Previous Year Question Papers NEET 2016 Phase II

13863.A person can see clearly object only when they lie between 50 cm and 400 cm from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens, the person has to use will be

convex, +0.15 diopter

convex, +2.25 diopter

concave, –0.25 diopter

concave, –0.2 diopter

Explanation:

$\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f}$

$\dfrac{1}{–4m} – \dfrac{1}{\infty} = \dfrac{1}{f}$

f = – 4m

power = $\dfrac{1}{f} = \dfrac{1}{–4}$ = – 0.25D

13864.A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 × 10^–5 cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is

0.15 cm

0.10 cm

0.25 cm

0.20 cm

Explanation:

Position of 1^st minima

$y = \dfrac{\lambda D}{a} = \dfrac{(5 × 10^{–8})(0.6)}{0.02 × 10^{–2}}$ = 0.15 cm.

13865.Electrons of mass m with de–Broglie wavelength λ fall on the target in an X–ray tube. The cutoff wavelength (λ₀) of the emitted X–ray is

$\lambda_0 = \lambda$

$\lambda_0 = \dfrac{2mc\lambda^2}{h}$

$\lambda_0 = \dfrac{2h}{mc}$

$\lambda_0 = \dfrac{2m^2c^2\lambda^3}{h^2}$

Explanation:

K.E. of electrons = $\dfrac{P^2}{2m}= \dfrac{\left(\dfrac{h}{\lambda}\right)^2}{2m} = \dfrac{h^2}{2m\lambda^2}$

So maximum energy of photon will also be this much.

$\dfrac{hc}{\lambda_0} = \dfrac{h^2}{2m\lambda^2}$

$\lambda_0 = \dfrac{2mc\lambda^2}{h}$

13866.Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is

–3 V

+3 V

+4 V

–1 V

Explanation:

$k_{max} = h\nu – \phi$

$2eV = 5eV – \phi$

$\phi = 3eV$

So V_st = 3 volt

V_cathode – V_anode = 3 volt

V_anode – V_cathode = – 3 volt

13867.If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength λ. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be

$\dfrac{20}{13}\lambda$

$\dfrac{16}{25}\lambda$

$\dfrac{9}{16}\lambda$

$\dfrac{20}{7}\lambda$

Explanation:

$\dfrac{1}{\lambda} = Re\left(\dfrac{1}{2^2}– \dfrac{1}{3^2}\right)$

$\dfrac{1}{\lambda^{}} = Re\left(\dfrac{1}{3^2}– \dfrac{1}{4^2}\right)$

Dividing $\lambda^{} = \dfrac{20}{7}\lambda$

13868.The half–life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is

Explanation:

N₁ = 0.6 N₀

N₂ = 0.15 N₀

$\dfrac{N_2}{N_1} = \left(\dfrac{1}{2}\right)^2$ so two half–life period has passed

So, time taken = 2t_1/2 = 2 × 30 = 60 minutes.

13869.For CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 4 V. If the current amplification factor of the transistor is 100 and the base resistance is 1 kΩ, then the input signal voltage is

15 mV

10 mV

20 mV

30 mV

Explanation:

A_v = $\beta \ \dfrac{R_{out}}{R_{in}}$

A_v = 100 × $\dfrac{2kΩ}{1kΩ}$

A_v = 200.

A_v = $\dfrac{(V_{out})_{AC}}{(V_{in})_{AC}}$

$(V_{in})_{AC} = \dfrac{4}{200}$ = 0.02 = 20 mV

13870.The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance R₁ will be Diagram on diodes D1 and D2

3.13 A

2.5 A

10.0 A

1.43 A

Explanation:

The diode D₁ will be be in reverse bias, so it will block the current and diode D₂ will be in forward bias, so it will pass the current

i = $\dfrac{10}{2 + 2}$ = 2.5 A

13871.What is the output Y in the following circuit, when all the three inputs A, B, C are first 0 and then 1? Circuit diagram - Find output Y

1, 1

0, 1

0, 0

1, 0

Explanation:
Circuit diagram – ABC all 0

13872.Plancks constant (h), speed of light in vacuum (c) and Newtons gravitational constant (G) are three fundamental constants. Which of the following combinations of these has dimension of length?

$\sqrt{\dfrac{Gc}{h^{3/2}}}$

$\dfrac{\sqrt{hG}}{c^{3/2}}$

$\dfrac{\sqrt{hG}}{c^{5/2}}$

$\sqrt{\dfrac{hc}{G}}$

Explanation:

L = (h)^a (c)^b (G)^c

m⁰L¹T⁰ = (m¹L²T^–1)^a (L¹T^–1)^b (m^–1C³T^–2) ^c

a – c = 0, 2a + b + 3c = 1, – a – b –2c = 0

solving b = –3/2, a = 1/2, c = 1/2

L = $\dfrac{\sqrt{hG}}{c^{3/2}}$

13873.Two cars P and Q start from a point at the same time in a straight line and their positions are represented by x_P(t) = at + bt² and x_Q(t) = ft – t². At what time do the cars have the same velocity?

$\dfrac{f – a}{2(1 + b)}$

$\dfrac{a – f}{1 + b}$

$\dfrac{a + f}{2(b – 1)}$

$\dfrac{a + f}{2(1 + b)}$

Explanation:

V_P = V_Q

a + 2bt = f – 2t

t = $\dfrac{f – a}{2(b + 1)}$

13874.In the given figure, a = 15 m / s² represents the total acceleration of a particle moving in the clockwise direction in a circle of radius R = 2.5 m at a given instant of time. The speed of the particle is Particle accelerating clockwise in a circle

Particle accelerating clockwise in a circle

6.2 m / s

4.5 m / s

5.0 m / s

5.7 m / s

Explanation:
Value of Particle accelerating clockwise in a circle

$a_c = \dfrac{V^2}{r}$

$15 \cos 30^\circ = \dfrac{V^2}{2.5}$

V² = 32.73

V = 5.7 m / sec

13875.A rigid ball of mass m strikes a rigid wall at 60º and gets reflected without loss of speed as shown in the figure below. The value of impulse imparted by the wall in the ball will be Ball hitting wall with impulse

$\dfrac{mV}{3}$

2mV

$\dfrac{mV}{2}$

Explanation:
J = 2mV cos 60º = mV

13876.A bullet of mass 10 g moving horizontally with a velocity of 400 m s^–1 strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be

160 m s^–1

100 m s^–1

80 m s^–1

120 m s^–1

Explanation:

During the collision, apply momentum conservation

(0.01)(400) + 0 = (2)V + (0.01)V

where V = $\sqrt{2gh}$

V = $\sqrt{2 × 10 × 0.1}$

V = $\sqrt{2}$

Solving V’ = 120 m / sec.

13877.Two identical balls A and B having velocities of 0.5 m / s and –0.3 m / s respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be

0.3 m / s and 0.5 m / s

– 0.5 m / s and 0.3 m / s

0.5 m / s and – 0.3 m / s

– 0.3 m / s and 0.5 m / s

Explanation:
Mass of balls are same and the collision is perfectly elastic, so their velocity will be interchanged.
So, V_A = – 0.3 m / s, V_B = 0.5 m / s

13878.A particle moves from a point (–2î + 5ĵ) to (4ĵ + 3k̂) when a force of (4î + 3ĵ) N is applied. How much work has been done by the force?

2 J

8 J

11 J

5 J

Explanation:

$\overrightarrow{S}= \overrightarrow{r_f}- \overrightarrow{r_i}$

= (4ĵ + 3k̂) – (–2î + 5ĵ)

= 2î – ĵ + 3k̂

$\overrightarrow{F}$ = 4î + 3ĵ

$\omega = \overrightarrow{F}. \overrightarrow{S}$ = (4î + 3ĵ) × (2î – ĵ + 3k̂)

= 8 – 3 = 5 J

13879.Two rotating bodies A and B of masses m and 2m with moments of inertia I_A and I_B (I_B > I_A) have equal kinetic energy of rotation. If L_A and L_B be their angular momenta respectively, then

$L_A > L_B$

$L_A = \dfrac{L_B}{2}$

$L_A > 2L_B$

$L_B > L_A$

Explanation:

$KE_A = KE_B$

$\dfrac{1}{2}I_A \omega_A^2 = \dfrac{1}{2}I_B \omega_B^2$ ⇒ since $I_B > I_A \ so \ \omega_B > \omega_A$

$\dfrac{1}{2}L_A \omega_A = \dfrac{1}{2}L_B \omega_B$ ⇒ $L_B > L_A$

13880.A solid sphere of mass m and radius R is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation (E_sphere / E_cylinder) will be

3 : 1

2 : 3

1 : 5

1 : 4

Explanation:

KE of sphere = $\dfrac{1}{2}\left(\dfrac{2}{5}mR^2\right)\omega^2 = \dfrac{1}{5}mR^2\omega^2$

KE of cylinder = $\dfrac{1}{2}\left(\dfrac{mR^2}{2}\right)(2\omega)^2 = mR^2\omega^2$

So, KE_sphere : KE_cylinder = 1 : 5

13881.A light rod of length l has two masses m₁ and m₂ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is

$\sqrt{m_1m_2}l^2$

$\dfrac{m_1m_2}{m_1 + m_2}l^2$

$\dfrac{m_1 + m_2}{m_1m_2}l^2$

$(m_1 + m_2)l^2$

Explanation:
Inertia of rod of with two masses