

P

V

C1 0.4

C

P

V

C1.4 r 1.4

C

As r 1.4 the gas is diatomic in nature. For example air molecules.

2 : (D)

21K.E mv

2

At lowest point in vertical circular motion. LV 5rg and at highest point hV rg

h

L

(K.E) 10.2

(K.E) 5

3 : (B)

Strain energy per unit volume U 1

u Stress Strainv 2

2U 1 F 1

u strainV 2 A 2

2 2

s s L

L L s

u Y Strain (S) A

u Y Strain (L) A

2 2

L2 2S

r 39 :1

r 1






4 : (A)

Total K.E of the rolling disc or ring is given by.

2 21 1K.E mv I

2 2

For ring and disk, translational kinetic energy 21mv

2 is constant.

Rolling K.E. of disc is 2 21mR

4

Rolling K.E of ring is 2 21mR

2

As for ring, 2 2 21 14J mv mR

2 2

2 2mR 4J

For disc

2 2 2 21 1 4 4mR mR J (2 1)J 3J

2 4 2 4

5 : (B)

1

2

f2

f, Speed of approach = Speed of leaving

The apparent frequency of sound level by observer when it is approaching source is given by

1 0

1

Vf f

V V … (1)

When observer is moving away from source

2 0

1

Vf f

V V … (2)

Here f0 is the frequency of sound main

Taking ratio’s of (1) with (2) we have






1 1 1

2 1 1

f V V V VV

f V V V V V

1

1

V V2

V V

1 12V 2V V V

12V V 3V

6 : (A)

For wire vibrating under tension, the fundamental frequency is given by

1

1 Tf

2 , Where T is tension & r site mass attached to the string.

It is given that, when it is sounded with a tuning fork of frequency x, 6 beats per seconds were heard

(x f) 6 12

1 1

1 1 225f f

2L 2L

2

1 256f

2L

1

2

f 15

f 16

1 1

16f f

15

2

16f (6 x)

15

2f (x 6) 12

1

V3

V






16

(x 6) (x 6)15

15x 90 16x 96

X 186 Hz

7 : (B)

The pressure exerted by the gas on the walls of container is 0 1 2P P P P

i.e. 20

1P P 8v 3gh

3

For a container, 2 21

1 1 m 2 2P v v K.E.

3 3 V 2 3

8 : (A)

P.E of the oscillating mass is given by

2 21 1E m x kx

2 2 Where k = 2m

When only black is oscillating, at x = A

E = Emax 21m A

2 and at mean position i.e. x = 0

E = 0

1

Am

When mass m2 is placed on top of the mass m1

Then, total mass is (m1 + m2) and E = 0 at this point, as x = 0.

When the combination reaches x = A,

Then

1

1 2

1A

m m






1 1

1 2

A m

A m m

9 : (A)

Using 2 21v u 2as and u u and s 4 r

2 2 22 v v v

2as v a2s 2 4 r 8 r

10 : (C)

The velocity of wave travelling on string is

T Tv n

2L

1 Tn

2L

2

T TV v

m m

TY

A L

T YA L

Y A LV

m

V A (A is area)

For string A, radius is 2r, and for string B, radius is r

2

A2

B

V 4r4 2

A r

Y is same for both and in is same for both.






11 : (A)

eff

T 2g

In air geff = g where as in water

3 3

eff3

910 10

8g g 9.89

108

91

1 889.8 9.89 8 98

9.8

9

2

eff

9T 2 2

g g

2 1 1T 9 T 3T

12 : (C)

2T cosh

g(R r)

For cos = 0 , 2T

hg(R r)

13 : (D)

Let 00

vf

2L be the fundamental frequency of the open pipe

is second overtone is 00

3v3f

2 2L

Let 0vf

4L be the fundamental frequency of close pipe.






And third overtone of the organ pipe closed at one end only is, (f3) closed = 0v7

4 L

As given 0 07v 3V150

4L 2L

0v7 6150

4 4 L

0v150

4L

0v300 Hz

2L

14 : (A)

Volume of disc is

3

2 R RA d R

6 6

Moment of inertia of disc is 21MR

2

When the disk is remolded in solid sphere of volume V lowing radius r, then

3 33 3R 4 R 3

r r6 3 6 4

3

3 R Rr r

8 2

Moment of inertia of sphere is given by 22m r

5

2 2 22 R MR MR 1m

5 4 10 2 5 5

15 : (B)

The sag of bending of beam is given by.

3WL

48 Y

Where W is load, L is length, Y is young’s modules and is

3bd

12 (area moment of inertia)






3 3

3 3

WL 12 WL

48Y bd 4bd Y

16 : (C)

is the initial phase. When oscillations of pendulum are damped, doesn’t change

17 : (B)

For open organ pipe

L 1.2 r

1 0.8 2

r L1.2 1.2 3

cm.

18 : (C)

y 12 m (5t 4y) .

Comparing in y = A sin (wt – kx)

We have A = 12, w = 5 and k = 4

(wt – kx) = phase difference

2

5t 4x

2

When t = 0,

4x2

x cm

8

19 : (C)

For the system described above






U GMU m V V

m X

Here M = 16 m and x = r

G

16 GMV

r

20 : (D)

4 40E e A(T T ) and A b

When and b changes to 3

is and b

3

A

A9

4

4

E' A' (327 373)

E A (27 273)

4E' 1 600

E 9 300

41E' (2) E

9

16 E

E'9

21 : (C)

The surface area of the liquid drop is 2A 4 R

Its surface energy is E

When the drop splits in 512 droplets, the surface area of each droplets is 22A 512 r

2512 4 r






The volume of bigger drop is 34R

3 and Volume of small droplets is 34

512 r3

3 34 4 RR 512 r r

3 3 8

2

22 1

RA 512 4 r 512 4 8A

8

Change in surface area is

22

2 1

512 RA A 4 R

64 2 24 (8R R ) 27R

Surface energy E = A . T (T is surface tension and A is area)

n 2 1

0 1 1

E A T 8 A8

E A T A

nE 8E

22 : (B)

M.I of rod whose axis of rotation is passing through center and perpendicular to the plane of rod is

2ML

12 and 2

1MK (where K.J radius of gravitation)

21 1

LMK K

2 3 …(1)

When axis of rotation of rod is passing through one and of rod, then

2

22 2

ML LMK K

3 3 …(2)

Taking ratios of (1) and (2) we get

1

2

K L 3 1

K L 22 3 1

2

K 1

K 2

23 : (C)






When bob is at rest, the pendulum has only potential energy which is given as P.E mg

When bob displaces by small angler displacement , the pendulum loses P.E of bob and it gets

converted to K.E.

Thus loss in P.E = gain in K.E

K.E mg mg cos mg (1 cos )

24 : (C)

c o2 360 1 1deg/ sec

t 12 3600 12 3600 12 10 120

25 : (A)

2R

g' gR h

When g

g'4

then.

2g R 1 R

g4 R h 2 R h

2R R h R h

26 : (A)

For potentiometer, when null point is obtained for a particular cell (EV) at L cm, (say).

Whose length is x cm E = L•V volts

When the length is increased from x to x’ then

E’ = L’ • V volts.

The balancing length increases when if the length of the potentiometer wire increases.

Hence (A) is correct choice.






27 : (C)

43

4 3

6 10 1B H H 10

A 3 10 2 10

28 : (B)

Peak value of current

0 rms

2I I 2 Amp

Co-efficient of mutual inductance is M = 1 horn

Induced emf in secondary is given by

2

die M•

dt where 0i i sin( t )

Here 2 n 100 cos n 50 )

2 0

de 1 (i sin t)

dt

0

2i cos( t) 2 50 cos(100 t)

For, t 0 , we have 2e 4 50 200V .

29 : (C)

When electron or any charged particle is accelerated through potential difference v, then kinetic energy

gained is given by E = eV … (1)

2 220 2

1 p hE mv

2 2m 2m … (2)

2

2

h hev

2m 2meV (3)

When proton of mass M is accelerated through P.d of gv, then the de-Broglie wavelength obtained is

h h m'

2M e gv 3 2MeV m






m'

3 M

30 : (C)

At point P, resultant intensity of interfering wave is

p res 1 2 1 2( ) 2 cos

For p res, ( ) 9 102

At point Q, resultant intensity of interfering wave is

Q res(I ) 210 2 9 10 6 4 for

p res g res( ) ( ) 10 10 6 6

31 : (B)

0A

C QV and Cd

As given

0 0 01 2 3

A A AC ; C ; C

3d 6d gd

eq 1 2 3C C C C C for parallel combination of capacitors

0 0eq

A 11 A1 1 1C

d 3 6 9 18d

32 : (A)

For sustained oscillations according to Barkhausen criteria. A 1

33 : (A)

According to photoelectric equation






hcE where

2pE (K.E)

2m

If E is constant then

1

If we decrease the wavelength , then stopping potential will increase such that

hc = constant.

34 : (B)

As at interface, velocity of light reduces by 20 % of C

20 i20% i i

100 5

35 : (C)

For ionosphere, the maximum frequency of radio waves that can be reflected back is given by g N .

Where N is maximum electron density of ionosphere and g is acceleration due to gravity.

36 : (A)

According to Bohr’s theory the wavelength of radiation emitted is given by

2 2

1 1 1R

ne ne

For ni = 5 and nf = 4, the wavelength emitted will be longest as energy difference 5 4E E E is very

small.

37 : (B)

When capacitors are connected is series

1 2eq

1 2

C CC

C C and the potential areas plates of capacitors is

given by Q

VC






If we enter the dielectric material of constant is in capacitor say C2 as shown then, the equivalent

capacitance of

02

AC k K.G.

d

1 2 1 1 1eq

1 2 1 1

C C C kC kCC

C C C C 2

The potential plates of C2 will be

2

2 1

Q QV

C kC

As there is only air between the plates of capacitor C1 and (R) = 1 for air

1

1

QV

C

2 eff

1 1

Q 1 f1 k 1 Q(V )

C k 1 k C

21

k 1V

k

1 2

kV V

k 1

If 2V V then

1

K.VV

k 1

38 : (A)

LC parallel resonant circuit has very high impedance.

39 : (C)

3

eff

V V 2i 1 10 A 1mA

R 1790 30 2000

This current provides full scale deflection (i.e. 20 division) In order to limit the deflection 10 divisions,

the resistance needed to connect such that the current reduces can be obtained as

1 1

2 2

iniAB

k i






1 2: 2

1 2i : i 2

341

2

i 1 10i 5 10 mA

2 2

eff

eff

v Vi Rs R

R Rs i

4

22000

5 10

4210 2000

5

34 10 2000

2000

The resistance of 1970 is to be replaced by 1970 + 2000 = 3970

40 : (B)

Here x2

For n = 4th dark bone (2n 1) (8 1) 9

7 69 9 9x 6 10 2.7 10 m

2 2 2

42.7 10 cm

41 : (B)

42 : (D)

The work done in increasing the potential is given by

2 22 1W dU (V V )






2 21 2 1

2 22 3 2

W (V V )

W (V V )

here 1 2 3V 5v, V 10V and V 15V

2

W 100 25 75

W 225 100 125

2W 1.67 W

43 : (C)

41 4 10 wb and 5

2 10.1 4 10 wb

0.72 Hz

t ?

Using

d ddt

dt

4 5 5

5

4 10 4 10 4 10 (9) 10.5 sec

72 10 72 2

44 : (A)

Resolving power of telescope is given by

dP

1.22

When decreases, resolving power decreases and vice versa

45 : (D)

Using hc

eV






When, wavelength is incident on metallic surface, the stopping potential required to stop most

energetic electron is V and when wavelength increases to 3 , stopping potential required as V

6

hceV ... (1)

And hc eV

... (2)3 6

Thus equation (2) can be rewritten as 2hc

eV 6 ...(3)

Subtracting equation (1) from (3) we get

hc5

and

o

hc

Thus o 5

46 : (A)

According to Brewster’s law, polarizing angle depends on wavelength and is different for different

colors.

47 : (A)

The force acting on the particle inside magnetic field is BF qvB sin

This provides the necessary centripetal force 2

C

mvF

r

2mv mv

qvB rr qB

x x x 1 x 1

y y y 2 y 2

r m v r m v

r m v r m v

For x 11 2

y 2

m rv : v 1,

m r






48 : (A)

For an electron revolving in stationary orbits it doesn’t radiate light through is velocity change.

49 : (D)

0B ni (1 )

50 : (A)

1 1

2 2

v2

3 v

Now 1 2v v v 5V QV IR

For potentiometer wire 1 1

2 2

R 2

R 3

x 2 x4

6 3 m

The resistance of wire is

0.1 1

cm m

eff(x 1) R 5

eff

V 5i 1 A

R 5





monksays studmonk the app facebook · pdf file4 monksays studmonk – the app facebook...

Documents