datavorous/entrance-exam-dataset huggingface dataset:
A bullet of mass $m$ hits a block of mass $M$ elastically. The transfer of energy is the maximum, when

Options:
A $M \ll m$
B $M \gg m$
C $M=m$
D $M=2 m$

The correct answer is:
$M=m$

Solution:
During elastic collision between two equal masses, the velocity of the two bodies gets interchanged so if one body is at rest, energy transfer will be maximum for $M=m$ $m_1=m_2=m$ (say)

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2023 (Manipur)
A bullet of mass M hits a block of mass M'. The transfer of energy is maximum, when

Options:
A M'=M
B M'=2M
C M'≪M
D M'â«M

The correct answer is:
M'=M

Solution:
If M=M' then bullet will transfer the whole of its velocity (and consequently 100% of its KE) to block and will itself come to rest as per the theory of collision.

Topic: Physics, Center of Mass Momentum and Collision, NEET
A bullet of mass m is fired with a velocity of 50 m s-1 at an angle Î¸ with the horizontal. At the highest point of its trajectory, it collides head on with a bob connected to massless string of length l and gets embedded in the bob of mass 3m. After the collision, the string moves to an angle of 120Â°. What is the angle Î¸ ?

Options:
A cos-1⁡45
B cos-1⁡54
C sin-1⁡45
D sin-1⁡54

The correct answer is:
cos-1⁡45

Solution:
Velocity of bullet at highest point of its trajectory = 50 cos⁡θ in horizontal direction.
As bullet of mass m collides with pendulum bob of mass 3m and two stick together, their common velocity
v′=50mcos⁡θm+3m=252cos⁡θ m s-1
As now under this velocity v′ pendulum bob goes up to an angle 120°, hence
v′22g=h=l1-cos⁡120°=1031-f-12=5
⇒ v′2=2×10×5=100 or v′=10
Comparing two answer of v′, we get
252cos⁡θ=10⇒cos⁡θ=45 or θ=cos-1⁡45

Topic: Physics, Center of Mass Momentum and Collision, NEET
A bullet of mass m leaves a gun of mass M kept on a smooth horizontal surface. If the speed of the bullet relative to the gun is v, the magnitude of recoil speed of the gun will be

Options:
A mMv
B

mM+mv

C

mM-mv

D Mmv

The correct answer is:

mM+mv

Solution:
Speed of the bullet relative to ground v→b=v→+v→r, where vr is recoil velocity of gun. Now for gun-bullet system applying the conservation law of momentum, we get
mv→b+Mv→r=0 or mv→+v→r+Mv→r=0
⇒ v→r=-mv→m+M or vr=mvm+M

Topic: Physics, Center of Mass Momentum and Collision, NEET
A bullet of mass m moving with velocity v1 strikes a suspended wooden block of mass M as shown in the figure and sticks to it. If the block rises to a height y, the initial velocity of the bullet is

Options:
A v1=m+Mm2gy
B v1=2gy
C v1=M+mM2gy
D v1=mm+M2gy

The correct answer is:
v1=m+Mm2gy

Solution:
As block raises to a height y, therefore speed just after collisions is
v1=2gy
As the collision is perfectly inelastic, using conservation of linear momentum,
mv1=m+M2gy
â´ v1=m+Mm2gy

Topic: Physics, Center of Mass Momentum and Collision, NEET
A bullet of mass m moving with velocity v strikes a block of mass M at rest and gets embedded into it. The kinetic energy of the composite block will be

Options:
A 12m2v2m+M
B 12 Mmv2m+M
C 12mv2 M+mM
D 12 mMv2M+m

The correct answer is:
12m2v2m+M

Solution:
mv=M + m vâ²
vâ²=mvM+m
K.E. of composite block=12M + mvâ²2
= 12 M + mÃ m2v2M+m2= m2v22M+m

Topic: Physics, Center of Mass Momentum and Collision, NEET
A bus is moving with a speed of $10 \mathrm{~ms}^{-1}$ on a straight road. A scooterist wishes to overtake the bus in $100 \mathrm{~s}$. If the bus is at a distance of $1 \mathrm{~km}$ from the scooterist, with what speed should the scooterist chase the bus?

Options:
A $20 \mathrm{~ms}^{-1}$
B $40 \mathrm{~ms}^{-1}$
C $25 \mathrm{~ms}^{-1}$
D $10 \mathrm{~ms}^{-1}$

The correct answer is:
$20 \mathrm{~ms}^{-1}$

Solution:
Let $y$ be the relative velocity of scooter(s) w.r.t. bus $(B)$, then
$$
\begin{aligned}
\therefore \quad v_S & =\mathrm{v}+\mathrm{v}_{\mathrm{B}} \\
\text { Relative velocity } & =\text { displacement } / \text { time } \\
& =\frac{1000}{100}=10 \mathrm{~ms}^{-1}
\end{aligned}
$$
Now, substituting the value of in Eq. (i), we get
$$
\mathrm{v}_{\mathrm{S}}=10+10=20 \mathrm{~ms}^{-1}
$$

Topic: Physics, Motion In One Dimension, NEET, NEET 2009 (Screening)
A bus is moving with a velocity of 10 m s-1â¡ on a straight road. A scooterist wishes to overtake the bus in 100 s. If the bus is at a distance of 1 km from the scooterist, the scooterist should chase the bus with a constant velocity of

Options:
A 50 m s-1â¡
B 40 m s-1â¡
C 30 m s-1â¡
D 20 m s-1â¡

The correct answer is:
20 m s-1â¡

Solution:
Let v be the speed of the scooterist then
v100=10100+1000
v=20 m s-1

Topic: Physics, Motion In One Dimension, NEET
A button cell used in watches functions as following
$\mathrm{Zn}(s)+\mathrm{Ag}_2 \mathrm{O}(s) & +\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons 2 \mathrm{Ag}(s) + \mathrm{Zn}^{2+}(\alpha q)+2 \mathrm{OH}^{-}(\alpha q)$
If half cell potentials are
$\begin{aligned}
\mathrm{Zn}^{2+}(a q)+2 e^{-} & \rightarrow \mathrm{Zn}(s) ; E^{\circ}=-0.76 \mathrm{~V} \\
\mathrm{Ag}_2 \mathrm{O}(s)+\mathrm{H}_2 \mathrm{O}(l) & +2 e^{-} \\
& \rightarrow 2 \mathrm{Ag}(s)+2 \mathrm{OH}^{-}(a q), \\
& E^{\circ}=0.34 \mathrm{~V}
\end{aligned}$
The cell potential will be

Options:
A $1.10 \mathrm{~V}$
B $0.42 \mathrm{~V}$
C $0.84 \mathrm{~V}$
D $1.34 \mathrm{~V}$

The correct answer is:
$1.10 \mathrm{~V}$

Solution:
Anode is always the site of oxidation thus anode half cell is
$\mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}(\mathrm{s}) ; \quad E^{\circ}=-0.76 \mathrm{~V}$
Cathode half cell is
$\begin{aligned}
& \mathrm{Ag}_2 \mathrm{O}(s)+ \mathrm{H}_2 \mathrm{O}(l)+2 e^{-} \longrightarrow 2 \mathrm{Ag}(s)+2 \mathrm{OH} \\
& E^{-}(a q) ; E^{\circ}=0.34 \mathrm{~V} \\
&= E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \\
&= 0.34-(-0.76)=+1.10 \mathrm{~V}
\end{aligned}$

Topic: Chemistry, Electrochemistry, NEET, NEET 2013 (All India)
A button cell used in watches functions as follows
Zn s + Ag 2 O s + H 2 O⇌ 2Ag s + Zn 2+ aq +2 OH −
If half cell potentials are
Zn2+(aq)+2e-→Zn(s)
E°=-0.76V
Ag 2 O (s) + H 2 O (l) +2 e − → 2Ag (s) +2 OH −
E°=0.34V
The cell potential will be

Options:
A 1.1 V
B 0.42 V
C 0.84 V
D 1.34 V

The correct answer is:
1.1 V

Solution:
As cell is
Zn/Zn2+||Ag+/Ag
Ecello=ECathodeo-EAnodeo
=0.34-(-0.76)=1.1V

Topic: Chemistry, Electrochemistry, NEET
A calorimeter contains 70.2 g of water at 15.3 Â°C. If 143.7 g of water at 36.5 Â°C is mixed with it, the common temperature becomes 28.7 Â°C. The water equivalent of the calorimeter is

Options:
A 15.6Â g
B 9.4Â g
C 6.3Â g
D 13.4 g

The correct answer is:
13.4 g

Solution:
Let m be the water equivalent of the calorimeter. Then,
70.2+m28.7-15.3=143.7 36.5-28.7
m=13.4 g

Topic: Physics, Thermal Properties of Matter, NEET
A camera objective has an aperture diameter d. If the aperture is reduced to diameter d2, the exposure time under identical conditions of light should be made

Options:
A 2 fold
B 2 fold
C 22 fold
D 4 fold

The correct answer is:
4 fold

Solution:
For camera, time of exposure ∝1Aperture2 ¸
Therefore
t2t1=d1d22
t2t1=dd22=4
t2t1=4
Therefore exposure time under identical conditions of light should be made 4 fold

Topic: Physics, Ray Optics, NEET
A candidate connects a moving coil ammeter A and a moving coil voltmeter V and a resistance R as shown in figure

If the voltmeter reads 20 V and the ammeter reads 4 A, then R is

Options:
A Equal to 5 Î©
B Greater than 5 Ω
C Less than 5 Î©
D Greater or less than 5 Î©Â depending upon its material

The correct answer is:
Greater than 5 Ω

Solution:
Here, i=4A;V=20 Volt; so,
R=VI=204=5Ω. Since, voltmeter is connected in parallel with resistance R, the effective resistance of this combination is 5 Ω only if the resistance R is greater than 5Ω, since total resistance in parallel combination becomes less than individual resistance.

Topic: Physics, Current Electricity, NEET
A cannon shell fired at an angle Î¸, with horizontal breaks into two equal parts at its highest point. One part retraces the path to the cannon with kinetic energy E1 and kinetic energy of the second part is E2 , the relation between E1 and E2 is

Options:
A E2=15E1
B E2=E1
C E2=4E1
D E2=9E1

The correct answer is:
E2=9E1

Solution:
A shell fired at an angle 'Î¸', velocity of the shell at the highest point is ucosÎ¸
Let its mass be m
The speed of one piece is ucosÎ¸ and let the speed of other piece be v.

Hence at the highest point, from conservation of momentum
mucosâ¡Î¸= -m2ucosâ¡Î¸+m2v
mucosâ¡Î¸+m2 ucosâ¡Î¸=m2v
2mucosâ¡Î¸+mucosâ¡Î¸2=m2v
mv=3mucosâ¡Î¸
v=3ucosâ¡Î¸ .......(i)
The kinetic energy
E1=12Ãm2u2cos2â¡Î¸
E1=14mu2cos2â¡Î¸ .......(ii)
Similarly,
E2=12Ãm2Ã9u2cos2â¡Î¸
E2=94mu2cos2â¡Î¸ ......(iii)
The relation between the E1 and E2
E2=9ÃE1 ( from the equation (ii) )
E2=9E1

Topic: Physics, Center of Mass Momentum and Collision, NEET
A capacitance of 2 Î¼F is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 Î¼F capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is:

Options:
A 32
B 2
C 16
D 24

The correct answer is:
32

Solution:
Let us assume that we connect n (whole number) capacitors in series such that the potential difference across the combination is 1000 V. This means the potential difference across each capacitor is

Vcapacitor=1000n V≤300 V

n⩾1000300=3.333...

nmin=4

This means we need to connect at least 4 capacitors in series to make sure that the potential across each one is less than 300 V.

Ceq=mnC=2 μF

m4=2⇒m=8

∴ Minimum no. of capacitors =8×4=32

Topic: Physics, Electrostatics, NEET
A capacitor is made of a flat plate of area A and a second plate having a stair-like structure as shown in the diagram. The capacitance of the arrangement is-

Options:
A ε0Ad
B 1811ε0Ad
C 1118ε0Ad
D ε0A3d

The correct answer is:
1118ε0Ad

Solution:
All the steps of stair makes three parellel capacitors.
Ceq=C1+C2+C3
=Îµ0A3d+Îµ0A3 Ã 2d+Îµ0A3Ã3 d=11Îµ0A18 d

Topic: Physics, Electrostatics, NEET
A capacitor of 2 μF is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is

Options:
A 0%.
B 20%.
C 75%.
D 80%.

The correct answer is:
80%.

Solution:
First position,

C 1 =2  μF
C 2 =8  μF
Q= C 1 V
U i = Q 2 2 C 1
Second position,
Q−q C 1 = q C 2
C 2 Q− C 2 q= C 1 q
q= C 2 Q C 1 + C 2
&Q−q= C 1 C 1 + C 2 Q
Energy on 2 μF=12(Q−q)2C1
(Uf)1=12C1(C1+C2)2Q2
Energy on 8 μF=12q2C
( U f ) 2 = 1 2 C 2 ( C 1 + C 2 ) 2 Q 2
Total energy =12C1(C1+C2)2Q2+12C2(C1+C2)2Q2=12Q2(C1+C2)
Energy dissipated =Uf−Ui=12C2C1(C1+C2)Q2 = 1 2 C 1 C 2 C 1 + C 2 V 2

Percentage of energy dissipated, = C 2 C 1 + C 2 ×100= 8 2+8 ×100=80%

Topic: Physics, Capacitance, NEET, NEET 2016 (Phase 1)
A capacitor of 8 F is connected as shown in the figure. Charge on the plates of the capacitor

Options:
A 32 C
B 40Â C
C 0Â C
D 80Â C

The correct answer is:
32 C

Solution:
When the capacitor completely charged, there is no current passes through it. Current exists only in other branch.
I=VReq=51+4= 1 A
The potential difference across capacitor =5-1=4 V
Charge stored in capacitor, q
=CÃV=8Ã4=32 C

Topic: Physics, Electrostatics, NEET
A capacitor of capacitance 100 μF is charged by connecting it to a battery of EMF 12 V and internal resistance 2 Ω. The time taken before 99% of the maximum charge is stored on the capacitor-

Options:
A 0.92 ms
B 0.4 ms
C 0.8 ms
D 0.1 ms

The correct answer is:
0.92 ms

Solution:
During charging of capacitor, charge at any instant is
q=q01-et/RC
0.99q0=q01-e-t/RC
0.99=1-e-t/RC
0.01=e-t/RC
et/RC=100
t=RClnâ¡100
t2100Ã10-62Ã2.303
t=0.92 ms

Topic: Physics, Current Electricity, NEET
A capacitor of capacitance $C_1$ charged up to $V$ volt and then connected to an uncharged capacitor $C_2$. Then, the final potential difference across each will be:

Options:
A $\frac{C_2 V}{C_1+C_2}$
B $\frac{C_1 V}{C_1+C_2}$
C $\left(1+\frac{C_2}{C_1}\right) V$
D $\left(1-\frac{C_2}{C_1}\right) V$

The correct answer is:
$\frac{C_1 V}{C_1+C_2}$

Solution:
$$
\begin{aligned}
& V_{\text {comma }}=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}\left(\because V_2=0\right) \\
& \Rightarrow V_{\text {comma }}=\frac{C_1 V_1}{C_1+C_2}
\end{aligned}
$$

Topic: Physics, Capacitance, NEET, NEET 2002
A capacitor of capacitance C=900 pF is charged fully by 100 V battery B as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C=900 pF as shown in figure (b). The electrostatic energy stored by the system (b) is

Options:
A 3.25×10-6 J
B 2.25×10-6 J
C 1.5×10-6 J
D 4.5×10-6 J

The correct answer is:
2.25×10-6 J

Solution:
Initial charge on 900 pF capacitor will be q=CV

When the two equal capacitors are joined together as shown in figure (b). The charge will be equally divided among them. Therefore,

q'=q2=CV2

Now the energy stored in them will be

E=q'22C+q'22C

⇒E=CV22×1C=CV24=900×10-12×10024

⇒E=2.25×10-6 J

Topic: Physics, Capacitance, NEET, NEET 2022 (Phase 1)
A capacitor of capacitance C is charged to a potential difference V and then connected in series with an open key and a pure resistor R. At time t=0, the key is closed. If I is current at time t=0, a plot of logâ¡I against t is shown in the graph (2). Later one of the parameters, i.e., V, R and C is changed, keeping the other two constant and graph (2) is recorded. Then

Options:
A C is reduced
B C is increased
C R is reduced
D R is increased

The correct answer is:
C is increased

Solution:
During discharging of a capacitor, the current is given by,
I=ERe- tRC
logâ¡I=logâ¡ERe- tRC
logâ¡I=-tRC+logâ¡ER
Intercept is constant â E & R constant
Magnitude of slope decreases â C is increased

Topic: Physics, Current Electricity, NEET
A capacitor of capacitance C is charged to a potential V. The flux of the electric field through a closed surface enclosing the capacitor is

Options:
A CVε0
B 2CVε0
C CV2ε0
D zero

The correct answer is:
zero

Solution:
As the net charge on a capacitor including both the plates is zero.
Therefore from Gauss's law, the flux of the electric field through a closed surface enclosing the capacitor is zero.

Topic: Physics, Electrostatics, NEET
A capacitor of capacitance C, is connected across an ac source of voltage V, given by V=V0sinωt. The displacement current between the plates of the capacitor would then be given by___

Options:
A Id=V0ωCsinωt
B Id=V0ωCcosωt
C Id=V0ωCcosωt
D Id=V0ωCsinωt

The correct answer is:
Id=V0ωCcosωt

Solution:
∵q=CV

And V=V0sinωt

Id=?

As we know,

∵Id=ε0dQdt

=ε0dEAdt∵ E=Vd

=ε0AdVddt

=ε0AdddtV0sinωt

Id=CV0ωcosωt

Topic: Physics, Electromagnetic Waves, NEET, NEET 2021
A capillary tube is dipped in water up to a depth l and the water rises to a height h<l in the capillary tube. The lower end of the tube is closed now and the tube is taken out of water. When the lower end of the tube is opened now the length of the liquid column in the tube will be

Options:
A l
B l+h
C h
D 2h

The correct answer is:
2h

Solution:
The force of surface tension of water (due to top meniscus) can lift a water column of length h<l. So with the help of two surfaces (top and bottom meniscus), we will be able to hold a length 2h

ΔP=2×2Tr=4Tr
2Tr=ρgh
and ΔP=ρgh'
∴ h'=2h

Topic: Physics, Mechanical Properties of Fluids, NEET
A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of the water in the capillary is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is:

Options:
A 5.0 g
B 10.0 g
C 20.0 g
D 2.5 g

The correct answer is:
10.0 g

Solution:
The relation between mass and radius of the capillary tube is given by,m∝r

⇒m2m1=r2r1

⇒m25=2rr

⇒m2=10 g

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2020 (Phase 1)
A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of water in the capillary tube is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is

Options:
A 2.5 g
B 5.0 g
C 10 g
D 20 g

The correct answer is:
10 g

Solution:
Height of liquid in column is h=2Tcos⁡θrρg
∴ Mass of liquid in column =π2hρ
∴ Mass=volume×density
∴ m=πr2 2Tcos⁡θrρgρ
∴ m= 2πTcos⁡θgr
∴ m∝r or m1m2= r1r2
or 5m2=r2r or m2=10 g

Topic: Physics, Mechanical Properties of Fluids, NEET
A capillary tube of the radius 0.5 mm is immersed in a beaker of mercury. The level inside the tube is 0.8 cm below the level in beaker and angle of contact is 120Â°. What is the surface tension of mercury, if the mass density of mercury is Ï=13.6Ã103 kg m-3 and acceleration due to gravity is g=10 m s-2 ?

Options:
A 0.225 N m-1
B 0.544 N m-1
C 0.285 N m-1
D 0.375 N m-1

The correct answer is:
0.544 N m-1

Solution:
Given,
The radius of capillary tube =0.5 mm
=0.5×10-3 m
Level inside tube =0.8 cm
=0.8×10-2 m
The angle of contract, θ=120o
The mass density of mercury,
ρ=13.6×103 kg m-2
Acceleration due to gravity, g=10 m s-2
h=2Tcos⁡θrρg
T=hr ρg2cos⁡θ
=0.8×10-2×0.5×10-3×13.6×103×102×cos⁡120
=0.8×10-2×0.5×10-3×13.6×103×102×12
=0.8×0.5×13.6×10-1
=0.544N m-1

Topic: Physics, Mechanical Properties of Fluids, NEET
A capillary tube of the radius r is immersed in water and water rise in it to a height H. Mass of water in the capillary tube is m. If the capillary of radius 2r is taken and dipped in water, the mass of water that will rise in the capillary tube will be

Options:
A m
B 2 m
C m2
D 4 m

The correct answer is:
2 m

Solution:
mgA=Â 2TR
âÂ Â Â Â Â Â Â Â mgÏr2=Â 2Â TcosÎ¸r
âÂ Â Â Â Â Â Â Â Â Â mr=Â 2ÏTcosÎ¸g=constant
âÂ Â Â Â Â Â Â Â Â Â Â mr=Â m′2r
âÂ Â Â Â Â Â Â Â Â m′=2m

Topic: Physics, Mechanical Properties of Fluids, NEET
A car A is moving with speed 40 km h-1 along a straight line 30Â° north of east and another car B is moving with same speed along a straight line 30Â° south of east. The relative velocity of car A as observed from the car B is

Options:
A 40 km h-1 north-east
B 40 km h-1 south
C 40 km h-1 north
D 40 km h-1 south-east

The correct answer is:
40 km h-1 north

Solution:
V→A=40cos⁡30o i^+20j^
V→B=40cos⁡30o i^-20j^
∴V→AB=V→A-V→B=40j^=40 km h-1 north

Topic: Physics, Motion In Two Dimensions, NEET
A car accelerates from rest at a constant rate Î± for some time, after which it decelerates at a constant rate Î² and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is

Options:
A ( Î± 2 â Î² 2 Î±Î² )t
B ( α 2 + β 2 αβ )t
C (α+β)t αβ
D Î±Î²t Î±+Î²

The correct answer is:
αβt α+β

Solution:
Let us draw the vât graph of the given situation. Initially it accelerates uniformly from rest. Hence, it is a straight line.

â v max =0+Î± t 1 âââ t 1 = v max Î± ââââââ0= v max âÎ² t 2 â t 2 = v max Î²

In Figure

t=t1+t2=vmaxÎ±+vmaxÎ²
t=vmax1Î±+1Î²=vmaxÎ±+Î²Î±Î²
vmax=tÎ±Î²Î±+Î²

Topic: Physics, Motion In One Dimension, NEET
A car is moving along a straight road with uniform acceleration. It passes through two points P and Q separated by a distance with velocity 30 km h-1 and 40 km h-1 respectively. The velocity of the car midway between P and Q is

Options:
A 33.3 km h-1
B 202 km h-1
C 252 km h-1
D 35 km h-1

The correct answer is:
252 km h-1

Solution:
Let PQ = x, then
a=402-3022x=350x ∵ v2=u2+2 as
Also, the velocity at the midpoint is given by
v2-302=2×350x×x2
This gives v=252 km h-1

Topic: Physics, Motion In One Dimension, NEET
A car is moving in a circular horizontal track of radius $10 \mathrm{~m}$ the roof of the car by a light wire of length $1.0 \mathrm{~m}$. The angle made by the wire with the vertical is:

Options:
A $\frac{\pi}{3}$
B $\frac{\pi}{6}$
C $\frac{\pi}{4}$
D $0^{\circ}$

The correct answer is:
$\frac{\pi}{4}$

Solution:
From the question $v=10 \mathrm{~m} / \mathrm{s}, \mathrm{r}=10$ $\mathrm{m}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$
If $\theta=$ angle made by wire with vertical then
$$
\begin{aligned}
\tan \theta & =\frac{V^2}{r g} \\
& =\frac{10 \times 10}{10 \times 10}=1 \\
\tan \theta=1 \Rightarrow \theta=\tan ^{-1} & (\mathrm{i})=\frac{\pi}{4}
\end{aligned}
$$

Topic: Physics, Laws of Motion, NEET, NEET 2013 (Karnataka)
A car is moving on a circular level road of the radius of curvature 300 m. If the coefficient of friction is 0.3 and acceleration due to gravity is 10 m s-2, the maximum speed the car can have is (in km h-1)

Options:
A 30 km h-1

B 81 km h-1

C 108 km h-1

D 162 km h-1

The correct answer is:
108 km h-1

Solution:
r = 300 m
μ = 0 . 3
g=10 m s-2
v max = μ rg
fs≤fs(max)
mv2r≤μmg
v2≤μrg
v≤μrg
vmax=μrg
= 0 . 3 3 0 0 1 0
=30 m s-1
= 108 km h-1

Topic: Physics, Laws of Motion, NEET
A car is moving towards a high cliff. The driver sounds a horn of frequency f. The reflected sound heard by the driver has frequency 2f. If v is the velocity of sound, then the velocity of the car, in the same velocity units, will be

Options:
A v2
B v3
C v4
D v2

The correct answer is:
v3

Solution:
1st the car is the source and at the cliff, one observes f'.
∴ f'=vv-vsf
2nd cliff is now source. It emits frequency f' and the observer is now the driver who observes f''.
∴ f''=v+v0vf' or 2f=v+v0v-vsf
⇒ 2v-2v0=v+v0⇒v=v0+2v0 as vs=v0
⇒ v0=v3

Topic: Physics, Waves and Sound, NEET
A car is moving towards a high cliff. The driver sounds a horn of frequency $f$. The reflected sound herd by the driver has frequency $2 f$. If $v$ is velocity of sound, then the velocity of the car, in the same velocity units, will be:

Options:
A $\frac{v}{\sqrt{2}}$
B $\frac{v}{3}$
C $\frac{v}{4}$
D $\frac{v}{2}$

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

Solution:
According to given condition
Cliff is stationary source of frequency $f$ where
$$
f=\left(\frac{V+V_{\text {car }}}{v}\right) f^{\prime}
$$
Frequency heard by the drives
$$
\begin{array}{rlrl}
f^{\prime \prime}=\left(\frac{V+V_{\mathrm{car}}}{v-v_{\mathrm{car}}}\right) f^1=\left(\frac{V+V_{\mathrm{car}}}{v-v_{\mathrm{car}}}\right) f \\
& =2 f \\
\Rightarrow & V+V_{\mathrm{car}} & =2 V-2 v_{\mathrm{car}} \\
\Rightarrow & 3 v_{\mathrm{car}} & =V \\
\Rightarrow & \quad V_{\mathrm{car}} & =\frac{v}{3}
\end{array}
$$

Topic: Physics, Waves and Sound, NEET, NEET 2004
A car is moving with 90 km h-1 blows a horn of 150 Hz, towards a cliff. The frequency of the reflected sound heard by the driver will be (speed of sound in air is 340 m s-1 )

Options:
A 150 Hz
B 140 Hz
C 180 Hz
D 174 Hz

The correct answer is:
174 Hz

Solution:
Given, vs=90 km h-1=90×518 m s-1=25 m s-1
νs=150 Hz, speed of sound, v=340 m s-1
Here, the source is an observer of reflected sound and moving towards the reflector (the cliff)
Thus, vo=25 m s-1
Since, Doppler's equation for sound,
ν=v+vsv-voνs=340+25340-25×150
=365315×150=173.88 Hz
∴ ν=174 Hz

Topic: Physics, Waves and Sound, NEET
A car is moving with speed 20ââm s-1 on a circular path of radius 100ââm. Its speed is increasing at a rate of 3 m s-2. The magnitude of the acceleration of the car at that moment is

Options:
A 1 m s-2
B 3 m s-2
C 4 m s-2
D 5 m s-2

The correct answer is:
5 m s-2

Solution:
Given v=20 m s-1,
R=100 m
dvdt= 3 m s-2
ac=v2R=(20)2100=4 m s-2
at=dvdt=3 m s-2
a=ac2+at2=42+32=5 m s-2

Topic: Physics, Laws of Motion, NEET
A car is moving with speed 20ââm/s on a circular path of radius 100ââm . Its speed is increasing at the rate of 3 m/s2. The magnitude of acceleration of the car at that moment is

Options:
A 1 m/s2
B 3 m/s2
C 4 m/s2
D 5 m/s2

The correct answer is:
5 m/s2

Solution:
Given v=20 m/s,
R=100 m
dv dt = 3  m/s 2
a c = v 2 R = ( 20 ) 2 100 =4  m/s 2
a t = dv dt =3  m/s 2
a= a c 2 + a t 2 = 4 2 + 3 2 =5  m/s 2

Topic: Physics, Laws of Motion, NEET
A car is negotiating a curved road of radius R . The road is banked at an angle θ. The coefficient of friction between the tyres of the car and the road is μs. The maximum safe velocity on this road is:

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

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

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

Topic: Physics, Laws of Motion, NEET, NEET 2016 (Phase 1)