datavorous/entrance-exam-dataset huggingface dataset:
A sagittal section of human brain is shown here. Identify at least two labels from a-d.

Options:
A c-Mid brain, d-Cerebellum
B a-Cerebrum, c-Pons
C b-Corpus callosum, d-Medulla
D a-Cerebral hemispheres, b-Cercbellum

The correct answer is:
a-Cerebrum, c-Pons

Solution:
Related Theory
$a$ - cerebrum, b- thalamus, $c$ - pons, $d$ cerebellum

Topic: Biology, Neural Control and Coordination, NEET, NEET 2013 (Karnataka)
A salt is soluble when its solubility is

Options:
A > 0.1M
B < 0.01M
C In between 0.01M and 0.1M
D Cannot say

The correct answer is:
> 0.1M

Solution:
Correct Option is : 1
> 0.1M

Topic: Chemistry, Ionic Equilibrium, NEET
A salt MA2 ionises as

MA 2 â M 2 + + 2 A -

It was found that a given solution of the salt had the same freezing point as solution of glucose of twice the molality. The apparent degree of ionization of the salt is

Options:
A 0.25
B 0.33
C 0.50
D 0.67

The correct answer is:
0.50

Solution:
MA 2 â M 2 + + 2 A - 1 - Î± Î± 2 Î±
i = Actual no. of particles in solution Number of particles taken
i = 1 + 2 Î±
Î T fâ¡ of MA 2 = Î T fâ¡ of glucose
i· K f ·m= K f ·2m×i
1 + 2 Î± = 2 â 2 Î± = 1 â Î± = 0.5

Topic: Chemistry, Solutions, NEET
A sample contains 10-2 kg each of the two substances A and B with half-lives 4 s and 8 s respectively. Their atomic weights are in the ratio of 1:2. Amounts of A and B after an interval 16 s are

Options:
A A=0.625×10-4 kg, B=0.25×10-3 kg
B A=0.25×10-3 kg, B=0.625×10-4 kg
C A=2.5×10-3 kg, B=6.25×10-4 kg
D A=6.25×10-3 kg, B=2.5×10-3 kg

The correct answer is:
A=6.25×10-3 kg, B=2.5×10-3 kg

Solution:
The amount left after N half - lives
N=N012n
For A,
n=tT12=16s4s=4
NA=10-2 kg124=6.25×10-4 kg
For B,
n=16s8s=2
NB=10-2 kg122=2.5×10-3 kg

Topic: Physics, Nuclear Physics, NEET
A sample contains a large number of nuclei. The probability that a nucleus in the sample will decay after four half-lives is-

Options:
A 14
B 34
C 1516
D 716

The correct answer is:
1516

Solution:
The probability that a nucleus will not decay is-
NN0=12n=q
When n is the number of half-lives
q=124=116
The probability that a nucleus will decay is
p=1-q=1-116=1516

Topic: Physics, Nuclear Physics, NEET
A sample contains large number of nuclei. The probability that a nucleus in sample will decay after four half lives is

Options:
A 14
B 34
C 1516
D 716

The correct answer is:
1516

Solution:
Probability that a nucleus will not decay is
NN0=12n=q
Where n is the number of half lives q=124=116
Probability that a nucleus will decay is
p=1-q=1-116=1516

Topic: Physics, Nuclear Physics, NEET
A sample of 0.1  g of water at 100 oC and normal pressure 1.013×105 N m-2 requires 54  cal of heat energy to convert to steam at 100 oC. If the volume of the steam produced is 167.1  cc, the change in internal energy of the sample, is

Options:
A 42.2  J
B 208.7ââJ
C 104.3  J
D 84.5  J

The correct answer is:
208.7  J

Solution:
The values are given in the above question, ∆Q=54 cal=54×4.18 joule=225.72 joule
∆W=PVsteam-VwaterFor water 0.1 gram=0.1 cc
∆W=1.013×105167.1×10-6-0.1×10-6 joule
⇒∆W=1.013×167×10-1=16.917 joule
. By first law of thermodynamics, the change in internal energy,
⇒∆U=∆Q-∆W=225.72-16.917=208.8 J

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2018
A sample of 1.0 mol of a monoatomic ideal gas is taken through a cyclic process of expansion and compression as shown in Fig. 6.1. What will be the value of ∆H for the cycle as a whole?

Options:
A ∆H(cycle) = 100 kJmol-1
B ∆H(cycle) = ∞
C ∆H(cycle) = 15 kJmol-1
D ∆H(cycle) = 0

The correct answer is:
∆H(cycle) = 0

Solution:
Correct Option is : (D)
∆H(cycle) = 0

Topic: Chemistry, Thermodynamics (C), NEET
A sample of 18 F is used internally as a medical diagnostic tool to look for the effects of the positron decay ( T 1 2 =110 min ) . How long does it take for 99% of the 18F to decay?

Options:
A 12.4 h
B 12.0 h
C 12.2 h
D 12.5 h

The correct answer is:
12.2 h

Solution:
Radioactive decay equation is
N=N0e-Î»t=N0e-lnâ¡2tT12
After decay of 99% of the initial sample only 1% will be left, and NN0=1% . We then have
NN0=1100=e-lnâ¡2tT12
If we take the natural logarithm, we have
-lnâ¡100=-lnâ¡2ÃtT12
Which on solving for t yields
â´ t=lnâ¡100lnâ¡2ÃT12
=logâ¡100logâ¡2ÃT12
=20.3010Ã110
=731minâ¡=12.2 h

Topic: Physics, Nuclear Physics, NEET
A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 ml of 0.5 M H2SO4.The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.

Options:
A % fo nitrogen = 95
B % fo nitrogen = 78
C % fo nitrogen = 44
D % fo nitrogen = 56

The correct answer is:
% fo nitrogen = 56

Solution:
Correct Option is : (D)
% fo nitrogen = 56

Topic: Chemistry, General Organic Chemistry, NEET
A sample of 210Po which is Î± - emitter with T12=138 days is observed by a student to have 2000 disintegration (2000 Bq). The activity in Î¼Ci for this source is

Options:
A 0.050 μCi
B 0.051 μCi
C 0.55 μCi
D 0.054 μCi

The correct answer is:
0.054 μCi

Solution:
The activity of a substance that has 2000 decays per second
A=20003.7 ×1010
=0.054 ×10-6Ci=0.054 μCi

Topic: Physics, Nuclear Physics, NEET
A sample of air contains only N2,O2 and H2O. It is saturated with water vapours and the total pressure is 640 torr. The vapour pressure of water is 40 torr and the molar ratio of N2:O2 is 3 : 1. The partial pressure of N2 in the sample is

Options:
A 480 torr
B 600 torr
C 525 torr
D 450 torr

The correct answer is:
450 torr

Solution:
Pressure of dry air =Pmoist air-Pwater vapours
Partial pressure = mole fraction Ã P T
â´ Partial pressure of N2 and O2 are also in the ratio of 3 : 1.
Partial pressure of N2=600Ã33+1=450 torr .

Topic: Chemistry, States of Matter, NEET
A sample of air which is free from CO2, is used to heat Na metal at 350Â°C to form X. X has the property to absorb CO2 and form Y. Identify "X" and "Y" respectively:

Options:
A Na2O and O2
B Na2O2 and O2
C NaO2 and O2
D Na2O2and O3

The correct answer is:
Na2O2 and O2

Solution:
Sodium is heated in
2Na+O2(air)⟶Na2O2(X)
2CO2+2Na2O2⟶2Na2CO3+O2(Y)

Topic: Chemistry, s Block Elements, NEET
A sample of an ideal gas is taken through the cyclic process ABCA in the diagram. It rejects 50 J of heat during the part AB, does not absorb or reject the heat during BC, and accepts 70 J of heat during CA. Forty joules of work is done on the gas during the part BC, The internal energies at B and C, respectively, will be
[Given that Internal energy at A i.e UA= 1500]

Options:
A 1450 J and 1410 J

B 1550 J and 1590 J
C 1450 J and 1490 J
D 1550 J and 1510 J

The correct answer is:
1450 J and 1490 J

Solution:
Along path AB (Isochoric)
Î Q = Î U + Î W = U B - U A + Î W
or - 50 = U B - U A = U B - 1 5 0 0
or U B = 1 4 5 0 J
Along path BC
or Î Q = U C - U B + Î W
or 0 = U C - U B + Î W = U C - U B + - 4 0
or 0 = U C - 1 4 5 0 - 4 or U C = 1 4 9 0 J

Topic: Physics, Thermodynamics, NEET
A sample of chloroform being used as anaesthetic is tested by

Options:
A Ammoniacal Cu2Cl2
B Fehling solution
C AgNO3 solution
D Adding Ag to the solution after boiling with alcoholic KOH solution

The correct answer is:
AgNO3 solution

Solution:
Before using the sample of chloroform as an anaesthetic, it is tested by treating with aq .
solution of AgNO3 . A pure sample does not give white ppt of AgCl with aq.AgNO3 .

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass). Determine the molality of chloroform in the water sample.

Options:
A 2.43 × 10-4m
B 1.25 × 10-4m
C 3.68 × 10-4m
D 0.52 × 10-4m

The correct answer is:
1.25 × 10-4m

Solution:
Correct Option is : (B)
1.25 × 10-4m

Topic: Chemistry, Solutions, NEET
A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass). Express this in percent by mass

Options:
A 6.4×10–3%
B 0.5×10–3%
C 3.4×10–3%
D 1.5×10–3%

The correct answer is:
1.5×10–3%

Solution:
Correct Option is : (D)
1.5×10–3%

Topic: Chemistry, Solutions, NEET
A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass). Express this in per cent by mass.

Options:
A 0.0019%
B 0.0034%
C 0.0015%
D 0.0024%

The correct answer is:
0.0015%

Solution:
Correct Option is : (C)
0.0015%

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass). Determine the molality of chloroform in the water sample.

Options:
A 1.255×10−4m
B 1.355×10−4m
C 1.465×10−4m
D 1.524×10−4m

The correct answer is:
1.255×10−4m

Solution:
Correct Option is : (A)
1.255×10−4m

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A sample of ferrous oxide has actual formula Fe0.93O1.00. In this sample what fraction of metal ions are Fe2+ions? What type of non stoichiometric defect is present in this sample?

Options:
A Fraction of Fe2+= 0.81 and Metal deficiency defect is present in this sample.
B Fraction of Fe2+= 0.81 and Metal excess defect is present in this sample.
C Fraction of Fe2+= 0.56 and Metal deficiency defect is present in this sample.
D Fraction of Fe2+= 0.95 and Metal deficiency defect is present in this sample.

The correct answer is:
Fraction of Fe2+= 0.81 and Metal deficiency defect is present in this sample.

Solution:
Correct Option is : (A)
Fraction of Fe2+= 0.81 and Metal deficiency defect is present in this sample.

Topic: Chemistry, Solid State, NEET
A sample of gas contracts by 1 litre against a constant pressure of 0.1 atm while 5.13 J heat is lost to surroundings. The change in internal energy, U of the system is

Options:
A 10.26 J
B 5.0 J
C 5.64 J
D 4.0 J

The correct answer is:
5.0 J

Solution:
ΔE=q+W
Wcont.=-PΔV
q=0 so ΔE=W=PΔV
W = 0.1 × 1 × 101.3 = 10.13 J
∵ 1 L-atm = 101.3 J
ΔE = −5.13 J + 10.13 J = + 5.0 J

Topic: Chemistry, Thermodynamics (C), NEET
A sample of HI(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium, the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium? 2HI(g) ⇌ H₂(g) + I₂(g)

Options:
A 4
B 7
C 0.2
D 5

The correct answer is:
4

Solution:
Correct Option is : (A)
4

Topic: Chemistry, Ionic Equilibrium, NEET
A sample of hydrogen gas in its ground state is irradiated with photons of 10.2 eV energies. The radiation from the above sample is used to irradiate two other samples of excited ionized He+ and excited ionized Li2+, respectively. Both the ionized samples absorb the incident radiation. How many spectral lines are obtained in the spectra of Li2+?

Options:
A 10
B 15
C 20
D 17

The correct answer is:
15

Solution:
Li+2 is excited from n=3 to n=6
no of spectral lines
So, N=nn-12
N=15

Topic: Physics, Atomic Physics, NEET
A sample of I51131 as iodine ion was administered to a patient in a carrier consisting of 0.10 mg of stable iodine ion. After 4 days 67.7% of initial radioactivity was detected in the thyroid gland of the patient. The percentage of stable iodine ion that had migrated to the thyroid gland, if t12 for iodide ion is 8 days is

Options:
A 65.8%
B 95.7%
C 85.8%
D 75.8%

The correct answer is:
95.7%

Solution:
Let us take x mg iodine migrated to thyroid gland.
if A is activity of complete sample of mass m. Then activity of x is given by
Ax=Amx
Ax is 67.7% of initial total sample at t=4 days.
Ax=67.7100A0 & A=A02t/T
T=half life time
So using above equation,
67.7100A0=A024/8 xm
xm×100=67.7×2=95.7%

Topic: Physics, Nuclear Physics, NEET
A sample of live microbial cells was presented to a scientist. He decides to stop the synthesis of a particular protein in the sample cells. Suggest a modification in the part of the cell for the scientist to enable the prevention of that particular protein synthesis.

Options:
A The DNA in the nucleus
B An enzyme in the lysosomes
C The polysaccharides in the cytoplasm
D A phospholipid in the plasma membrane

The correct answer is:
The DNA in the nucleus

Solution:
DNA is considered the blue print of an organism. If the DNA is altered, it will alter the organism's characteristic. Also, DNA contains genes which are responsible for synthesis of proteins. Thus, altering/modifying the DNA can result in transcription or no transcription for the particular protein. In this case, the scientist needs to modify the DNA using genetic engineering to disable protein synthesis.

Topic: Biology, Molecular Basis of Inheritance, NEET
A sample of milk splits after 60 min. at 300 K and after 40 min. at 400 K when the population of lactobacillus acidophilus in it doubles. The activation energy (in kJ/mol) for this process is closest to
(Given, R = 8.3 J mol-1K-1,ln23=0.4,e-3=4.0)

Options:
A 39.8
B 19.9
C 3.98
D 7.96

The correct answer is:
3.98

Solution:
lnk2k1=EaR1T1-1T2
ln6040=Ea8.31300-1400
Ea=3.98 kJ/mole

Topic: Chemistry, Chemical Kinetics, NEET
A sample of NaNO3 weighing 0.38 g is placed in a 50.0 mL volumetric flask. The flask is then filled with water to the mark on the neck. What is the molarity of the solution?

Options:
A 8.94 Ã 10-6M
B 8.94 Ã 10-2M
C 1.94Â ÃÂ 10-2M
D 8.94Â ÃÂ 10-4M

The correct answer is:
8.94 Ã 10-2M

Solution:
Molarity Ã Volume in mL = Milli mole =wMÃ1000
â´ Molarity Ã 50 =0.3885Ã1000
[Mol. wt. of NaNO3 = 85]
â´ Molarity = 8.94 Ã 10-2

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A sample of pure \(\mathrm{PCI} 5\) was introduced into an evacuated vessel at \(473 \mathrm{~K}\). After equilibrium was attained, the concentration of \(\mathrm{PCI} 5\) was found to be \(0.5 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1}\). If the value of \(K_c\) is \(8.3 \times 10^{-3}\), what are the concentrations of \(\mathrm{PCl} 3\) and \(\mathrm{Cl} 2\) at equilibrium?
\(P C l 5(g) \rightleftharpoons P C l 3(g)+C l 2(g)\)

Options:
A \([P C l 3]=0.035 \mathrm{~mol} \mathrm{~L}^{-1},[C l 2]=0.035 \mathrm{~mol} \mathrm{~L}^{-1}\)
B \([P C l 3]=0.015 \mathrm{~mol} \mathrm{~L}^{-1},[C l 2]=0.015 \mathrm{~mol} \mathrm{~L}^{-1}\)
C \([P C l 3]=0.030 \mathrm{~mol} \mathrm{~L}^{-1},[C l 2]=0.030 \mathrm{~mol} \mathrm{~L}^{-1}\)
D \([P C l 3]=0.040 \mathrm{~mol} \mathrm{~L}^{-1},[C l 2]=0.040 \mathrm{~mol} \mathrm{~L}^{-1}\)

The correct answer is:
\([P C l 3]=0.035 \mathrm{~mol} \mathrm{~L}^{-1},[C l 2]=0.035 \mathrm{~mol} \mathrm{~L}^{-1}\)

Solution:
Correct Option is : (A)
\([P C l 3]=0.035 \mathrm{~mol} \mathrm{~L}^{-1},[C l 2]=0.035 \mathrm{~mol} \mathrm{~L}^{-1}\)

Topic: Chemistry, Ionic Equilibrium, NEET
A sample of radioactive element contains $4 \times 10^{16}$ active nuclei. The half-life of the element is 10 decays, then number of decayed nuclei after 30 days is:

Options:
A $0.5 \times 10^{16}$
B $2 \times 10^{16}$
C $3.5 \times 10^{16}$
D $1 \times 10^{16}$

The correct answer is:
$3.5 \times 10^{16}$

Solution:
Half life number is given by $n=\frac{1}{T}$ $=\frac{30}{10}=3$
Nuclear of active nuclei left after $\mathrm{n}$ half-life is
$$
\begin{aligned}
& N=N_0\left(\frac{1}{2}\right)^n \\
= & \frac{4 \times 10^{16}}{(2)^3}=0.5 \times 10^{16} \\
\therefore & \text { Decay nuclei }=N_0-N \\
= & 4 \times 10^{16}-0.5 \times 10^{16} \\
= & 3.5 \times 10^{16}
\end{aligned}
$$

Topic: Physics, Nuclear Physics, NEET, NEET 2002
A sample of radioactive element has a mass of $10 \mathrm{~g}$ at an instant $t=0$. The approximate mass of this element in the sample after two mean lives is:

Options:
A $1.35 \mathrm{~g}$
B $2.50 \mathrm{~g}$
C $3.70 \mathrm{~g}$
D $6.30 \mathrm{~g}$

The correct answer is:
$1.35 \mathrm{~g}$

Solution:
\(\begin{aligned} & \mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}} \Rightarrow \mathrm{m}=\mathrm{m}_0 \mathrm{e}^{-\lambda \mathrm{t}}=\mathrm{m}_0 \mathrm{e}^{-\lambda(2 / \lambda)} \\ & =\frac{10}{\mathrm{e}^2}=1.35 \mathrm{gm}\end{aligned}\)

Topic: Physics, Nuclear Physics, NEET, NEET 2003
A sample of radioactive material decays simultaneously by two processes A and B with half-lives 12 and 14 hours respectively. For first half-hour, it decays with process A, next one hour with process B and for a further half an hour with both A and B. If originally there were N0 nuclei, find the number of nuclei after 2 hours of such decay-

Options:
A N028
B N024
C N026
D N025

The correct answer is:
N028

Solution:
After first half hour with the process A ,
N=N02
for next one hour with the process B , i.e. t=12 hr to t=112 hr
N=N02124=N0125
Half life for both A and B ,
1t1/2=11/2=11/4
1t1/2=2+4
t1/2=16hr
for further half an hour with both A and B , i.e. t=112 hr to t=2 hr
N=N0125 123=N0128

Topic: Physics, Nuclear Physics, NEET
A sample of radioactive material has mass m, decay constant λ and molecular weight M. Avogadro's constant =NA. The initial activity of the sample is

Options:
A λm
B λmM
C Î»mNAM
D mNAeλ

The correct answer is:
λmNAM

Solution:
In a radioactive sample, activity is directly proportional to the number of active nuclei.
Activity =λN, where N=NAmM.

Activity=λmNAM

Topic: Physics, Nuclear Physics, NEET
A sample of the radioactive element has a mass of 10 g at an instant t=0. The approximate mass of this element in the sample after two mean lives is

Options:
A 1.35 g

B 2.50 g
C 3.70 g
D 6.30 g

The correct answer is:
1.35 g

Solution:
Let, t = 0, M0 = 10 g
t = 2 τ = 2 1 λ given
Then from, M⁡ = M⁡ 0 e⁡ - λ t = 1 0 e⁡ - λ 2 λ
= 1 0 1 e⁡ 2 = 1 · 3 5 g.

Topic: Physics, Nuclear Physics, NEET
A sample was given to the lab technician for figuring out whether the cells of the sample come under the category of prokaryote or eukaryote. Four of the dyes are described in the table below are available to the lab tech for the determining process.

Dye	Test

Acridine orange	stains DNA and RNA

Osmium tetroxide	stains lipids

Eosin	stains cell cytoplasm

Nile blue	stains cell nuclei

Predict the stain that was used by the lab technician that help her identify whether sample cells will come under the group of prokaryote or eukaryote.

Options:
A Acridine orange
B Eosin
C Nile blue
D Osmium tetroxide

The correct answer is:
Nile blue

Solution:
The eukaryotic cell has a defined nucleus where it is absent in the prokaryotic cells. Thus, best would be for the lab tech to use the Nile blue dye which will stain the nuclei. Since nuclei are absent in prokaryote, there will no successful staining whereas the eukaryote will get stained and help determine the type of cell sample. DNA, lipids, and cytoplasm is commonly found in both the cell types.

Topic: Biology, Cell - The Unit of Life, NEET
A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius (1.02)R. The period of the second satellite is larger than the first one by approximately

Options:
A 0.7%
B 1.0%
C 1.5%
D 3.0%

The correct answer is:
3.0%

Solution:
from kepler's law,
T2∝R3 ⇒T∝R32
⇒T= k R 3 2 ;k k is some constant . for radius =R
T=k R 3 2
dT dR = 3 2 k R 1 2
when radius =1.02R;dR=1.02R−R=0.02R
dT dR = 3 2 k R 1 2
⇒dT= 3 2 k R 1 2 ×dR
⇒dT=1.5kR12×0.02R
% change = dT T ×100= 0.03T T ×100=3

Topic: Physics, Gravitation, NEET
A satellite is orbiting just above the surface of the earth with period T. If d is the density of the earth and G is the universal constant of gravitation, the quantity 3πGd represents :

Options:
A T
B T
C T2
D T3

The correct answer is:
T2

Solution:
The formula to calculate the mass of the Earth considering it a sphere is given by

M=43πR3d ...1

The formula to calculate the time period of revolution of the satellite can be written as

T=2πR3GM ...2

Substitute the expression for the mass from equation (1) into equation (2) and simplify to obtain the expression for the required quantity.

T=2πR3G43πR3d=2π34πGd=12π24πGd=3πGd⇒3πGd=T2

Topic: Physics, Gravitation, NEET, NEET 2023 (All India)
A satellite is orbiting the Earth in a circular orbit of radius R . Which one of the following statements is true?

Options:
A Angular momentum varies as 1R
B Linear momentum varies as R
C Frequency of revolution varies as 1R2
D Kinetic energy varies as 1R

The correct answer is:
Kinetic energy varies as 1R

Solution:
∵ Kinetic energy of satellite,
KE=GMm2R
∴ KE∝1R
∴ Option Kinetic energy varies as 1R is correct.

Topic: Physics, Gravitation, NEET
A satellite is revolving in a circular orbit at a height h above the surface of the earth of radius R. The speed of the satellite in its orbit is one fourth the escape velocity from the surface of the earth. The relation between h and R is

Options:
A h=2R
B h=3R
C h=5R
D h=7R

The correct answer is:
h=7R

Solution:
Ï 0 = 1 4 Ï e
GM ( R+h ) = 1 4 2GM R
GM ( R+h ) = 1 16 Ã 2GM R
R+h=8( R )
R+h=8R
7R=h

Topic: Physics, Gravitation, NEET
A satellite moves around the earth in a circular orbit of radius R making one revolution per day. A second satellite moving in a circular orbit moves around the earth once in 8 days. The radius of the orbit of the second satellite is

Options:
A 8 R

B 4 R

C 2 R

D R

The correct answer is:
4 R

Solution:
Given that, T 1 =1 day and T 2 = 8 days
∴ T 2 T 1 = ( r 2 r 1 ) 3/2 ⇒ r 2 r 1 = ( T 2 T 1 ) 2/3 = ( 8 1 ) 2/3 =4 ⇒ r 2 =4 r 1 =4R

Topic: Physics, Gravitation, NEET
A satellite of mass m, initially at rest on the earth, is launched into a circular orbit at a height equal to the radius of the earth. The minimum energy required is

Options:
A 34mgR
B 12mgR
C 14mgR
D 34mgR

The correct answer is:
34mgR

Solution:
We know
V 0 = GM r & g = GM R 2
From energy conservation
U i + K i = U f + K f
- GM m R + K i = - GM m 2 R + 1 2 mv 0 2
K i = GM m 2 R + 1 2 m GM 2 R 2 ⇒ K i = 3 GM m 4 R ⇒ K i = 3 4 mgR

Topic: Physics, Gravitation, NEET
A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches height R ( R= radius of the earth), it ejects a rocket of mass m10 so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is ( G is the gravitational constant; M is the mass of the earth):

Options:
A m20u2+113200GMR
B 5mu2-119200GMR
C 3m8u+5GM6R2
D m20u-2GM3R2

The correct answer is:
5mu2-119200GMR

Solution:
-GMemR+12mu2=-GMem2R+12mv2
v=u2-GMeR
M10vr=Mu2-GMeR
Kinetic energy =12M10VT2+Vr2=M2081GM2R+100u2-100GMR
= M 20 100 u 2 − 119GM 2R
=5mu2-119GM200R

Topic: Physics, Gravitation, NEET
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 g o , the value of acceleration due to gravity at the earth's surface, is

Options:
A mg0R22R+h
B -mg0R22R+h
C 2mg0R2R+h
D -2mg0R2R+h

The correct answer is:
-mg0R22R+h

Solution:
The total energy of a satellite in the orbit is

E=-GMm2r

r=R+h

g0=GMR2
⇒E=-mg0R22R+h

Topic: Physics, Gravitation, NEET, NEET 2016 (Phase 2)
A satellite of mass m revolves around the earth of radius R at a height x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is

Options:
A gx
B gRR-x
C gR2R+x
D gR2R+x1/2

The correct answer is:
gR2R+x1/2

Solution:
For a satellite
centripetal force = Gravitational force
â´ mÎ½02R+x=GMmR+x2
or Î½02=GMR+x=gR2R+x [âµ g=GMR2]
or Î½0=gR2R+x.

Topic: Physics, Gravitation, NEET
A satellite which is geostationary in a particular orbit is taken to another orbit, the distance of which is twice that of earlier orbit. The time period of the satellite in the second orbit is :

Options:
A 24 h
B 48 h
C 482 h
D 482 h

The correct answer is:
482 h

Solution:
T2∝ r3
T1T2=r1r23/2; 24 hT=r2r3/2
T=2422; T=482 h

Topic: Physics, Gravitation, NEET
A satellite which is geostationary in a particular orbits is taken to another orbit, the distance of which is twice that of earlier orbit. The time period of the satellite in the second orbit is

Options:
A 24 h
B 48 h
C 482 h
D 482 h

The correct answer is:
482 h

Solution:
T2∝r3
T1T2=r1r232;
24T=r2r32
T=2422
T=482 h

Topic: Physics, Gravitation, NEET