datavorous/entrance-exam-dataset huggingface dataset:
An $L C$ circuit contains a $20 \mathrm{mH}$ inductor and a $50 \mu \mathrm{F}$ capacitor with initial charge of $10 \mathrm{mC}$. The resistance of the circuit is negligible. Let the instant the circuit is closed be $t=0$.
(a) What is the total energy stored initially? Is it conserved during $L C$ oscillations?
(b) What is the natural frequency of the circuit?
(c) At what time is the energy stored
(i) completely electrical (i.e. stored in the capacitor)?
(ii) completely magnetic (i.e. stored in the inductor)?
(d) At what times is the total energy shared equally between the inductor and capacitor?
(e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?

Options:

None

Solution:
(a) Total energy is initially in the form of electric field within the plates of charged capacitor.
$$
U_E=\frac{Q^2}{2 C}=\frac{\left(10 \times 10^{-3}\right)^2}{2 \times 50 \times 10^{-6}}=1 \mathrm{~J}
$$
If we neglect the losses due to resistance of connecting wires, the total energy remain consumed during $L C$ oscillations.
(b) Natural frequency of the circuit
$f=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \times \sqrt{20 \times 10^{-3} \times 50 \times 10^{-6}}} \mathrm{~Hz}=\frac{500}{\pi}$
(c) Instantaneous electrical energy
$$
U_E=\frac{q_0^2 \cos ^2 \omega t}{2 C}
$$
(i) At $\omega t=0, \pi, 2 \pi, 3 \pi \ldots$ the energy is completely electrical.
$$
t=\frac{n \pi}{2 \pi f}=\frac{n}{2 f}=\frac{n \pi}{1000} \sec n=0,1,2,3,4 \ldots
$$
or $t=0, T / 2, T, 3 T / 2 \ldots$
(ii) Instantaneous magnetic energy
$$
U_B=\frac{1}{2} L q_0^2 \omega^2 \sin ^2 \omega t
$$
or $U_B=\frac{q_0^2}{2 C} \sin ^2 \omega t$
so at $\omega \mathrm{t}=\pi / 2,3 \pi / 2,5 \pi / 2 \ldots$
The energy is completely magnetic
$$
t=\frac{(2 n+1) \pi}{2(2 \pi f)}=\frac{(2 n+1)}{4 f}=\frac{(2 n+1) \pi}{2000} \mathrm{sec}
$$
where $n=0,1,2,3,4 \ldots \quad$ or $t=T / 4,3 T / 4,5 T / 4 \ldots$
(d) timings for energy shared equally between inductor and capacitor.
$$
\begin{aligned}
&U_B=U_E \\
&\frac{q_0^2}{2 C} \sin ^2 \omega t=\frac{q_0^2}{2 C} \cos ^2 \omega t \\
&\tan ^2 \omega t=1 \text { or } \tan \omega t=\tan \pi / 4 \\
&t=\frac{\pi}{4 \omega}, \frac{3 \pi}{4 \omega}, \frac{5 \pi}{4 \omega} \ldots \text { or } t=\frac{T}{8}, \frac{3 T}{8}, \frac{5 T}{8} \ldots
\end{aligned}
$$
(e) When a resistor in inserted in the circuit, eventually all the energy will be lost as heat across resistence. The oscillation will be damped.

Topic: Physics, Alternating Current
An L.C.R. circuit contains resistance of 110 Ω and a supply of 220 V at 300 rad s-1 angular frequency. If only capacitance is removed from the circuit, current lags behind the voltage by 45°. If on the other hand, the only the inductor is removed the current leads by 45° with the applied voltage. The R.M.S. current flowing in the circuit will be:

Options:
A 2.5 A
B 2 A
C 1 A
D 1.5 A

The correct answer is:
2 A

Solution:
tan45°=1ωCR=ωLR⇒XL=XC

⇒ resonance

i=VR

=220110=2 A

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (25 Feb Shift 2)
An LCR circuit as shown in the figure is connected to a voltage source $\mathrm{V}_{\mathrm{ac}}$ whose frequency can be varied.
The frequency, at which the voltage across the resistor is maximum, is :

Options:
A
$902 \mathrm{~Hz}$

B
$143 \mathrm{~Hz}$

C
$23 \mathrm{~Hz}$

D
$345 \mathrm{~Hz}$

The correct answer is:
$23 \mathrm{~Hz}$

Solution:
Frequency $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$
$$
\begin{aligned}
& =\frac{1}{2 \times 3.4 \sqrt{24 \times 2 \times 10^{-6}}} \\
& \simeq 23 \mathrm{~Hz}
\end{aligned}
$$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2013 (22 Apr Online)
An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q0 and then connected to the L and R as shown below:
If a student plots graphs of the square of maximum charge QMax2 on the capacitor with time (t) for two different values L1 and L2(L1>L2) of L then which of the following represents this graph correctly? (plots are schematic and not drawn to scale)

Options:
A
B
C
D

The correct answer is:

Solution:
Comparing to the damped pendulum. We write, mdvdt=-kx-bv, bv is resistive force, ⇒ Amplitude, A=A0e-bt2m. Comparing results, we can write
qc=+Ldidt+iR, as charge decreasing,
qc=Ld2qdt2-dqdtR
⇒A=q; R=b, m=L
⇒q=q0e-Rt2L⇒q2=q02e-RtL⇒q2=q02e-RtL.
Exponentially decreasing function, L1 graph has higher values than L2.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2015 (04 Apr)
An LCR series circuit is connected to an oscillator (AC supply) having RMS voltage 200 V. If the RMS voltages across resistor, inductor and capacitor are equal, then RMS voltage across the resistor would be

Options:
A 100 V
B 200 V2
C 200 V
D 1002V

The correct answer is:
200 V

Solution:
∵ V=VR2+VL-VC2=VR∵VL=VC
VR=V=200 volt

Topic: Physics, Alternating Current, JEE Main
An LCR series circuit of capacitance 62.5 nF and resistance of 50 Ω, is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is ____mH.(Take π2=10

Options:

The correct answer is:
100

Solution:
Given here, f=2 kHz=2000 Hz, C=62.5 nF=62.5×10-9 F and R=50 Ω

Resonance frequency (amplitude of the current will be maximum in this case) is given by f=12πLC, here, L is the inductance.

Putting the values,

2000 Hz=12πL×62.5×10-9

L=14π2×20002×62.5×10-9=0.1 H=100 mH

Topic: Physics, Alternating Current, JEE Main, JEE Main 2023 (25 Jan Shift 1)
An LPG cylinder contains gas at a pressure of 300 kPa at 27°C. The cylinder can withstand the pressure of 1.2×106 Pa. The room in which the cylinder is kept catches fire. The minimum temperature at which the bursting of cylinder will take place is __________ C∘. (Nearest integer)

Options:

The correct answer is:
927

Solution:
P1 T1=P2 T2⇒300×103300=1.2×106 T2

⇒T2=1200 K

T2=927°C

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2021 (25 Jul Shift 2)
An L-shaped tube with a small orifice is held in a water stream as shown in fig. The upper end of the tube is $10.6 \mathrm{~cm}$ above the surface of water. What will be the height of the jet of water coming from the orifice? Velocity of water stream is $2.45 \mathrm{~m} / \mathrm{s}$

Options:
A Zero
B $20.0 \mathrm{~cm}$
C $10.6 \mathrm{~cm}$
D $40.0 \mathrm{~cm}$

The correct answer is:
$20.0 \mathrm{~cm}$

Solution:
According to continuity equation \(A_1 V_1=A_2 V_2\)
\(\Rightarrow A_1=A_2 ; V_1=V_2\)
Equation of motion: \(v^2=u^2+2 a s \Rightarrow(0)=(2.45)^2-2(10) \times 5\)
\(\begin{aligned}
& S=\frac{(2.45)^2}{2 \times 10}=30.6 \mathrm{~cm} \\
& H_{\text {jet }}=30.6-10.6=20 \mathrm{~cm}
\end{aligned}\)

Topic: Physics, Mechanical Properties of Fluids, JEE Main
An massless equilateral triangle EFG of side 'a' (As shown in figure) has three particles of mass m situated at its vertices. The moment of inertia of the system about the line EX perpendicular to EG in the plane of EFG is N20ma2 where N is an integer. The value of N is ___________ .

Options:

The correct answer is:
25

Solution:
I=m×O2+ma2+ma22

=54ma2=2520ma2

N=25

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2020 (03 Sep Shift 2)
An $n-p-n$ transistor is connected in common-emitter configuration as shown in the figure. If the collector current is $5 \mathrm{~mA}$, $V_{B E}=0.6 \mathrm{~V}, V_{C E}=3 \mathrm{~V}$ and common-emitter current amplification factor is 50 , then the values of $R_1$ and $R_2$ are respectively.

Options:
A $1 \mathrm{k} \Omega, 74 \mathrm{k} \Omega$
B $74 \mathrm{k} \Omega, 1 \mathrm{k} \Omega$
C $37 \mathrm{k} \Omega, 2 \mathrm{k} \Omega$
D $2 \mathrm{k} \Omega, 37 \mathrm{k} \Omega$

The correct answer is:
$74 \mathrm{k} \Omega, 1 \mathrm{k} \Omega$

Solution:
In given circuit, $V_{C C}=i_B R_B+V_{B E}$
$$
\Rightarrow \quad R_B=R_1=\frac{V_{C C}-V_{B E}}{i_B}
$$
As, $i_B=\frac{i_C}{\beta}=\frac{5 \times 10^{-3}}{50}=1 \times 10^{-4} \mathrm{~A}$
$$
\Rightarrow \quad R_1=\frac{8-0.6}{1 \times 10-4}=7.4 \times 10^4=74 \times 10^3 \Omega=74 \mathrm{k} \Omega
$$
and by KVL in closed collector loop, we get
$$
\Rightarrow \quad \begin{aligned}
V_{C C} & =i_C R_L+V_{C E} \\
R_L & =\frac{V_{C C}-V_{C E}}{i_C} \\
& =\frac{8-3}{5 \times 10^{-3}}
\end{aligned}
$$
So,
$$
R_2=R_L=1.0 \mathrm{k} \Omega
$$

Topic: Physics, Semiconductors, JEE Main
An NPN transistor is used in common emitter configuration as an amplifier with 1 kΩ load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and 15 μA change in the base current of the amplifier. The input resistance and voltage gain are:

Options:
A 0.33 kΩ, 1.5
B 0.67 kΩ, 300
C 0.67 kΩ, 200
D 0.33 k Ω, 300

The correct answer is:
0.67 kΩ, 300

Solution:
load resistace=output resistaceRL=1 kΩ=1000 Ω
input voltage=ΔVi=10 mV=10×10-3 V
input current=base current=ΔiB=15 μA=15×10-6 A
output current=collector current=ΔiC=3 mA=3×10-3 A
ΔiBRi=ΔVi=0.67 kΩ
Ri=10×10-315×10-6=0.67 kΩ
Voltage gain =ΔICRLΔVi=3×10-3×100010×10-3=300

Topic: Physics, Semiconductors, JEE Main, JEE Main 2019 (09 Apr Shift 1)
An npn transistor operates as a common emitter amplifier with a power gain of 106. The input circuit resistance is 100 Ω and the output load resistance is 10 kΩ. The common emitter
current gain β will be (Round off to the Nearest Integer)

Options:

The correct answer is:
100

Solution:
106=β2×R0Ri
106=β2×104102
β2=104⇒β=100

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (18 Mar Shift 1)
An n-p-n transistor has three leads A, B and C. Connecting B and C by moist fingers, A to the positive lead of an ammeter, and C to the negative lead of the ammeter, one finds large deflection. Then, A, B and C refer respectively to :

Options:
A Base, emitter and collector
B Base, collector and emitter
C Emitter, base and collector
D Collector, emitter and base

The correct answer is:
Base, collector and emitter

Solution:
Now, according to the question, A lead of the transistor is connected to the positive lead of the ammeter and the ammeter gives a deflection. We know that in an n-p-n transistor, the emitter and the collector are negatively doped, while the base is positively doped. So, in this type of transistor the electric field is from the base to the other terminals. Thus, the lead A must be the base of the transistor.
Now, the lead C is connected to the negative lead of the ammeter. Also, the leads B and C are connected by moist fingers. So, some conduction is there in between these two terminals. Therefore, both the emitter and the collector are connected to the negative end of the terminal. But the moist fingers are less conductive than the lead. So the lead B is not directly connected to the negative end.
We are having a large deflection in the ammeter. This means that there is a huge amount of current from the base to the other terminal. This is possible only when the terminal adjacent to the base terminal has a high number of charge carriers. As the emitter is most highly doped, so the lead C must correspond to the emitter.
Now the only terminal left for the third lead B is the collector. So, the lead B must be the collector.
Thus, A is the Base, B is the collector, and C is the emitter.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2014 (09 Apr Online)
An N-P-N transistor is being used in CE mode for which the current transfer ratio is, Î±=2526. The input resistance is 1000 Î© and amplitude of A.C. input voltage is 10 mV. The amplitude of the amplified output collector current is

Options:
A 25Â Î¼A
B 250 Î¼A
C 125Â Î¼A
D 50Â Î¼A

The correct answer is:
250 Î¼A

Solution:
Î²=Î±1-Î±
or 25261-2526=Î²=25

and Input current Ib=ViRi=10 mV1000 Î©=10 Î¼A
we know, Î²=IcIb=25

â´ Output current Ic=25Ib=250 Î¼A

Topic: Physics, Semiconductors, JEE Main
An NPN transistor operates as a common emitter amplifier, with a power gain of 60 dB . The input circuit resistance is 100 Ω and the output load resistance is 10 kΩ . The common emitter current gain β is:

Options:
A 6×102
B 102
C 104
D 60

The correct answer is:
102

Solution:
Av×β=Pgain

60=10log10PP0
P=106=β2×Rout Rin =β2×104100
β2=104
β=100=102

Topic: Physics, Semiconductors, JEE Main, JEE Main 2019 (10 Apr Shift 1)
An NPN-transistor circuit is arranged as shown in fiaure. It is

Options:
A A common base amplifier circuit
B A common emitter amplifier circuit
C A common collector amplifier circuit
D Neither of the above

The correct answer is:
A common emitter amplifier circuit

Solution:
Because emitter $(M)$ is common to both, base (P) and collector (M).
Correct option is B. A common emitter amplifier circuit
An $N P N$ transistor circuit is arranged as shown in figure. It is A common emitter amplifier circuit because emitter $(N)$ is common to both, base ( $P$ ) and collector $(N)$.

Topic: Physics, Semiconductors, JEE Main
An object $1 \mathrm{~cm}$ tall is placed $4 \mathrm{~cm}$ infront of a mirror. In order to produce an upright image of $3 \mathrm{~cm}$ height one needs a

Options:
A Convex mirror of radius of curvature $12 \mathrm{~cm}$
B Concave mirror of radius of curvature $12 \mathrm{~cm}$
C Concave mirror of radius of curvature $4 \mathrm{~cm}$
D Plane mirror of height $12 \mathrm{~cm}$

The correct answer is:
Concave mirror of radius of curvature $12 \mathrm{~cm}$

Solution:
Erect and enlarged image can produced by concave mirror.
$\frac{I}{O}=\frac{f}{f-u} \Rightarrow \frac{+3}{+1}=\frac{f}{f-(-4)} \Rightarrow f=-6 \mathrm{~cm}$
$\Rightarrow R=2 f=-12 \mathrm{~cm}$

Topic: Physics, Ray Optics, JEE Main
An object 2.4 m in front of a lens forms a sharp image on a screen 12 cm behind the lens. A glass plate 1 cm thick, of refractive index 1.50 is interposed between lens and screen with its plane faces parallel to screen. At what distance from the lens, should object be placed so as to get a sharp image on the screen?

Options:
A 2.4 m

B 3.2 m

C 5.6 m

D 7.2 m

The correct answer is:
5.6 m

Solution:
According to the thin lens formula
1 fâ¡ = 1 vâ¡ - 1 uâ¡
Here, u = -2.4 m = - 240 cm, v = 12 cm
â´ 1 fâ¡ = 1 1 2 - 1 - 2 4 0 = 1 1 2 + 1 2 4 0
1fâ¡=21240 or fâ¡=24021 cm
When a glass plate is interposed between lens and film, so shift produced by it will be
Shift =t 1-1Î¼=11-11.5=11-23=13 cm
To get an image at the film, the lens should form an image at a distance
vâ¡'=12-13=353 cm
Again using lens formula
â´ 2 1 2 4 0 = 3 3 5 - 1 uâ¡ â² or 1 uâ¡ â² = 3 3 5 - 2 1 2 4 0 = 1 5 3 7 - 2 1 4 8
1 uâ¡ â² = 1 5 1 4 4 - 1 4 7 3 3 6 or 1 uâ¡ â² = - 3 1 6 8 0
uâ¡'=-560 cm=-5.6 m
uâ¡'=5.6 m

Topic: Physics, Ray Optics, JEE Main
An object $5 \mathrm{~cm}$ tall is placed $1 \mathrm{~m}$ from a concave spherical mirror which has a radius of curvature of $20 \mathrm{~cm}$ The size of the image is

Options:
A $0.11 \mathrm{~cm}$
B $0.50 \mathrm{~cm}$
C $0.55 \mathrm{~cm}$
D $0.60 \mathrm{~cm}$

The correct answer is:
$0.55 \mathrm{~cm}$

Solution:
$\frac{I}{O}=\frac{f}{(f-u)} \Rightarrow \frac{I}{+5}=\frac{-10}{-10-(-100)} \Rightarrow I=0.55 \mathrm{~cm}$

Topic: Physics, Ray Optics, JEE Main
An object $2.4 \mathrm{~m}$ in front of a lens forms a sharp image on a film $12 \mathrm{~cm}$ behind the lens. A glass plate $1 \mathrm{~cm}$ thick, of refractive index $1.50$ is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object be shifted to be in sharp focus on film?

Options:
A
$7.2 \mathrm{~m}$

B
$2.4 \mathrm{~m}$

C
$3.2 \mathrm{~m}$

D
$5.6 \mathrm{~m}$

The correct answer is:
$5.6 \mathrm{~m}$

Solution:
Case $\mathrm{I}: \mathrm{u}=-240 \mathrm{~cm}, \mathrm{v}=12$, by Lens formula
$\frac{1}{f}=\frac{7}{80}$
Case $\mathrm{II}$: $\mathrm{v}=12-\frac{1}{3}=\frac{35}{3}$ (normal shift $=1-\frac{2}{3}=\frac{1}{3}$ )
$f=\frac{7}{80}$
$u=5.6$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2012 (Offline)
An object and a concave mirror are approaching each other with velocities 10 m/s and 5 m/s as shown in figure. The velocity of image of object at the instant shown in figure is

Options:
A 65 m/s i^
B 65 m/s -i^
C 40 m/s i^
D 40 m/s -i^

The correct answer is:
65 m/s -i^

Solution:
The formula for velocity of image is given by :
=VIm=ff-u=-10-10+15=-2
â´VIm=-4 VOm
Therefore, VâI-5-i^=-410i^+5i^
âVâI=-65 i^

Topic: Physics, Ray Optics, JEE Main
An object as shown in the figure consists of a sphere of mass $1 \mathrm{~kg}$ and radius $5 \mathrm{~cm}$ attached to a rod of length $50 \mathrm{~cm}$ and mass $500 \mathrm{~g}$. If this has to be balanced horizontally on a knife-edge, at what distance from point A must the knife-edge be placed?

Options:
A 20 cm
B 30 cm
C 45 cm
D More than one of the above

The correct answer is:
45 cm

Solution:
CONCEPT:
- Center of mass: Center of the mass of a body is the weighted average position of all the parts of the body with respect to mass.
- The Center of mass is used in representing irregular objects as point masses for the ease of calculation.
- For simple shaped objects, its center of mass lies at the centroid.
- For irregular shapes, the center of mass is found by the vector addition of the weighted position vectors.
The position coordinates for the center of mass can be found by:
$C_x=\frac{m_1 x_1+m_2 x_2+\ldots m_n x_n}{m_1+m_2+\ldots m_n} \quad C_y=\frac{m_1 y_1+m_2 y_2+\ldots m_n y_n}{m_1+m_2+\ldots m_n}$
CALCULATION:
The center of mass of an object is the point at which the object can be balanced. Hence, the given object can be balanced by placing a knife edge under the center of mass.
Given that:
Mass of the rod, $m_A=500 \mathrm{~g}=0.5 \mathrm{~kg}$
Mass of the sphere, $m_B=1 \mathrm{~kg}$
$\begin{aligned} & x_1=25 \mathrm{~cm} \\ & x_2=50+5=55 \mathrm{~cm}\end{aligned}$
To find the coordinates of the centre of mass of the object from $A$,
$C_x=\frac{m_1 x_1+m_2 x_2+\ldots m_n x_n}{m_1+m_2+\ldots m_n}=\frac{m_A x_A+m_B x_B}{m_A+m_B}=\frac{(0.5 \times 25)+(1 \times 55)}{0.5+1}=45 \mathrm{~cm}$
Hence, the knife-edge must be placed $45 \mathrm{~cm}$ from point $A$.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
An object falling through a fluid is observed to have acceleration given by $a=g-b v$ where $g=$ gravitational acceleration and $b$ is constant. After a long time of release, it is observed to fall with constant speed. What must be the value of constant speed?

Options:

None

Solution:
After long time of released the speed becomes constant acceleration, $a=\frac{d v}{d t}=0$
As given that, acceleration, $a=g-b v$ where, $g=$ gravitational acceleration
As the given acceleration equation when speed increases acceleration will decrease. At a certain speed say $v_0$, acceleration will be zero and speed will remain constant.
So, $a=g-b v_0=0$
$$
\Rightarrow v_0=\frac{g}{b}
$$
Hence the constant speed after long time of release is $\left(\frac{g}{b}\right)$.

Topic: Physics, Motion In One Dimension
An object flying in air with velocity $(20 \hat{i}+25 \hat{j}-12 \hat{k})$ suddenly breaks in two pieces whose masses are in the ratio $1: 5$. The smaller mass flies off with a velocity $(100 \hat{i}+35 \hat{j}+8 \hat{k})$.
The velocity of the larger piece will be

Options:
A $4 \hat{\mathrm{i}}+23 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}$
B $-100 \hat{\mathrm{i}}-35 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}$
C $20 \hat{i}+15 \hat{j}-80 \hat{k}$
D $-20 \hat{i}-15 \hat{j}-80 \hat{k}$

The correct answer is:
$4 \hat{\mathrm{i}}+23 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}$

Solution:
Let $m$ be the mass of an object flying with velocity $v$ in air. When it split into two pieces of masses in ratio $1: 5$, the mass of smaller piece is $\mathrm{m} / 6$ and of bigger piece is $\frac{5 \mathrm{~m}}{6}$.
This situation can be interpreted diagrammatically as below
As the object breaks in two pieces, so the momentum of the system will remains conserved i.e. the total momentum (before breaking) = total momentum (after breaking)
$$
\begin{aligned}
& \mathrm{mv}=\frac{\mathrm{m}}{6} \mathrm{v}_1+\frac{5 \mathrm{~m}}{6} \mathrm{v}_2 \Rightarrow \mathrm{v}=\frac{\mathrm{v}_1}{6}+\frac{5 \mathrm{v}_2}{6} \\
& \text { Here, } \quad \begin{aligned}
\mathrm{v} & =20 \hat{\mathrm{i}}+25 \hat{\mathrm{j}}-12 \hat{\mathrm{k}} \\
\mathrm{v}_1 & =100 \hat{\mathrm{i}}+35 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}
\end{aligned}
\end{aligned}
$$
$$
\begin{array}{rlrl}
\Rightarrow & 20 \hat{\mathrm{i}}+25 \hat{\mathrm{j}}-12 \hat{\mathrm{k}} & =\frac{(100 \hat{\mathrm{i}}+35 \hat{\mathrm{j}}+8 \hat{\mathrm{k}})}{6}+\frac{5 \mathrm{v}_2}{6} \\
\Rightarrow & (120 \hat{\mathrm{i}}+150 \hat{\mathrm{j}}-72 \hat{\mathrm{k}}) & =(100 \hat{\mathrm{i}}+35 \hat{\mathrm{j}}+8 \hat{\mathrm{k}})+5 \mathrm{v}_2 \\
\Rightarrow & \mathrm{v}_2 & =\frac{1}{5}(20 \hat{\mathrm{i}}+115 \hat{\mathrm{j}}-80 \hat{\mathrm{k}}) \\
& =4 \hat{\mathrm{i}}+23 \hat{\mathrm{j}}-16 \hat{\mathrm{k}}
\end{array}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
An object flying in air with velocity $(20 \hat{i}+25 \hat{j}-12 \hat{k})$ suddenly breaks in two pieces whose masses are in the ratio $1: 5$. The smaller mass flies off with a velocity $(100 \hat{i}+35 \hat{j}+8 \hat{k})$.
The velocity of the larger piece will be

Options:
A $4 \hat{\mathrm{i}}+23 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}$
B $-100 \hat{\mathrm{i}}-35 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}$
C $20 \hat{i}+15 \hat{j}-80 \hat{k}$
D $-20 \hat{i}-15 \hat{j}-80 \hat{k}$

The correct answer is:
$4 \hat{\mathrm{i}}+23 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}$

Solution:
Let $m$ be the mass of an object flying with velocity $v$ in air. When it split into two pieces of masses in ratio $1: 5$, the mass of smaller piece is $\mathrm{m} / 6$ and of bigger piece is $\frac{5 \mathrm{~m}}{6}$.
This situation can be interpreted diagrammatically as below
As the object breaks in two pieces, so the momentum of the system will remains conserved i.e. the total momentum (before breaking) = total momentum (after breaking)
$$
\begin{aligned}
& \mathrm{mv}=\frac{\mathrm{m}}{6} \mathrm{v}_1+\frac{5 \mathrm{~m}}{6} \mathrm{v}_2 \Rightarrow \mathrm{v}=\frac{\mathrm{v}_1}{6}+\frac{5 \mathrm{v}_2}{6} \\
& \text { Here, } \quad \begin{aligned}
\mathrm{v} & =20 \hat{\mathrm{i}}+25 \hat{\mathrm{j}}-12 \hat{\mathrm{k}} \\
\mathrm{v}_1 & =100 \hat{\mathrm{i}}+35 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}
\end{aligned}
\end{aligned}
$$
$$
\begin{array}{rlrl}
\Rightarrow & 20 \hat{\mathrm{i}}+25 \hat{\mathrm{j}}-12 \hat{\mathrm{k}} & =\frac{(100 \hat{\mathrm{i}}+35 \hat{\mathrm{j}}+8 \hat{\mathrm{k}})}{6}+\frac{5 \mathrm{v}_2}{6} \\
\Rightarrow & (120 \hat{\mathrm{i}}+150 \hat{\mathrm{j}}-72 \hat{\mathrm{k}}) & =(100 \hat{\mathrm{i}}+35 \hat{\mathrm{j}}+8 \hat{\mathrm{k}})+5 \mathrm{v}_2 \\
\Rightarrow & \mathrm{v}_2 & =\frac{1}{5}(20 \hat{\mathrm{i}}+115 \hat{\mathrm{j}}-80 \hat{\mathrm{k}}) \\
& =4 \hat{\mathrm{i}}+23 \hat{\mathrm{j}}-16 \hat{\mathrm{k}}
\end{array}
$$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2019 (Odisha)
An object is allowed to fall from a height R above the earth, where R is the radius of earth. Its velocity when it strikes the earth’s surface, ignoring air resistance, will be :

Options:
A 2gR
B gR
C gR2
D 2gR

The correct answer is:
gR

Solution:
As gravitational field is a conservative field and therefore mechanical energy will be conserved. Therefore,
Loss in PE = Gain in KE

-GMm2R--GMmR=12mv2

⇒v2=GMR=gRas g=GMR2

⇒v=gR

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2023 (30 Jan Shift 2)
An object is at a distance of $0.5 \mathrm{~m}$ in front of a plane mirror. Distance between the object and image is

Options:
A $0.5 \mathrm{~m}$
B $1 \mathrm{~m}$
C $0.25 \mathrm{~m}$
D $1.5 \mathrm{~m}$

The correct answer is:
$1 \mathrm{~m}$

Solution:
in plane mirror distance of object= distance of image from mirror

Topic: Physics, Ray Optics, JEE Main
An object is dropped from a height h from the ground. Every time it hits the ground it loses 50% of its kinetic energy. The total distance covered as t→∞ is:

Options:
A 3h
B ∞
C 53h
D 83h

The correct answer is:
3h

Solution:
Since the object is dropped from height, thus its potential energy is equal to its kinetic energy.

As per question, kinetic energy after hitting the ground is 12mv′2=12×12 mv2
v′=v2
v=eu
e=12
H=h1+e21-e2
The total distance covered=h1+121-12=3h

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2017 (08 Apr Online)