Page : 2 - Previous Year Question Papers NEET 2016 Phase II

13883.A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g₀, the value of acceleration due to gravity at the earths surface, is

$–\dfrac{2mg_0R^2}{R + h}$

$\dfrac{mg_0R^2}{2(R + h)}$

$–\dfrac{mg_0R^2}{2(R + h)}$

$\dfrac{2mg_0R^2}{R + h}$

Explanation:
TE = $–\dfrac{GMm}{2(R + h)} = –\dfrac{GMm}{2(R + h)}\dfrac{R^2}{R^2} = –\dfrac{g_0mR^2}{2(R + h)}$

13884.A rectangular film of liquid is extended from (4 cm × 2 cm) to (5 cm × 4 cm). If the work done is 3 × 10^–4 J, the value of the surface tension of the liquid is

8.0 Nm^–1

0.250 Nm^–1

0.125 Nm^–1

0.2 Nm^–1

Explanation:
Increase in surface area = (20 cm² – 8 cm²) × 2
= 12 × 2 cm
= 24 cm² (film has two surfaces)

So work done = T.ΔS = T × 24 × 10^–4 = 3 × 10^–4

So T = $\dfrac{3}{24}N / m = \dfrac{1}{8}N m^{–1}$ = 0.125 N/m

13885.Three liquids of densities ρ₁, ρ₂ and ρ₃ (with ρ₁ > ρ₂ > ρ₃), having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact θ₁, θ₂ and θ₃ obey

$\pi > \theta_1 > \theta_2 > \theta_3 > \dfrac{\pi}{2}$

$\dfrac{\pi}{2}> \theta_1 > \theta_2 > \theta_3 \geq 0$

$0 \leq \theta_1 < \theta_2 < \theta_3 < \dfrac{\pi}{2}$

$\dfrac{\pi} \theta_1 < \theta_2 < \theta_3 < \pi$

Explanation:

$h = \dfrac{2T\cos \theta}{\rho gr}$

$\dfrac{\cos \theta_1}{\rho_1}= \dfrac{\cos \theta_2}{\rho_2}= \dfrac{\cos \theta_3}{\rho_3}$

$\cos \theta_1 > \cos \theta_2 > \cos \theta_3 \ as\ \rho_1 > = \rho_2 > \rho_3$

$0 \leq \theta_1 < \theta_2 < \theta_3 < \dfrac{\pi}{2}$

13886.Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100 ºC, while the other one is at 0 ºC. If the two bodies are brought into contact, then assuming no heat loss, the final common temperature is

0 ºC

50 ºC

more than 50 ºC

less than 50 ºC but greater than 0 ºC

Explanation:
Body at 100°C temperature has greater heat capacity than body at 0°C so final temperature will be closer to 100°C. So, T_c > 50°C

13887.A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newtons law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be

$\dfrac{7}{4}$T

$\dfrac{3}{2}$T

$\dfrac{4}{3}$T

Explanation:

$\triangle T = \triangle T_0e^{–\lambda t}$

$T = 2Te^{–\lambda(10 \ min.)}$

$\triangle T = 2Te^{–\lambda(20 \ min.)} = 2T\left(\dfrac{1}{2}\right)^2 = \dfrac{T}{2}$

$T_f = T + \dfrac{T}{2} = \dfrac{3T}{2}$

13888.One mole of an ideal monatomic gas undergoes a process described by the equation PV³ = constant. The heat capacity of the gas during this process is

$\dfrac{3}{2}$R

$\dfrac{5}{2}$R

Explanation:

PV³ = constant.

for a polytropic process, PV^α = constant

$C = C_v + \dfrac{R}{1 – \alpha} = \dfrac{3}{2}R + \dfrac{R}{1 – 3} = \dfrac{3R}{2} – \dfrac{R}{2} = R$

13889.The temperature inside a refrigerator is t₂ ºC and the room temperature is t₁ ºC. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be

$\dfrac{t_1 + t_2}{t_1 + 273}$

$\dfrac{t_1}{t_1 – t_2}$

$\dfrac{t_1 + 273}{t_1 – t_2}$

$\dfrac{t_2 + 273}{t_1 – t_2}$

Explanation:

$\dfrac{Q_{more}}{W} = \dfrac{Q_{more}}{Q_{more} – Q_{less}} = \dfrac{T_{more}}{T_{more} – T_{less}} = \dfrac{t_1 + 273}{t_1 – t_2}$

13890.A given sample of an ideal gas occupies a volume V at a pressure P and absolute temperature T. The mass of each molecule of the gas is m. Which of the following gives the density of the gas?

mkT

P / (kT)

Pm / (kT)

P / (kTV)

Explanation:

$n = \dfrac{PV}{RT} = \dfrac{mass}{Molar \ mass}$

$density = \dfrac{mass}{volume} = \dfrac{(Molar \ mass)P}{RT} = \dfrac{(mN_A)P}{RT} = \dfrac{mP}{KT}$

13891.A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3 s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5 s. The value of m in kg is

$\dfrac{9}{16}$

$\dfrac{3}{4}$

$\dfrac{4}{3}$

$\dfrac{16}{9}$

Explanation:

$T = 2\pi\sqrt{\dfrac{m + 1}{k}}$ = 5 sec

Dividing and squaring, $\dfrac{m}{m + 1} = \left(\dfrac{3}{5}\right)^2 = \dfrac{9}{25}$

25m = 9m + 9

So, m = $\dfrac{9}{16}$ kg

13892.The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe L meter long. The length of the open pipe will be

$\dfrac{L}{2}$

Explanation:
overtones of open and closed organ pipes

$\dfrac{3\lambda}{2} = l_0$

$\lambda = \dfrac{3l_0}{3}$

$f = \dfrac{3V}{2l_0}$

comparison of overtones of open and closed organ pipes

$\dfrac{3\lambda}{4} = L_c$

$ \lambda = \dfrac{4L_e}{3}$

$ f = \dfrac{3V}{4L_e} = \dfrac{3V}{4L} = \dfrac{3V}{2l_0}$

$l_0 = 2L$

13893.Three sound waves of equal amplitudes have frequencies (n – 1), n, (n + 1). They superimpose to give beats. The number of beats produced per second will be

Explanation:
no. of beats = 1
(HCF of beat frequencies)

13894.An electric dipole is placed at an angle of 30º with an electric field intensity 2 ×10⁵ N / C. It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2 cm, is

7 μC

8 mC

2 mC

5 mC

Explanation:

τ = PE sin θ

4 = P × 2 × 10⁵ × $\dfrac{1}{2}$

P = 4 × 10^–5 cm = q × 2 × 10^–2

So, q = $\dfrac{4 \times 10^{–5}}{2 \times 10^{–2}}$ = 2 × 10^–3 coulomb

13895.A parallel– plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constant k₁, k₂, k₃ and k₄ as shown in the figure below. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by parallel plate capacitor

$\dfrac{1}{k} = \dfrac{1}{k_1} + \dfrac{1}{k_2} +\dfrac{1}{k_3}+ \dfrac{3}{2k_4}$

$k = k_1 + k_2 + k_3 + 3k_4$

$k = \dfrac{2}{3}(k_1 + k_2 + k_3) + 2k_4$

$\dfrac{2}{k} = \dfrac{3}{(k_1 + k_2 + k_3)} + \dfrac{1}{k_4}$

Explanation:
Ans: 4 or bonus parallel plate capacitor explanation

$\dfrac{1}{C_1} + \frac{3}{C_4} = \frac{3d}{2k_1\epsilon_0A} + \dfrac{3d}{2k_4\epsilon_0A} = \dfrac{3d}{2\epsilon_0A}\left\{\dfrac{1}{k_1} + \dfrac{1}{k_4}\right\}$

$C_{eq} = \dfrac{K\epsilon_0A}{d} = \dfrac{2\epsilon_0A}{3d}\left\{\dfrac{k_1k_4}{k_1 + k_4} + \dfrac{k_2k_4}{k_2 + k_4} + \dfrac{k_3k_4}{k_3 + k_4}\right\}$

$k = \dfrac{2}{3}\left\{\dfrac{k_1k_4}{k_1 + k_4} + \dfrac{k_2k_4}{k_2 + k_4} + \dfrac{k_3k_4}{k_3 + k_4}\right\}$

Alter:

Wrong solution seems to be correct as per the options given.

parallel plate capacitor explanation part 2

$\dfrac{2}{3}\dfrac{A}{D}\epsilon_0 = \dfrac{2A}{3}\dfrac{\epsilon_0}{d}(k_1 + k_2 + k_3) = C_A$

$\dfrac{2}{3}\dfrac{A}{D}\epsilon_0 = \dfrac{A\epsilon_0}{d}k_4 = C_B$

$\dfrac{1}{C_{eq}} = \dfrac{1}{C_A} + \dfrac{1}{C_B}$

$\dfrac{d}{A\epsilon_0K_{eq}} = \dfrac{3d}{2A\epsilon_0}(k_1 + k_2 + k_3) + \dfrac{d}{2A\epsilon_0k_4}$

$\dfrac{1}{K_{eq}} = \dfrac{3}{2}(k_1 + k_2 + k_3) + \dfrac{1}{2k_4}$

13896.The potential difference (V_A – V_B) between the points A and B in the given figure is potential difference in voltages

+ 9 V

– 3V

+ 3 V

+ 6 V

Explanation:
potential difference in voltages explanation

V_A – V_B = 9 volt

13897.A filament bulb (500 W, 100 V) is to be used in a 230 V main suply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is :

13 Ω

230 Ω

46 Ω

26 Ω

Explanation:
filament bulb current flow explanation

i = $\dfrac{500}{100}$ = 5A

so, 130 = 5R

R = 26

13898.A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be

2n²B

n²B

2nB

Explanation:
circular loop diagram

$B = \dfrac{\mu_0i}{2R} = \dfrac{\mu_0i(2\pi)}{2l} = \dfrac{\mu_0\pi i}{2l}$

$B = \dfrac{\mu_0ni}{2r} = \dfrac{\mu_0ni}{2\left(\dfrac{l}{2n\pi}\right)} = \dfrac{n^2\mu_0\pi i}{2l} = n^2B$

13899.A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60º is W. Now the torque required to keep the magnet in this new position is

$\dfrac{2W}{\sqrt{3}}$

$\dfrac{W}{\sqrt{3}}$

$\sqrt{3}W$

$\dfrac{\sqrt{3}W}{2}$

Explanation:

W_ext = U_f – V_i

= – MB cos 60° – (–MB)

= MB(1 – cos 60°) = MB/2 = W

r = MB sin 60° = MB $\dfrac{\sqrt{3}}{2}$ =$ \sqrt{3}$W

13900.An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57 × 10^–2 T. If the value of e / m is 1.76 × 10¹¹ C / kg, the frequency of revolution of the electron is

6.28 MHz

1 GHz

100 MHz

62.8 MHz

Explanation:

$R = \dfrac{mV}{qB}$

$\omega = \dfrac{V}{R} = \dfrac{qB}{m}$

$f\omega = \dfrac{\omega}{2\pi} = \dfrac{1}{2\pi}. \dfrac{qB}{m} = \dfrac{1.76 \times 10^{11} \times 3.57 \times 10^{–2}}{2 \times 3.14} = 10^9 Hz$

13901.Which of the following combinations should be selected for better tuning of an L–C–R circuit used for communication?

R = 25 Ω, L = 1.5 H, C = 45 μF

R = 20 Ω, L = 1.5 H, C = 35 μF

R = 25 Ω, L = 2.5 H, C = 45 μF

R = 15 Ω, L = 3.5 H, C = 30 μF

Explanation:

Option with highest quality factor sould be chosen as most appropriate answer.

$Q = \dfrac{1}{R}\sqrt{\dfrac{L}{C}}$

13902.A uniform magnetic field is restricted within a region of radius r. The magnetic field changes with time at a rate $\dfrac{d\overrightarrow{B}}{dt}$. Loop 1 of radius R > r encloses the region r and loop 2 of radius R is outside the region of magnetic field as shown in the figure below. Then the e.m.f. generated is magnetic field loop

$–\dfrac{\text{d}\overrightarrow{B}}{\text{d}t}\pi r^2$ in loop 1 and zero in loop 2

zero in loop 1 and zero in loop 2

$–\dfrac{\text{d}\overrightarrow{B}}{\text{d}t}\pi r^2$ in loop 1 and $–\dfrac{\text{d}\overrightarrow{B}}{\text{d}t}\pi r^2$ in loop 2

$–\dfrac{\text{d}\overrightarrow{B}}{\text{d}t}\pi R^2$ in loop 1 and zero in loop 2

Explanation:

$e = – \dfrac{\text{d}\phi}{\text{d}t}= –\dfrac{\text{d}}{\text{d}t}{\pi r^2B} = – \pi r^2\dfrac{\text{d}B}{\text{d}t}$ in loop 1 & zero in loop 2.

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