datavorous/entrance-exam-dataset huggingface dataset:
A cylindrical block of wood (density = 650 kg m-3), of base area 30 cm2 and height 54 cm, floats in a liquid of density 900 kg m-3. The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length (nearly) :

Options:
A 52 cm
B 26 cm
C 39 cm
D 65 cm

The correct answer is:
39 cm

Solution:
Case-I
In equilibrium
Fup=mg
ViÏLg=mg
AhÏLg=ALÏsgâ¦â¦â¦1
Case-II
Let block is pressed by x more in liquid
Fnet=Fup1-mg
=Vi1ÏLg-mg
Fnet=A h+xÏLg-ALÏsg
Fnet=AhÏLg+AxÏLg-ALÏsg. . . . . . . . . . (2)
from equation (1) & (2)
Fnet=AxÏLg restoring force in upward direction
Î±=Fnetm=AÏLgAÏsLx
Î±=Ï2x for SHM
and Ï2=gl
so gl=ÏLgÏsL
so l=ÏsLÏL=650Ã54900=39cm

Topic: Physics, Oscillations, JEE Main, JEE Main 2015 (11 Apr Online)
A cylindrical capacitor has charge $\mathrm{Q}$ and length L. If both the charge and length of the capacitor are doubled, by keeping other parameters fixed, the energy stored in the capacitor

Options:
A remains same
B increases two times
C decreases two times
D increases four times

The correct answer is:
increases two times

Solution:
The energy stored in a capacitor is given by
$E=\frac{Q^{2}}{2 C}$
Here, $C=\frac{2 \pi \in_{0} L}{\ln \left(\frac{R_{2}}{R_{1}}\right)}$, where
$\begin{array}{l}
\mathrm{L}=\text { length of the cylindrical conductor } \\
\mathrm{R}_{1}=\text { inner radius } \\
\mathrm{R}_{2}=\text { outer radius } \\
\therefore \mathrm{C} \propto \mathrm{L} \Rightarrow \mathrm{E} \propto \frac{\mathrm{Q}^{2}}{\mathrm{~L}}
\end{array}$
When both $Q$ and $L$ are doubled, by keeping other parameters fixed, the energy stored (E') becomes
$\begin{array}{l}
E^{\prime} \propto \frac{(2 Q)^{2}}{2 L} \\
\propto \frac{2 Q^{2}}{L} \\
\therefore E^{\prime}=2 E
\end{array}$

Topic: Physics, Semiconductors, VITEEE, VITEEE 2008
A cylindrical capacitor has charge Q and length L. If both the charge and the length of the capacitor are doubled by keeping the other parameters fixed, then the energy stored in the capacitor

Options:
A remains same
B increases two times
C decreases two times
D increases four times

The correct answer is:
increases two times

Solution:
Energy of a charged capacitor,
E=12Q2C
For a cylindrical capacitor,
C= 2ÏÎµ0Llogeâ¡ba
E= 12Q22ÏÎµ0Llogeâ¡ba ...(i)
where L= length of the cylinder
a and b = radii of two concentric cylinder
C'= 2ÏÎµ0(2L)logeâ¡ba
E'=122Q2C'
= 1 2 2Q 2 2Ï Îµ 0 2L log e b a ...(ii)
From Eqs. (i) and (ii), we get
E'= 2E

Topic: Physics, Electrostatics, NEET
A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a as shown in the figure. Both the cylinder and the cavity are infinity long. A uniform current density J flows along the length. If the magnitude of the magnetic field at the point P is given by N12Î¼0aJ, then the value of N is

Options:

The correct answer is:
5

Solution:
B = μ 0 J π a 2 2 π a - μ 0 J π a 2 / 4 2 π 3 a 2
B = 5 μ 0 J a 1 2 = μ 0 N J a 1 2
So N = 5

Topic: Physics, Magnetic Effects of Current, JEE Main
A cylindrical conductor of radius R carrying current i along the axis such that the magnetic field inside the conductor varies as B=B0r2 0<râ¤R, then which of the following is incorrect?

Options:
A Current density at distance ârâ from central axis of the conductor is J(r) =2B0rÎ¼0
B Current density at distance ârâ from central axis of the conductor is J(r) =3B0rÎ¼0
C Half of the total current would be confined within the radius of R213 of the conductor
D For r > R magnetic field would vary as B(r)=B0R3r

The correct answer is:
Current density at distance ârâ from central axis of the conductor is J(r) =2B0rÎ¼0

Solution:
Hint: Use Ampere’s law
Sol: As Br=B0r2
∴ Total current (upto R)
B0R22πR=μ0I0 ⇒ I0=2πR3B0μ0
And let upto 'r' it confines half of the current.
Then B0r'2·2πr'=μ0I02
⇒ B0·2πr'3=μ02·2πR3B0μ0
⇒ r'=R213
Now let J→(r) is the current density at ‘r’
Then B0r2·2πr=μ0∫0rJ(r) 2πrdr
∵ ∫Bdl = μ0I
⇒ 2πB0r3=μ0·2π∫0rJ(r)rdr
∴ Differentiating both side w.r.t(r)
2πB0·3r2·dr=μ02π·J(r) rdr
⇒ J(r)=3B0rμ0

For r>R;

B×2πr=μ0I0

⇒ B×2πr=μ0×2πR3B0μ0

⇒B=B0R3r

Topic: Physics, Magnetic Effects of Current, JEE Main
A cylindrical container of volume 4.0×10-3 m3 contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is 400 K. The pressure of the mixture of gases is :
[Take gas constant as 8.3 J mol-1 K-1

Options:
A 24.9×103 Pa
B 249×101 Pa
C 24.9×105 Pa
D 24.9 Pa

The correct answer is:
24.9×105 Pa

Solution:
For gas A

PAV=nART

PA=(1)RTV

for gas B

PBV=nBRT

PB=(2)RTV

Now according to the Dalton's law of mixture,

P=PA+PB

=3RTV=3×8.314(400)4×10-3=24.9×105 Pa

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (26 Aug Shift 2)
A cylindrical copper rod has length $\mathrm{L}$ and resistance $\mathrm{R}$. If it is melted and formed into another rod of length $2 \mathrm{~L}$, the resistance will be

Options:
A $\mathrm{R}$
B $2 \mathrm{R}$
C $4 \mathrm{R}$
D $8 \mathrm{R}$.

The correct answer is:
$4 \mathrm{R}$

Solution:
Applying volume conservation
$\begin{array}{l}
\mathrm{A} \times \mathrm{L}=\mathrm{A}^{\prime} \times 2 \mathrm{~L} \\
\mathrm{~A}^{\prime}=\frac{\mathrm{A}}{2} \\
\mathrm{R}=\frac{\rho \mathrm{L}}{\mathrm{A}} \\
\mathrm{R}^{\prime}=\frac{\rho \times 2 \mathrm{~L}}{\mathrm{~A}^{\prime}}=\frac{\rho \times 4 \mathrm{~L}}{\mathrm{~A}} \\
\mathrm{R}^{\prime}=4 \mathrm{R}
\end{array}$

Topic: Physics, Current Electricity, JEE Main
A cylindrical metal rod is shaped into a ring with a small gap as shown. On heating the system :

Options:
A \(x\) decreases, \(r\) and \(d\) increase
B \(x\) and \(r\) increase, \(d\) decreases
C \(x, r\) and d all increase
D \(x\) and \(r\) decreased, \(d\) remains constant

The correct answer is:
\(x\) and \(r\) increase, \(d\) decreases

Solution:
Material expands outward and so \(x, r\) increases. Due to linear expansion diameter of rod will increase.

Topic: Physics, Thermal Properties of Matter, VITEEE, VITEEE 2022
A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two end conduct an amount of heat ' $Q_1$ ' in time ' $t$ ' the metallic rod is melted and the material is formed into a rod of length four times the length of original rod. The amount of heat conducted by the new rod when placed in thermal contact with the same two reservoirs in time $t$ is ' $Q_2$ '. Then $\frac{Q_1}{Q_2}$ is

Options:
A $16$
B $\frac{1}{16}$
C $\frac{1}{4}$
D $4$

The correct answer is:
$16$

Solution:
The heat conduction rate is given by:
$\frac{Q}{t}=\frac{k A\left(T_1-T_2\right)}{l}$
Considering volume conservation:
$\begin{aligned} & A_1 l_1=A_2 l_2 \\ & A_1 l_1=A_2\left(4 l_1\right) \\ & \Rightarrow A_2=\frac{A_1}{4} \\ & \frac{Q_2}{t}=\frac{k\left(\frac{A_1}{4}\right)\left(T_1-T_2\right)}{4 l_1}=\frac{1}{16} \frac{k A_1\left(T_1-T_2\right)}{l_1}=\frac{1}{16} \frac{Q_1}{t} \\ & \Rightarrow Q_2=\frac{1}{16} Q_1\end{aligned}$

Topic: Physics, Thermal Properties of Matter, JEE Main
A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat $Q$ in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod when placed in thermal contact with the two reservoirs in time $\mathrm{t}$ ?

Options:
A $Q / 4$
B $Q / 16$
C $2 \mathrm{Q}$
D $Q / 2$

The correct answer is:
$Q / 16$

Solution:
In steady state the amount of heat flowing from one face to the other face in time $t$ is given by $Q=\frac{K A\left(\theta_1-\theta_2\right) t}{l}$ where $K$ is coefficient of thermal conductivity of material of rod $\Rightarrow \quad \frac{\mathrm{Q}}{\mathrm{t}} \propto \frac{\mathrm{A}}{\mathrm{l}} \propto \frac{\mathrm{r}^2}{\mathrm{l}}$
As the metallic rod is melted and the material is formed into a rod of half the radius
$$
\begin{aligned}
\mathrm{V}_1 & =\mathrm{V}_2 \\
\Rightarrow \quad \pi \mathrm{r}_1^2 \mathrm{l}_1 & =\pi \mathrm{r}_2^2 \mathrm{l}_2 \\
\quad \quad \quad \mathrm{l}_1 & =\frac{\mathrm{l}_2}{4}
\end{aligned}
$$
Now, from Eqs. (i) and (ii)
$$
\begin{aligned}
\Rightarrow \quad & \frac{Q_1}{Q_2}=\frac{r_1^2}{l_1} \times \frac{l_2}{r_2^2}=\frac{r_1^2}{l_1} \times \frac{4 l_1}{\left(r_1 / 2\right)^2} \\
\Rightarrow \quad Q_1 & =16 Q_2
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main
A cylindrical metallic wire is stretched to increase its length in such a way that the metallic wire changes its resistance by 6%. The percentage increase in its length is

Options:
A 2%
B 4%
C 3%
D 12%

The correct answer is:
3%

Solution:
Resistance of a uniform wire is given by,

R=ρLA.

As volume will remain constant, therefore,

AL=V⇒A=VL

So we can write resistance as,

R=ρLVL=ρVL2

Now for small changes we can write,

∆RR=2∆LL⇒∆LL×100=12∆RR×100=12×6=3%

Topic: Physics, Current Electricity, TS EAMCET, TS EAMCET 2022 (18 Jul Shift 2)
A cylindrical piston of mass M slides smoothly inside a long cylinder closed at one end, enclosing a certain mass of gas. The cylinder is kept with its axis horizontal. If the piston is disturbed from its equilibrium position, it oscillates simple harmonically. The period of oscillation will be (Assume isothermal process)

Options:
A T=2ÏMhPA
B T=2πMAPh
C T=2πMPAh
D T=2πMPhA

The correct answer is:
T=2πMhPA

Solution:
Let the piston be displaced through distance x towards left, then volume decreases, pressure increases. If ÎP is an increase in pressure and ÎV is a decrease in volume, then considering the process to take place slowly (i.e. Isothermal)
P1V1=P2V2âPV=(P+ÎP)(V-ÎV)
âPV=PV+ÎPV-PÎV-ÎPÎV
âÎP.V-P.ÎV=0 (neglecting ÎP.ÎV)
ÎP(Ah)=P(Ax)âÎP=Pâxh

F=-PAhx

Ï=PAMh

T=2ÏMhPA

Topic: Physics, Oscillations, JEE Main
A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency ω. If the radius of the bottle is 2.5 cm then ω is close to: ( density of water =103kg/m3)

Options:
A 5.00 rad sec-1â¡Â
B 2.50Â radÂ sec-1â¡
C 7.9 rad sec-1â¡
D 3.75Â radÂ sec-1â¡

The correct answer is:
7.9 rad sec-1â¡

Solution:
Let at equilibrium, y length of bottle is immerged in water:
So, mg=fbuo
mg=ρAyg ..... (1) [Equilibrium Condition]
Now it is further pushed by distance x
Frestoring=ρ[Ax+y)g-mg
mω2x=ρAxg+ρAyg-mg
ω=ρAgm=ρAgρV=πr2gV=7.9 rad/sec

Topic: Physics, Oscillations, JEE Main, JEE Main 2019 (10 Jan Shift 2)
A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of 0.1 g of ice per second. If the rod is replaced by another with half the length and double the radius of the first and if the thermal conductivity of the material of the second rod is 14 that of the first, the rate at which ice melts in g s-1 will be

Options:
A 3.2
B 1.6
C 0.2
D 0.1

The correct answer is:
0.2

Solution:
As dQdt=KA dTdx, therefore, when
dt â12, Aâ22=4, Kâ14
dQdt becomes twice ; m would become twice
Mass of ice melted/s= 2Ã0.1g=0.2g

Topic: Physics, Thermal Properties of Matter, JEE Main
A cylindrical steel rod of length $0.10 \mathrm{~m}$ and thermal conductivity $50 \mathrm{~W} \cdot \mathrm{m}^{-1} \mathrm{~K}^{-1}$ is welded end to end to copper rod of thermal conductivity $400 \mathrm{~W} \cdot \mathrm{m}^{-1} . \mathrm{K}^{-1}$ and of the same area of cross section but $0.20 \mathrm{~m}$ long. The free end of the steel rod is maintained at $100^{\circ} \mathrm{C}$ and that of the copper and at $0^{\circ} \mathrm{C}$. Assuming that the rods are perfectly insulated from the surrounding, the temperature at the junction of the two rods-

Options:
A $20^{\circ} \mathrm{C}$
B $30^{\circ} \mathrm{C}$
C $40^{\circ} \mathrm{C}$
D $50^{\circ} \mathrm{C}$

The correct answer is:
$20^{\circ} \mathrm{C}$

Solution:
$\frac{50 \times \mathrm{A} \times(100-\mathrm{T})}{0.1}=\frac{400 \times \mathrm{A} \times(\mathrm{T}-0)}{0.2}$

Topic: Physics, Thermal Properties of Matter, KVPY, KVPY 2012 (SB/SX)
A cylindrical tank has a hole of $1 \mathrm{~cm}^2$ in its bottom. If the water is allowed to flow into the tank from a tube above it at the rate of $70 \mathrm{~cm}^3 / \mathrm{sec}$. Then the maximum height up to which water can rise in the tank is

Options:
A $2.5 \mathrm{~cm}$
B $5 \mathrm{~cm}$
C $10 \mathrm{~cm}$
D $0.25 \mathrm{~cm}$

The correct answer is:
$2.5 \mathrm{~cm}$

Solution:
The height of water in the tank becomes maximum when the volume of water flowing into the tank per second becomes equal to the volume flowing out per second.
Volume of water flowing out per second
$=A v=A \sqrt{2 g h}$...(i)
Volume of water flowing in per second
$=70 \mathrm{~cm}^3 / \mathrm{sec}$ ...(ii)
From (i) and (ii) we get
$A \sqrt{2 g h}=70 \Rightarrow 1 \times \sqrt{2 g h}=70$
$\Rightarrow 1 \times \sqrt{2 \times 980 \times h}=70$
$\Rightarrow 1 \times \sqrt{2 \times 980 \times h}=70$
$\therefore h=\frac{4900}{1960}=2.5 \mathrm{~cm}$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A cylindrical tube filled with water Î¼w=43 is closed at its both ends by two thin, silvered plano-convex lenses, as shown in the figure. Refractive index of lenses L1 and L2 are 2.0 and 1.5, while their radii of curvatures are 5 cm and 9 cm, respectively. A point object is placed somewhere at a point O on the axis of the cylindrical tube. If it is found that all the images formed by multiple refractions and reflections coincide with the object, then the distance between both the lenses is

Options:
A 8Â cm
B 18 cm
C 10Â cm
D 14Â cm

The correct answer is:
18 cm

Solution:
For the lens L1, the ray must move parallel to the axis after refraction, for the image to coincide with the object.

μ1∞+μwx=μ1 -μwR1, where x is the distance of the object from the lens L1

43x=2-435

⇒x=10 cm

For the lens L2, the ray must appear to come from the centre of curvature after refraction, for the image to coincide with the object.

μ2-R2+μwy=μ1 -μw∞, where y is the distance of the object from the lens L2

y=R24332=8 cm

Distance between the two lenses = 10 cm + 8 cm = 18 cm

Topic: Physics, Ray Optics, JEE Main
A cylindrical tube of length $0.2 \mathrm{~m}$ and radius $R$ with sugar solution of concentration $C$ produce a rotation of $\theta$ in the plane of vibration of a plane polarized light. The same sugar solution is transferred to
another tube of length $0.3 \mathrm{~m}$ of same radius. The remaining gap is filled by distilled water. Now the optical rotation produced is

Options:
A $\theta$
B $2 \frac{\theta}{3}$
C $3 \frac{\theta}{2}$
D $9 \frac{\theta}{4}$

The correct answer is:
$\theta$

Solution:
We know the specific rotation,
$S=\frac{\theta}{l \cdot C}=\frac{\theta \cdot v}{l \cdot m}$
where $\theta=$ rotation produced
$I=$ length of the tube
$m=$ mass of sugar dissolved
$V=$ volume of sugar solution
$\Rightarrow \quad \theta=\frac{S \cdot 1 \cdot m}{V}$
For first case, $\quad \theta=\frac{S / m}{V}...(i)$
For second case, $\quad \theta_{1}=\frac{S I_{1} m_{1}}{V_{1}}...(ii)$
Here, $\quad I=0.2 \mathrm{~m}$ and $\quad l_{1}=0.3 \mathrm{~m}$ also, $\quad V_{1}=V+\frac{V}{2}=1.5 V=\frac{3}{2} V$
On dividing Eq. (ii) by Eq. (i), we get
$\begin{aligned}
\quad \frac{\theta_{1}}{\theta}=& \frac{\frac{S l_{1} m}{V_{1}}}{\frac{S I m}{V}}=\frac{l_{1}}{l} \times \frac{V}{V_{1}}=\frac{l_{1}}{l} \times \frac{2 V}{3 V}=\frac{0.3}{0.2} \times \frac{2}{3}=1 \\
\Rightarrow \quad \theta_{1} &=\theta
\end{aligned}$

Topic: Physics, Wave Optics, KCET, KCET 2013
A cylindrical tube of uniform cross-sectional area $A$ is fitted with two air tight frictionless pistons. The pistons are connected to each other by a metallic wire. Initially the pressure of the gas is $P_0$ and temperature is $T_0$, atmospheric pressure is also $P_0$. Now the temperature of the gas is increased to $2 T_0$, the tension in the wire will be

Options:
A $2 P_0 A$
B $P_0 A$
C $\frac{P_0 A}{2}$
D $4 P_0 A$

The correct answer is:
$P_0 A$

Solution:
Volume of the gas is constant $V=$ constant $\therefore P \propto T$
i.e., pressure will be doubled if temperature is doubled
$\therefore P=2 P_0$
Now let $F$ be the tension in the wire. Then equilibrium of any one piston gives
$F=\left(P-P_0\right) A=\left(2 P_0-P_0\right) A=P_0 A$

Topic: Physics, Thermodynamics, JEE Main
A cylindrical tube, open at both ends, has a fundamental frequency, $\mathrm{f}$, in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now

Options:
A
$f$

B
$\frac{f}{2}$

C
$\frac{3 f}{4}$

D
$2 f$

The correct answer is:
$f$

Solution:
$f_o=\frac{v}{2 \ell}$
$f_c=\frac{v}{2 \ell}$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2012 (Offline)
A cylindrical tube open at both ends, has a vibrating air column of fundamental frequency ' $\mathrm{f}$ ' in air. The tube is dipped vertically in water so that half of its is in water. The fundamental frequency of the vibrating air column is now

Options:
A f
B $\frac{\mathrm{f}}{2}$
C $\frac{3 \mathrm{f}}{2}$
D 2f

The correct answer is:
f

Solution:
$v=\lambda f$
Fundamental frequency's $\lambda=\frac{v}{f} \Rightarrow f_0=\frac{v}{2 L}=f$
If the tube is dipped half into liquid then it acts as if closed at $\frac{\mathrm{L}}{2}$.
See figure
The frequency can be written as $\mathrm{f}_{\mathrm{c}}=\frac{\mathrm{v}}{\lambda_1}=\frac{\mathrm{v}}{2 \mathrm{~L}}=\mathrm{f}$
$\therefore$ Fundamental frequency would remain the same ' $\mathrm{f}$ '.

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2022 (08 Aug Shift 2)
A cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure. The radius of vessel is 5 cm and the angular speed of rotation is ω rad s-1. The difference in the height, h in cm of liquid at the Centre of vessel and at the sides of the vessel will be :

Options:
A 2ω225g
B 5ω22g
C 25ω22g
D 2ω25g

The correct answer is:
25ω22g

Solution:
ρdrω2r=ρgdh

ω2∫0Rr dr=g∫0hdh

ω2R22=gh

h=ω2R22g=25ω22g cm.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2020 (02 Sep Shift 1)
A cylindrical vessel of $90 \mathrm{~cm}$ height is kept filled upto the brim. It has four holes $1,2,3,4$ which are respectively at heights of $20 \mathrm{~cm}, 30 \mathrm{~cm}, 45 \mathrm{~cm}$ and $50 \mathrm{~cm}$ from the horizontal floor $P Q$. The water falling at the maximum horizontal distance from the vessel comes from

Options:
A Hole number 4
B Hole number 3
C Hole number 2
D Hole number

The correct answer is:
Hole number 3

Solution:
Horizontal range will be maximum when $h=\frac{H}{2}=\frac{90^{\circ}}{2}$
$=45 \mathrm{~cm}$ i.e hole 3

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A cylindrical vessel of cross-section A contains water to a height $\mathrm{h}$. There is a hole in the bottom of radius ' $a$ '. The time in which it will be emptied is:

Options:
A
$\frac{2 \mathrm{~A}}{\pi \mathrm{a}^2} \sqrt{\frac{\mathrm{h}}{\mathrm{g}}}$

B
$\frac{\sqrt{2} \mathrm{~A}}{\pi \mathrm{a}^2} \sqrt{\frac{\mathrm{h}}{\mathrm{g}}}$

C
$\frac{2 \sqrt{2} \mathrm{~A}}{\pi \mathrm{a}^2} \sqrt{\frac{\mathrm{h}}{\mathrm{g}}}$

D
$\frac{\mathrm{A}}{\sqrt{2} \pi \mathrm{a}^2} \sqrt{\frac{\mathrm{h}}{\mathrm{g}}}$

The correct answer is:
$\frac{\sqrt{2} \mathrm{~A}}{\pi \mathrm{a}^2} \sqrt{\frac{\mathrm{h}}{\mathrm{g}}}$

Solution:
$$
\text { Let the rate of falling water level be }-\frac{d h}{d t}
$$
Initially at $t=0 ; h=h$
$$
t=t ; h=0
$$
Then, $A\left(-\frac{d h}{d t}\right)=\pi a^2 \cdot v$
$$
d t=-\frac{A}{\pi a^2 \sqrt{2 g h}} d h
$$
$[\because$ velocity of efflux of liquid $v=\sqrt{2 g h}]$ Integrating both sides
$$
\int_0^t d t=-\frac{A}{\sqrt{2 g} \pi a^2} \int_h^0 h^{-1 / 2} d h
$$
$$
[t]_0^t=-\frac{A}{\sqrt{2 g} \pi a^2} \cdot\left[\frac{h^{1 / 2}}{1 / 2}\right]_h^0
$$
$$
t=\frac{\sqrt{2} A}{\pi a^2} \sqrt{\frac{h}{g}}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2014 (12 Apr Online)
A cylindrical vessel of height 500 mm has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it upto height H . Now the top is completely sealed with a cap and the orifice at the bottom is opened and some water is allowed to comes out slowly from the orifice. Eventually, the water level in the vessel becomes steady with a height of water column being 200 mm . Find the fall in height ( in mm ) of water level due to opening of the orifice.
[ Take atmospheric pressure =1.0Ã105âN m-2, density of water =1000âkgâ m -3 and g=10ââ ms -2 . Assume temperature to be constant and neglect any effect of surface tension.]

Options:

The correct answer is:
6

Solution:
In this question, we will have to assume that temperature is constant for enclosed air above water (or pV = constant) (âµ Boyle's Law)
p = p0 - Ïgh ...(i)
p0 [A(500 - H)] = p[A(300)] ...(ii)
Solving these two equations, we get H = 206 mm .
Therefore, Level fall = (206 - 200) mm = 6 mm

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A cylindrical wire of mass (0.4 ± 0.01) g has length (8 ± 0.04) cm and radius (6 ± 0.03) mm. The maximum error in its density will be

Options:
A 3.5%
B 5%
C 1%
D 4%

The correct answer is:
4%

Solution:
The formula to calculate the density of the material of the wire is given by

ρ=MπR2L ...1

Hence, the formula to calculate the percentage error in calculating the density of the material can be written as

dρρ×100=dMM+2dRR+dLL×100 ...2

Substitute the values of the known parameters into equation (2) to calculate the required percentage error.

=0.010.4+2×0.036+0.048×100=4%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2023 (08 Apr Shift 1)
A cylindrical wire of radius 0.5 mm and conductivity 5×107 S m-1 is subjected to an electric field of 10 mV m-1. The expected value of current in the wire will be x3π mA. The value of x is _________.

Options:

The correct answer is:
5

Solution:
Conductivity σ=5×107 S m-1

Radius r=0.5 mm=5×10-4 m

E=10×10-3 V m-1

J=σE=10×10-3×5×107

⇒J=5×105

⇒iA=5×105

i=5×105×πr2

=5×105×π×5×10-42

=125π×10-3 A

Therefore, i=125 π mA

Hence, x=5

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (24 Feb Shift 2)
A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 11000 of the original amplitude is close to:

Options:
A 50 s
B 100 s
C 20 s
D 10 s

The correct answer is:
20 s

Solution:
f=5
So, T=15
Time taken for amplitude to drop to half: 10T=105=2 s
Now,
A01000=A0(2)t2
(2)t2=1000
t2log2=3
t=6log2≈20 s

Topic: Physics, Oscillations, JEE Main, JEE Main 2019 (08 Apr Shift 2)
A data consists of n observations: x1, x2, … , xn. If ∑i=1nxi+12=9n and ∑i=1nxi-12=5n, then the standard deviation of this data is

Options:
A 5
B 7
C 5
D 2

The correct answer is:
5

Solution:
Given:

∑i=1nxi+12=9n

⇒∑i=1nxi2+2xi+1=9n ...1

And, ∑i=1nxi-12=5n

⇒∑i=1nxi2-2xi+1=5n ...2

Form 1+2, we get

∑i=1n2xi2+2=14n

⇒∑i=1nxi2=6n ...3

From 1-2, we get

⇒∑i=1nxi=n ...4

∴ Variance, σ2=Ex2-Ex2

⇒σ2=∑x2n-∑xn

⇒σ2=6nn-nn2=6-1=5

∴ Standard deviation, σ=5.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2019 (09 Jan Shift 2)
A data consists of n observations: x1,x2,…,xn. If ∑i=1n(xi+1)2=11n and ∑i=1nxi-12=7n, then the variance of this data is

Options:
A 5
B 8
C 6
D 7

The correct answer is:
7

Solution:
We have, ∑i=1nxi+12=11n…….i
and ∑i=1nxi-12=7n…ii
Adding i and ii , we get
2∑i=1nxi2+1=18n⇒∑i=1nxi2+1=9n
⇒∑i=1nxi2+n=9n⇒∑i=1nxi2=8n
⇒∑i=1nxi2n=8
Subtracting i and ii, we get
4∑i=1nxi=4n⇒∑i=1nxi=n⇒∑i=1nxin=1
Now, variance =1n∑i=1nxi2-∑i=1nxin2=8-1=7

Topic: Mathematics, Statistics, JEE Main
A data consists of \( n \) observations: \( x_{1}, x_{2}, \ldots, x_{n} . \) If \( \sum_{i=1}^{n}\left(x_{i}+1\right)^{2}=11 n \) and \( \sum_{i=1}^{n}\left(x_{i}-1\right)^{2}=7 n \), then the variance of this data is

Options:
A \( 5 \)
B \( 8 \)
C \( 6 \)
D \( 7 \)

The correct answer is:
\( 7 \)

Solution:
We have, ∑i=1nxi+12=11n…….i
and ∑i=1nxi-12=7n…ii
Adding i and ii , we get
2∑i=1nxi2+1=18n⇒∑i=1nxi2+1=9n
⇒∑i=1nxi2+n=9n⇒∑i=1nxi2=8n
⇒∑i=1nxi2n=8
Subtracting i and ii, we get
4∑i=1nxi=4n⇒∑i=1nxi=n⇒∑i=1nxin=1
Now, variance =1n∑i=1nxi2-∑i=1nxin2=8-1=7

Topic: Mathematics, Statistics, JEE Main