datavorous/entrance-exam-dataset huggingface dataset:
A car is negotiating a curved road of radius R . The road is banked at an angle θ. The coefficient of friction between the tyres of the car and the road is μs. The maximum safe velocity on this road is:

Options:
A gR2μs+tan⁡θ1-μstan⁡θ
B gRμs+tan⁡θ1-μstan⁡θ
C gRμs+tan⁡θ1-μstan⁡θ
D gR2μs+tanθ1-μstan⁡θ

The correct answer is:
gRμs+tan⁡θ1-μstan⁡θ

Solution:
N=mgcos⁡θ+mv2Rsin⁡θ
f max =μN ⇒ fmax= μ s mgcos⁡θ+ μ s mv2Rsin⁡θ
mg⁡sinθ+fmax=mv2R⁡cosθ
mgsin⁡θ+ μ s mgcos⁡θ+ μ s mv2Rsin⁡θ=mv2Rcos⁡θ
gsin⁡θ+ μ s gcos⁡θ=v2Rcos⁡θ- μ s sin⁡θ
gRtan⁡θ+ μ s 1- μ s tan⁡θ=v2

Topic: Physics, Laws of Motion, NEET
A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m s-2 and the car has acceleration 4 m s-2 . The car will catch up with the bus after time :

Options:
A 120 s
B 15 s
C 110 s
D 102 s

The correct answer is:
102 s

Solution:
aCB=2 m sec-2
200=12×2t2
t=102 s

Topic: Physics, Motion In One Dimension, NEET
A car moves from $\mathrm{X}$ to $\mathrm{Y}$ with a uniform speed $v_u$. The average speed for this round trip is :

Options:
A $\sqrt{v_u v_d}$
B $\frac{v_d v_u}{v_d+v_v}$
C $\frac{v_u+v_d}{2}$
D $\frac{2 v_d v_u}{v_d+v_u}$

The correct answer is:
$\frac{2 v_d v_u}{v_d+v_u}$

Solution:
Average speed
$\begin{aligned}
& =\frac{\text { Total distance }}{\text { Total time }} \\
& =\frac{2 d}{\frac{d}{V_u}+\frac{d}{V_d}}=\frac{2 V_d V_u}{V_d+V_u}
\end{aligned}$

Topic: Physics, Motion In One Dimension, NEET, NEET 2007
A car of mass 1000 kg moves on a circular path with a constant speed of 16 m s-1. It is turned by 90Â°after travelling 628 m on the road. The centripetal force acting on the car is

Options:
A 160Â N
B 320Â N
C 640 N
D 1280Â N

The correct answer is:
640 N

Solution:
2πr4=628
r=62823.14=400 m
Centripetal Force, F=mv2r
F=1000162400
F=640 N

Topic: Physics, Motion In Two Dimensions, NEET
A car of mass $1000 \mathrm{~kg}$ negotiates a banked curve of radius $90 \mathrm{~m}$ on a frictionless road. If the banking angle is $45^{\circ}$, the speed of the car is

Options:
A $20 \mathrm{~ms}^{-1}$
B $30 \mathrm{~ms}^{-1}$
C $5 \mathrm{~ms}^{-1}$
D $10 \mathrm{~ms}^{-1}$

The correct answer is:
$30 \mathrm{~ms}^{-1}$

Solution:
The angle of banking
$\tan \theta=\frac{v^2}{r g}$
Given, $\theta=45^{\circ}, r=90 \mathrm{~m}$
and $g=10 \mathrm{~m} / \mathrm{s}^2$
$\begin{aligned}
\tan 45^{\circ} & =\frac{v^2}{90 \times 10} \\
v & =\sqrt{90 \times 10 \times \tan 45^{\circ}}
\end{aligned}$
Speed of $\operatorname{car} v=30 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Laws of Motion, NEET, NEET 2012 (Screening)
A car of mass 1500 kg is lifted up a distance of 30 m by crane A in 0.5 minutes. The second crane B does the same job in 1 minute. The ratio of their powers is

Options:
A 1Â :2
B 2 :1
C 1Â :4
D 4Â :1

The correct answer is:
2 :1

Solution:
Given, mass of the car m=1500 kg
Height of car h1=30 m
Time taken by crane A in lifting the car t1=0.5 min.
Time taken by crane B in lifting the car t2=1 min.
We know that, P=mght
â´Pâ1t
Now, P1P2=t2t1
â P1P2=6030
â´ P1P2=21

Topic: Physics, Work Power Energy, NEET
A car round on unbanked curve of radius 92 m without skidding at a speed of 26 m s-1. The smallest possible coefficient of static friction between the tyres and the road is

Options:
A 0.75
B 0.60
C 0.45
D 0.30

The correct answer is:
0.75

Solution:
fs≤fsmax
Centripetal force is equal to maximum static friction
mv2r≤μmg
μ≥v2rg
μmin=v2rg

μmin=26292×9.8

μmin=0.75

Topic: Physics, Laws of Motion, NEET
A car runs at a constant speed on a circular track of radius $100 \mathrm{~m}$, taking 62.8 seconds for every circular lap. The average velocity and average speed for each circular lap respectively is:

Options:
A $10 \mathrm{~m} / \mathrm{s}, 0$
B 0,0
C $0,10 \mathrm{~m} / \mathrm{s}$
D $10 \mathrm{~m} / \mathrm{s}, 10 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$10 \mathrm{~m} / \mathrm{s}, 10 \mathrm{~m} / \mathrm{s}$

Solution:
Distance trevelled in one revolution
$\begin{aligned}
& =2 \pi r \\
& \therefore \text { Average speed }=\frac{\text { Totaldistance }}{\text { Total time }} \\
& \Rightarrow \quad \frac{2 \pi r}{t}=\frac{2 \times 3.14 \times 100}{62.8} \\
& =10 \mathrm{~m} / \mathrm{s}
\end{aligned}$
Displacement in one revolution is zero.
$\begin{aligned}
\therefore \text { Average Velocity }= & \frac{\text { Net displacement }}{\text { True }} \\
& =\frac{0}{t}=0
\end{aligned}$

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2006
A car starts from rest and accelerates at 5 m s-2, at t=4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t=6 s ? (g=10 m s-2

Options:
A 20 m s-1,5 m s-2
B 20 m s-1,0
C 202 m s-1 ,10 m s-2
D 202 m s-1,0

The correct answer is:
202 m s-1 ,10 m s-2

Solution:
velocity of car at t=4sec is

v=u+at

v=0+5(4)

=20 m s-1

At t=6sec

acceleration is due to gravity ∴a=g=10 m s-2

vx=20 m s-1 (due to car)

vy=u+at

=0+g(2) (downward)

=20 m s-1 (downward)

v=vx2+vy2

=202+202

=202 m s-1

Topic: Physics, Motion In One Dimension, NEET, NEET 2021
A car starts from rest and moves with uniform acceleration a on a straight road from time t=0 to t=T. After that, a constant deceleration of magnitude a brings it to rest. In this process the average speed of the car is

Options:
A aT4
B 3aT2
C aT2
D aT

The correct answer is:
aT2

Solution:
Distance while accelerating
X1=12aT2 ......(1)
Velocity v=aT .......(2)
Time taken during deceleration
0=aT-at
t=T
∴ Distance while decelerating
x2=aTT-12aT2
x2=12aT2 .......(3)
Average Speed =x1+x2T+T
vav=12aT2+12aT22T
vav=aT2

Topic: Physics, Motion In One Dimension, NEET
A car travelling on a straight track moves with a uniform velocity of v1 for some time and with uniform velocity v2 for the next equal time. The average velocity of the car is

Options:
A v1v2
B 2v1v2v1+v2
C v1+v22
D v1-v22

The correct answer is:
v1+v22

Solution:
Average velocity is equal to ratio of total displacement to total time. Velocity is a vector quantity and average velocity is defined as the total displacement undergone by the body per unit time.
Displacement in first instance =v1t
Displacement in second instance =v2t
â´ Average velocity =v1t+v2tt+t=v1+v22

Topic: Physics, Motion In One Dimension, NEET
A car turns a corner on a slippery road at a constant speed of 10 m s-1. If the coefficient of friction is 0.5, the minimum radius of the arc in m in which the car turns is

Options:
A 20
B 10
C 5
D 4

The correct answer is:
20

Solution:
Given, v=10 m s-1 ,μ=0.5
We know that vmax=μgR
On squaring both sides, we get
R=v2μg
=1000.5×10=20 m

Topic: Physics, Laws of Motion, NEET
A car weighing 1000 kg is going up an incline with a slope of 2 in 25 at a steady speed of 18 kmph. If g=10 ms-2 , the power of its engine is

Options:
A 4 kW
B 50Â kW
C 625Â kW
D 25Â kW

The correct answer is:
4 kW

Solution:
P=mgsinθ×V=4000 W

Topic: Physics, Work Power Energy, NEET
A carbon resistor of 47±4.7 kΩ is to be marked with rings of different colours for its identification. The colour code sequence will be:

Options:
A Yellow - Green - Violet - Gold
B Yellow - Violet - Orange - Silver
C Violet - Yellow - Orange - Silver
D Green - Orange - Violet - Gold

The correct answer is:
Yellow - Violet - Orange - Silver

Solution:
Given resistance of a carbon resistor is

R=47±4.7×103

Where according to color code of carbon resistor given digits are digit 4 for yellow color, digit 7 for violet color, the multiplier 103 for orange color and tolerance 4·747=10% for silver color.

That's why as per color code, the sequence of colors rings are

Yellow - Violet - Orange - Silver

Topic: Physics, Current Electricity, NEET, NEET 2018
A carbonyl compound reacts with hydrogen cyanide to form cyanohydrine which on hydrolysis forms a racemic mixture of $\alpha$-hydroxy acid. The carbonyl compound is:

Options:
A formaldehyde
B acetaldehyde
C acetone
D diethyl ketone

The correct answer is:
acetaldehyde

Solution:
Caution
In racemic mixture, $\alpha$-C-atoms must have asymmetric carbon atoms.

Topic: Chemistry, Aldehydes and Ketones, NEET, NEET 2006
A Carnot engine absorbs 1000 J of heat energy from a reservoir at 127 Â°C and rejects 600 J of heat energy during each cycle. The efficiency of the engine and temperature of the sink will be

Options:
A 70% and -10°C
B 50% and -20°C
C 40% and -33°C
D 20% and -43°C

The correct answer is:
40% and -33°C

Solution:
Q1=1000 J
Q2=600
Î·=1-Q2Q1
=40%
T1T2=Q1Q2
400T2=1000600
T2=240K-273=-33 Â°C

Topic: Physics, Thermodynamics, NEET
A Carnot engine efficiency is equal to 17. If the temperature of the sink is reduced by 65 K, the efficiency becomes 14. The temperature of the source and the sink in the first case are respectively

Options:
A 620 K, 520 K
B 520 K, 606.67 K
C 606.67 K, 520 K
D 520 K, 610 K

The correct answer is:
606.67 K, 520 K

Solution:
âµ1-T2T1=17 (in case 1)
â´ T2T1=67 ... (i)
1-T2-65T1=14 (in case 2)
â´ T2-65T1=34 ...(ii)
From Eqs. (i) and (ii),
T2T1T2-65T1=6734=87
âT2T2-65=87
â7T2=8T2-8Ã65
âT2=520 K
â´T2T1=67
â´T1=T2Ã76=520Ã76
=606.67 K

Topic: Physics, Thermodynamics, NEET
A Carnot engine has an efficiency of 16. When the temperature of the sink is reduced by 62 â, its efficiency is doubled. The temperature of source and sink are,

Options:
A 99 ℃, 37 ℃
B 124 ℃, 62 ℃
C 37 ℃, 99 ℃
D 62 ℃, 124 ℃

The correct answer is:
99 ℃, 37 ℃

Solution:
η=1-T2T1 ⇒T2T1=1-η=1-16=56 ...(i)
in the second case T2-62T1=1-η′=1-26=23 ...(ii)
from Eqs. (i) and (ii)
now T2-62=23T1=23×65 T2
⇒ T2=310 K=310-273=37 ℃
T1=65 T2=65 ×310=372 K=372-273=99 ℃

Topic: Physics, Thermodynamics, NEET
A Carnot engine has an efficiency of 50% when its source is at a temperature 327 °C. The temperature of the sink is:

Options:
A 200 °C
B 27 °C
C 15 °C
D 100 °C

The correct answer is:
27 °C

Solution:
Efficiency of Carnot engine is given by,

η=1-TsinkTsource⇒50100=1-Tsink327+273 K⇒Tsink=12×600=300 K=300-273 °C=27 °C

Topic: Physics, Thermodynamics, NEET, NEET 2023 (All India)
A Carnot engine has the same efficiency between (i) 100 K and 500 K, (ii) T K and 900 K. The value of T is

Options:
A 180 K
B 90 K
C 270 K
D 360 K

The correct answer is:
180 K

Solution:
η=1-T2T1=1-100500=1-T900
∴ T900=15 or T=180 K

Topic: Physics, Thermodynamics, NEET
A Carnot engine, having an efficiency of Î·=1/10 as a 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 a lower temperature is

Options:
A 99 J
B 90 J
C 1 J
D 100 J

The correct answer is:
90 J

Solution:
Relation between coefficient of performance and efficiency of refrigerator is
β=1-ηη
∴ β=1-110110=9
Coefficient of performance, β=Heat absorbed(Q2)Work done (W)
⇒9=Q210
Q2=90 J

Topic: Physics, Thermodynamics, NEET
A Carnot engine, having an efficiency of η=110 as 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 a lower temperature is:

Options:
A 100 J
B 99 J
C 90 J
D 1 J

The correct answer is:
90 J

Solution:
For engine and refrigeration operating between two same temperature, the relationship between efficiency and coefficient of performance β,

η=11+β

110=11+β⇒β=9

From the principle of the refrigerator, coefficient of performance is given by the ratio of the amount of energy absorbed from the reservoir to work, i.e.,

Q2W=β

Q210=9

Q2=90 J.

Topic: Physics, Thermodynamics, NEET, NEET 2015 (Phase 1)
A Carnot engine having an efficiency of 110 is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at a lower temperature is

Options:
A 99 J
B 100 J
C 1 J
D 90 J

The correct answer is:
90 J

Solution:
For Carnot engine using as refrigerator
W=Q2T1T2-1
It is given η=110⇒η=1-T2T1⇒T2T1=910
So ,Q2=90 J as W=10 J

Topic: Physics, Thermodynamics, NEET
A Carnot engine, having an efficiency of a heat engine (Î·=110), is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at a lower temperature is

Options:
A 99 J

B 90 J

C 1 J

D 100 J

The correct answer is:
90 J

Solution:
Relation between coefficient of performance and efficiency of refrigerator is
Î²=1-Î·Î·
â´ Î²=1-110110=9
Coefficient of performance, Î²=Heat absorbed(Q2)Work done (W)
â 9=Q210
or Q2=90 J

Topic: Physics, Thermodynamics, NEET
A Carnot engine operates with a source at 500 K and sink at 375 K. Engine consumes 600 kcal of heat per cycle. The heat rejected to sink per cycle is

Options:
A 250 kcal
B 350 kcal
C 450 kcal
D 550 kcal

The correct answer is:
450 kcal

Solution:
The efficiency of the engine is
Î· = 1 - T 2 T 1 = 1 - 375 500 = 25 %
Heat consumed per cycle Q 1 = 600 kcal
Now, as, [ Î·= Q 1 â Q 2 Q 1 ]
â Heat rejected to the sink is Q 2 = Q 1 âÎ·. Q 1
Q 1 - Î· Q 1 = 6 0 0 - 0.25 Ã 6 0 0 = 450 kcal

Topic: Physics, Thermodynamics, NEET
A Carnot engine operates with a source at 500 K and sinks at 375 K. Engine consumes 600 kcal of heat per cycle. The heat rejected to sink per cycle is

Options:
A 250 kcal
B 350 kcal
C 450 kcal
D 550 kcal

The correct answer is:
450 kcal

Solution:
The efficiency of the engine is
Î· = 1 - T 2 T 1 = 1 - 375 500 = 25 %
Heat consumed per cycle Q 1 = 600 kcal
Now, as, [ Î·= Q 1 â Q 2 Q 1 ]
â Heat rejected to the sink is Q 2 = Q 1 âÎ·. Q 1
Q 1 - Î· Q 1 = 6 0 0 - 0.25 Ã 6 0 0 = 450 kcal

Topic: Physics, Thermodynamics, NEET
A Carnot engine takes 3Ã106 cal of heat from a reservoir at 627 â and gives it to a sink at 27 â. The work done by the engine is

Options:
A 4.2Ã106Â J
B 8.4Ã106 J
C 16.8Ã106Â J
D Zero

The correct answer is:
8.4Ã106 J

Solution:
Given, T1=627+273=900K
Q1=3Ã106 cal
T2=27+273=300 K
â´ Q1T1=Q2T2
â Q2=T2T1ÃQ1
=300900Ã3Ã106
=1Ã106 cal
Work done =Q1-Q2
=3Ã106-1Ã106=2Ã106cal
Work done =( 2Ã 10 6 âcal )( 4.2âJ 1âcal )=8.4Ã 10 6 âJ

Topic: Physics, Thermodynamics, NEET
A Carnot engine takes 300 calories of heat from a source at 500 K and rejects 150 calories of heat to the sink. The temperature of the sink is

Options:
A 125K
B 250 K
C 750Â K
D 1000Â K

The correct answer is:
250 K

Solution:
Given, Q1=300 cal, Q2=150 cal
T1=500 K, T2= ?
Q2Q1=T2T1
150300=T2500
T2=250 K

Topic: Physics, Thermodynamics, NEET
A Carnot engine whose sink is at $300 \mathrm{~K}$ has an efficiency of $40 \%$. By how much should the temperature of source be increased so as to increase its efficiency by $50 \%$ of original efficiency?

Options:
A $380 \mathrm{~K}$
B $275 \mathrm{~K}$
C $325 \mathrm{~K}$
D $250 \mathrm{~K}$

The correct answer is:
$250 \mathrm{~K}$

Solution:
Efficiency of carnot engine =
$$
\begin{aligned}
& \eta=1-\frac{T_2}{T_1} \\
& \text { or } \quad \frac{T_2}{T_1}=1-\eta \\
& =1-\frac{40}{100}=\frac{3}{5} \\
& \therefore \quad T_1=\frac{50}{100} n=\frac{3}{2} n \\
& =500 \mathrm{~K} \\
& \begin{array}{l}
\text { New efficiency } \\
\eta^{\prime}=\eta+\frac{50}{100} \eta=\frac{3}{2} \eta=60 \%
\end{array} \\
& \therefore \quad \frac{T_2}{T_1}=1-\frac{60}{100}=\frac{2}{5} \\
& \therefore T_1^{\prime}=\frac{5}{2} \times T_2=\frac{5}{2} \times 300=750 \text {. } \\
&
\end{aligned}
$$
Increase in temperature of source
$$
\begin{aligned}
& =T_1^{\prime}-T_1 \\
& =750-300=250 \mathrm{~K} .
\end{aligned}
$$

Topic: Physics, Thermodynamics, NEET, NEET 2006
A Carnot engine working between 300 K and 400 K has 800 J of useful work. The amount of heat energy supplied to the engine from the source is

Options:
A 2400 J

B 3200 J

C 1200 J

D 3600 J

The correct answer is:
3200 J

Solution:
Î·=WorkdoneHeatsuppliedâ¡
=T1-T2T1=WQ1
W=800 J,T1=400K,T2=300K
On substituting the value
800Q1=400-300400=100400=14
Q1=800Ã4=3200 J

Topic: Physics, Thermodynamics, NEET
A Carnot's engine having an efficiency of 110 as 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 lower temperature is

Options:
A 1  J
B 90  J
C 99  J
D 100  J

The correct answer is:
90  J

Solution:
Î²= coefficient of performance of refrigerator
Î·= efficiency of heat engine
Î²=1-Î·Î·
=1-110110=910110
Î²=9
Î²=Q2W
âµQ2= heat absorbed from sink and W= work done in a cycle
Q2=9Ã10=90 J

Topic: Physics, Thermodynamics, NEET, NEET 2017
A Carnot's engine works between a source at a temperature of 27 â and a sink at -123 â. Its efficiency is

Options:
A 0.5
B 0.25
C 0.75
D 0.4

The correct answer is:
0.5

Solution:
T1=27 ℃=27+273 K=300 K
T2=-123+273=150 K
η=1-T2T1=1-150300=0.5

Topic: Physics, Thermodynamics, NEET
A cart loaded with sand having total mass m0 = 1800 kg moves on a straight horizontal road starting from rest under the action of a force of 120 N. The sand spills through a small hole in the bottom at a rate of K = 0.5 kg/s. What will be the velocity of the cart after 20 min ?

Options:
A 120 m/s
B 60 m/s
C 80 m/s
D 40 m/s

The correct answer is:
120 m/s

Solution:
In this case, sand falls on the ground vertically and comes to rest. Hence, horizontal velocity of ejected sand with respect to ground is zero; so
Fâ¡ ext = dâ¡ dâ¡ t mv or Fâ¡ ext dt = dâ¡ mv
Which on integration gives
Fâ¡ ext ∫ t 1 t 2 dt = ∫ dâ¡ mv
Fâ¡ ext t 2 - t 1 = mâ¡ 2 v 2 - mâ¡ 1 v 1 … i
The rate of leakage of sand, K = (–dm / dt) is constant which on integration gives
∫ mâ¡ 1 mâ¡ 2 dâ¡ mâ¡ = - Kâ¡ ∫ t 1 t 2 dâ¡ t
(m2 – m1) = –K(t2 – t1) ...(ii)
Here, t1 = 0, t2 = t, v1 = 0, v2 = v
From Eqns. (i) and (ii), Fext . t = (m0 – Kt)v
v = Fâ¡ ext · t m 0 - K t
Substituting for Fext, t, K and m0, we get
v = 1 2 0 × 2 0 × 6 0 1 8 0 0 - 0 . 5 × 2 0 × 6 0 = 1 2 0 m/s

Topic: Physics, Center of Mass Momentum and Collision, NEET
A cart of mass M is tied to one end of a massless rope of length 10 m. The other end of the rope is in the hands of a man of mass M, the entire system is on a smooth horizontal surface. The man is at x=0 and the cart at x=10 m. If the man pulls the cart by a rope, the man and the cart will meet at the point

Options:
A xÂ =Â 0
B x=5 m
C x=10 m
D They will never meet

The correct answer is:
x=5 m

Solution:
They meet at the centre of mass
xCOM=m1x1+m2x2m1+m2
=M×0+M×102M=5 m

Topic: Physics, Center of Mass Momentum and Collision, NEET
A cation is in the centre touches three anions. Assume that the anions also touch each other. The limiting radius ratio, r+/r- is

Options:
A 0.1547
B 0.4141
C 0.7322
D 0.2252

The correct answer is:
0.1547

Solution:
r++r-r-=23
r+r++1=23
r+r-+1=1.1547
r+r-=0.1547

Topic: Chemistry, Solid State, NEET
A cavity of the radius R2 is made inside a solid sphere of radius R. The centre of the cavity is located at a distance R2 from the centre of the sphere. The gravitational force on a particle of mass m at a distance R2 from the centre of the sphere on the line joining both the centres of sphere and cavity is (opposite to the centre of the cavity).
[Here g=GMR2, where M is the mass of the sphere]

Options:
A mg2
B 3 mg8
C mg16
D None of these

The correct answer is:
3 mg8

Solution:
Gravitation field at mass m due to a full solid sphere
E→1=-g0r→R=-g0r^2

Gravitational field due to the cavity at mass m

M'=R23R3M ⇒M'=M8
E→2=--GM' r^R2,
=GMr^8R2=g0r^8
E→net=-g02+g08r^
=-3g0r^8
⇒ F→net=-3mg0r^8

Topic: Physics, Gravitation, NEET
A cell at the telophase stage is observed by a student in a plant brought from the field. He tells his teacher that this cell is not like other cells at the telophase stage. There is no formation of the cell plate and thus the cell is containing more chromosomes as compared to other dividing cells. This would result in:

Options:
A Aneuploidy
B Polyploidy
C Somaclonal variation
D Polyteny

The correct answer is:
Polyploidy

Solution:
Polyploid cells have a chromosome number that is more than double the haploid number. Polyploidy is common in nature and provides a major mechanism for adaptation and speciation. The cells of these plants do not form cell plate at the end of cell division thus the cells have an extra number of chromosomes as compared to other dividing cells.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2016 (Phase 1)
A cell can be balanced against $110 \mathrm{~cm}$ and $100 \mathrm{~cm}$ of potentiometer wire, respectively with and without being short circuited through a resistance of $10 \Omega$ Its internal resistance is

Options:
A $1.0 \Omega$
B $0.5 \Omega$
C $2.0 \Omega$
D zero

The correct answer is:
$1.0 \Omega$

Solution:
Key Idea : This problem is based on the application of potentiometer in which we find the internal resistance of a cell.
In potentiometer experiment in which we find internal resistance of a cell, let $E$ be the emf of the cell and $V$ the terminal potential difference, then
$\frac{E}{V}=\frac{l_1}{l_2}$
where $l_1$ and $l_2$ are lengths of potentiometer wire with and without short circuited through a resistance.
Since, $\frac{E}{V}=\frac{R+r}{R}[\because E=I(R+r)$ and $V=I R]$
$\therefore \frac{R+r}{R}=\frac{l_1}{l_2}$
$\begin{aligned}
1+\frac{r}{R} & =\frac{110}{100} \\
\frac{r}{R} & =\frac{10}{100} \\
r & =\frac{1}{10} \times 10=1 \Omega
\end{aligned}$
$r=\frac{1}{10} \times 10=1 \Omega$

Topic: Physics, Current Electricity, NEET, NEET 2008 (Screening)
A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short-circuited through a resistance of 10 Î©. Its internal resistance is

Options:
A 1 Ω
B 0.5 Ω
C 2 Ω
D Zero

The correct answer is:
1 Ω

Solution:
In potentiometer experiment in which we find internal resistance of a cell, let E be the emf of the cell and V the terminal potential difference, then
EV=l1l2
Where l1 and l2 are lengths of potentiometer wire with and without short circuited through a resistance.
Since,
EV=R+rR âµE=IR+rand V=IR
â´ R+rR=l1l2
or 1+rR=110100 or rR=10100
or r=110Ã10=1 Î©

Topic: Physics, Current Electricity, NEET
A cell divides every one minute. At this rate of cell division it can fill a 100ml of beaker in one hour. How much time does it take to fill a 50ml of beaker?

Options:
A 29 minutes
B 30 minutes
C 59 minutes
D 60 minutes

The correct answer is:
59 minutes

Solution:
Bacteria divide every 1 minute. It means the generation time of bacteria is 60 seconds. It is given that it takes 60 minutes to fill a beaker. Thus after one hour the beaker will be completely filled. Hence half of the beaker i.e. 50 ml will be filled in 59 minutes.

Topic: Biology, Organisms and Populations, NEET
A cell during the equational division of genetic material will show which of the following events?

I. interphase

II. anaphase

III. cytokinesis

Options:
A I, II and III
B I and III only
C I only
D II only

The correct answer is:
II only

Solution:
Mitosis is one of the two divisional phase in cell-cycle. It occurs after interphase and is succeeded by cytokinesis (cytoplasm division). Mitosis has prophase, metaphase, anaphase and telophase as the stages of karyokinesis in the same order.

Topic: Biology, Cell Cycle and Cell Division, NEET
A cell has 24 chromosomes at the end of metaphase in mitosis. What will be the number of chromosomes and chromatids at each pole at the beginning of telophase?

Options:
A 12 chromosomes, 24 chromatids
B 24 chromosomes, 12 chromatids
C 12 chromosomes, 12 chromatids
D 24 chromosomes, 24 chromatids

The correct answer is:
24 chromosomes, 24 chromatids

Solution:
In the metaphase of mitosis, the condensation of chromosomes have already taken place and they move to spindle equator to get themselves aligned along with the metaphase plate. In the anaphase, daughter chromosomes are formed, the centromeres of each chromosome splits and chromatids separate, therefore at the beginning of telophase there are 24 chromosomes and 24 chromatids at each pole.

Topic: Biology, Cell Cycle and Cell Division, NEET
A cell increases in volume, if the external medium is

Options:
A Hypotonic
B Hypertonic
C Isotonic
D None of these

The correct answer is:
Hypotonic

Solution:
If the plasmolysed cell (flaccid cell) is placed in a hypotonic solution then endosmosis occurs, which makes the cell again turgid (volume increases).

Topic: Biology, Transport in Plants, NEET