datavorous/entrance-exam-dataset huggingface dataset:
Two rods OA and OB of equal length and mass are lying on xy plane as shown in figure. Let Ix,Iy and Iz be the moment of inertias of both the rods about x,y and z axis respectively. Then moment of inertia of the combined rod system is

Options:
A I x = I y > I z
B I x = I y < I z
C I x > I y > I z
D I z > I y > I x

The correct answer is:
I x = I y < I z

Solution:
I x = I y = 2 ml 2 3 sin 2 4 5 ο = ml 2 3
I z = 2 ml 2 3 = 2 3 ml 2
Hence, I x = I y < I z
NOTE : for lx and ly you can remember the
following formula which is obtained by integration.
I xx = ml 2 3 sin 2 θ

Topic: Physics, Rotational Motion, NEET
Two rods of the same length and diameter having thermal conductivities K1 and K2 are joined in parallel. The equivalent thermal conductivity of the combination is

Options:
A K1K2K1 + K2
B K1 + K2
C K1+K22
D K1K2

The correct answer is:
K1+K22

Solution:
Let n slabs each of length l, area
A and thermal conductivities
K1, K2, K3,â¦â¦.,Knare connected in parallel, then,
Keq=K1+K2+K3+â¦+Knn
For two slabs of equal area Keq=K1+K22

Topic: Physics, Thermal Properties of Matter, NEET
Two rods one made of copper and other made of steel of the same length and same cross sectional area are joined together. The thermal conductivity of copper and steel are $385 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~K}^{-1} \mathrm{~m}^{-1}$ and $50 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1}$ respectively. The free ends of copper and steel are held at $100^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$ respectively. The temperature at the junction is, nearly:

Options:
A $12^{\circ} \mathrm{C}$
B $50^{\circ} \mathrm{C}$
C $73^{\circ} \mathrm{C}$
D $88.5^{\circ} \mathrm{C}$

The correct answer is:
$88.5^{\circ} \mathrm{C}$

Solution:
As per question, we have;
$\begin{aligned}
& \theta=\text { ? } \\
& \text {As } \\
& \mathrm{K}=\frac{Q d}{\mathrm{~A} \Delta t}, \\
& \text {where, } \mathrm{K}=\text { thermal conductivity } \\
& Q=\text { amount of heat transferred } \\
& d=\text { distance between two planes } \\
& \mathrm{A}=\text { area of surface, and } \\
& \Delta \mathrm{T}=\text { difference in Temp. } \\
& \frac{\Delta \mathrm{Q}}{\Delta t}=\frac{385 \times \mathrm{A} \times\left(100^{\circ} \mathrm{C}-\theta\right)}{l} \\
& =\frac{50 \times \mathrm{A} \times\left(\theta-0^{\circ} \mathrm{C}\right)}{l} \\
& \Rightarrow 385\left(100^{\circ} \mathrm{C}-\theta\right)=50\left(\theta-0^{\circ} \mathrm{C}\right) \\
& \Rightarrow \theta=88.5^{\circ} \mathrm{C}
\end{aligned}$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2022 (Phase 2)
Two rods P and Q of equal lengths have thermal conductivities K1 and K2 and cross-sectional areas A1 and A2 respectively. If the temperature difference across the ends of each rod is the same, then the condition for which the rate of flow of heat through each of them will be equal, is

Options:
A A1A2=K2K1
B A1A2=K1K2
C A1A2= K1K2
D A1A2= K2K12

The correct answer is:
A1A2=K2K1

Solution:
âQâtP=âQâtQ
K1A1T1-T2l=K2A2T1-T2l
Or K1A1=K2A2 or A1A2=K2K1

Topic: Physics, Thermal Properties of Matter, NEET
Two rods P and Q of the same length and same diameter having thermal conductivity in the ratio 2:3 are joined end to end. If the temperature at one end of P is 100 Â°C and at one end of Q is 0 Â°C, then the temperature of the interface is

Options:
A 40 Â°C
B 50 °C
C 60 °C
D 70 °C

The correct answer is:
40 Â°C

Solution:
KPA[100−θ]l=KQA[θ]l
∴KpKQ=23=θ100−θ or θ=40 °C

Topic: Physics, Thermal Properties of Matter, NEET
Two rods P and Q of the same length and same diameter having thermal conductivity in the ratio 2:3 are joined end to end. If the temperature at one end of P is 100 Â°C and at one end of Q is 0 Â°C, then the temperature of the interface is

Options:
A 40 °C
B 50 °C
C 60 °C
D 70 °C

The correct answer is:
40 °C

Solution:
KPA[100−θ]l=KQA[θ]l
∴KpKQ=23=θ100−θ or θ=40 °C

Topic: Physics, Thermal Properties of Matter, NEET
Two rotating bodies, A and B of masses, m and 2m with moments of inertia. IA and IB IB>IA have equal kinetic energy of rotation. If, LA and LB be their angular momenta, respectively, then,

Options:
A LA=LB2
B LA=2LB
C LB>LA
D LA>LB

The correct answer is:
LB>LA

Solution:
E=L22I.

⇒EA=EB

⇒LA22IA=LB22IB.

IB>IA⇒LB>LA.

Topic: Physics, Rotational Motion, NEET, NEET 2016 (Phase 2)
Two satellite A and B go round a planet P in circular orbits having radii 4R and R respectively. If the speed of the satellite A is 3v , the speed of the satellite B will be

Options:
A 12v
B 6v
C 4v3
D 3v2

The correct answer is:
6v

Solution:
The speed of satellite is called orbital speed
v0=GMeReâ vâ 1Râ vBvA=RARB
vB3v=4RRâ vB=6v

Topic: Physics, Gravitation, NEET
Two satellite are revolving around the earth with velocities v1 & v2 in radii r1 and r2r1> r2 respectively then -

Options:
A v1=v2
B v1>v2
C v1< v2
D v1r1=v2r2

The correct answer is:
v1< v2

Solution:
orbital velocity of satellite
v 0 = Gm r
â v01v02 = r02r01
if r 1 > r 1 â v 1 < v 2

Topic: Physics, Gravitation, NEET
Two satellites A and B go round a planet P in circular orbits having radii 4R and R respectively. If the speed of satellite A is 3V, the speed of the satellite B will be.

Options:
A 12 V

B 6 V

C 4 3 V
D 3 2 V

The correct answer is:
6 V

Solution:
v= GM R ⇒ v A v B = R B R A = R 4R = 1 2
∴ v A v B = 3V v B = 1 2 ∴ v B =6 V

Topic: Physics, Gravitation, NEET
Two satellites, A and B, have masses m and 2m respectively. A is in a circular orbit of radius R, and B is in a circular orbit of radius 2R around the earth. The ratio of their kinetic energies, KAKB is:

Options:
A 2
B 12
C 1
D 12

The correct answer is:
1

Solution:
K=12mV2=12mGMR2=GMm2R∝mR
KAKB=mAmB×RBRA=m2 m×2 RR=1

Topic: Physics, Gravitation, NEET
Two satellites A and B, have the ratio of their masses as mA:mB=3:1, are in the circular orbits of radii r and 4r respectively. The ratio of total mechanical energy of A to that of B is

Options:
A 13
B 3
C 34
D 12

The correct answer is:
12

Solution:
E=-GMm2r
or Eâmr
∴ E1E2=m1m2×r2r1=31×41=121=12

Topic: Physics, Gravitation, NEET
Two satellites are in the parking orbits around the earth. Mass of one is 10 times of the other. The ratio of their periods of revolution is

Options:
A 1
B 10
C 10
D 100

The correct answer is:
1

Solution:
Both the satellites will have the same time period as the time period is independent of the mass of the satellite

Topic: Physics, Gravitation, NEET
Two satellites of earth, $\mathrm{S}_1$ and $\mathrm{S}_2$ are moving in the same orbit. The mass of $S_1$ of four times the mass of $S_2$. Which one of the following statements is true?

Options:
A The potential energies of earth and satellite in the two cases are equal.
B $S_1$ and $S_2$ are moving with the same speed
C The kinetic energies of the two satellites are equal
D The time period of $S_1$ is four times that of $S_2$.

The correct answer is:
$S_1$ and $S_2$ are moving with the same speed

Solution:
$\mathrm{K} . \mathrm{E} .=\frac{G M m}{2 r}$
$\Rightarrow$ Kinetic energies are unequal
$\begin{aligned}
& \mathrm{T}=\frac{2 \pi r^{3 / 2}}{\sqrt{G M}} \\
& \Rightarrow \text { Time period are equal } \\
& \text {P.E. }=-\frac{G M m}{r} \\
& \Rightarrow \text { Potential energies are unequal } \\
& V=\sqrt{\frac{G M}{r}} \\
& \Rightarrow \text { orbital speeds are equal. }
\end{aligned}$

Topic: Physics, Gravitation, NEET, NEET 2007
Two satellites of masses M and 9M are orbiting a planet in a circular orbit of radius r. Their frequency of revolution will be in the ratio of

Options:
A 1:9
B 1:3
C 1:1
D 3:1

The correct answer is:
1:1

Solution:
Using Kepler's IIIrd law of planetary motion,
T=2Ïr3GM or T2âr3
Frequency, f=1T=12ÏGMr3
Frequency is independent of the mass of the satellite.
Hence, required ratio =1:1

Topic: Physics, Gravitation, NEET
Two separate bulbs contain ideal gases A and B at the same temperature. The molecular mass of A is half that of gas B and the density of gas A is twice that of gas B . The ratio of the pressure of A to that of gas B is

Options:
A 1/4
B 4
C 2
D 1/2

The correct answer is:
4

Solution:
da=2db ; 2Ma=Mb
PV=nRT=mMRT ; P=mV.RTM=dRTM
PaPb=dadbMbMa=2dbdb×2MaMa=4

Topic: Chemistry, States of Matter, NEET
Two SHM are represented by the equations x1=20sin⁡5πt+π4 and x2=10sin⁡5πt+3cos⁡5πt. The ratio of the amplitudes of the two motions is

Options:
A 0.5
B 1
C 0.25
D 32

The correct answer is:
1

Solution:
For SHM 1 ,
x1=20sin⁡5πt+π4
Its amplitude, A1=20
For SHM 2,
x2=10sin⁡5πt+3cos⁡5πt
=10×2sin⁡5πt2+32cos⁡5πt
20sin⁡5πt.cos⁡π3+sin⁡π3cos⁡5πt
=20sin⁡5πt+π3
It's amplitude, A2=20
∴A1A2=2020=1

Topic: Physics, Oscillations, NEET
Two short bar magnets with magnetic moments 400 ab-A cm2 and 800 ab-A cm2 are placed with their axis in the same straight line with similar poles facing each other and with their centres at 20 cm from each other. Then the force of repulsion is

Options:
A 12 dyne
B 6 dyne
C 800 dyne
D 150 dyne

The correct answer is:
12 dyne

Solution:
F=Î¼04Ï.6M1M2r4=6M1M2r4
In CGS system, Î¼04Ï=1
=6Ã800Ã40020Ã20Ã20Ã20= 12 dyne.

Topic: Physics, Magnetic Effects of Current, NEET
Two similar coils of radius $R$ are lying concentrically with their planes at right angles to each other. The currents flowing in them are $I$ and $2 I$, respectively. The resultant magnetic field induction at the centre will be

Options:
A $\frac{\sqrt{5} \mu_0 I}{2 R}$
B $\frac{3 \mu_0 I}{2 R}$
C $\frac{\mu_0 I}{2 R}$
D $\frac{\mu_0 I}{R}$

The correct answer is:
$\frac{\sqrt{5} \mu_0 I}{2 R}$

Solution:
The magnetic field $(B)$ at the centre of circular current carrying coil of radius $R$ and current $I, B=\frac{\mu_0 I}{2 R}$
Similarly, if current $=2 I$, then
$\text {Magnetic field }=\frac{\mu_0 2 I}{2 R}=2 B$
So, resultant magnetic field
$\begin{aligned}
& =\sqrt{B^2+(2 B)^2}=\sqrt{5 B^2}=\sqrt{5 B} \\
& =\frac{\mu_0 I \sqrt{5}}{2 R}
\end{aligned}$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2012 (Screening)
Two similar point charges q1 and q2 are placed at a distance r apart in the air. The force between them is F1. A dielectric slab of thickness t(< r) and dielectric constant K is placed between the charges. Then the force between the same charges is F2. The ratio is

Options:
A 1
B K
C r–t+tKr2
D rr–t+tK2

The correct answer is:
r–t+tKr2

Solution:
F1= kq1q2r2=q1q24πε0r2 , F2 = q1q24πε0(r–t+tK)2

Topic: Physics, Electrostatics, NEET
Two similar springs P and Q have spring constants KP and KQ, such that KP>KQ. They are stretched, first by the same amount (case a), then by the same force (case b). the work done by the springs WP and WQ are related as respectively, in case

Options:
A WP=WQ;WP>WQ
B WP=WQ;WP=WQ
C WP>WQ;WQ>WP
D WP<WQ;WQ<WP

The correct answer is:
WP>WQ;WQ>WP

Solution:
KP>KQ
Case (a): Stretched by same amount
WP=12KPx2
WQ=12KQx2 ∵as xP=xQ=x
∴WP>WQ
Case (b): Stretched by same force,
F=KPxP=KQxQ
xP=FKP;xQ=FKQ
WP=12KPxP2=12KPF2KP2=12F2KP
WQ=12KQxQ2=12KQF2KQ2=12F2KQ
∴WP<WQ

Topic: Physics, Work Power Energy, NEET, NEET 2015 (Phase 1)
Two similar thin equi-convex lenses of focal length f each are kept coaxially in contact with each other, such that the focal length of the combination is F1. When the space between the two lenses is filled with glycerin (which has the same refractive index (μ=1.5) as that of glass), then the equivalent focal length is F2. The ratio F1:F2 will be

Options:
A 2:1
B 1:2
C 2:3
D 3:4

The correct answer is:
1:2

Solution:
Case I:
1Feq+1F+1F
F1=Feq=F2
Case II:
F2=Feq=F
F1:F2=1:2

Topic: Physics, Ray Optics, NEET, NEET 2019 (All India)
Two simple harmonic motions are represented by the equations y1=0.1sin100πt+π3 and y2=0.1cos100πt. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is

Options:
A -Ï3
B π6
C -π6
D π3

The correct answer is:
-π6

Solution:
ν1=ddty1=0.1×100πcos100πt+π3
ν2=ddty2=-0.1×πsinπt
=0.1×πcosπt+π2
∴ Δϕ=π3-π2=-π6

Topic: Physics, Oscillations, NEET
Two simple pendulums A and B are made to oscillate simultaneously and it is found that A completes 10 oscillations in 20 s and B completed 8 oscillations in 10 s. The ratio of the length of A and B is

Options:
A 2564
B 6425
C 85
D 54

The correct answer is:
6425

Solution:
According to questions,
For pendulum A,
TA=2πlAg
2010=2πlAg ...(i)
For pendulum B,
TB=2πlBg
108=2πlBg ...(ii)
Dividing Equation. (i) by Equation. (ii), we get
lAlB=1601002
lAlB=6425

Topic: Physics, Oscillations, NEET
Two slits in Young's double slit experiment are 1.5 mm apart and the screen is placed at a distance of 1 m from the slits. If the wavelength of light used is $600 \times 10^{-9} \mathrm{~m}$ then the fringe separation is

Options:
A $4 \times 10^{-5} \mathrm{~m}$
B $9 \times 10^{-8} \mathrm{~m}$
C $4 \times 10^{-7} \mathrm{~m}$
D $4 \times 10^{-4} \mathrm{~m}$

The correct answer is:
$4 \times 10^{-4} \mathrm{~m}$

Solution:
Fringe width $=$ Fringe separation $(\beta)=\frac{\lambda D}{d}$
$\Rightarrow \quad \beta=\frac{600 \times 10^{-9} \times 1}{1.5 \times 10^{-3}}=\frac{6 \times 10^{-7}}{1.5 \times 10^{-3}}=4 \times 10^{-4} \mathrm{~m}$

Topic: Physics, Wave Optics, NEET, NEET 2024 (Re-NEET)
Two slits in Young's experiment have widths in the ratio 1 : 25. The ratio of intensity at the maxima and minima in the interference pattern, ImaxImin is:

Options:
A 12149
B 49121
C 49
D 94

The correct answer is:
94

Solution:
I1I2=251⇒A1A2=51
AmaxAmin=5+15-1=64=32
ImaxImin=322=94

Topic: Physics, Wave Optics, NEET, NEET 2015 (Phase 2)
Two slits, S1 and S2, illuminated by a white light source, give a white central maximum. A transparent sheet, of refractive index 1.25 and thickness t1, is placed in front of S1. Another transparent sheet, of refractive index 1.50 and thickness t2, is placed in front of S2. If the central maxima are not affected, then the ratio t1:t2 will be

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

The correct answer is:
2 : 1

Solution:
Since there is no shift in central maxima, the path difference introduced by the two sheets are equal.
i.e. μ1-1t1=μ2-1t2
Where μ1 and μ2 are refraction index
i.e. t1t2=(μ2-1)(μ1-1)=(1.5-1)(1.25-1)=0.50.25=2

Topic: Physics, Wave Optics, NEET
Two small identical spheres having charges + 10μC and - 90μC attract each other with a force of F newton. If they are kept in contact and then separated by the same distance, the new force between them is -

Options:
A F6
B 16F
C 16F9
D 9F

The correct answer is:
16F9

Solution:
F=k1090r2 ……..i
charge on each when contact is made
q1′=q2′=10+–902=-40 μC
∴ New force F′=k4040r2
∴   F  ′F=40×4010×90=169
∴ F′=169F

Topic: Physics, Electrostatics, NEET
Two small satellites are moving in circular orbits around the earth at a distance R and R+△R from the centre of the earth. If their time period of rotation are T and T+△T respectively, then (ΔR<K)

Options:
A ΔT=TΔRR
B ΔT=3TΔRR
C ΔT=32TΔRR
D ΔT=23TΔRR

The correct answer is:
ΔT=32TΔRR

Solution:
T∝R32
(T+ΔT)T=R+ΔRR3/2
1+ΔTT=1+ΔRR3/2
∴ΔR<<R
1+ΔTT=1+32ΔRR
ΔTT=32ΔRR
ΔT=32T·ΔRR

Topic: Physics, Gravitation, NEET
Two small spherical metal balls, having equal masses, are made from materials of densities $\rho_1$ and $\rho_2\left(\rho_1=8 \rho_2\right)$ and have radii of $1 \mathrm{~mm}$ and $2 \mathrm{~mm}$, respectively. They are made to fall vertically (from rest) in viscous medium whose coefficient of viscosity equals $\eta$ and whose density is $0.1 \rho_2$. The ratio of their terminal velocities would be

Options:
A $\frac{79}{72}$
B $\frac{19}{36}$
C $\frac{39}{72}$
D $\frac{79}{36}$

The correct answer is:
$\frac{79}{36}$

Solution:
The terminal velocity achieved by ball in a viscous fluid is
$$
v_{\mathrm{t}}=\frac{2(\rho-\sigma) \mathrm{r}^2 \mathrm{~g}}{9 \eta}
$$
where, $\rho=$ density of metal of ball,
$\sigma=$ density of viscous medium,
$\mathrm{r}=$ radius of ball and
$\eta=$ coefficient of viscosity of medium
Terminal velocity of first ball,
$$
\begin{aligned}
\mathrm{v}_{\mathrm{t}_1} & =\frac{2\left(\rho_1-\sigma\right) \mathrm{r}_1^2 \mathrm{~g}}{9 \eta} \\
& =\frac{2\left(8 \rho_2-\sigma\right) \mathrm{r}_1^2 \mathrm{~g}}{9 \eta}
\end{aligned}
$$
...(i) $\left[\because \rho_1=8 \rho_2\right]$
Similarly, for second ball
$$
\mathrm{v}_{\mathrm{t}_2}=\frac{2\left(\rho_2-\sigma\right) \mathrm{r}_2^2 \mathrm{~g}}{9 \eta}
$$
From Eq. (i) and (ii), we get
$$
\begin{aligned}
\frac{\mathrm{v}_{\mathrm{t}_1}}{\mathrm{v}_{\mathrm{t}_2}} & =\frac{2\left(8 \rho_2-\sigma\right) \mathrm{r}_1^2 \mathrm{~g}}{2\left(\rho_2-\sigma\right) \mathrm{r}_2^2 \mathrm{~g}} \times \frac{9 \eta}{9 \eta} \\
& =\left(\frac{8 \rho_2-0.1 \rho_2}{\rho_2-0.1 \rho_2}\right)\left(\frac{r_1}{r_2}\right)^2
\end{aligned}
$$
..(iii) $\left[\because \sigma=0.1 \rho_2\right]$
Here, $\mathrm{r}_1=1 \mathrm{~mm}$ and $\mathrm{r}_2=2 \mathrm{~mm}$
Substituting these values in Eq. (iii), we get
$$
\Rightarrow \quad \frac{\mathrm{v}_{\mathrm{t}_1}}{\mathrm{v}_{\mathrm{t}_2}}=\left(\frac{7.9 \rho_2}{0.9 \rho_2}\right)\left(\frac{1}{2}\right)^2=\frac{79}{36}
$$

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2019 (Odisha)
Two smooth and similar right-angled prisms are arranged on a smooth horizontal plane as shown in the figure. The lower prism has a mass 3 times that of the upper prism. The prisms are held in an initial position as shown and are released. As the upper prism touches the horizontal plane, the distance moved by the lower prism is

Options:
A a-b
B a-b3
C b-a2
D a-b4

The correct answer is:
a-b4

Solution:
The displacement of the centre of mass of the system along the horizontal direction is zero. This means

xcm=m1x1+m2x2m1+m2=0

Let x be the displacement of the lower prism, then
xcm=ma-b-x+3m-xm+3m=0

⇒a-b-x-3x=0⇒x=a-b4

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two soap bubbles of radii 3 mm and 4 mm confined in vacuum coalesce isothermally to form a new bubble. The radius of the bubble formed in mm is

Options:
A 3
B 4
C 5
D 7

The correct answer is:
5

Solution:
P1V1+P2V2=PV
P1-0=4SR1⇒P1R1=4S
Similarly P2R2=4S=PR
P143πR1 3+P243πR2 3=P43πR3
R12+R22=R2
R=5 mm

Topic: Physics, Mechanical Properties of Fluids, NEET
Two solid balls A and B whose masses are 200 g and 400 g respectively, are moving in mutually opposite directions. If the velocity A is 0.3 m s-1 and both the balls come to rest after collision, then the velocity of the ball B is

Options:
A 0.15 m s-1
B -0.15 m s-1
C 1.5 m s-1
D none of these

The correct answer is:
-0.15 m s-1

Solution:
From the conservation of linear momentum
Initial momentum = Final momentum
2000.3+400vB=0
vB=-0.15 m s-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two solid conductors are made up of same material, have same length and same resistance. One of them has a circular cross section of area A1 and the other one has a square cross section of area A2. The ratio A1A2 is

Options:
A 1.5
B 1
C 0.8
D 2

The correct answer is:
1

Solution:
First, remember the relation between resistance and specific resistivity in terms of length and cross-section area, R=ρlA, given, the resistance of the conductors is the same, R1=R2 and they are made from the same material, so ρ1=ρ2

⇒1A1=1A2⇒A1A2=1.

Topic: Physics, Current Electricity, NEET, NEET 2020 (Phase 2)
Two solid cylinders A and B of the same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. The cylinder A has most of its mass concentrated near the surface, while B has most of its mass concentrated near the axis. Which statement is correct?

Options:
A Both cylinder A and B reach the ground at the same time
B Cylinder A has larger linear acceleration than cylinder B
C Cylinder B reaches the ground with larger angular speed
D Both cylinder A and B reach the ground with the same translational kinetic energy.

The correct answer is:
Cylinder B reaches the ground with larger angular speed

Solution:
We can treat cylinders A as hollow and B as solid. For the hollow cylinder A,
K2R2=1 and
Solid cylinder B, K2R2=12
∴ VA=2gh1+1=2gh
VB=2gh1+12=4gh3
VB>VA
∴ ωB>ωA

Topic: Physics, Rotational Motion, NEET
Two solid spheres of radii R1 and R2 are made of the same material and have similar surfaces. These are raised to the same temperature and then allowed to cool under identical conditions. The ratio of their initial rates of loss of heat is

Options:
A R1R2
B R2R1
C R12R22
D R22R12

The correct answer is:
R12R22

Solution:
Rate of loss of heat per second,
dQdt=σ AT4-T04=σ4πR2T4-T04
∴dQdt1=σ4πR12 T4-T04 and
dQdt2=σ4πR22(T4-T04)
∴dQdt1dQdt2=R12R22

Topic: Physics, Thermal Properties of Matter, NEET
Two solutions, A and B, each of 100 L was made by dissolving 4 g of NaOH and 9.8 g of H2SO4 in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is

Options:
A 5.3
B 10.6
C 9.4
D 7.5

The correct answer is:
10.6

Solution:
M(H2SO4)=9.898Ã100=10-3m
MNaOH=440Ã100=10-3m
Equivalents of resulting solution [OHâ]=40Ã10-3-10Ã10-3Ã250=25Ã10-3
pOH=3.39, pH=10.6

Topic: Chemistry, Ionic Equilibrium, NEET
Two solutions of a substance (non-electrolyte) are mixed in the following manner.
480 ml of 1.5 M (first solution) +520 mL of 1.2 M (second solution). What is the molarity of the final mixture?

Options:
A 1.50 M
B 1.20 M
C 2.70 M
D 1.344 M

The correct answer is:
1.344 M

Solution:
Mmix=M1V1+M2V2V1+V2
Mmix=1.5Ã480+1.2Ã5201000=1.344 M

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Two sound sources are placed along the diameter of a circle of radius R Râ«4Î». How many minima will be heard as one moves along the perimeter of the circle?

Options:
A 16
B 12
C 4
D 8

The correct answer is:
16

Solution:
In this figure S1 and S2 are sound sources.
S1S2=4Î»
â´Îx=S2P-S1P=4Î»sinÎ¸
For minima Îx=(2n+1)Î»2
â´4Î»sinÎ¸=(2n+1)Î»2
â sinÎ¸=(2n+1)8
n=0; sinÎ¸=18âÎ¸=sin-118
n=1;ââsinÎ¸=38âÎ¸=sin-138
n=2;ââsinÎ¸=58âÎ¸=sin-158
n=3;ââsinÎ¸=78âÎ¸=sin-178
For n=4;sinÎ¸>1 it is impossible
In the quarter circle, 4 minima are heard, in the entire circle total number of minima will be 16.

Topic: Physics, Waves and Sound, NEET
Two sound waves with wavelengths 5.0 $\mathrm{m}$ and $5.5 \mathrm{~m}$ respectively, each propagate in a gas with velocity $330 \mathrm{~m} / \mathrm{s}$. We expect the following number of beats per second:

Options:
A 6
B 12
C 0
D 1

The correct answer is:
6

Solution:
$$
\begin{gathered}
\text { Frequency }=\frac{\text { Velocity }}{\text { Wave length }} \\
V_1=\frac{V}{\lambda_1}=\frac{330}{5}=66 \mathrm{H}_2
\end{gathered}
$$
and $V_2=\frac{V}{\lambda_2}=\frac{330}{5 \cdot 5}=60 \mathrm{H}_2$
$\therefore$ Number of beats per second $=V_1-V_2=60-60=6$.

Topic: Physics, Waves and Sound, NEET, NEET 2006
Two source $\mathrm{P}$ and $\mathrm{Q}$ produce notes of frequency $660 \mathrm{~Hz}$ each. A listener moves from $\mathrm{P}$ to $\mathrm{Q}$ with a speed of $1 \mathrm{~ms}^{-1}$. If the speed of sound is $330 \mathrm{~m} / \mathrm{s}$, then the number of beats heard by the listener per second will be:

Options:
A 4
B 8
C 2
D zero

The correct answer is:
4

Solution:
$$
\text { According to the question }
$$
Speed of listener $V_L=1 \mathrm{~ms}^{-1}$
Speed of sound $V=330 \mathrm{~ms}^{-1}$
Frequency of each source $v=600 \mathrm{~Hz}$
$\therefore$ Apparent frequency due to $\mathrm{P}=\mathrm{V}^{\prime}$
$=\frac{v\left(\mathrm{~V}-\mathrm{V}_{\mathrm{L}}\right)}{\mathrm{V}}$
Apparent frequency due to $\mathrm{Q}=\mathrm{V}^{\prime \prime}=$ $\frac{v\left(\mathrm{~V}-\mathrm{V}_{\mathrm{L}}\right)}{\mathrm{V}}$
$\therefore$ No. of heat heard by the listener per second.
$$
\begin{aligned}
& \mathrm{V}^{\prime \prime}-\mathrm{V}^{\prime}=\frac{v\left(\mathrm{~V}-\mathrm{V}_{\mathrm{L}}\right)}{\mathrm{V}}-\frac{v\left(\mathrm{~V}_0-\mathrm{V}_{\mathrm{L}}\right)}{\mathrm{V}} \\
& =\frac{2 v_{\mathrm{L}}}{v}=\frac{2 \times 660 \times 1}{330}=4
\end{aligned}
$$

Topic: Physics, Waves and Sound, NEET, NEET 2013 (Karnataka)
Two sources of equal E.M.F. are connected to an external resistance R. The internal resistances of the two sources are R1 and R2 R2>R1. If the potential difference across the source having internal resistance R2 is zero, then

Options:
A R = R 1 R 2 R 1 + R 2
B R = R 1 R 2 R 2 - R 1
C R = R 2 R 1 + R 2 R 2 - R 1
D R=R2-R1

The correct answer is:
R=R2-R1

Solution:
I = 2 E R 1 + R 2 + R
âµ E - IR 2 = 0 (Given)

â´ E=IR2
or E = 2 ER 2 R 1 + R 2 + R
or R1+R2+R=2R2
or R=R2-R1

Topic: Physics, Current Electricity, NEET