datavorous/entrance-exam-dataset huggingface dataset:
A bar magnet is oscillating in the Earth's magnetic field with a period $T$. What happens to its period and motion if its mass is quadrupled?

Options:
A motion remains simple harmonic with time period $=\frac{T}{2}$
B motion remains S.H.M with time period $=2 T$
C motion remains S.H.M. with time period $=4 T$
D motion remains S.H.M and period nearly constant

The correct answer is:
motion remains S.H.M with time period $=2 T$

Solution:
Time period of bar magnet $T$
$$
=2 \pi \sqrt{\frac{I}{14 H}}
$$
Where $\mathrm{I}=$ M.I. $=\frac{m \mathrm{~L}^2}{12}$
$m=$ mass of magnet
$\therefore T=2 \pi \sqrt{\frac{m L^2}{12 \mathrm{MH}}}$
$\Rightarrow T \propto \sqrt{m}$
$\therefore$ If mass is quadruplet then $T$ becomes twice.

Topic: Physics, Magnetic Properties of Matter, JEE Main
A bar magnet is oscillating in the earth's magnetic field with a period $T$. What happens to its period and motion, if its mass is quadrupled?

Options:
A Motion remains simple harmonic with time period $4 T$.
B Motion remains simple harmonic with time period nearly constant.
C Motion remains simple harmonic and time period becomes $T / 2$
D Motion remains simple harmonic with time period $2 T$.

The correct answer is:
Motion remains simple harmonic with time period $2 T$.

Solution:
Time period of simple harmonic motion,
$$
T=2 \pi \sqrt{\left|\frac{\theta}{\alpha}\right|}=2 \pi \sqrt{\frac{I}{m H}}
$$
$\Rightarrow \quad T \propto \sqrt{I} \Rightarrow T \propto \sqrt{m} \quad\left[\therefore=\frac{m l^{2}}{12}\right]$
$\frac{T_{1}}{T_{2}}=\sqrt{\frac{m_{1}}{m_{2}}}=\sqrt{\frac{m}{4 m}}=\frac{1}{2}$
$\Rightarrow \quad T_{2}=2 T_{1}=2 T \quad\left[\therefore T_{1}=T_{1}\right.$
So, motion remains simple harmonic with time period $2 T$.

Topic: Physics, Magnetic Properties of Matter, COMEDK, COMEDK 2021
A bar magnet is released from rest along the axis of a very long vertical copper tube. After some time the magnet will

Options:
A Move down with an acceleration equal to g
B Oscillate inside the tube
C Move down with almost constant speed
D Move down with an acceleration greater than g

The correct answer is:
Move down with almost constant speed

Solution:
The flux connected to the copper tube will fluctuate when the magnet moves under the influence of gravity. Eddy currents will be created in the copper tube's body as a result, these induced currents oppose the collapse of the magnet, in accordance with Lenz's law.

A retarding force will consequently act on the magnet. This force will keep rising with the magnet's rising velocity until it equals the force of gravity. The net force acting on the magnet will then vanish. The magnet will therefore reach a steady speed.

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2023 (10 Apr Shift 2)
A bar magnet of length $6 \mathrm{~cm}$ has a magnetic moment of $4 \mathrm{~J} \mathrm{~T}^{-1}$. Find the strength of magnetic field at a distance of $200 \mathrm{~cm}$ from the centre of the magnet along its equatorial line.

Options:
A
$4 \times 10^{-8}$ tesla

B
$3.5 \times 10^{-8}$ tesla

C
$5 \times 10^{-8}$ tesla

D
$3 \times 10^{-8}$ tesla

The correct answer is:
$5 \times 10^{-8}$ tesla

Solution:
Along the equatorial line, magnetic field strength
$$
\left.B=\frac{\mu_0}{4 \pi} \frac{M}{\left(r^2+\ell^2{ }^{3 / 2}\right.}\right)
$$
Given: $M=4 \mathrm{~J} T^{-1}$
$$
r=200 \mathrm{~cm}=2 \mathrm{~m}
$$
$\ell=\frac{6 \mathrm{~cm}}{2}=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}$
$$
\therefore B=\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{4}{\left[2^2+\left(3 \times 10^{-2}\right)^2\right]^{3 / 2}}
$$
Solving we get, $B=5 \times 10^{-8}$ tesla

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2012 (07 May Online)
A bar magnet of length 10 cm and pole strength 2 A m makes an angle 60o with a uniform magnetic field of induction 50 T. The couple acting on it is

Options:
A 53 N m
B 3 N m
C 103 N m
D 203 N m

The correct answer is:
53 N m

Solution:
C=2lmBsinâ¡Î¸
= 1 0 - 1 Ã 2 Ã 5 0 Ã 3 2
= 5 3 N m

Topic: Physics, Magnetic Properties of Matter, JEE Main
A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the center of the magnet. If BH=0.4 G, the magnetic moment of the magnet is 1 G=10-4 T

Options:
A 2.880×103 J T-1
B 2.880×102 J T-1
C 2.880 J T-1
D 28.80 J T-1

The correct answer is:
2.880 J T-1

Solution:
i.e. 2μ04πmr2×7r=0.4×10-4

⇒2×10-7×m×772+1823/2×104=0.4×10-4

m=4×10-2×3733/214

⇒M=m×14 cm=m×14100=0.04×3733/214×14100

⇒M=4×10-4×7203.82=2.88 J T-1.

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2021 (16 Mar Shift 1)
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, JEE Main
A bar magnet of magnetic moment $M$ and moment of inertia $I$ is freely suspended such that the magnetic axial line is in the direction of magnetic meridian. If the magnet is displaced by a very small angle $(\theta)$, the angular acceleration is (Magnetic induction of earth's horizontal field $=B_H$ )

Options:
A $\frac{M B_H \theta}{I}$
B $\frac{I B_H \theta}{M}$
C $\frac{M \theta}{I B_H}$
D $\frac{I \theta}{M B_H}$

The correct answer is:
$\frac{M B_H \theta}{I}$

Solution:
When magnet is displaced by a very small angle $\theta$, then restoring couple acting on the magnet is
$\tau=-M B_H \sin \theta$
Negative sign shows the restoring nature of torque. Now since $\tau=I \alpha$ and $\sin \theta \approx \theta$ for small angular displacement
$\therefore \quad I \alpha=M B_H \theta$
or
$\begin{aligned} \alpha & =\text { angular acceleration } \\ & =\frac{M B_H \theta}{I}\end{aligned}$

Topic: Physics, Magnetic Effects of Current, JEE Main
A bar magnet of magnetic moment \( M \) is bent as a part of the circle shown in the figure. The new magnetic moment is

Options:
A \( M \)
B \( \frac{3 M}{5 \pi} \)
C \( \frac{5 M}{3 \pi} \)
D \( \frac{2 M}{\pi} \)

The correct answer is:
\( \frac{3 M}{5 \pi} \)

Solution:
According to the question:
Initially, M=qm56×2πR

Now it is bent
M′=qmR
⇒M′=3M5π

Topic: Physics, Magnetic Effects of Current, JEE Main
A bar magnet with a magnetic moment 5.0 A m2 is placed in parallel position relative to a magnetic field of 0.4 T. The amount of required work done in turning the magnet form parallel to antiparallel position relative to the field direction is ________.

Options:
A 4 J
B 1 J
C 2 J
D Zero

The correct answer is:
4 J

Solution:
For a bar magnet, the work done in turning from parallel to antiparallel position relative to the field direction is

W=MBcosθ1-cosθ2, here, θ1=0° and θ1=180°.

W=-MBcos180°-mBcos0°

W=2 MB=2×5×0.4=4 J

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (31 Jan Shift 1)
A bar of mass m and length l is in pure translational motion with its centre of mass moving with a velocity v. It collides and sticks to another identical bar at rest as shown in the figure. Assuming that after the collision the system of the two bars becomes one composite bar of length 2l, then the angular velocity of the composite bar will be

Options:
A 3 v 4 l , anti-clockwise
B 4 v 3 l , anti-clockwise
C 3 v 4 l , clockwise
D 4 v 3 l , clockwise

The correct answer is:
3 v 4 l , anti-clockwise

Solution:
By law of conservation of angular momentum
Î£ mvr = I system Ï
â mv l 2 = 2 m 2 l 2 1 2 Ï = 2 m 4 l 2 1 2 Ï
â Ï = 3 v 4 l anti-clockwise
Note that clockwise or anti-clockwise rotation can only be determined here by the given diagram.

Topic: Physics, Rotational Motion, JEE Main
A bar of mass $m$ resting on a smooth horizontal plane starts moving due to a constant force $F$. In the process of its rectilinear motion the angle $\theta$ between the direction of this force and the horizontal varies as $\theta=k x$, where $k$ is a constant and $x$ is the distance traversed by the bar from its initial position. The velocity $(v)$ of the bar as a function of the angle $\theta$ is

Options:
A $v=\sqrt{\frac{2 F \sin \theta}{m k}}$
B $v=\sqrt{\frac{2 F}{m k \sin \theta}}$
C $v=\frac{2 F \sin \theta}{m k}$
D $v=\frac{2 F}{m k \sin \theta}$

The correct answer is:
$v=\sqrt{\frac{2 F \sin \theta}{m k}}$

Solution:
Component of force responsible for horizontal motion is $F \cos \theta$.
So, acceleration of bar, $a=\frac{F \cos \theta}{m}$
As acceleration, $a=\frac{d v}{d t}$, we have
$\Rightarrow \quad \frac{d v}{d t}=\frac{F \cos \theta}{m}$
we can also express this as,
$\begin{array}{ll}\Rightarrow & \frac{d v}{d x} \cdot \frac{d x}{d t}=\frac{F \cos \theta}{m} \\ \Rightarrow & v d v=\frac{F \cos \theta}{m} d x\end{array}$
Integrating both sides and putting $\theta=k x$, we get
$\int v d v=\int \frac{F \cos k x}{m} d x$
$\begin{array}{ll}\Rightarrow & \frac{v^2}{2}=\frac{F}{m} \int \cos k x \cdot d x \\ \Rightarrow & \frac{v^2}{2}=\frac{F}{m} \cdot\left(\frac{\sin k x}{k}\right) \\ \Rightarrow & v^2=\frac{2 F \sin k x}{m k} \\ \Rightarrow & v=\sqrt{\frac{2 F \sin \theta}{m k}}\end{array}$

Topic: Physics, Laws of Motion, JEE Main
A bar of mass m resting on a smooth horizontal plane starts moving due to force |F→|=mg9. The magnitude of the force remains constant with time. The force vector makes an angle θ with the horizontal which varies with the distance covered as θ=Cx. If the constant C=10 degree meter , then the speed of the bar, when θ becomes equal to 30° for the first time, is (Assume g=10 m s-2)

Options:
A 0.33 m s-1
B 0.50 m s-1
C 1.0 m s-1
D 0.8 m s-1

The correct answer is:
0.33 m s-1

Solution:
Given,

The magnitude of a force, |F→|=mg9 and

Force vector makes a varying angle θ with the horizontal as θ=Cx,

where, C=10 degree meter .

Thus, Horizontal component of force,

Fx=F cos θ=F cos Cx.

Applying work energy theorem,

Wnet=∆K.E.

⇒∫0xFxdx=12mv2

(when θ becomes 30°, x=θC=30°10° m-1=3 m)

⇒ ∫0xmg9 cos Cx dx=12mv2

⇒ mg9∫03cos Cxdx=12mv2

⇒ mg9C sinC 3-sin 0=12mv2

⇒ 109 10° m-1 sin 30°=12v2

⇒ 1912=12v2

⇒ v=13=0.33 m s-1.

Topic: Physics, Laws of Motion, TS EAMCET, TS EAMCET 2019 (03 May Shift 2)
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, JEE Main
A barometer kept in a stationary elevator reads $76 \mathrm{~cm}$. If the elevator starts accelerating up the reading will be

Options:
A Zero
B Equal to $76 \mathrm{~cm}$
C More than $76 \mathrm{~cm}$
D Less than $76 \mathrm{~cm}$

The correct answer is:
Less than $76 \mathrm{~cm}$

Solution:
$h=\frac{P}{\rho g} \therefore h \propto \frac{1}{g}$. If lift moves upward with some acceleration then effective $g$ increases. So the value of $h$ decreases i.e. reading will be less than $76 \mathrm{~cm}$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A barometer tube reads $76 \mathrm{~cm}$ of mercury. If the tube is gradually inclined at an angle of $60^{\circ}$ with vertical, keeping the open end immersed in the mercury reservoir, the length of the mercury column will be

Options:
A $152 \mathrm{~cm}$
B $76 \mathrm{~cm}$
C $38 \mathrm{~cm}$
D $38 \sqrt{3} \mathrm{~cm}$

The correct answer is:
$152 \mathrm{~cm}$

Solution:
$\cos 60^{\circ}=\frac{h}{l}$
$\Rightarrow l=\frac{h}{\cos 60^{\circ}}=\frac{76}{1 / 2}$
$\therefore l=152 \mathrm{~cm}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
(a) Baryte, (b) Galena, (c) Zinc blende and (d) Copper pyrites. How many of these minerals are sulphide based?

Options:

The correct answer is:
3

Solution:
Baryte is a mineral consisting of barium sulphate BaSO4

Galena, also called lead glance, is the natural mineral of lead (II) Sulphate PbS

Zinc blende, is a mineral consisting of ZnS.

Copper pyrites is a copper iron sulphide mineral.

It has the chemical formula CuFeS2.

Number of minerals which are sulphide based are 3.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2022 (24 Jun Shift 2)
A basic communication system consists of:
A. Transmitter, B. Information source, C. User of information, D. Channel, E. receiver
Choose the correct sequence in which these are arranged in a basic communication system.

Options:
A A,B,C,D,E
B B, A, D, E, C
C B, D, A, C, E
D B,E,A,D,C

The correct answer is:
B, A, D, E, C

Solution:

Topic: Physics, Communication System, JEE Main
A bat moving at 10 m s−1 towards a wall sends a sound signal of 8000 Hz towards it. On reflection, it hears a sound of frequency f. The value of f in Hz is close to
(speed of sound=320 m s−1)

Options:
A 8258
B 8424
C 8000
D 8516

The correct answer is:
8516

Solution:
We can assume that reflected wave is due to image of B coming with same speed in opposite direction
Observer
f= v+10v-10 ×f0
= 320+10320-10 ×8000
= 330310 ×8000
= 3331 ×8000
= 8516 Hz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2015 (10 Apr Online)
A bat moving at 10 msâ1 towards a wall sends a sound signal of 8000 Hz towards it. On reflection it hears a sound of frequency f. The value of f in Hz is close to
(speed of sound=320âmsâ1)

Options:
A 8258
B 8424
C 8000
D 8516

The correct answer is:
8516

Solution:
We can assume that reflected wave is due to image of B coming with same speed in opposite direction
Observer
f= v+10v-10 Ãf0
= 320+10320-10 Ã8000
= 330310 Ã8000
= 3331 Ã8000
= 8516 Hz

Topic: Physics, Waves and Sound, JEE Main
A batsman hits back a ball of mass 0.4 kg straight in the direction of the bowler without changing its initial speed of 15 m s-1. The impulse imparted to the ball is _____ N s.

Options:

The correct answer is:
12

Solution:
Impulse, J=∫Fdt=Change in momentum

Therefore,
Impulse, Δp=mv--mv=2mv=2×0.4×15 =12 N s

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2022 (26 Jun Shift 2)
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, JEE Main
A battery is charged at a potential of 15 V in 8 h when the current flowing is 10A. The battery on discharge supplies a current of 5A for 15 h. The mean terminal voltage during discharge is 14V. What is the watt-hour efficiency of the battery?

Options:

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, JEE Main
A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle Î¸ at the centre. The value of the magnetic induction at the centre due to the current in the ring is

Options:
A proportional to (180Â°-Î¸)
B inversely proportional to r
C zero, only if (Î¸=180Â°)
D zero for all values of Î¸

The correct answer is:
zero for all values of Î¸

Solution:
For a current flowing into a circular arc, the magnetic induction at the centre is
B = μ 0 i⁡ 4 π θ
or B∝i⁡θ
In the given problem, the total current is divided into two arcs
i⁡∝1resistance of arc
∝ 1 length of arc
∝ 1 angle subtended at centre θ

or iθ=constant
i.e., magnetic field at centre due to arc AB is equal and opposite to the magnetic field at centre due to arc ACB. Or the net magnetic field at the centre is zero.

Topic: Physics, Magnetic Effects of Current, JEE Main
A battery is constructed of $\mathrm{Cr}$ and $\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7$. The unbalanced chemical equation when such a battery discharges is following:
$$
\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{Cr}+\mathrm{H}^{+} \rightarrow \mathrm{Cr}^{3+}+\mathrm{H}_2 \mathrm{O}+\mathrm{Na}^{+}
$$
If one Faraday of electricity is passed through the battery during the charging, the number of moles of $\mathrm{Cr}^{3+}$ removed from the solution is

Options:
A
$\frac{4}{3}$

B
$\frac{1}{3}$

C
$\frac{3}{3}$

D
$\frac{2}{3}$

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

Solution:
Reduction half reaction :
$$
\mathrm{Cr}_2 \mathrm{O}_7^{2-}+6 e^{-}+14 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}
$$
Oxidation half reaction :
$$
\mathrm{Cr} \longrightarrow \mathrm{Cr}^{3+}+3 e^{-}
$$
Overall reaction :
$$
\begin{aligned}
& \mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{Cr}+14 \mathrm{H}^{+}+3 e^{-} \\
& 3 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}
\end{aligned}
$$
$3 \mathrm{~F}$ of electricity $=3$ moles of $\mathrm{Cr}^{3+}$
$1 \mathrm{~F}$ of electricity $=\frac{3}{3}$ moles of $\mathrm{Cr}^{3+}$

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2012 (07 May Online)
A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be

Options:
A
$1$

B
$2$

C
$\frac{1}{4}$

D
$\frac{1}{2}$

The correct answer is:
$\frac{1}{2}$

Solution:
$\frac{\frac{1}{2} q v}{q v}=\frac{1}{2}$

Topic: Physics, Capacitance, JEE Main, JEE Main 2007
A battery of 6 V is connected to the circuit as shown below. The current I drawn from the battery is

Options:
A 1 A
B 2 A
C 611 A
D 43 A

The correct answer is:
1 A

Solution:
The given circuit in the question is a case of a balanced Wheatstone bridge. Therefore, the circuit can be simplified as given below.

The equivalent resistance for the given circuit will be,

Req=3+3×6+63+3+6+6+2=6 Ω

Therefore, the current supplied by the battery is

i=VReq=66=1 A

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (26 Jul Shift 2)
A battery of 10 V is connected to a resistance of 20 Î© through a variable resistance R. If the variable resistance R is increased at the rate of 5 Î© min-1, then the amount of charge that passes through the battery in 4 min is

Options:
A 120 C
B 120loge2 C
C 120loge2 C
D 60loge2 C

The correct answer is:
120loge2 C

Solution:
Q= â«i dt = â«VRdt= â«02401020+560tdt t converted in seconds
=10560 lnâ¡560t+200240
=6005lnâ¡4020=120lnâ¡2

Topic: Physics, Current Electricity, JEE Main
A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of the battery is 2.5 V, the power dissipated within the internal resistance is:

Options:
A 0.50 W
B 0.072 W
C 0.10 W
D 0.125 W

The correct answer is:
0.10 W

Solution:
Given,

Emf of the battery, E=3 V

Potential difference across resister R is VR=2.5 V

Power dissipation in the resister is P=0.5 W

By using Kirchhoff's voltage law is given by,

Vr+VR=E ...(1)

Vr is voltage across internal resister.

Using equation (1) and substitute the given values,

⇒Vr+2.5=3

⇒Vr=0.5

Now the ratio of potential across R and r is given by,

VRVr=IRIr=2.50.5=5

⇒Rr=5

Now the power dissipation across the internal resistance is given by

PRPr=I2RI2r=Rr

⇒PRPr=5

⇒Pr=PR5

⇒Pr=0.55=0.1W

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (04 Sep Shift 1)
A battery of emf $\mathrm{E}$ and internal resistance $\mathrm{r}$ is connected with an external resistance $\mathrm{R}$ as shown in the figure. The battery will act as a constant voltage source if

Options:
A $r< < R$
B $r>>R$
C $r=R$
D It will never act as a constant voltage source

The correct answer is:
$r< < R$

Solution:
$V_R=\frac{E}{(R+r)} \times R, \quad V_R=\frac{E}{(1+r / R)}$
$r < R, V_R \approx E$

Topic: Physics, Current Electricity, JEE Main
A battery of internal resistance 4 Î© is connected to the network of resistances as shown in the figure. In order that the maximum power be delivered to the network, the value of R (in Î©) should be

Options:

The correct answer is:
2

Solution:
The given circuit is a balanced Wheatstone's bridge.

Thus, no current will flow across 6R. The given circuit will now be equivalent to:-
For maximum power, net external resistance = Total internal resistance
or 2R=4 Î©
or R=2 Î©

Topic: Physics, Current Electricity, JEE Main
A battery of internal resistance 4 Î© is connected to the network of resistances, as shown in the figure. In order to deliver the maximum power to the network, what should be the value of R in ohm?

Options:

The correct answer is:
2

Solution:
The given figure is a balanced form of Wheatstone bridge therefore, 6R resistance can be removed.
â´ 1RP=1R1+1R2=13R+16R=12R
â RP=2R.
The power delivered to the network is maximum when internal resistance = External resistance.
4=2R
â R=2 Î©

Topic: Physics, Current Electricity, JEE Main
$A B C D$ is parallelogram. The position vectors of $A$ and $C$ are respectively, $3 \hat{i}+3 \hat{j}+5 \hat{k}$ and $\hat{i}-5 \hat{j}-5 \hat{k}$. If $M$ is the midpoint of the diagonal $D B$, then the magnitude of the projection of $\overrightarrow{O M}$ on $\overrightarrow{O C}$, where $O$ is the origin, is

Options:
A
$7 \sqrt{51}$

B
$\frac{7}{\sqrt{50}}$

C
$7 \sqrt{50}$

D
$\frac{7}{\sqrt{51}}$

The correct answer is:
$\frac{7}{\sqrt{51}}$

Solution:
In a parallelogram, diagonals bisect each other. So, mid point of $D B$ is also the midpoint of $A C$.
Mid-point of $M=2 \hat{i}-\hat{j}$
Direction ratio of $O C=(1,-5,-5)$
Direction ratio of $O M=(2,-1,0)$
Angle $\theta$ between $O M$ and $O C$ is given by
$$
\left.\cos \theta=\frac{(1 \times 2+)-5(-1)(+-5) 0(}{\sqrt{2^2+(-1)^2} \sqrt{()^2+(-5)^2+(-5)^2}}\right)()
$$
$$
=\frac{2+5}{\sqrt{5} \sqrt{51}}=\frac{7}{\sqrt{5} \sqrt{51}}
$$
Projection of $\overrightarrow{O M}$ on $\overrightarrow{O C}$ is given by
$$
|O M| \cdot \cos \theta=\sqrt{5} \times \frac{7}{\sqrt{5} \times \sqrt{51}}=\frac{7}{\sqrt{51}}
$$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2012 (07 May Online)