Grass-G/jee-main-questions huggingface dataset:
Subject: Physics — Mechanical property of solids — Young's modulus
Difficulty: Moderate

Question: Two wires of the same length and same area of cross-section are
suspended as shown. Their Young's moduli are in the ratio 2: 3. Blocks
are in equilibrium in each arrangement.

1) At equilibrium strain will be in ratio 3:4
2) At equilibrium stress will be in ratio 2: 3
3) When stretched and released, they will oscillate with time periods
in the ratio 3: 2
4) When stretched and released, they will oscillate with time periods
in the ratio [IMAGE] leku yEckbZ,oa leku vuqizLFk dkV {kS=kQy ds nks rkj fp=k vuqlkj vkyfEcr gSA mu xq.kkadksa

Correct option: 4

Solution:
[IMAGE] Ratio of time period vkorZdky
Subject: Physics — Kinetic theory of gases — Root mean squared speed
Difficulty: Moderate

Question: Two moles of hydrogen are mixed with n moles of helium. The root
mean square speed of gas molecules in the mixture is $\sqrt{2}$ times
the speed of sound in the mixture. Then n is.
gkbMªkstu ds nks eksyksa n eksyksa ds lkFk


Answer: 2

Solution:
$V_{rms} = \sqrt{\frac{3RT}{M_{mix}}}$
$V_{sound\ } = \sqrt{\frac{\gamma RT}{M_{mix}}}$
$V_{rms} = \sqrt{2}V_{sound\ }$
$\sqrt{\frac{3RT}{M_{mix}}} = \sqrt{2}\sqrt{\frac{\gamma RT}{M_{mix}}}$
$\gamma_{mix} = \frac{3}{2}$
$\gamma_{mix} = \frac{n_{1}C_{P_{1}} + n_{2}C_{P_{2}}}{n_{1}C_{v_{1}} + n_{2}C_{v_{2}}}$
$\frac{3}{2} = \frac{2 \times \frac{7R}{2} + n \times \frac{5R}{2}}{2 \times \frac{5R}{2} + n \times \frac{3R}{2}} \Rightarrow \frac{3}{2} = \frac{14 + 5n}{10 + 3n} \Rightarrow 30 + 9n = 28 + 10n \Rightarrow n = 2$
Subject: Physics — Thermodynamics — Expansion of gas
Difficulty: Moderate

Question: An ideal gas may expands from V_0 to 2V_0 according to
following three processes. Molar specific heat for processes b will be
izlkfjr gks ldrh

1) positive if a is isothermal and c is adiabatic processes
2) negative if a is isothermal and c is adiabatic processes
3) positive if c is isothermal and a is adiabatic processes
4) negative if c is isothermal and a is adiabatic processes

Correct option: 2

Solution:
If a is isothermal and c is adiabatic processes then for b
$\Delta U$ will be negative and Q will be positive so specific heat will
be negative.
 rFkk c:)ks"e izØe gks rks bds fy, $\Delta U$_.kkRed
gksxk rFkk _.kkRed gksxhA
Subject: Physics — Thermal properties of matter — Thermal expansion
Difficulty: Moderate

Question: A composite bar of length
[IMAGE] is made up from a rod of material 1 and
of length [IMAGE] attached to a rod of material 2 and of
length [IMAGE] as shown. If
[IMAGE] and [IMAGE] ar their respective coefficients of linear
expansion, then equivalent coefficient of linear expansion for the
composite rod is:,d la;qDr NM+ dh yEckbZ L = L_1+ L_2gS tks inkFkZ 1 dh,d NM+ dh
yEckbZ L_1gS,oa inkFkZ 2 dh,d NM+ dh yEckbZ L_2gS tks fp=kkuqlkj
la;qDr NM+


Correct option: 3

Solution:
When it has been given a temperature change ∆T, the new length
of the composite bar is the sum of the individual final lengths (Let
 be required coefficient of linear expansion)
tc b kUrj ifjorZuΔT, la;qDr NM+ dh ubZ yEckbZ vfUre yEckbZ gksxk (ekuk 'α izlkj
xq.kkad gS)
Subject: Physics —  — 
Difficulty: Moderate

Question: A ball of mass 1 kg is at rest in position P by means of two light
strings OP and RP as shown in the figure. The string RP is now cut and
the ball swings upto a position Q. If θ = 60° and tension in the string
in position OP and OQ are T1 (Just before cutting) and T2 respectively,
then [IMAGE] is equal to

1) 1
2) 2
3) 3
4) 4
1 kg OP RP 
 P RP Q 
 
θ = 60° OP OQ T_1 ( )
 T_2, [IMAGE]

Correct option: 4

Solution:
T1 cos θ = mg........(i)
T2 = mg cos θ........(ii)
[IMAGE] = cos2 θ = 4
TOPIC: CIRCULAR MOTION
SUB TOPIC: DYNAMICS
Subject: Physics — Electrostatics — Electric field due to infinite sheet of charge
Difficulty: Moderate

Question: Two infinite planes each with uniform surface charge density
$+ \sigma$ are kept in such a way that the anglebetween them is 30°. The
electric field in the region shown between them is given by

1) $\frac{\sigma}{2\epsilon_{0}}\left\lbrack (1 + \sqrt{3})\overset{\hat{}}{y} - \frac{\overset{\hat{}}{x}}{2} \right\rbrack$
2) $\frac{\sigma}{2\epsilon_{0}}\left\lbrack \left( 1 - \frac{\sqrt{3}}{2} \right)\overset{\hat{}}{y} - \frac{\overset{\hat{}}{x}}{2} \right\rbrack$
3) $\frac{\sigma}{2\epsilon_{0}}\left\lbrack (1 + \sqrt{3})\overset{\hat{}}{y} + \frac{\overset{\hat{}}{x}}{2} \right\rbrack$
4) $\frac{\sigma}{\epsilon_{0}}\left\lbrack \left( 1 + \frac{\sqrt{3}}{2} \right)\overset{\hat{}}{y} + \frac{\overset{\hat{}}{x}}{2} \right\rbrack$

Correct option: 2

Solution:
$\overset{\rightarrow}{E} = \frac{\sigma}{2\varepsilon_{0}}\left\lbrack \left( 1 - \frac{\sqrt{3}}{2} \right)\overset{\hat{}}{y} - \frac{1}{2}\overset{\hat{}}{x} \right\rbrack$.
Subject: Physics — Mechanical Properties of Fluids — Hydrostatic pressure
Difficulty: Moderate

Question: Two immiscible liquids are poured in a U-tube having densities
$\rho_{1} = 1.0 \times 10^{3}\ kg/m^{3}$ and
$\rho_{3} = 3.0 \times 10^{3}\ kg/m^{3}$. If the ratio of heights (of
the liquids above their interface) $\frac{h_{1}}{h_{2}}$ is x, then find
the value of x.


Answer: 3

Solution:
$p_{0} + \rho_{1}{gh}_{1} = p_{0} + \rho_{2}{gh}_{2}$
$\Rightarrow \frac{h_{1}}{h_{2}} = \frac{\rho_{2}}{\rho_{1}} = 3$
Subject: Physics —  — 

Question: A small 1 kg mass is launched from the top of a cliff with
speed V at an angle of 30° above the horizontal. When the mass reaches
the ground, its velocity is directed at 45° below the horizontal. Which
one of the following choices is the magnitude of the total impulse that
was imparted to the mass during its flight ? Ignore air resistance.


Correct option: 1

Solution:
V cos 30° = V1 cos 45°
V [IMAGE] = [IMAGE] V1 = [IMAGE] J = pf - pi
=
Subject: Physics —  — 

Question: A conducting loop of area 10 cm^2 and resistance 1 W is kept
in a time varying magnetic field, perpendicular to its plane, which is
zero at t = 0. The current I induced in the loop as a function of time t
is given as I = (8t + 1) mA. The magnetic field B as a function of time
t is

1) (4t^2 + t + 1) T
2) (4t^2 + t + 2) T
3) (4t^2) T
4) (4t^2 +
t) T

Correct option: 4

Solution:
[IMAGE] = (8t + 1) ´ 10^-3 A
⇒ ε = R(8t + 1) × 10^-3
[IMAGE] = R (8t + 1) × 10^-3 and [IMAGE] ∴ [IMAGE] (8t + 1) ´ 10^-3 ⇒ [IMAGE] ⇒ [IMAGE] ⇒ [IMAGE] ⇒ [IMAGE] ⇒ [IMAGE]
Subject: Physics — Nuclear Physics — Radioactivity
Difficulty: Moderate

Question: The activity of a radioactive sample falls from
[IMAGE] in 30 minutes. Its half life is close to

1) 62 min
2) 66 min
3) 72 min
4) 52 min

Correct option: 1

Solution:
[IMAGE] (i) (ii)
$\Rightarrow \text{\quad}\frac{ln2}{ln(7/5)} = \frac{t_{\frac{1}{2}}}{(30\min)} \Rightarrow \text{\quad}(2.06004)30 = t_{\frac{1}{2}} = 61.8\text{min}$
Subject: Physics —  — 

Question: The mass of an object is 0.074 kg and its volume is 290 cm^3.
Its density expressed to the correct significant figures (with rounding
off) will be



Solution:
B
= 0.2551724 (up to two significant figures)
Subject: Physics — Rigid Body Dynamics — Moment of inertia
Difficulty: Moderate

Question: Mass per unit area of a circular disc of radius a depends on the
distance r from its centre as [IMAGE] The moment of
inertia of the disc about the axis, perpendicular to the plane and
passing through its centre is


Correct option: 2

Solution:
Subject: Physics — Kinetic Theory of Gases — Ideal Gas
Difficulty: Moderate

Question: 70 calories of heat is required to raise the temperature of 2
moles of an ideal gas at constant pressure from 30°C to 35°C (R= 2 cal
/mol-°C) The gas may be:
ls] bl gks tkrk
gs rks xSl gks ldrh

2) He

Correct option: 1

Solution:
[IMAGE] So gas is
Subject: Physics — Ray Optics — Lens maker's formula
Difficulty: Moderate

Question: A point object in air is in front of the curved surface of a
plano-convex lens. The radius of curvature of thecurved surface is 30 cm
and the refractive index of the lens material is 1.5, then the focal
length of thelens (in cm) is_______.

1) 50
2) 40
3) 60
4) 30

Correct option: 3

Solution:
$\frac{1}{f} = (\mu - 1)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$
$R_{1} = \infty$
$R_{2} = - 30cm$
$\frac{1}{f} = (1.5 - 1)\left( \frac{1}{\infty} - \frac{1}{- 30} \right)$
$\begin{aligned}
 & \frac{1}{f} = \frac{0.5}{30} \\
 & f = 60cm
\end{aligned}$
Subject: Physics —  — 

Question: A wire of circular cross section has inner portion of radius R
made of material of resisitivity ρ and is surrounded by an outer portion
of thickness R made of a material of double resisitivity. Find the
resistance of length L of such a wire.


Correct option: 4

Solution:
R = [IMAGE] = [IMAGE] =
Subject: Physics — Mechanical Properties of Fluids — Archimedes principle
Difficulty: Moderate

Question: The densities of wood and benzene at 0°C
are$\ 880{kgm}^{- 3}$and $\ 900\ {kgm}^{- 3}$, respectively. The
coefficients of volume expansion of wood is
$1.2 \times 10^{- 3^{{^\circ}}}C^{- 1}$ and that of benzene is
$1.5 \times 10^{- 3^{{^\circ}}}C^{- 1}$. Value of temperature (in °C) at
which a piece of this wood would just sink in benzene at the same
temperature is T. Then find value of [IMAGE] (where T is
temperature in °C).


Answer: 2

Solution:
For just sink condition
$\rho_{ow\ }\left( 1 - \gamma_{w\ }\Delta T \right) = \rho_{ob\ }\left( 1 - \gamma_{b}\Delta T \right)$
$880\left( 1 - 1.2 \times 10^{3}(T - 0) \right) = 900\left( 1 - 1.5 \times 10^{- 3}(T - 0) \right)$
$T10^{- 3}(1.5 - 1.2) = 900 - 880$
$T \times 10^{- 3} \times 0.3 = 20$
$T = \frac{20 \times 10^{4}}{3}{^\circ}C$
Subject: Physics — Electronic device — Gates
Difficulty: Moderate

Question: In the given circuit, value of Y is

1) toggles between 0 and 1
2) 0
3) will not execute
4) 1

Correct option: 2

Solution:
Subject: Physics — Thermodynamic — Adiabatic process
Difficulty: Hard

Question: A litre of dry air at STP expands adiabatically to a volume of 3
litres. If [IMAGE] the work done by air is:
 air to be an ideal gas]
1 yhVj gok tks fd ekud rki o nkcSTP ij gS]:)ks"e
çfØ çlkfjr gksdj 3 yhVjvk;ru

1) 88.75 J
2) 48 J
3) 60.7 J
4) 100.8 J

Correct option: 1

Solution:
Subject: Physics — Heat — Specific heat
Difficulty: Tough

Question: Temperature of 100 g of water in a thermoflaskremains fixed for
a pretty long time at 50ºC. Anequal mass of sand at 20ºC is poured in
the flaskand shaken for some time so that the temperatureof the mixture
is 40ºC. Now the experiment is repeated with 100 g of a liquid at 50ºC
and an equalamount of sand at 20ºC when the temperature ofthe mixture is
found to be 30ºC. The specific heat of the liquid (in
$kJ{kg}^{- 1}K^{- 1}$ ) is

1) 1.05
2) 2.01
3) 1.55
4) 1.95

Correct option: 1

Solution:
$m_{\omega}\ S_{\omega}\ \Delta T_{\omega} = m_{S}\ S_{S}\ \Delta T_{S}$
$100 \times 1 \times 10 = 100 \times S_{S} \times 20$
$\Rightarrow S_{S} = \frac{1}{2}cal/{gm}^{\circ}C$
$m_{S}\ S_{S}\ \Delta T_{S} = m_{\mathcal{l}}{\ S}_{\mathcal{l}}\ \Delta T_{\mathcal{l}}$
$100 \times \frac{1}{2} \times 10 = 100 \times {\ S}_{\mathcal{l}} \times 20$
$\Rightarrow S_{\mathcal{l}} = \frac{1}{4}cal/ºC\ gm$
$S_{\mathcal{l}} = \frac{4.2kJ}{4}/^{\circ}cgm = 1.05kJ/ºC\ gm$
Subject: Physics — Gravitation — Binding energy
Difficulty: Easy

Question: A Comet is observed as it passes our solar system. Its closest
approach to the sun is r = 6.67 × 10^10 m,at that time it has velocity
v = 2 × 10^6 m/s. If G = 6.67 × 10^-11 N m^-2 kg^-2 and mass of
the sun is 2 × 10^30kg. Then which of the following is true in respect
with this comet?

1) The comet will return after 20 years
2) The comet will never return to the solar system
3) The comets path is circular
4) The comets total energy is negative

Correct option: 2

Solution:
Total energy of the comet = U+K
$= \frac{- GM_{s}m}{r} + \frac{1}{2}m\left( 2 \times 10^{6} \right)^{2}$ where m mass
of the Comet
$= m\left\lbrack \left( - 2 \times 10^{9} \right) + \left( 2 \times 10^{12} \right) \right\rbrack$>0
$\therefore$Comets will never returns to the solar system because total
energy is +ve.
Subject: Physics — Mechanical properties of liquid — Buoyancy
Difficulty: Moderate

Question: For a fluid which is flowing steadily, the level in the vertical
tubes is best represented by
LFkk;h:i ls izokfgr nzo


Correct option: 1

Solution:
From continuity equation, velocity at cross-section (1) is more
than that at cross-section (2).
lkarR; lehdj.k ls vuqizLFk dkV(1) ij osx ] vuqizLFk dkV(2) ij osx
Subject: Physics — Collision — elastic collision
Difficulty: Moderate

Question: A particle of mass m is dropped from a height h above the ground.
At the same time another particle of the same mass is thrown vertically
upwards from the ground with a speed of[IMAGE]. If they
collide head-on completely in elastically, the time taken for the
combined mass to reach the ground, in units of [IMAGE] is


Correct option: 2

Solution:
[IMAGE] time for collision [IMAGE] After [IMAGE] And [IMAGE] at the time of collision
Subject: Physics — Electronic devices — Zener diode
Difficulty: Moderate

Question: The circuit shown below is working as a 8V dc regulated voltage
source. When 12V is used as input,the power dissipated (in mW) in each
diode is 8n. Find n. (considering both zener diodes are identical)
__________.
 A tc blesa 12V oksYVrk yxk;h tkrh gS rks + esa gksus okyh mW esa 8n gksxh (nksuks T+khuj
Mk;ksM,d leku gSa) __________.


Answer: 5

Solution:
$i = \frac{(12 - 8)}{(200 + 200)}A = \frac{4}{400} = 10^{- 2}A$
Power loss in each diode gkfu = (4)(10^-2 ) W
= 40 mW
Subject: Physics — Current Electricity — KVL
Difficulty: Moderate

Question: The current i in the network is

1) 0.2A
2) 0.3A
3) 0A
4) 0.6A

Correct option: 2

Solution:
Both diodes are in reverse biased
Subject: Physics — Digital electronics — Lopic pates
Difficulty: Moderate

Question: In the given circuit, value of Y is

1) toggles between 0 and 1
2) 0
3) will not execute
4) 1

Correct option: 2

Solution:
$$
[IMAGE] = 0
Subject: Physics — Modern physics — Photoelectric emission
Difficulty: Moderate

Question: A beam of electromagnetic radiation of intensity
[IMAGE] is comprised of wavelength,
[IMAGE] It falls normally on a metal (work function
[IMAGE] ) of surface area of [IMAGE]. If
one in [IMAGE] photons ejects an electron, total number
of electrons ejected in[IMAGE] is
[IMAGE] then x is....


Answer: 6

Solution:
Energy of photon. [IMAGE] (so photoelectric
effect will take place)
[IMAGE] No. of photons falling per second
[IMAGE] No. of photoelectron emitted per second
Subject: Physics —  — 

Question: An electron starting from rest and accelerated through V =
20,000 V in an x-ray tube loses half the energy it acquired as it
approaches an atom of the target material. The remaining KE is lost in
subsequent interactions with other target atoms. Find the wavelength of
the radiation emitted in the first interaction.

1) 6.63
2) 4.14
3) 1.24
4) 12.4

Correct option: 3

Solution:
∆E = 1 × 104 eV = [IMAGE] λ = [IMAGE] nm = 0.124 nm
Subject: Physics — Rigid Body Dynamics — Conservation of angular momentum
Difficulty: Tough

Question: A circular platform is free to rotate in a horizontal plane about
a vertical axis passing through its centre. A tortoise is sitting at the
edge of the platform. Now the platform is given an angular velocity
$\omega_{0}$. When the tortoise moves along a chord of the platform with
a constant velocity (with respect to the platform) the angular velocity
of the platform $\omega(t)$ will vary with time t as


Correct option: 2

Solution:
since, there is no external torque, angular momentum will
remain conserved. The moment of inertia will first decrease till the
tortoise moves from A to C and then increase as it moves from C and D.
The re fore $\omega$ will initially increase and then decrease. Let R be
the radius of platform m the mass of disc and M is the mass of platform.
Moment of inertia when the tortoise is at A
and moment of inertia when the tortoise is at B
$I_{2} = {mr}^{2} + \frac{{MR}^{2}}{2}$
here $r^{2} = a^{2} + {\lbrack\sqrt{R^{2} - a^{2}} - vt\rbrack}^{2}$
From conservation of angular momentum
$\omega_{0}I_{1} = \omega(t)I_{2}$
substituting the values we can see that variation of $\omega(t)$ is
nonlinear.
Subject: Physics — Mechanical Properties of Fluids — Archimedes principle
Difficulty: Moderate

Question: A vessel contains two immiscible liquids of densities
[IMAGE]. A solid block of volume
[IMAGE] and density [IMAGE] is tied to one end of a string and other
is tied to the bottom of the vessel as shown in figure. The block is
immersed with [IMAGE] of its volume in the liquid of higher density and
[IMAGE] in the liquid of lower density. The entire system is kept in an elevator
which is moving upwards with an acceleration of a = g/2. Find the
tension in the string. (Take g = 10 m
[IMAGE] ) (in N),d ik=k esa nks ρ_1 = 1000 kg
m^-3rFkk ρ_2 = 1500 kg m^-3 Hkjs gSaV = 10^-3 m^3vk;ru rFkk d
= 800 kg m^-3?kuRo dk,d Bksl CykWd jLlh ca
dk nqljk fljk fp=kkuqlkj ik=k ds ry ij ca
 $\frac{2^{th}}{5}$Hkkx mPp ?kuRo nzo esa rFkk$\frac{3^{th}}{5}$Hkkx
fuEu ?kuRo ds nzo esa Mqck gSA lEiw.kZ Roj.k


Answer: 6

Solution:
[IMAGE] We analysed this problem From the reference frame of Elevator. Total
buoyant force On the block,
CykWd ij dqy mRIykou cy
[IMAGE] For equilibrium,
Subject: Physics — Work Power & Energy — Work energy theorem
Difficulty: Moderate

Question: A particle of mass m is released from point A and it stops at B
find out the coefficient of friction

1) $\mu = \frac{1}{4}$
2) $\mu = \frac{1}{3}$
3) $\mu = \frac{2}{3}$
4) None of these

Correct option: 1

Solution:
$W_{g} + W_{f} = K_{f} + K_{i}$
$mg\left( \frac{3h}{4} \right) - \mu mg3h = 0 - 0 \Rightarrow \mu = \frac{1}{4}$
Subject: Physics —  — 

Question: Four functions given below describe motion of a particle. (I) y
= sin ωt - cos ωt, (II) y = sin3 wt, (III) y = 5 cos
[IMAGE], (IV) y = 1 + ωt + ω2t2. Therefore, simple
harmonic motion is represented by

1) only I
2) I, II, III and IV
3) I and III
4) I and II

Correct option: 3

Solution:
I and III are showing S.H.M. on the basis of superposition
principle.
Subject: Physics — Rigid Body Dynamics — Moment of inertia
Difficulty: Easy

Question: Find the moment of inertia of a uniform rectangular plate of
mass M, edges of length
 and about itsaxis passing through
centre and perpendicular to

1) $\frac{M\left( \mathcal{l}^{2} + b^{2} \right)}{6}$
2) $\frac{M\left( \mathcal{l}^{2} + b^{2} \right)}{12}$
3) $\frac{M\left( \mathcal{l}^{2} + b^{2} \right)}{3}$
4) None

Correct option: 2

Solution:
Using perpendicular axis theorem l_3 = I_1 + I_2
$I_{2} = \frac{M\mathcal{l}^{2}}{12}$
$I_{3} = \frac{M\left( \mathcal{l}^{2} + b^{2} \right)}{12}$
$I_{3} = \frac{M\left( \mathcal{l}^{2} + b^{2} \right)}{12}$
Subject: Physics — Moving charges and magnetism — Magnetic field due to current carrying conducting
Difficulty: Moderate

Question: The radius of a coil of wire with 21 turns is 0.22 m, and i_2
current flows clockwise in the coil as shown. A long straight wire
carrying a current is toward the left is located 0.05 m from the edge of
the coil. Themagnetic field at the centre of the coil is zero tesla. The
ratio $\frac{i_{1}}{9i_{2}}$ is (Use$\pi = 22/7$)


Answer: 9

Solution:
$\frac{\mu_{0} \cdot i_{1}}{2\pi(d + r)} = \frac{N \times \mu_{0}i_{2}}{2r}$
$\Rightarrow N = \frac{i_{1}}{i_{2}} \cdot \frac{r}{\pi(d + r)}$
$\frac{i_{1}}{i_{2}} = 81$
Subject: Physics —  — 

Question: A uniform solid sphere of radius R and mass M purely rolls down
an inclined plane. The coefficient of friction between the sphere and
the inclined plane is[IMAGE]. If the maximum value of
angle of inclination for this to be possible is
[IMAGE] then n is: (Take[IMAGE] )


Answer: 7

Solution:
[IMAGE] put in (1)
[IMAGE] Now [IMAGE] TOPIC:Rotaional dynamics
SUB TOPIC: Rolling down the incline
LEVEL: Moderate
Subject: Physics — Centre of Mass — Collision
Difficulty: Moderate

Question: A small sphere of mass 1 kg is moving with a velocity
$(6\widehat{i} + \widehat{j})ms^{- 1}$. It hits a fixed smooth wall and
rebound with velocity $(4\widehat{i} + \widehat{j})ms^{- 1}$. The
coefficient of restitution between the sphere and the wall is


Correct option: 2

Solution:
Impulse = Change in momentum
$= 1(4\widehat{i} + \widehat{j}) - 1(6\widehat{i} + \widehat{j}) = - 2\widehat{i}$
 
Which is perpendicular to the wall. Componentof initial velocity along
$\widehat{i} = 6\widehat{i}$
$\Rightarrow$ Speed of approach =6 
Therefore $e = \frac{4}{6} = \frac{2}{3}$
Subject: Physics — Wave Optics — YDSE
Difficulty: Easy

Question: The distance between two slits in a Young's double slit
experiment is 3 mm. The distance of the screenfrom the slits is 1 m.
Waves of wavelength 1 mm are incident on the plane of the slits
normally.The distance of the first maxima on the screen from the central
maxima will be: (in cm)

1) 36.6
2) 35.4
3) 34.4
4) 35.4

Correct option: 4

Solution:
At first maxima d sinθ = λ
$3sin\theta = 1 \Rightarrow sin\theta = \frac{1}{3}$
$\tan\theta = \frac{y}{D} \Rightarrow y = D \times \tan\theta = 1 \times \frac{1}{\sqrt{8}}m = \frac{100}{\sqrt{8}}cm = 35.4cm$
Subject: Physics — Mechanical Properties of Fluids — Archimedes principle
Difficulty: Moderate

Question: The densities of wood and benzene at 0 ºC are 880 kg m^-3 and
900 kg m^-3, respectively. The coefficientsof volume expansion of wood
is 1.2 × 10^-3 ºC^-1.and that of benzene is
1.5 × 10^-3 ºC^-1. Value oftemperature (in ºC) at which a piece of
this wood would just sink in benzene at the same temperature isT. Then
find value of 3×10^-4T (where T is tempture in ºC).
ydM+h rFkk csathu dk ?kuRo 0ºC ij Øe'k 880 kg m^-3rFkk 900 kg
m^-3gSA ydM+h rFkk csathu izlkj xq.kkad Øe'k 1.2 × 10^-3
C^-1rFkk 1.5 × 10^-3 ºC^-1gSA ftl rkieku ij ydM+h csathu esa
iw.kZ:Ik ls Mqcrh gS rks og rkieku T (ºC esa) gSA rc 3×10^-4T

1) 40
2) 20
3) 60
4) 80

Correct option: 2

Solution:
For just sink condition
Mqcus
Subject: Physics —  — 
Difficulty: Moderate

Question: Two blocks of masses 2 kg and 4 kg are connected through a massless
inextensible string. The
co-efficient of friction between 2 kg block and ground is 0.4 and the
coefficient of friction between 4 kg block and ground is 0.6. Two forces
F1 = 10 N and F2 = 20 N are applied on the blocks as shown in the
figure. Calculate the frictional force ( in N )between 4 kg block and
ground (Assume initially the tension in the string was just zero before
forces F1 and F2 were applied)
[IMAGE] 2 kg 4 kg 2 kg
 0.4 4 kg 
0.6 F1 = 10 N F2 = 20 N 
4 kg (, F1 F2 )


Answer: 18

Solution:
System will not move (Equilibrium)
For equilibrium of 2 kg block
[IMAGE] 10 = 8 + T
⇒ T = 2N
For equilibrium of 4 kg block
[IMAGE] T + f2 = 20
⇒ f2 = 18 N
TOPIC: NLM
SUB TOPIC: FRICTION
Subject: Physics —  — 

Question: A ball is thrown horizontal from the top of a tower 40 m high. The
ball strikes the ground at a point 80 m from the bottom of the tower.
Find the angle that the velocity vector makes with the horizontal just
before the ball hits the ground.

1) 45°
2) 90°
3) 37°
4) 53°

Correct option: 1

Solution:
Subject: Physics — Heat — Thermal expansion
Difficulty: Moderate

Question: If the length of steel rod is 5cm longer than copper rod at all
temperatures. ($\alpha$ for copper $= 1.7 \times 10^{- 5}ºC^{- 1}$ and
$\alpha$ for steel $\left. \ = 1.1 \times 10^{- 5}C^{- 1} \right)$

1) Length of steel rod is 14.17cm and length of copper rod is 9.17cm
2) Length of steel rod is 15.17cm and length of copper rod is 6.17cm
3) Length of steel rod is 14.17cm and length of copper rod is 10.17cm
4) Length of steel rod is 15.17cm and length of copper rod is 9.17cm

Correct option: 1

Solution:
$\mathcal{l}_{1}\left( 1 + \alpha_{1}\Delta T \right) - \mathcal{l}_{2}\left( 1 + \alpha_{2}\Delta T \right) = \mathcal{l}_{1} - \mathcal{l}_{2}$
$\mathcal{l}_{1}\alpha_{1}\Delta T = \mathcal{l}_{2}\alpha_{2}\Delta T$
$\mathcal{l}_{1}\alpha_{1} = \mathcal{l}_{2}\alpha_{2}$
$\Rightarrow \mathcal{l}_{1} \times 1.7 = \mathcal{l}_{2} \times 1.1$
$17\mathcal{l}_{1} = 11\mathcal{l}_{2}$
As $\left( \mathcal{l}_{2} - \mathcal{l}_{1} = 5 \right)$
$\Rightarrow 17\mathcal{l}_{2} - 11\mathcal{l}_{2} = 5 \times 17$
$6\mathcal{l}_{2} = 85$
$\Rightarrow \mathcal{l}_{2} = \frac{85}{6} = 14.17cm$
$\mathcal{l}_{1} = 9.17cm$