datavorous/entrance-exam-dataset huggingface dataset:
A ball of mass (m) 0.5 kg is attached to the end of a string having a length (L) 0.5 m. The ball is rotated on a horizontal circular path about the vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of the angular velocity of ball (in rad s-1) is

Options:
A 9
B 18
C 27
D 36

The correct answer is:
36

Solution:
r=l sin θ T cos θ component will cancel mg.
T sin θ component will provide the necessary centripetal force to the ball towards centre C.
Therefore Tsin θ=mrω2=m (l sin θ) ω2 or T= mlω2

Therefore ω=Tml

or ωmax=Tmaxml=3240.5=36 rad s-1

Topic: Physics, Laws of Motion, NEET
A ball of mass m=0.5 kg is attached to the end of a string having length L= 0.5 m is rotated on a horizontal circular path about the vertical axis. The maximum tension that the string can bear is 324 N. What is the maximum possible value of angular velocity of ball (in radian/s) ?

Options:
A 9
B 18
C 27
D 36

The correct answer is:
36

Solution:
T sin Î¸ = m Ï 2 r
T sin Î¸ = m Ï 2 L sin Î¸
T = m Ï 2 L
3 2 4 = 1 2 Ï 2 1 2
Therefore Ï = 3 6

Topic: Physics, Laws of Motion, NEET
A ball of mass m=0.5 kg is attached to the end of a string having length L=0.5 m. The ball is rotated on a horizontal circular path about the vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of the angular velocity of ball (in rad s-1) is

Options:
A 9
B 18
C 27
D 36

The correct answer is:
36

Solution:
T sinÎ¸=mÏ2r
T sinÎ¸=mÏ2LsinÎ¸

âT=mÏ2L
â 324=12Ï212
Therefore Ï=36 rad s-1

Topic: Physics, Laws of Motion, NEET
A ball of mass m=0.5 kg is attached to the end of a string having length L=0.5 m. The ball is rotated on a horizontal circular path about the vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of the angular velocity of ball (in rad s-1) is

Options:
A 9
B 18
C 27
D 36

The correct answer is:
36

Solution:
Tsinθ=mω2r
Tsinθ=mω2Lsinθ

⇒T=mω2L
⇒ 324=12ω212
Therefore ω=36 rad s-1

Topic: Physics, Laws of Motion, NEET
A ball of mass m approaches a wall of mass M M>>m, with speed 2 m s-1 along the normal to the wall. The speed of wall is 1 m s-1 towards the ball. The speed of the ball after an elastic collision with the wall is -

Options:
A 2 m s-1 away from the wall
B 3 m s-1 away from the wall
C 4 m s-1 away from the wall
D 5 m s-1 away from the wall

The correct answer is:
4 m s-1 away from the wall

Solution:
Let v be the velocity of ball after collision, collision is perfectly elastic
â´ e = 1
Or relative velocity of separation = relative velocity of approach
â´v â 1 = 2 + 1
Or v = 4m/s (away from the wall)

Topic: Physics, Center of Mass Momentum and Collision, NEET
A ball of mass m approaches a wall of mass M >>m with speed 4 m s-1 along the normal to the wall. The speed of the wall is 1 m s-1 towards the ball. The speed of the ball after an elastic collision with the wall is -

Options:
A 5 m s-1 away from the wall
B 9 m s-1 away from the wall
C 3 m s-1 away from the wall
D 6 m s-1 away from the wall

The correct answer is:
6 m s-1 away from the wall

Solution:
Let v be the velocity of ball after collision, collision is elastic
â´ e =1
or
relative velocity of separation=relative velocity of approach
â´ v- 1=4+1
or v=6 m/s (away from the wall)

Topic: Physics, Center of Mass Momentum and Collision, NEET
A ball of mass m approaches a wall of mass M >>m with speed 4 m s-1 along the normal to the wall. The speed of wall is 1 m s-1 towards the ball. The speed of the ball after an elastic collision with the wall is

Options:
A 5 m s-1 away from the wall
B 9 m s-1 away from the wall
C 3 m s-1 away from the wall
D 6 m s-1 away from the wall

The correct answer is:
6 m s-1 away from the wall

Solution:
Let v be the velocity of the ball after the collision.
As collision is elastic
â´ e=1 or
The relative velocity of separation = relative velocity of approach
â´ v-1=4+1
or v=6 m s-1 (away from the wall)

Topic: Physics, Center of Mass Momentum and Collision, NEET
A ball of mass m is dropped onto a smooth fixed wedge of inclination Î¸. The collision is perfectly elastic. If after the collision, the x component of the velocity of the block is 2gh, then Î¸ is

Options:
A 30°
B 45Â°
C 60°
D 15°

The correct answer is:
45°

Solution:
uâ=-2gh j^
vâ=uââvâ= 2gh i^
â ucosâ¡Î¸=vsinâ¡Î¸
tanâ¡Î¸=1
â Î¸=45o

Topic: Physics, Center of Mass Momentum and Collision, NEET
A ball of mass m moving at a speed v makes a head-on collision with an identical ball at rest. The kinetic energy of the balls after the collision is 34th of the original. What is the coefficient of restitution?

Options:
A 13
B 12
C 2
D 3

The correct answer is:
12

Solution:
Since we know that in the case, when m1=m2 and v1=0
then, v1â²=1+e2v2 and v2â²=1-e2v2
Given that, KF=34Ki
Using energy conservation
â 12mvâ²12+12mvâ²22=3412mv2
Substituting the values, we get
1+e22+1-e22=34
â 1+e2+1-e2=3
â 2+2e2=3âe2=12âe=12

Topic: Physics, Center of Mass Momentum and Collision, NEET
A ball of mass m moving with a speed u undergoes a head-on elastic collision with a ball of mass nm initially at rest. The fraction of the incident energy transferred to the heavier ball is

Options:
A n1+n
B n1+n2
C 2n1+n2
D 4n1+n2

The correct answer is:
4n1+n2

Solution:
Speed of heavier ball
v=0+2mm+nmu
v=2un+1
The fraction of energy transferred
=12nmv212mu2
=nv2u2
=nu22un+12
=4nn+12

Topic: Physics, Center of Mass Momentum and Collision, NEET
A ball of radius r and density ρ falls freely under gravity through a distance h before entering water. Velocity of ball does not change even on entering water. If viscosity of water is η, the value of h is given by

Options:
A 2 9 r 2 ( 1 - ρ η ) g
B 2 81 r 2 ( ρ - 1 η ) g
C 2 8 1 r 4 ( ρ - 1 η ) 2 g
D 2 9 r 4 ( ρ - 1 η ) 2 g

The correct answer is:
2 8 1 r 4 ( ρ - 1 η ) 2 g

Solution:
Velocity of ball when the water surface
v = 2 gh ............ i
Terminal velocity of ball inside the water
v = 2 9 r 2 g ρ - 1 η ............ i i
Equating (i) and (ii)we get 2 gh = 2 9 r 2 g η ρ - 1 ⇒ h = 2 8 1 r 4 ( ρ - 1 η ) 2 g

Topic: Physics, Mechanical Properties of Fluids, NEET
A ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet. In which direction will it move?

Options:
A Towards the magnet.
B Away from the magnet.
C Parallel to the magnet.
D No movement will occur.

The correct answer is:
Away from the magnet.

Solution:
Correct Option is : (B)
Away from the magnet.

Topic: Chemistry, Magnetic Properties of Matter, NEET
A ball reaches a racket at 60 m s-1 along +x direction and leaves the rocket in the opposite direction with the same speed. Assuming that the mass of the ball as 50 g and the contact time is 0.02 second, the force exerted by the racket on the ball is

Options:
A 300 N along + X direction

B 300 N along - X direction

C 300 kN along + X direction

D 300 kN along - X direction

The correct answer is:
300 N along - X direction

Solution:
The change in momentum of the ball, Δp=mv--mv=2mv=2×0·05×60=6 kg m s-1
Force exerted on the bell by the rocket, F = Δ p Δ t = 6 0 · 0 2 = 3 0 0 N

Topic: Physics, Center of Mass Momentum and Collision, NEET
A ball rolls off the top of stair-way with a horizontal velocity of magnitude 1.8 m s-1. The steps are 0.20ââm high and 0.20ââm wide. Which step will the ball hit first?

Options:
A First
B Second
C Third
D Fourth

The correct answer is:
Fourth

Solution:
Given, x=0.20 m, y=0.20 m, u=1.8 m s-1
Let the ball strike the nth step of stairs,
Vertical distance travelled =ââny
Using equation of motion in y-direction
s=ut+12at2
â¹-ny=0-12gt2 ...(1)
Horizontal distance travelled =âânx
â¹nx=ut
â¹t=nxu ...(2)
using (1) and (2)
ny=12gnxu2
or n=2u2gyx2 =3.3â¹4th stair

Topic: Physics, Motion In Two Dimensions, NEET
A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $K$. If radius of the ball be $R$, then the fraction of total energy associated with its rotational energy will be:

Options:
A $\frac{K^2+R^2}{R^2}$
B $\frac{K^2}{R^2}$
C $\frac{K^2}{K^2+R^2}$
D $\frac{R^2}{K^2+R^2}$

The correct answer is:
$\frac{K^2}{K^2+R^2}$

Solution:
Total energy $=\frac{1}{2} I \omega^2+\frac{1}{2} m v^2$
$=\frac{1}{2} m v^2\left(1+\frac{K^2}{R^2}\right)$
$\begin{aligned}
& \text {Rotational energy }=\frac{1}{2} I \omega^2 \\
& =\frac{K^2+R^2}{1+\frac{K^2}{R^2}}
\end{aligned}$
$\begin{aligned}
& \text {Required fraction }=\frac{K^2 / R^2}{1+K^2 / R^2} \\
& =\frac{K^2}{R^2+K^2}
\end{aligned}$

Topic: Physics, Rotational Motion, NEET, NEET 2003
A ball whose density is 0.4 x 103 kg m-3 falls into water from a height of 9 cm. To what depth does the ball sink ?

Options:
A 9 cm
B 6 cm
C 4.5 cm
D 2.25 cm

The correct answer is:
6 cm

Solution:
The velocity of ball before entering the water surface
v = 2 gh = 2 g × 9
when ball enters into water, due to up thrust of water the velocity of ball decreases (or retarded)
The retardation,
a = apparent weight mass of ball
= V ρ - σ g V ρ = ρ - σ g ρ
= 0 . 4 - 1 0 . 4 g = - 3 2 g
If h be the depth upto which ball sin k, then
0 - v 2 = 2 × - 3 2 g × h
⇒ 2 g × 9 = 3 gh
∴ h = 6 cm.

Topic: Physics, Motion In One Dimension, NEET
A balloon has 5.0 g moles of helium at 7°C. Calculate the number of atoms of helium in the balloon.

Options:
A 30 × 10²³
B 35 × 10²³
C 25 × 10²³
D 40 × 10²³

The correct answer is:
30 × 10²³

Solution:
Correct Option is : (A)
30 × 10²³

Topic: Chemistry, Kinetic Theory of Gases, NEET
A balloon is been blown up by a healthy person with no known respiratory disorder. Select the option that correctly provides information about the proportion of gases of air inside the balloon compared to the proportion of gases of air outside the balloon.

No.	water vapour	oxygen	carbon dioxide

1	More	More	Less

2	Less	More	Less

3	More	Less	More

4	less	Less	More

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Outside the balloon the air is atmospheric air, that has the following composition - About 21% oxygen, about 0.04% CO2 and less than 1% H2O
Exhaled air has the following composition- About 16% O2 , About 4% CO2 , About 5% H2O .
Inside the balloon exhaled air is present. Therefore, column 3 is the right answer.

Topic: Biology, Breathing and Exchange of Gases, NEET
A balloon is filled with hydrogen at room temperature. It will burst if pressure exceeds 0.2 bar. If at 1 bar pressure the gas occupies 2.27 L volume, upto what volume can the balloon be expanded?

Options:
A 11.95 L
B 10.65 L
C 11.35 L
D 10.54 L

The correct answer is:
11.35 L

Solution:
Correct Option is : (C)
11.35 L

Topic: Chemistry, States of Matter, NEET
A balloon weighing 50 kg is filled with 685.2 kg of helium at 1 atm pressure and 25oC. What will be its pay load if it displaced 5108 kg of air ?

Options:
A 4372.8 kg
B 4422.8 kg
C 5793.2 kg
D 5843.2 kg

The correct answer is:
4372.8 kg

Solution:
Mass of the filled balloon = 50 + 685.2
= 735.2 kg
Pay load = mass of displaced air - mass of balloon
= 5108 - 735.2 = 4372.8 kg

Topic: Chemistry, States of Matter, NEET
A balloon with mass m is descending down with an acceleration a (where a<g). How much mass should be removed from it so that it starts moving up with an acceleration a?

Options:
A 2mag+a
B 2mag-a
C mag+a
D mag-a

The correct answer is:
2mag+a

Solution:
Let the upward force be F.

When balloon descending down, then mg-F=ma …(i)
Let the mass removed be m'.

When balloon ascending upward, then F-m-m'g=m-m'a
from (i)
F-mg+m'g=ma-m'a
mg-ma-mg+m'g=ma-m'a
m'g+a=2ma
m'=2mag+a

Topic: Physics, Laws of Motion, NEET, NEET 2014
A bar magnet 8 cm long is placed in the magnetic meridian with the N pole pointing towards geographical north. Two neutral points separated by a distance of 6 cm are obtained on the equatorial axis of the magnet. If the horizontal component of the earth's field = 3.2Ã10-5 T then pole strength of the magnet is

Options:
A 0.5 A m
B 1 A m
C 0.25 A m
D 2 A m

The correct answer is:
0.5 A m

Solution:
Field due to a bar magnet on equatorial position.
B=μ0m2l4πr2+l23/2
Given
2l=8 cm
r=62=3 cm
At the null point,
B=BH
μ04πm2lr2+l23/2=BH
10-7 m8×10-23×10-22+4×10-223/2=3.2×10-5
10-7m8×10-25×10-23=3.2×10-5
m=1253.28×10-2
m=0.5 A m

Topic: Physics, Magnetic Properties of Matter, NEET
A bar magnet has a length 3 cm, cross-sectional area 2 cm2 and magnetic moment 3 A m2. The intensity of magnetization of the bar magnet is

Options:
A 2Ã105 A m-1
B 3×105 A m-1
C 4×105 A m-1
D 5×105 A m-1

The correct answer is:
5×105 A m-1

Solution:
L=3 cm=3× 10 −2 m
A=2 cm 2 =2× 10 −4 m 2
M=3 A m2
Mz→ Intensity of magnetization
=ML×A=3 A m23×2×10−6 m3=12×106 A m-1=5×105A m-1

Topic: Physics, Magnetic Properties of Matter, NEET
A bar magnet has a magnetic moment of 10âAâm2. If its magnetic length is 5âcm, Its pole strength is

Options:
A 200âAâm
B 100 A m
C 300 A m
D 50 A m

The correct answer is:
200 A m

Solution:
m⁡=Ml=100.05=200 A m

Topic: Physics, Magnetic Properties of Matter, NEET
A bar magnet has coercivity 4Ã103 A m-1. It is desired to demagnetise it by inserting it inside a solenoid 12 cm long and having 60 turns. The current that should be sent through the solenoid is

Options:
A 2 A

B 4 A

C 6 A

D 8 A

The correct answer is:
8 A

Solution:
The bar magnet coercivity 4Ã103 A m-1 i.e., it requires a magnetic intensity H=4Ã103 A m-1to get demagnetised. Let i be the current carried by solenoid having n number of turns per metre length, then by definition H = ni. Here H=4Ã103 A m-1
n=Nl=600.12=500 m-1
âi=Hn=4Ã103500=8.0 A

Topic: Physics, Magnetic Effects of Current, NEET
A bar magnet having a magnetic moment of $2 \times 10^4 \mathrm{JT}^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $\mathrm{B}=6 \times 10^{-4} \mathrm{~T}$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ}$ from the field is

Options:
A $0.6 \mathrm{~J}$
B $12 \mathrm{~J}$
C $6 \mathrm{~J}$
D $2 \mathrm{~J}$

The correct answer is:
$6 \mathrm{~J}$

Solution:
Key Idea The work done in rotating a magnetic dipole against the torque acting on it, when placed in magnetic field is stored inside it in the form of potential energy.
When magnetic dipole is rotated from initial position $\theta=\theta_1$ to final position $\theta=\theta_2$, then work done $=\mathrm{MB}\left(\cos \theta_1-\cos \theta_2\right)$
$$
\begin{aligned}
& =\operatorname{MB}\left(1-\frac{1}{2}\right) \\
& =\frac{2 \times 10^4 \times 6 \times 10^{-4}}{2}=6 \mathrm{~J}
\end{aligned}
$$

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2009 (Screening)
A bar magnet is held at a right angle to a uniform magnetic field. The couple acting on the magnet is to be halved by rotating it from this position. The angle of rotation is

Options:
A 60°
B 45°
C 30°
D 75°

The correct answer is:
60°

Solution:
Torque, τ=MB sin⁡θ
Where θ=90°
∴ τ=MB
Given, τ2=12 τ1
∴ MB sin⁡θ=12 MB
∴ sin θ⁡=12
⇒ θ=30°
Angle of rotation =90°-30°=60°

Topic: Physics, Magnetic Properties of Matter, NEET
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

Options:
A W3
B 3W
C 3W2
D 2W3

The correct answer is:
3W

Solution:
W=PEcos⁡θ1-cos⁡θ2

W=PEcos⁡0-cos⁡60o

=PE2.

⇒PE=2W

τ=PEsin⁡θ=2Wsin⁡60o=3W.

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2016 (Phase 2)
A bar magnet is released into a copper ring directly below it. The acceleration of the magnet will be

Options:
A equal to the acceleration due to gravity at that place
B less than the acceleration due to gravity at that place
C greater than the acceleration due to gravity at that place
D twice the acceleration due to gravity at that place

The correct answer is:
less than the acceleration due to gravity at that place

Solution:
According to Lenz's law, acceleration of the magnet will be less than the acceleration due to gravity, due to the fact that, emf induced in the ring opposes the motion of magnet.

Topic: Physics, Electromagnetic Induction, NEET
A bar magnet is suspended freely in a uniform magnetic field of induction B. When will the rate of change of torque Ï with deflection Î¸ be maximum? [Here Î¸ is the angle between the magnetic field and the magnetic moment of the bar magnet]

Options:
A Î¸=0Â°
B θ=45°
C θ=60°
D Î¸=90Â°

The correct answer is:
θ=0°

Solution:
Ï=MBsinÎ¸
dÏdÎ¸=MBcosÎ¸
For dÏdÎ¸ to be maximum,

cosÎ¸=1âÎ¸=0Â°

Topic: Physics, Magnetic Properties of Matter, NEET
A bar magnet of length $l$ and magnetic dipole moment $M$ is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be

Options:
A $M$
B $\frac{3}{\pi} M$
C $\frac{2}{\pi} M$
D $\frac{M}{2}$

The correct answer is:
$\frac{3}{\pi} M$

Solution:
Given, \(\theta=60^{\circ}=\frac{\pi}{3} \mathrm{rad}\)
The new magnetic length of the dipole will be, \(I^{\prime}=r\)
Now, \(M=m \cdot I\) and \(M^{\prime}=m \cdot I^{\prime}\)
\(\therefore \frac{\mathrm{M}}{\mathrm{M}^{\prime}}=\frac{\mathrm{I}}{\mathrm{I}^{\prime}} \ldots \quad (1)\)
Now, angle in radians \(=\frac{\text { arc length }}{\text { radius }}\)
\(\therefore \frac{\pi}{3}=\frac{1}{r} \Rightarrow 1=\frac{\pi r}{3}\)
In the equilateral triangle, marked by dotted lines, \(I^{\prime}=r\)
\(\therefore \frac{1}{I^{\prime}}=\frac{\pi}{3} \ldots \quad (2)\)
From and, we get,
\(\begin{aligned}
& \frac{M}{M^{\prime}}=\frac{\pi}{3} \\
& \therefore M^{\prime}=\frac{3 M}{\pi}
\end{aligned}\)
Hence, \((C)\) is the correct answer.
Alternative method:
Using the short cut formula about bending of a bar magnet, \(M=\frac{2 M \sin \left(\frac{\theta}{2}\right)}{\theta}\)
\(=\frac{2 M \sin 30^{\circ}}{\frac{\pi}{3}}=\frac{3 M}{\pi}\)
Hence, (C) is the correct answer.

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2013 (All India)
A bar magnet of magnetic moment 1.5 J T–1lies aligned with the direction of a uniform magnetic field of 0.22 T. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction?

Options:
A 0.15 J
B 0.22 J
C 0.33 J
D 0.44 J

The correct answer is:
0.33 J

Solution:
Correct Option is : (C)
0.33 J

Topic: Chemistry, Magnetic Properties of Matter, NEET
A bar magnet of magnetic moment M1 is axially cut into two equal parts. If these two pieces are arranged perpendicular to each other, the resultant magnetic moment is M2. Then the value of M1M2 is

Options:
A 122
B 1
C 12
D 2

The correct answer is:
2

Solution:
M2=2·M2⇒ M1M2=MM2 = 2

Topic: Physics, Magnetic Properties of Matter, NEET
A bar magnet of magnetic moment $M$ placed at right angles to a magnetic induction $B$. If a force $F$ is experienced by each pole of the magnet. The length of the magnet will be:

Options:
A $M B / F$
B $B F / M$
C $M F / B$
D $F / M B$

The correct answer is:
$M B / F$

Solution:
Here
$$
\begin{aligned}
\tau & =\vec{M} \times \vec{B} \\
|\tau| & =M B \sin 90^{\circ}
\end{aligned}
$$
and we know that $\tau=F l$
$$
\therefore F l=\mathrm{MB} \quad \Rightarrow l=\frac{M B}{F}
$$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2013 (Karnataka)
A bar magnet of the moment M→=i^+j^ is placed in a magnetic field induction B→=3i^+4j^+4k^. The torque acting on the magnet is

Options:
A 4i^-4j^+k^
B i^+k^
C i^-j^
D i^+j^+k^

The correct answer is:
4i^-4j^+k^

Solution:
When bar magnet is palced in external magnetic field it experiences torque.
τ→=M→×B→
τ→=4i^-4j^+k^

Topic: Physics, Magnetic Properties of Matter, NEET
A bar of mass m and length l is hanging from point A as shown in the figure. If the Young's modulus of elasticity of the bar is Y and area of cross-section of the wire is A, then the increase in its length due to its own weight will be

Options:
A mgL2 AY
B mgA2Â LY
C mg2Â LAY
D 2Â LYmgA

The correct answer is:
mgL2 AY

Solution:
Consider a small section dx of the bar at a distance x from B. The weight of the bar for a length x is, W=mgLx.
Elongation in x will be
de=WAYdx=mgLAYdxâx .
Total elongation of the bar can be obtained by integrating this expression for x=0 to x=L.
â´ Îl= â«x=0x=Lde=mgLAY â«0Lxâdx
â Îl=mgL2AY

Topic: Physics, Mechanical Properties of Solids, NEET
A barometer is constructed using a liquid (density =760 kg m-3) What would be the height of the liquid column, when a mercury barometer reads 76 cm?

(density of mercury =13600 kg m-3)

Options:
A 1.36 m
B 13.6 m
C 136 m
D 0.76 m

The correct answer is:
13.6 m

Solution:
Recall the formula of liquid pressure due to column in terms of density of the liquid, the height of the liquid column, and the acceleration due to gravity, P=hρg

⇒76 cm×ρHg×g=h×ρL×g

⇒h=76 cm×ρHgρL=76 cm×13600760=13.6 m.

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2020 (Phase 2)
A battery consists of a variable number 'n' of identical cells (having internal resistance 'r' each) which are connected in series. The terminals of the battery are short-circuited and the current I is measured. Which of the graphs shows the correct relationship between I and n?

Options:
A

B

C

D

The correct answer is:

Solution:
In series, ε net =nε
and r net =nr

current in circuit
I= nε nr = ε​ r = constant.
⇒I independent of n

Topic: Physics, Current Electricity, NEET, NEET 2018
A battery has an open circuit potential difference 10 V between the terminals. When loads 9 Î© and 4 Î© are connected one by one across the battery, the power in the load resistance is the same. The amount of heat approximately generated in one second in the load when a load of 5 Î© is connected across the battery will be

Options:
A 10.4 J
B 9 J
C 6.8 J
D 4.13 J

The correct answer is:
4.13 J

Solution:
Given: v29+r29=v24+r24
⇒ r=6
∴ H=10112×5×1=4.13 J

Topic: Physics, Current Electricity, NEET
A battery is charged at a potential of 15 $\mathrm{V}$ for 8 hours when the current flowing is $10 \mathrm{~A}$. The battery on discharge supplies a current of $5 \mathrm{~A}$ for 15 hours. The mean terminal voltage during discharge is $14 \mathrm{~V}$. The "Watt-hour" efficiency of the battery is:

Options:
A $82.5 \%$
B $80 \%$
C $90 \%$
D $87.5 \%$

The correct answer is:
$87.5 \%$

Solution:
When the battery is charged then Watt hour $=V I t=15 \times 10 \times 8=1200$ W/hr.
During discharge the watt hour $u=$ $1050 \mathrm{~W} / \mathrm{hr}$.
$$
\begin{gathered}
\therefore \text { Efficiency of battery }=\frac{1050}{1200} \times 100 \\
=87.5 \%
\end{gathered}
$$

Topic: Physics, Current Electricity, NEET, NEET 2004
A battery is charged at a potential of 15 V in 8 h when the current flowing is 10 A. The battery on discharge supplies a current of 5 A for 15 h. The mean terminal voltage during discharge is 14 V. The watt-hour efficiency of battery is

Options:
A 80%
B 90%
C 87.5%
D 82.5%

The correct answer is:
87.5%

Solution:
The power of the battery, when charged is given by
P1=V1I1
The electrical energy dissipated is given by
Et=P1t1
E1=V1I1T1=15Ã10Ã8
=1200 Wh
Similarly, the electrical energy dissipated during the discharge of battery is given by
E2=V2I2t2
=14Ã5Ã15
=1050 Wh
Hence, watt-hour efficiency of the battery is given by
Î·=E2E1Ã100
=0.875Ã100=87.5%

Topic: Physics, Current Electricity, NEET
A battery is charged to a voltage of 15 V for 8 hours with 10 A current. On discharge, the battery supplies a current of 5 A for 15 hours at a mean terminal voltage of 14 V. The efficiency of the battery is

Options:
A 80%
B 90%
C 87.5%
D 82.5%

The correct answer is:
87.5%

Solution:
The power of the battery, when charged, is given by P1=V1I1 The electrical energy dissipated is given by E1=P1t1
i. e., E1=P1I1t1=15Ã10Ã8=1200Wh
Similarly, the electrical energy dissipated during the discharge of battery is given byE2=V2I2t2=14Ã5Ã15=1050Wh

Hence, watt-hour efficiency of the battery is given by Î·=E2E1Ã100=0.875Ã100=87.5%

Topic: Physics, Current Electricity, NEET