datavorous/entrance-exam-dataset huggingface dataset:
A 5 kg mass bounces in simple harmonic motion at the end of a spring. At which point the acceleration of the mass is the greatest?

Options:
A When the spring is fully compressed and when the spring is fully extended
B When the spring is at its rest length
C When the spring is halfway between its rest length and is fully extended or compressed length
D The acceleration is constant.

The correct answer is:
When the spring is fully compressed and when the spring is fully extended

Solution:
When the spring is fully compressed or fully extended is when the Hook's law forces are the greatest, and this indicates that the acceleration is the greatest. F=ma .

Topic: Physics, Oscillations, NEET
A $0.5 \mathrm{~kg}$ ball moving with a speed of 12 $\mathrm{m} / \mathrm{s}$ strikes a hard wall at an angle of $30^{\circ}$ with the wall. It is reflected with the same speed at the same angle. If the ball is in contact with the wall for 0.25 seconds, the average force acting on the wall is:

Options:
A $96 \mathrm{~N}$
B $48 \mathrm{~N}$
C $24 \mathrm{~N}$
D $12 \mathrm{~N}$

The correct answer is:
$24 \mathrm{~N}$

Solution:
Applying law of conservation of momentum.

Change in final momentum = initial momentum
$m v \sin \theta$ after collision $-(-m v \sin$ $\theta$) before collision
$\begin{aligned}
& f \times t=\text { change in momentum }=2 m v \sin \theta \\
& \Rightarrow \quad f=\frac{2 m v \sin \theta}{t}
\end{aligned}$
Putting the values we get,
$\begin{aligned}
f= & \frac{2 \times 0.5 \times 12 \times \sin 30^{\circ}}{0.25} \\
& =24 \mathrm{~N} .
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2006
A 5 molar solution of H2SO4 is diluted from 1 L to 10 L. What is the normality of the solution?

Options:
A 0.25 N
B 1 N
C 7 N
D 2 N

The correct answer is:
1 N

Solution:
Molarity of H2SO4=5 M
Normality of H2SO4=2×5=10 N
N1V1=N2V2
10×1=N2×10
N2=1 N

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A 5 molar solution of H 2 S O 4 â is diluted from 1 litre to a volume of 10 litres, the normality of the resulting solution will be :

Options:
A 0.1N
B 5N
C 0.5N
D 1N

The correct answer is:
1N

Solution:
Dilation formula
M 1 V 1 = M 2 V 2 ⇒ M 2 = 0 . 5
N 2 = 0 . 5 X Factor = 0 . 5 × 2 = 1 . 0 N

Topic: Chemistry, Redox Reactions, NEET
A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

Options:
A 352.12 K
B 269.07 K
C 112.24 K
D 411.03 K

The correct answer is:
269.07 K

Solution:
Correct Option is : (B)
269.07 K

Topic: Chemistry, Solutions, NEET
A 5 watt source emits monochromatic light of wavelength $5000 Å$. When placed $0.5 \mathrm{~m}$ away, it liberates photoclectrons from a photosensitive metallic surface. When the source is moved to a distance of $1.0 \mathrm{~m}$, the number of photoelectrons liberated will be reduced by a factor of :

Options:
A 8
B 16
C 2
D 4

The correct answer is:
4

Solution:
Intensity is given by
$I=\frac{p}{4 \pi d^2}$
where $p=$ power of source
$d=$ distance
We know (assume) that light to spread out uniformly is all direction i.e. spherical source.
$\begin{aligned}
& \therefore I \propto \frac{1}{d^2} \text { or } \frac{I_1}{I_2}=\frac{d_2^2}{d_1^2} \\
& \text {or } \frac{I_1}{I_2}=\left(\frac{1}{0.5}\right)^2 \\
& \Rightarrow \frac{I_1}{I_2}=4 \\
& \text {or } I_2=\frac{I_1}{4}
\end{aligned}$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2007
A 5-ampere fuse wire can withstand a maximum power of 1 watt in the circuit. The resistance of the fuse wire is :

Options:
A $0.04 \mathrm{ohm}$
B $0.2 \mathrm{ohm}$
C $5 \mathrm{ohm}$
D $0.4 \mathrm{ohm}$

The correct answer is:
$0.04 \mathrm{ohm}$

Solution:
$$
\begin{aligned}
P & =I^2 R \\
\Rightarrow \quad & I=25 \times R \\
\Rightarrow \quad & R=\frac{1}{25}=0.04 \Omega
\end{aligned}
$$

Topic: Physics, Current Electricity, NEET, NEET 2005
A 6 V battery is connected to the terminals of the three-metre long wire of uniform thickness and resistance of 100 Î©. The difference of potential between two points on the wire separated by a distance of 50 cm will be

Options:
A 1.5 V
B 3.0 V
C 1.0 V
D 2.0 V

The correct answer is:
1.0 V

Solution:
R∝ℓ
For 300 cm, R =100 Ω
For 50 cm, R'=100300×50=503 Ω
∴IR=6
⇒IR'=6R×R'=6100×503=1 V

Topic: Physics, Current Electricity, NEET
A 6 volt battery is connected to the terminal of a three metre long wire of uniform thickness and resistance of 100 $\mathrm{ohm}$. The difference of potential between two points on the wire separated by a distance of $50 \mathrm{~cm}$ will be:

Options:
A 2 volt
B 3 volt
C 1 volt
D 1.5 volt

The correct answer is:
1 volt

Solution:
$$
\begin{gathered}
\therefore \text { Voltage on } 50 \mathrm{~cm}=\frac{6}{300} \times 50 \\
=1 \text { volt }
\end{gathered}
$$

Topic: Physics, Current Electricity, NEET, NEET 2004
A 10.0 L flask contains 64 g of oxygen at 27°C. (Assume O2 gas is behaving ideally). The pressure inside the flask in bar is

(Given R=0.0831 L bar K-1 mol-1)

Options:
A 498.6
B 49.8
C 4.9
D 2.5

The correct answer is:
4.9

Solution:
From the ideal gas equation,

P=nRTVP=pressure of oxygenV=Volume of oxygenT=temperature of oxygen in Kn=Number of moles of oxygen gas

n=6432=2

P=2×0.0831×30010=4.986 bar

Topic: Chemistry, States of Matter, NEET, NEET 2022 (Phase 1)
A $10 \mu \mathrm{F}$ capacitor is connected to a $210 \mathrm{~V}, 50 \mathrm{~Hz}$ source as shown in figure. The peak current in the circuit is nearly $(\pi=3.14)$ :

Options:
A 0.93 A
B 1.20 A
C $\quad 0.35 \mathrm{~A}$
D 0.58 A

The correct answer is:
0.93 A

Solution:
Capacitive Reactance
$\begin{aligned}
X_C & =\frac{1}{\omega C}=\frac{1}{2 \pi f C}=\frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} \\
& =\frac{1000}{3.14}
\end{aligned}$
$\begin{aligned}
& V_{\mathrm{rms}}=210 \mathrm{~V} \\
& i_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C}=\frac{210}{X_C}
\end{aligned}$
$\begin{aligned}
\text { Peak current } & =\sqrt{2} i_{\mathrm{ms}}=\sqrt{2} \times \frac{210}{1000} \times 3.14=0.932 \\
& \simeq 0.93 \mathrm{~A}
\end{aligned}$

Topic: Physics, Alternating Current, NEET, NEET 2024
A 10 g bullet moving directly upward at 1000 m s-1 strikes and passes through the centre of mass of a 10ââkg block initially at rest. The bullet emerges from the block moving directly upward at 400 m s-1. What will be the velocity of the block just after the bullet comes out of it?

Options:
A 1 m s-1
B 0.4 m s-1
C 1.4 m s-1
D 0.6 m s-1

The correct answer is:
0.6 m s-1

Solution:
Given,
Weight of bullet mB=10 g
=10×10-3 kg
Weight of block M=10 kg
The initial velocity of the bullet vi=1000 m s-1
The final velocity of the bullet vf=400 m s-1
By the law of conservation of momentum,
mBvi=mBvf+10v [where, v=velocity of block]
10×10-31000=10×10-3400+10v
10=4+10v
10v=10-4
v=610=0.6 m s-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
A 1.0 g sample of substance A at 100 oC is added to 100 mL of H2O at 25 oC . Using separate 100 mL portions of H2O , the procedure is repeated with substance B and then with substance C. How will the final temperatures of the water compare?

Substance	Specific Heat

A	0.60 Jg-1oC-1

B	0.40 Jg-1oC-1

C	0.20 Jg-1oC-1

Options:
A TC>TB>TA
B TB>TA>TC
C TA>TB>TC
D TA=TB=TC

The correct answer is:
TA>TB>TC

Solution:
Heat absorbed or evolved, ΔQ=msΔt
Where, m = mass of substance, s = specific heat
Δt= temperature difference
It me Solution: s∝1Δt
Higher the temperature of given solution, lesser is the temperature difference. So higher is the specific heat. As, specific heat order is A > B > C. So, order of temperature of solution is A > B > C.

Topic: Chemistry, Thermodynamics (C), NEET
A 1.0 kg ball drops vertically into a floor from a height of 25 cm. It rebounds to a height of 4 cm. The coefficient of restitution for the collision is

Options:
A 0.16
B 0.32
C 0.40
D 0.56

The correct answer is:
0.40

Solution:
Initial velocity before colliding floor u=2gh
u=5
final velocity after colliding floor v=2gh′
v=0.8
We know that coefficient of restitution e=vu
e=0.85=0.4

Topic: Physics, Work Power Energy, NEET
A 10 kg mass travelling 2 m/s meets and collides elastically with a 2 kg mass travelling 4 m/s in the opposite direction. Find the final velocities of both objects.

Options:
A VAf = 2 m/s ,âVBf = 3 m/s

B VAf = 0 m/s ,âVBf = 6 m/s
C VAf = 5 m/s ,âVBf = 8 m/s
D VAf = 4 m/s ,âVBf = 2 m/s

The correct answer is:
VAf = 0 m/s ,âVBf = 6 m/s

Solution:
We know the following
mA=10 kg .
VAi=2 m/s
mB=2 kg
VBi=−4 m/s . The negative sign is because the velocity is in the negative direction.
Now we need to find V Af and V Bf . Use the equations from above. Let’s start with V Af .
VAf=mA−mBmA+mBVAf+(mB−mA)(mA−mB)VBi
Plug in our known values.
VAf=(10−2)kg(10+2)kg.(2 m/s)+2(2kg)(10+2kg).(−4 m/s)
VAf=812.(2 m/s)+412.(−4 m/s)
VAf=1612 m/s+−1612 m/s
VAf=0 m/s
The final velocity of the larger mass is zero. The collision completely stopped this mass.
Now for V Bf
VAf=2mA(mA+mB)VAf+(mB−mA)(mA−mB)VBi
Plug in our known values
VBf=2(10kg)(10+2)kg.2 m/s+(2−10)kg(10+2kg).−4 m/s
VBf=4012 m/s+3212 m/s
VBf=72 kg12 kg m/s
VBf=6 m/s

Topic: Physics, Center of Mass Momentum and Collision, NEET
A 10 kg stone is suspended with a rope of breaking strength 30 kg-wt. The minimum time in which the stone can be raised through a height 10 m starting from rest is
(Taking g = 10 m/s2)

Options:
A 0.5 s
B 1.0 s
C 2 3 s
D 2.0 s

The correct answer is:
1.0 s

Solution:
Maximum tension in the string T = mg
= 30 x 10

= 300 N

T - mg = ma
From the figure
300 - 10 x 10 = 10a
â´ a = 20 m s-2
Thus, the maximum acceleration with which the stone can be raised is 20 m s-2.
Given, s = 10 m
and u = 0
â´ 1 0 = 1 2 2 0 t 2
â t = 1 s

Topic: Physics, Laws of Motion, NEET
A $12 \mathrm{~cm}$ wire is given a shape of a right angled triangle $A B C$ having sides $3 \mathrm{~cm}$, $4 \mathrm{~cm}$ and $5 \mathrm{~cm}$ as shown in the figure. The resistance between two ends (AB, $B C, C A)$ of the respective sides are measured one by one by a multi-meter. The resistance will be in the ratio:

Options:
A $9: 16: 25$
B $27: 32: 35$
C $21: 24: 25$
D $3: 4: 5$

The correct answer is:
$27: 32: 35$

Solution:
Resistance of side $A B=R_1=\frac{3 \rho}{A}$
Resistance of side $B C=R_2=\frac{4 \rho}{A}$
Resistance of side $\mathrm{AC}=\mathrm{P}_3=\frac{5 \rho}{\mathrm{A}}$
(Where $A=$ Area of cross-section and $\rho=$ resistivity of wire)
$\therefore$ The resistance between $\mathrm{A}$ and $\mathrm{B}$ is
$$
\begin{aligned}
& \mathrm{R}_{\mathrm{AB}}=\frac{R_1\left(R_2+R_3\right)}{R_1+R_2+R_3}=\frac{\frac{3 \rho}{A}\left(\frac{4 \rho}{A}+\frac{5 \rho}{A}\right)}{\frac{3 \rho}{A}+\frac{4 \rho}{A}+\frac{5 \rho}{A}} \\
& =\frac{27 \rho}{12 A} \\
& \text { Similarly } R_{B C}=\frac{R_2\left(R_1+R_3\right)}{R_1+R_2+R_3} \\
& \frac{\frac{4 \rho}{A}\left(\frac{3 \rho}{A}+\frac{5 \rho}{A}\right)}{\frac{3 \rho}{A}+\frac{4 \rho}{A}+\frac{5 \rho}{A}}=\frac{32}{12} \frac{\rho}{A} \\
& \text { And } R_{A C}=\frac{R_3\left(R_1+R_2\right)}{R_1+R_2+R_3} \\
& =\frac{\frac{5 \rho}{A}\left(\frac{3 \rho}{A}+\frac{3 \rho}{A}\right)}{\frac{5 \rho}{A}+\frac{3 \rho}{A}+\frac{4 \rho}{A}}=\frac{35}{12} \frac{\rho}{A} \\
& \therefore R_{A B}: R_{B C}: R_{A C}=\frac{27}{12}: \frac{32}{12}: \frac{35}{12} \\
& =27: 32: 35 \\
&
\end{aligned}
$$

Topic: Physics, Current Electricity, NEET, NEET 2013 (Karnataka)
A 12 pF capacitor is connected to a 50 V battery, the electrostatic energy stored in the capacitor in nJ is

Options:
A 15
B 7.5
C 0.3
D 150

The correct answer is:
15

Solution:
Electrostatic energy stored $U=\frac{1}{2} C V^2$
$=\frac{1}{2} \times 12 \times 10^{-12} \times(50)^2$
$=6 \times 25 \times 10^{-10}$
$=15 \times 10^{-9} \mathrm{~J}$
$=15 \mathrm{~nJ}$

Topic: Physics, Capacitance, NEET, NEET 2024 (Re-NEET)
A 0.016 M of an acid solution in benzene is dropped on a water surface, the benzene evaporates and the acid forms a monomolecular film of solid type. What volume of the above solution would be required to cover a 500 cm2 surface area of water with a monomolecular layer of acid? The area covered by a single acid molecule is 0.2 nm.

Options:
A 0.94
B 0.0239
C 9.45
D 96.5

The correct answer is:
0.0239

Solution:
0.0239 cm3

Topic: Chemistry, Surface Chemistry, NEET
A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.

Options:
A 5
B 3
C 2
D 1

The correct answer is:
5

Solution:
l=20 cm; m=1 g; T=0.5 N; f=100 Hz
μ=1×10-320×10-2=12×10-2
V=0.50.5×10-2=10 m s-1 m/s
λ=0.1
λ2=0.05 m=5 cm

Topic: Physics, Waves and Sound, NEET
A 20 litre container at 400 K contains CO2 (g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO2 attains its maximum value, will be
(Given that: SrCO3(s)⇌SrO(s)+CO2(g) , Kp = 1.6 atm)

Options:
A 10 litre
B 4 litre
C 2 litre
D 5 litre

The correct answer is:
5 litre

Solution:
As here maximum press of CO2=Kp=1.6 atm.
P1V1=P2V2
V2=P1V1P2=0.4×201.6=5 litre

Topic: Chemistry, States of Matter, NEET
A 20 litre container at 400 K contains CO2g at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO ). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when the pressure of CO2 attains its maximum value, will be:
Given that:SrCO3s⇌SrOs+CO2g, Kp=1.6 atm

Options:
A 5 litre
B 10 litre
C 4 litre
D 2 litre

The correct answer is:
5 litre

Solution:
Max. pressure of CO2=Pressure of CO2 at equilibrium.
For reaction,
SrCO3s ⇌SrOs+CO2
Kp=PCO2=1.6 atm = maximum pressure of CO2.
The volume of the container at this stage.
V=nRTP ......(i)
Since the container is sealed and the reaction was not earlier at equilibrium
∴ n=constant
n=PVRT=0.4×20RT ......(ii)
Put equation (ii) in equation (i)
V=0.4×20RT RT1.6=5 L

Topic: Chemistry, States of Matter, NEET, NEET 2017
A $0.0020 \mathrm{~m}$ aqueous solution of an ionic compound $\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right) \mathrm{Cl}$ freezes at $-0.00732^{\circ} \mathrm{C}$. Number of moles of ions which $1 \mathrm{~mol}$ of ionic compound produces on being dissolved in water will be $\left(\mathrm{k}_{\mathrm{f}}=-1.86^{\circ} \mathrm{C} / \mathrm{m}\right)$

Options:
A 2
B 3
C 4
D 1

The correct answer is:
2

Solution:
Given,
$$
\begin{aligned}
\text { molality, } \mathrm{m} & =0.0020 \mathrm{~m} \\
\Delta \mathrm{T}_{\mathrm{f}} & =0^{\circ} \mathrm{C}-0.00732^{\circ} \mathrm{C}=0.00732^{\circ} \mathrm{C} \\
\mathrm{k}_{\mathrm{f}} & =1.86^{\circ} \mathrm{C} / \mathrm{m} \\
\Delta \mathrm{T}_{\mathrm{f}} & =\mathrm{i} \cdot \mathrm{k}_{\mathrm{f}} \times \mathrm{m} \\
\mathrm{i} & =\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{f}} \times \mathrm{m}} \\
& =\frac{0.00732}{1.86 \times 0.0020} \\
& =1.96 \approx 2
\end{aligned}
$$
Since, the compound is ionic, so number of moles produced is equal to vant' Hoff factor, $i$. Hence, 2 moles of ions are produced.

Topic: Chemistry, Solutions, NEET, NEET 2009 (Screening)
A 20 N metal block is suspended by a spring balance. A beaker containing some water is placed on a weighing machine which reads 40 N. The spring balance is now lowered so that the block gets immersed in the water. The spring balance now reads 16 N. The reading of the weighing machine will be

Options:
A 36Â N
B 60Â N
C 44 N
D 56Â N

The correct answer is:
44 N

Solution:
Let Reading of spring balance =R1
And reading of weighing machine be =R2
∴ R1+R2=w1+w2
16+R2=20+40
R2=44 N

Topic: Physics, Laws of Motion, NEET
A 20 V AC is applied to a circuit consisting of a resistance and a coil with negligible resistance. If the voltage across the resistance is 12 V, the voltage across the coil is

Options:
A 16 V
B 10 V
C 8 V
D 6 V

The correct answer is:
16 V

Solution:
The voltage across LR combination is given by
V2=VR2+VL2
VL=V2-VR2
VL=400-144=256
VL=16 V

Topic: Physics, Alternating Current, NEET
"A 20 W bulb emits light of wavelength 5000 Å shining on a metal surface at a distance of 2 m. Assuming the metal surface has a work function of 2 eV, and each atom on the surface can be treated as a circular disk with a radius of 1.5 Å. What is the time required for the atom to receive energy equal to the work function of 2 eV?"

Options:
A 30 s
B 25 s
C 20 s
D 28.4 s

The correct answer is:
28.4 s

Solution:
Correct Option is : (D)
28.4 s

Topic: Chemistry, Dual Nature of Matter, NEET
A 24 kg block resting on a floor has a rope tied to its top. The maximum tension, the rope can withstand without breaking is 310 N. The minimum time in which the block can be lifted a vertical distance of 4.6 m by pulling on the rope is

Options:
A 1.2 s

B 1.3 s
C 1.7 s
D 2.3 s

The correct answer is:
1.7 s

Solution:
Effective upward force = 310 âmg
=310-24Ã9.8=74.8 N
Upward acceleration
â´ a = 74.8/24 = 3.12 m s - 2
As s=ut+12at2
4.6 =0+12Ã3.12Ãt2
or t2=4.61.56=2.95
or t=2.95=1.7 s

Topic: Physics, Laws of Motion, NEET
A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57µm. What is the rate of emission of quanta per second?

Options:
A 7.18 × 1019s–1
B 6.25 × 1019s–1
C 8.69 × 1019s–1
D 5.12 × 1019s–1

The correct answer is:
7.18 × 1019s–1

Solution:
Correct Option is : (A)
7.18 × 1019s–1

Topic: Chemistry, Structure of Atom, NEET
A 30 Î¼F capacitor is connected to a 150 V, 60 Hz AC supply. The RMS value of current in the circuit is

Options:
A 17 A
B 1.7 A
C 1.7 mA
D 1.7 Î¼A

The correct answer is:
1.7 A

Solution:
Given that
C=30 Î¼F=30Ã10-6 F
V=150 V, f=60 Hz
Irms=VXC=1501ÏC
=150ÃÏC
=150Ã2ÏfÃC
=150Ã2Ã3.14Ã60Ã30Ã10-6
=1.6956 Aâ1.7 A

Topic: Physics, Alternating Current, NEET
A 30 V, 90 W lamp is to be operated on a 120 V DC line. For proper glow, a resistor of

Î© should be connected in series with the lamp.

Options:
A 40
B 10
C 20
D 30

The correct answer is:
30

Solution:
Resistance of the lamp,
R0=V2P=30290=10 Ω
Current in the lamp,
i=3010=3 A
When lamp is operated on a 120 V, then resistance,
R′=V′i=1203=40 Ω
Thus, for proper glow, the resistance required to the put in series will be
R=R′-R0=40-10=30 Ω

Topic: Physics, Current Electricity, NEET
A 35 year old man has extremely long limbs and digits. He also has severe disfigurement of the facial features. Which of the following could be a probable cause for it?

Options:
A Hypersecretion of a hormone from neurohypophysis
B Hypersecretion of a hormone from adenohypophysis
C Hypersecretion of a hormone from thyroid gland
D Hypersecretion of a hormone from adrenal gland

The correct answer is:
Hypersecretion of a hormone from adenohypophysis

Solution:
The person is suffering from acromegaly. It is a condition seen in middle aged adults as a result of hypersecretion of growth hormone from the anterior pituitary, i.e., adenohypophysis. It is characterized by excessive growth of body, extremely long limbs and digits and severe disfigurement of facial features. It can lead to premature death.

Topic: Biology, Chemical Coordination and Integration, NEET
A 40 Î¼F capacitor in a defibrillator is charged to 3000 V. The energy stored in the capacitor is sent through the patient during a pulse of duration 2 ms. The power delivered to the patient is

Options:
A 45 kW
B 90 kW
C 180 kW
D 360 kW

The correct answer is:
90 kW

Solution:
Power, P=Energytime=12CV2t
P=CV22t
P=40×16-6×9×1062×2×10-3
P=90×1000
P=90 kW

Topic: Physics, Electrostatics, NEET
A 40 μF capacitor is connected to a 200 V, 50 Hz AC supply. The rms value of the current in the circuit is, nearly:

Options:
A 2.05 A
B 2.5 A
C 25.1 A
D 1.7 A

The correct answer is:
2.5 A

Solution:
Current flowing through the capacitor is given by,Irms=VXC=V1/Cω=VCω

⇒Irms=200×40×10−6×2π×50

⇒Irms=2.5 A

Topic: Physics, Alternating Current, NEET, NEET 2020 (Phase 1)
A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic masses: Na = 23, Hg = 200, 1 Faraday = 96500 coulombs). The total number of moles of chlorine gas evolved is

Options:
A 0.5
B 1.0
C 2.0
D 3.0

The correct answer is:
1.0

Solution:
NaCl→Na++Cl-
At anode: 2Cl-→Cl2
Moles of Cl-=2 in 500 ml.
Therefore 1 mole of Cl2 evolves.

Topic: Chemistry, Electrochemistry, NEET
A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter of 1.0 cm. What is the pressure exerted on the horizontal floor?

Options:
A 6.24 × 106 Pa

B 5.63 × 107 Pa

C 4.81 × 106 Pa

D 7.19 × 107 Pa

The correct answer is:
6.24 × 106 Pa

Solution:
Given, mass of girl (m) = 50 kg
Diameter of circular heel (2r) = 1.0 cm
Radius (r) = 0.5 cm
= 5 × 10-3 m
Area of the circular heel (A) = πr2
= 3.14 × (5 × 10-3)2 m2
= 78.50 × 10-6 m2
The pressure exerted on the horizontal floor
p = F A = mg A
= 50 × 9.8 78.50 × 10 - 6
= 6.24 × 106 Pa

Topic: Physics, Mechanical Properties of Solids, NEET
A 60kg body is pushed with just enough force to start it moving across a floor and the same force continues to act afterwards. The coefficient of static friction and sliding friction are 0.5 and 0.4 respectively. The acceleration of the body is

Options:
A 6 m s-2
B 4.9 m s-2
C 3.92 m s-2
D 1 m s-2

The correct answer is:
1 m s-2

Solution:
Force required to just start the motion is,
fL=Î¼smg
Once body starts sliding friction becomes kinetic,
â´ a=fL-fkm=Î¼smg-Î¼kmgm
a=Î¼s-Î¼kg=0.5-0.410 m s-2
a=1 m s-2

Topic: Physics, Laws of Motion, NEET
A 60W, 120 V bulb is connected to a 240 V, 60 Hz supply with an inductance in series. Find the value of inductance so that bulb gets correct voltage.

Options:
A 2.3πH
B 23 H
C πH
D 23πH

The correct answer is:
23πH

Solution:
R=120260=240 Ω we require i=0.5 A
Or Z=480 Ω
XL=4802-2402=2403 Ω
L= 240360×2π=2.3πH

Topic: Physics, Alternating Current, NEET
A 60 W bulb is placed at a distance of 4 m from you. The bulb is emitting light of wavelength 600 nm uniformly in all directions. In 0.1 s, how many photons will enter your eye, if the pupil of the eye has a diameter of 2 mm? [Take hc=1240 eV nm]

Options:
A 2.84 x 1012
B 2.84 x 1011
C 9.37 x 1011
D 6.84 x 1011

The correct answer is:
2.84 x 1011

Solution:
The intensity of light at the location of your eye is
I = P 4 π r 2 = 6 0 4 π × 4 2 W m - 2
The energy entering into your eye per second is
P 1 = I × π d 2 4
where d is the diameter of pupil
P 1 = 6 0 4 π × 4 2 × π × 2 × 1 0 - 3 2 4
= 9.375 × 1 0 - 7 Js - 1
Let n be the number of photons entering into the eye per second, then
P 1 = n × hc λ
9.375 × 1 0 - 7 = n × 1 2 4 0 × 1.6 × 1 0 - 1 9 6 0 0
n = 2.84 × 1 0 1 2 photon s - 1
So, the number of photons entering the eye in 0.1 s = 0.1 n = 2.84 x 1011

Topic: Physics, Wave Optics, NEET
A 70 kg man standing on ice throws a 3 kg body horizontally at 8 m s-1. The friction coefficient between the ice and his feet is 0.02. The distance, through which the man slip is

Options:
A 0.3 m
B 2 m
C 1 m
D â

The correct answer is:
0.3 m

Solution:
At the time of throwing body, velocity gained by man is
v0=3Ã 870=2470 m/s
â´ Stopping distance s=v022Î¼g
Or s=24Ã2470Ã70Ã2Ã0.02Ã10
=57649Ã40=144490=0.3 m

Topic: Physics, Center of Mass Momentum and Collision, NEET
A 90° corner is made from a transparent optical material with a refractive index such that A cannot see B when he is standing behind the corner. Minimum value of refractive index is

Options:
A 3
B 2
C 5
D 7

The correct answer is:
2

Solution:
rmin>θc
r=90-β
r is minimum when β is maximum.
The maximum value of β is θc
∴90o-θc>θc
45o>θc
⇒12>1μ
⇒μ>2

Topic: Physics, Ray Optics, NEET
A 100 Ω resistance and a capacitor of 100 Ω reactance are connected in series across a 220 V source. When the capacitor is 50% charged, the peak value of the displacement current is

Options:
A 2.2ââA
B 11  A
C 4.4  A
D 112 A

The correct answer is:
2.2  A

Solution:
Given, R=100 Ω, XC=100 Ω.

Z=XC2+R2=1002 Ω.

Vmax=2202 V.

I max = V max Z =22021002=2.2 A.

Topic: Physics, Alternating Current, NEET, NEET 2016 (Phase 2)