Physics NEET-2017 With Solution - C7 REPORTturningpointkota.com/pdf/NEET2017QuestionPaperAnswer... · 2017-05-07 · Code-W Answer & Solution Physics NEET-2017 TURNING POINT Corporate

Code-W Answer & Solution Physics NEET-2017

TURNING POINT

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1

91. Two blocks A and B of masses 3m and m respectively are connected by a massless andinextensible string. The whole system is suspended by a massless spring as shown in figure.The magnitudes of acceleration of A and B immediately after the string is cut, arerespectively.

A

B

3m

m

(1)g

g,3

(2)g

,g3

(3) g, g (4)g g

,3 3

Sol. [2]

3m

m

Before

kx = 4 mg

3mgT = mg

T = mg

mg

3m

m

After

kx = 4 mg

3mgT = 0

T = 0

mg

3m

4mg 3mga g / 3

3m

−⇒ = = ↑

m

mga g

m⇒ = = ↓

92. The acceleration due to gravity at a height 1 km above the earth is the same as at a depth dbelow the surface of earth. Then:

(1)1

d km2

= (2) d = 1 km (3)3

d km2

= (4) d = 2 km

Sol. [4]

Outside the earth, 0

2hg g 1

R

= −

Inside the earth, 0

dg g 1

R

= −

for same value of g,2h d

R R=

⇒ d 2h 2km= =

93. A particle executes linear simple harmonic motion with an amplitude of 3 cm. When theparticle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of itsacceleration. Then its time period in seconds is:

(1)5

π(2)

5

2π(3)

4

5

π(4)

2

3

π

Sol. [3]At distance x from mean position,

2 2 2A x xω − = ω


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2

⇒ 2 2 2 2A x x− = ω

⇒2 2 2 2

2 2

A x 3 2 5

x 2 2

− −ω = = =

⇒ 2 4T sec

5

π π= =ω

94. The resistance of a wire is 'R' ohm. If it is melted and stretched to 'n' times its original length,its new resistance will be:

(1) nR (2)R

n(3) n2R (4)

2

R

nSol. [3]

Volume of wire remain same, thusV

A =

Or,2

RA V

= ρ = ρ

⇒ 2R ∝ or 2R ' n R=

95. A capacitor is charged by a battery. The battery is removed and another identical unchargedcapacitor is connected in parallel. The total electrostatic energy of resulting system:(1) increases by a factor of 4 (2) decreases by a factor of 2(3) remains the same (4) increase by a factor of 2

Sol. [2]

Initial energy of system, Ui = 21CV

2

V

C+++

–––

when another capacitor is connected in parallel,C+

++

–––

C+++

–––

V'

V'

CV 0 VV '

C C 2

+= =+

final energy of system, 2f

1U (2C)V '

2= 21

CV4

=


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96. Two rods A and B of different materials are welded together as shown in figure. Theirthermal conductivities are K1 and K2. The thermal conductivity of the composite rod will be:

A K1

B K2

T2T1

d

(1) 1 2K K

2

+(2) 1 23(K K )

2

+(3) 1 2K K+ (4) 1 22(K K )+

Sol. [1]Since rods are in parallel, then

1 2

1 1 1

R R R= +

⇒ 1 2K(2A) K A K A

d d d= +

⇒ 1 2K KK

2

+=

97. The two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and260 Hz. What is the fundamental frequency of the system.(1) 10 Hz (2) 20 Hz (3) 30 Hz (4) 40 Hz

Sol. [2]For closed organ pipe, difference in successive frequencies is equals to twice of fundamentalfrequency. Then, 0 02f 40 f 20Hz= ⇒ =

98. The bulk modulus of a spherical object is 'B'. If it is subjected to uniform pressure 'p', thefractional decrease in radius is:

(1)P

B(2)

B

3p(3)

3p

B(4)

P

3B

Sol. [4]

Since,P V P P

B orV / V V B B

∆ −∆ ∆= = =

−∆

r 1 V P

r 3 V 3B

∆ ∆− = − =

99. A physical quantity of the dimensions of length that can be formed out of c, G and2

0

e

4π ∈is

[c is velocity of light, G is universal constant of gravitation and e is charge]:

(1)

1/ 22

20

1 eG

c 4

π ∈

(2)

1/ 22

2

0

ec G

4

π ∈

(3)

1/ 22

20

1 e

c G4

π ∈

(4)2

0

1 eG

c 4π ∈Sol. [1]

r2

p q

0

eL K[c] [G]

4

= πε

⇒p q r0 1 0 1 1 1 3 2 1 3 2M LT LT M L T M L T− − − − =

⇒ q r 0− + = ⇒ q r=p 3q 3r 1+ + = ⇒ 6q = –p + 1

and p 2q 2r 0− − − = ⇒ p 4q= −


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By solving them, p = –2, q =1

2and

1r

2=

Thus,2

20

1 GeL K

C 4=

πεwhere K is a dimensionless constant

100. Figure shows a circuit that contains three identical resistors with resistance R = 9.0 each,two identical inductors with inductance L = 2.0 mH each, and an ideal battery with emf =18 V. The current 'i' through the battery just after the switch closed is....

+

–

R

CL

R

R

L

(1) 2 mA (2) 0.2 A (3) 2 A (4) 0 ampereSol. [Bonus]

+

–

R

CL

R

R

L

when switch is closed, inductor provides infinite resistance and capacitor provides zeroresistance.

+

–

RR

Thus, current through battery,2E

IR / 2 R

ε= = = 4A

101. One end of string of length l is connected to a particle of mass 'm' and the other end isconnected to a small peg on a smooth horizontal table. If the particle moves in circle withspeed '', the net force on the particle (directed towards center) will be (T represents thetension in the string)

(1) T (2)2m

Tυ+

(3)2m

Tυ−

(4) Zero

Sol. [1]

Net force on the particle =2mv

T =

102. The photoelectric threshold wavelength of silver is 3250 × 10–10 m. The velocity of theelectron ejected from a silver surface by ultraviolet light of wavelength 2536 × 10–10 m is:(Given h = 4.14 × 10–15 eVs and c = 3 × 108 ms–1)

(1) 5 16 10 ms−≈ × (2) 6 10.6 10 ms−≈ × (3) 3 161 10 ms−≈ × (4) 6 10.3 10 ms−≈ ×Sol. [1], [2]


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5

0

hc 124003.81 eV

3250φ = = =

λKinetic energy of electrons,

hcK = − φ

λ1912400

3.31 1.08 eV 1.73 10 J2536

−= − = = ×

⇒19

6 5

31

2 1.73 10V 0.6 10 m/s 6 10 m/s

9.1 10

−

−

× ×= = × = ××

103. Radioactive material 'A' has decay constant '8 ' and material 'B' has decay constant ''.Initially they have same number of nuclei. After what time, the ratio of number of nuclei of

material 'B' to that 'A' will be1

e?

(1)1

λ(2)

1

7λ(3)

1

8λ(4)

1

9λSol. [2]

After time t, AtA 0N N e−λ=

& BtB 0N N e−λ=

Since, B

A

Ne

N=

⇒B

A

t

t

ee

e

−λ

−λ =

⇒ ( )A B t 1λ − λ =

⇒A B

1 1t

7= =

λ − λ λ

104. A rope is would around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angularacceleration of the cylinder if the rope is pulled with a force of 30 N?(1) 25 m/s2 (B) 0.25 rad/s2 (C) 25 rad/s2 (D) 5 m/s2

Sol. [3]30 N

2

2

FR F 3025rad/s

I mR mR 3 0.4

τα = = = = =×

105. Two cars moving in opposite directions approach each other with speed of 22 m/s and 16.5m/s respectively. The driver of the first car blows a horn having a frequency 400 Hz. Thefrequency heard by the driver of the second car is [velocity of sound 340 m/s]:(1) 350 Hz (2) 361 Hz (3) 411 Hz (4) 448 Hz

Sol. [4]VS VO

O

S

V V 340 16.5f ' f 400

V V 340 22

+ + = = − −


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356.5400 448Hz

318= × =

106. A 250 - Turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 Aand subjected to a magnetic field of strength 0.85 T. Work done for rotating the coil by 180°against the torque is:(1) 9.1 J (2) 4.55 J (3) 2.3 J (4) 1.15 J

Sol. [1]

( )W MB cos0 cos180 2 MB= ° − ° =6 42(NBIA) 2 250 0.85 85 10 2.1 1.25 10− −= = × × × × × × ×

69.48 10 J 9.5 J−= × = µ

107. A long solenoid of diameter 0.1 m has 2 × 104 turns per meter. At the centre of the solenoid, acoil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis.The current in the solenoid reduces at a constant rate to 0A from 4A in 0.05 s. If theresistance of the coil is 10 2, the total charge flowing through the coil during this time is:(1) 32 C (2) 16 C (3) 32 C (4) 16 C

Sol. [3]Magnetic field inside the solenoid, B = 0nI

7 4 34 10 2 10 I 8 10 I− −= π × × × × = π × ×Charge flowing through the ring,

21 2N(B B ) r

qR R

∆φ − π∆ = =

3 4

2

100 8 10 (4 0) 10

10

− −× π × − × π ×=π

32 C= µ

108. Suppose the charge of a proton and an electron differ slightly. One of them is –e, the other is(e + e). If the net of electrostatic force and gravitational force between two hydrogen atomsplaced at a distance d (much greater than atomic size) apart is zero, then e is of the order of[Given mass of hydrogen mh = 1.67 × 10–27 kg]

(1) 2010 C− (2) 2310 C− (3) 3710 C− (4) 4710 C−

Sol. [3]

–ee+e

–ee+e

2 2 2

2 2 2 2

2Ke(e e) Ke K(e e) Gm

r r r r

+ ∆ + ∆− + + =

⇒2

2 2e eKe 2 1 1 1 Gm

e e

∆ ∆ − + + + + =

⇒2

2 2e e 2 eKe 2 2 1 1 Gm

e e e

∆ ∆ ∆ − − + + + + =

⇒2

2 2

2

eKe Gm

e

∆× =


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⇒( )211 272

379

6.67 10 6.67 10Gme 5.7 10 C

K 9 10

− −−

× × ×∆ = = = ×

×109. Two astronauts are floating in gravitational free space after having lost contact with their

spaceship. The two will:(1) keep floating at the same distance between them(2) move towards each other(3) move away from each other(4) will become stationary

Sol. [1]They have same velocity when separates from spaceship, thus distance will remain same.

110. The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is:(1) 2 (2) 1 (3) 4 (4) 0.5

Sol. [3]

Last line of Balmer series,2 2

1

1 1 1R

2

= − λ ∞

Last line of Lyman series,2 2

2

1 1 1R

1

= − λ ∞

1

2

4λ⇒ =λ

111. The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at atemperature T (Kelvin) and mass m, is:

(1)h

mkT(2)

h

3mkT(3)

2h

3mkT(4)

2h

mkTSol. [2]

2m(KE) 3mKT32m KT

2

λ λ λλ = = =

112. A thin prism having refracting angle 10° is made of glass of refractive index 1.42. This prismis combined with another thin prism of glass of refractive index 1.7. This combinationproduces dispersion without deviation. The refracting angle of second prism should be:(1) 4° (2) 6° (3) 8° (4) 10°

Sol. [2]for dispersion without deviation,( 1)A ( ' 1)A ' 0µ − + µ − =

⇒ 1 0.42A ' A 10 6

' 1 0.7

µ −= − = × ° = ° µ −

113. Thermodynamic processes are indicated in the following diagram.


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8

300 K500 K700 K

fi IV

IIIfII

I

f

f

P

V

Match the following :Column-1 Column-2

P. Process I a. AdiabaticQ. Process II b. IsobaricR. Process III c. IsochoricS. Process IV d Isothermal(1) P a, Q c, R d, S b (2) P c, Q a, R d, S b(3) P c, Q d, R b, S a (4) P d, Q b, R a, S c

Sol. [2]Theory based

114. A U tube with both ends open to the atmosphere, is partially filled with water. Oil, which isimmiscible with water, is poured into one side until it stands at a distance of 10 mm above thewater level on the other side. Meanwhile the water rises by 65 mm from its original level (seediagram). The density of the oil is:

10 mmFinal water level

Initial water level

C

Water

B

Oil

A

Pa Pa

F

E

D65 mm

65 mm

(1) 3650kg m− (2) 3425kg m− (3) 3800kg m− (4) 3928kg m−

Sol. [4]Pressure at B = Pressure at C

0 0 0 wP g(140) P g(130)+ ρ = + ρ

30

131000 928kg / m

14⇒ ρ = × =

115. A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. They are connected in seriesand the new force constant is k'. Then they are connected in parallel and force constant is k".Then k' : k" is:(1) 1 : 6 (2) 1 : 9 (3) 1 : 11 (4) 1 : 14

Sol. [3]

1 2 3K '(6L) K (L) K (2L) K (3L)= = =⇒ 1 2 3K 6K ', K 3K ' & K 2K '= = =when connected in parallel,

1 2 3K" K K K 11K '= + + =


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⇒ K ' 1

K" 11=

116. Which of the following statements are correct?(a) Centre of mass of a body always coincides with the centre of gravity of the body.(b) Centre of mass of a body is the point at which the total gravitational torque on the body iszero.(c) A couple on a body produce both translational and rotational motion in a body.(d) Mechanical advantage greater than one means that small effort can be used to lift a largeload.(1) (b) and (d) (2) (a) and (b) (3) (b) and (c) (4) (c) and (d)

Sol. [1]Theory based

117. A beam of light from a source L is incident normally on a plane mirror fixed at a certaindistance x from the source. The beam is reflected back as a spot on a scale placed just abovethe source L. When the mirror is rotated through a small angle the spot of the light is foundto move throught a distance y on the scale. The angle is given by:

(1)y

2x(2)

y

x(3)

x

2y(4)

x

y

Sol. [1]

2

y

Reflected beam rotates by angle 2 when mirror is rotated by angle .

Since,y y

2x 2x

θ = ⇒ θ =

118. A gas mixture consists of 2 moles of O2 and 4 moles of Ar at temperature T. Neglecting allvibrational modes, the total internal energy of the system is:(1) 4 RT (2) 15 RT (3) 9 RT (4) 11 RT

Sol. [4]

Since,f

E nRT2

=

5 3E (2)(R)T (4)(R)T

2 2= +

11RT=

119. Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits theground with a speed of 50 m/s. Take 'g' constant with a value 10 m/s2. The work done by the(i) gravitational force and the (ii) resistive force of air is:(1) (i) –10 J (ii) –8.25 J (2) (i) 1.25 J (ii) –8.25 J(3) (i) 100 J (ii) 8.75 J (4) (i) 10 J (ii) –8.75J

Sol. [4]


TURNING POINT


10

Work done by gravity,3 3

gw mgh 10 10 10 10J−= = × × =by work energy theorem,

2g R

1w w mv

2+ =

⇒ 3 2R

1w (10 )(50) 10

2−= −

1.25 10 8.75J= − = −

120. A carnot engine having an efficiency of1

10as heat engine, is used as a refrigerator. If the

work done on the system is 10 J, the amount of energy absorbed from the reservoir at lowertemperature is:(1) 1 J (2) 90 J (3) 99 J (4) 100 J

Sol. [2]

Since,

111 10 9

1

10

−− ηβ = = =η

2 22

Q Q9 Q 90J

work done 10β = ⇒ = ⇒ =

121. An arrangement of three parallel straight wires placed perpendicular to plane of papercarrying same current 'I' along the same direction is shown in Fig. Magnitude of force perunit length on the middle wire 'B' is given by:

d CB

A

d

90°

(1)2

0i

2 d

µπ

(2)2

02 i

d

µπ

(3)2

02 i

d

µπ

(4)2

0i

2 d

µπ

Sol. [4]F CB

A

F

totalF 2 F=2

0I22 d

µ=π

20I

2 d

µ=π


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11

122. The x and y coordinates of the particle at any time are x = 5t – 2t2 and y = 10t respectively,where x and y are in meters and t in seconds. The acceleration of the particle at t = 2s is:(1) 0 (2) 5 m/s2 (3) –4 m/s2 (4) –8 m/s2

Sol. [3]dx dyˆ ˆ ˆ ˆV i j (5 4t)i 10 jdt dt

= + = − +

anddv ˆa 4idt

= = −

or 2a 4 m/s=

123. The ratio of resolving powers of an optical microscope for two wavelengths 1 = 4000 Å and2 = 6000 Å is:(1) 8 : 27 (2) 9 : 4 (3) 3 : 2 (4) 16 : 81

Sol. [3]Resolving power of microscope ∝ λ

the ratio of resolving power = 2

1

6000 3

4000 2

λ = =λ

124. Preeti reached the metro station and found that the escalator was not working. She walked upthe stationary escalator in time t1. On other days, if she remains stationary on the movingescalator, then the escalator takes her up in time t2. The time taken by her to walk up on themoving escalator will be:

(1) 1 2t t

2

+(2) 1 2

2 1

t t

t t−(3) 1 2

2 1

t t

t t+(4) t1 – t2

Sol. [3]When escalator is not working

1

1

dV

t= ...(1)

when she stand and escalator in moving.

2

2

dV

t= ...(2)

when they both moves upward

( )2 1

dV V

t '+ = ...(3)

from eq. (1), (2) and (3)

⇒2 1

d d d

t t t '+ =

⇒ 1 2

2 1

t tt '

t t=

+

125. A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radiuswere halved and the temperature doubled, the power radiated in watt would be :(1) 225 (2) 450 (3) 1000 (4) 1800

Sol. [4]


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12

As power ∝ 4 2T r×4 2

1 1 14 2

2 2 2

P T r

P T r=

∴4 2

2

2 4

450 T r

P r(2T)

2

=

2P 1800 watt=

126. A potentiometer is an accurate and versatile device to make electrical measurements ofE.M.F. because the method involves:(1) cells(2) potential gradients(3) a condition of no current flow through the galvanometer(4) a combination of cells, galvanometer and resistances

Sol. [3]Conceptual

127. The given electrical network is equivalent to:YA

B

(1) AND gate (2) OR gate (3) NOR gate (4) NOT gateSol. [3]

AB

y1 y1 y1 = y1

1y A B= +

128. In a common emitter transistor amplifier the audio signal voltage across the collector is 3 V.The resistance of collector is 3 K. If current gain is 100 and the base resistance is 2 k, thevoltage and power gain of the amplifier is:(1) 200 and 1000 (2) 15 and 200 (3) 150 and 15000 (4) 20 and 2000

Sol. [3]

Voltage gain Av = C

B

R

R

β

Av =3

100 1502

× = Power gain = Voltage gain × Current gain

= C

B

R

R

β × β

=3

10000 150002

× =

129. Two discs of same moment of inertia rotating about their regular axis passing through centreand perpendicular to the plane of disc with angular velocities 1 and 2. They are brought


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13

into contact face to face coinciding the axis of rotation. The expression for loss of energyduring this process is:

(1) 21 2

1I( )

2ω + ω (2) 2

1 2

1I( )

4ω − ω (3) 2

1 2I( )ω − ω (4) 21 2

1I( )

8ω − ω

Sol. [2]As angular momentum in conserved therefore

i fL L=

1 2I I 2I 'ω + ω = ω

∴ 1 2'2

ω + ωω =

Intial KE is 2 21 2

1 1I I

2 2ω + ω

Final KE is2

1 21(2I)

2 2

ω + ω

loss = intial KE – final KE2

2 2 1 21 2

1I( ) I

2 2

ω + ω = ω + ω −

21 2

1I( )

4= ω − ω

130. Young's double slit experiment is first performed in air and then in a medium other than air. Itis found that 8th bright fringe in the medium lies where 5th dark fringe lies in air. Therefractive index of the medium is nearly:(1) 1.25 (2) 1.59 (3) 1.69 (4) 1.78

Sol. [4]

m a

98

2β = β

⇒ m aD 9 D8

d 2 d

λ λ× = ⋅

⇒ a

m

16

9

λ =λ

or a

m

161.78

9

λµ = = =λ

131. Which one of the following represents forward bias diode?

(1)R –2V0V

(2)R –3V–4 V

(3)R +2V–2V

(2)R 5V3V

Sol. [3]Zero volt is higher potential than –2 volt.

132. Two Polaroids P1 and P2 are placed with their axis perpendicular to each other. Unpolarisedlight I0 is incident on P1. A third polaroid P3 is kept in between P1 and P2 such that its axismakes an angle 45° with that of P1. The intensity of transmitted light through P2 is :


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14

(1) 0I

2(2) 0I

4(3) 0I

8(4) 0I

16Sol. [3]

45°

I0 I1 I2 I3

20 01 2 1

I II & I I cos 45

2 4= = ° =

and 2 03 2

II I cos 45

8= ° =

133. In an electromagnetic wave in free space that root mean square value of the electric field isErms = 6V/m. The peak value of the magnetic field is:(1) 1.41 × 10–8 T (2) 2.83× 10–8 T (3) 0.70 × 10–8 T (4) 4.23 × 10–8 T

Sol. [2]

8rmsrms 8

E 6B 2 10 T

C 3 10−= = = ×

×8

0 rmsB 2 B 2.83 10 T−= = ×

134. If 1θ and 2θ be the apparent angles of dip observed in two vertical planes at right angles to

each other, then the true angle of dip θ is given by:

(1) 2 2 21 2cot cot cotθ = θ + θ (2) 2 2 2

1 2tan tan tanθ = θ + θ(3) 2 2 2

1 2cot cot cotθ = θ − θ (4) 2 2 21 2tan tan tanθ = θ − θ

Sol. [1]

1 2

tan tantan & tan

cos cos(90 )

θ θθ = θ =

α − αwhere, is actual dip angle

⇒ 2 2cos sin 1α + α =

⇒2 2

2 21 2

tan tan1

tan tan

θ θ+ =θ θ

⇒ 2 2 21 2cot cot cotθ + θ = θ

135. The diagrams below show regions of equipotentials40 V20 V

BA

40 V20 V

BA

30 V10 V

30 V10 V

BA

40 V20 V30 V

10 V

20 V40 V

BA

10 V 30 V

A positive charge is moved from A to B in each diagram.(1) Maximum work is required to move q in figure (c)


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15

(2) In all the four cases the work done is the same(3) Minimum work is required to move q in figure (a)(4) Maximum work is required to move q in figure (b)

Sol. [2]Since, B AW q V q(V V ) q(40 10) 30q= ∆ = − = − =work done is same in all the four cases.

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Physics NEET-2017 With Solution - C7 REPORTturningpointkota.com/pdf/NEET2017QuestionPaperAnswer... · 2017-05-07 · Code-W Answer & Solution Physics NEET-2017 TURNING POINT Corporate