datavorous/entrance-exam-dataset huggingface dataset:
A stream of water flowing horizontally with a speed of 15 ms-1 gushes out of a tube of cross-sectional area 10-2 m2 and hits a vertical wall nearby. The force exerted on the wall by the impact of water (assuming it does not rebound) is

Options:
A 2.25×103 N
B 2.5Ã103Â N
C 3.0Ã103Â N
D 3.5Ã103Â N

The correct answer is:
2.25×103 N

Solution:
If the velocity of the stream be v (say) . In 1 second , the distance travelled =v.
â´ The volume of water hitting the wall per second, V=va
Where, a is the cross-sectional area.
â V=15 m/sÃ10-2 m2
[ â´ Given, v=15 m/s,âa=10-2â m2 ]
â V=15Ã10-2 m3/s
Mass of water hitting the wall per second,
m=volumeÃDensity
=15Ã10-2Ã103
=150âkg/s
[âµ Density of water =1000âkg/m3 ]
Initial momentum of the water hitting the wall per second =mÃv
=150Ã15
=2250 kgâm/s2
=2250 N
Final momentum per second =0
â´ Force exerted on the wall = Change in the momentum of water stream per second
=2250â N
=2.25Ã103â N

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A stretched wire of a material whose young's modulus $Y=2 \times 10^{11} \mathrm{Nm}^{-2}$ has poisson's ratio 0.25 . Its lateral strain $\varepsilon_l=10^{-3}$. The elastic energy density of the wire is

Options:
A $16 \times 10^5 \mathrm{Jm}^{-3}$
B $1 \times 10^5 \mathrm{Jm}^{-3}$
C $4 \times 10^5 \mathrm{Jm}^{-3}$
D $8 \times 10^5 \mathrm{Jm}^{-3}$

The correct answer is:
$16 \times 10^5 \mathrm{Jm}^{-3}$

Solution:
Given, Young's modulus, $Y=2 \times 10^{11} \mathrm{Nm}^{-2}$
Poisson ratio, $\sigma=0.25$
Lateral strain, $\varepsilon_1=10^{-3}$
Elastic potential energy density is given by, $\mathrm{PE}$ $=\frac{1}{2} \times Y \times(\text { strain })^2$
Poisson ratio
$\begin{aligned} & =\frac{\text { Lateral strain }}{\text { longitudinal strain }}=\frac{\varepsilon_1}{\text { longitudinal strain }} \\ & \Rightarrow \sigma=\frac{\varepsilon_1}{\text { longitudinal strain }} \\ & \Rightarrow \text { longitudinal strain }=\frac{\varepsilon_1}{\sigma}=\frac{10^{-3}}{0.25}=c \times 10^{-3}\end{aligned}$
Elastic potential energy density $=\frac{1}{2} \times Y \times\left(\frac{E_l}{\sigma}\right)^2$
$\begin{aligned} & =\frac{1}{2} \times 2 \times 10^{11} \times\left(4 \times 10^{-3}\right)^2 \\ & =10^{11} \times 16 \times 10^{-6}=16 \times 10^5 \mathrm{Jm}^{-3}\end{aligned}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A string 2.0 m long and fixed at its ends is driven by a 240 Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is

Options:
A 320  m s-1,  120  Hz
B 320â m s-1,ââ80 âHz
C 180  m s-1,  80  Hz
D 180  m s-1,  120  Hz

The correct answer is:
320â m s-1,ââ80 âHz

Solution:
For a string fixed at both the ends,
f=n2Lv
For third harmonic, 240=32(2)v
⇒v=240×43
=320  m s-1
Fundamental frequency f0=12Lv
=14×320 = 80 Hz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2019 (09 Apr Shift 2)
A string fixed at both ends has a standing wave mode for which the distances between adjacent nodes is 18 cm. For the net consecutive standing wave mode distance between adjacent nodes is 16 cm. The minimum possible length of the string is

Options:
A 144  cm
B 204  cm
C 288  cm
D 72  cm

The correct answer is:
144  cm

Solution:
Let the vibration takes place in the nth mode
So, for 1st case, nÎ»2=L ......(i)
And for 2nd case, n+1Î»â²2=L .....(ii)
From Equations. (i) and (ii), we get
nÎ»2=(n+1)Î»â²2 âµ Î»2=18 cm andÎ»â²2=16 cm
â 18n=n+116
â n=8
So, minimum possible length l=nÎ»2
â l=8Ã18=144 cm

Topic: Physics, Waves and Sound, JEE Main
A string fixed at both ends oscillates in \( 5 \) segments, length \( 10 \mathrm{~m} \) and velocity of wave is \( 20 \mathrm{~ms}^{-1} \). What is the frequency?

Options:
A \( 5 \mathrm{~Hz} \)
B \( 15 \mathrm{~Hz} \)
C \( 10 \mathrm{~Hz} \)
D \( 2 \mathrm{~Hz} \)

The correct answer is:
\( 5 \mathrm{~Hz} \)

Solution:
Frequency of vibration is given by
v=p2lTm where v=Tm
∴v=p2lv=5×202×10=5Hz

Topic: Physics, Waves and Sound, JEE Main
A string fixed at both the ends is vibrating in two segments.The wavelength of the corresponding wave is

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

The correct answer is:
$l$

Solution:
\(\begin{aligned} & \ell=\frac{2 \lambda}{2} \\ & \Rightarrow \ell^{=}=\lambda\end{aligned}\)

Topic: Physics, Waves and Sound, JEE Main
A string having length \( 0.5 \mathrm{~m} \) fixed at one end and other end is connected to a block of mass \( m=2 \mathrm{~kg} \) as shown in figure. The string is set into vibrations which is represented by \( y=4 \sin \left(\frac{\pi x}{5}\right) \cos (50 \pi t) \), where \( x \) and \( y \) are in \( \mathrm{cm} \) and \( \mathrm{t} \) is in second. The number of anti-nodes between point \( \mathrm{A} \) and the fixed pulley is \( 2 n \), find \( n \).

Options:

The correct answer is:
5

Solution:
k=π5
⇒2πλ=π5
⇒λ=10 cm
mλ2=0.5
m×5×10-2=0.5
m=102n=10

Topic: Physics, Waves and Sound, JEE Main
A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is y=0.3sin⁡0.157x cos(200πt). The length of the string is
(All quantities are in SI units.)

Options:
A 20Â m
B 60Â m
C 40Â m
D 80 m

The correct answer is:
80 m

Solution:
From the equation, k=0.157
2πλ=0.157
λ=40 m
But, length of the wire in 4th harmonic, l=4×λ2=2λ=80 m.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2019 (09 Apr Shift 1)
A string is clamped at both the ends and it is vibrating in its \( 4^{\text {th }} \) harmonic. The equation of the stationary wave is
\( Y=0.3 \sin (0.157 x) \cos (200 \pi t) \). The length of the string is,

Options:
A \( 60 \mathrm{~m} \)
B \( 40 \mathrm{~m} \)
C \( 80 \mathrm{~m} \)
D \( 20 \mathrm{~m} \)

The correct answer is:
\( 80 \mathrm{~m} \)

Solution:
We are given an equation of stationary wave,

Y=0.3sin0.157xcos200πt.

Comparing it with the general equation of stationary wave,

Y=asinkxcosωt.

We get,

k=2πλ=0.157

⇒λ=2π0.157=4π2 ∵12π≈0.157.

As the possible wavelength associated with the nth harmonics of a vibrating string fixed at both ends is given as,

λ=2ln ⇒l=nλ2.

Now, according to the question, string is vibrating in the 4th harmonic, so,

4λ2=l⇒2λ=l

⇒l=2×4π2=8π2 [using equation 1]

Now,

∴π2≈10

⇒l≈80 m.

Topic: Physics, Waves and Sound, JEE Main
A string is stretched between fixed points separated by 75.0 cm. it is observed to have resonant frequency of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

Options:
A 105 Hz
B 52.5 Hz
C 140 Hz
D 65 Hz

The correct answer is:
105 Hz

Solution:
For string fixed at both the ends, resonant frequency are given by
v=nv2L
Where symbols have their meaning. It is given that 315 Hz and 420 Hz are two consecutive resonant frequency, let these nth and (n+1)th harmonics.
315=nv2L â¦i
420=n+1v2L â¦ii
â¹Eq.(i)Ã·Eq.(ii)
â¹315450=nn+1â¹n=3
From Eq. (i), lowest resonant frequency
v0=v2L=3153=105 Hz

Topic: Physics, Waves and Sound, JEE Main
A string is stretched between fixed points separated by $75 \mathrm{~cm}$. It is observed to have resonant frequencies of $420 \mathrm{~Hz}$ and $315 \mathrm{~Hz}$. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

Options:
A
$10.5 \mathrm{~Hz}$

B
$105 \mathrm{~Hz}$

C
$1.05 \mathrm{~Hz}$

D
$1050 \mathrm{~Hz}$

The correct answer is:
$105 \mathrm{~Hz}$

Solution:
$\frac{\mathrm{n}}{2 \ell}(\mathrm{v})=315, \frac{(\mathrm{n}+1)}{2 \ell} \mathrm{v}=420$
Solving $\frac{\mathrm{v}}{2 \ell}=105$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2006
a string is wound around a hollow cylinder of mass $5 \mathrm{~kg}$ and radius $0.5 \mathrm{~m}$. If the string is now pulled with a horizontal force of $40 \mathrm{~N}$, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string)

Options:
A $20 \mathrm{rad} / \mathrm{s}^{2}$
B $16 \mathrm{rad} / \mathrm{s}^{2}$
C $12 \mathrm{rad} / \mathrm{s}^{2}$
D $10 \mathrm{rad} / \mathrm{s}^{2}$

The correct answer is:
$16 \mathrm{rad} / \mathrm{s}^{2}$

Solution:
From newton's second law

$40+\mathrm{f}=\mathrm{m}(\mathrm{R} \alpha)$ ...(i)

Taking torque about 0 we get

$40 \times \mathrm{R}-\mathrm{f} \times \mathrm{R}=\mathrm{I} \alpha$

$40 \times \mathrm{R}-\mathrm{f} \times \mathrm{R}=\mathrm{mR}^{2} \alpha$

$40-\mathrm{f}=\mathrm{mR} \alpha$ ...(ii)

Solving equation (i) and (ii) $\alpha=\frac{40}{\mathrm{mR}}=16 \mathrm{rad} / \mathrm{s}^{2}$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (11 Jan Shift 2)
A string is wrapped around the rim of a wheel of moment of inertia $0.20 \mathrm{~kg}-\mathrm{m}^2$ and radius $20 \mathrm{~cm}$. The wheel is free to rotate about its axis initially the wheel is rest. The string is now pulled by a force of $20 \mathrm{~N}$. The angular velocity of the string after 5 seconds will be :

Options:
A $70 \mathrm{rad} / \mathrm{s}$
B $90 \mathrm{rad} / \mathrm{s}$
C $95 \mathrm{rad} / \mathrm{s}$
D $100 \mathrm{rad} / \mathrm{s}$

The correct answer is:
$100 \mathrm{rad} / \mathrm{s}$

Solution:
Angular impulse $=$ change in angular momentum $\tau \mathbf{t}=\mathbf{I} \omega-0$
$\omega=\mathrm{F} \frac{\mathrm{Rt}}{\mathrm{I}}=\frac{20 \times 0.2}{0.2}=100 \mathrm{rad} / \mathrm{sec}$

Topic: Physics, Rotational Motion, JEE Main
A string is wrapped around the rim of a wheel of moment of inertia $0.40 \mathrm{kgm}^2$ and radius $10 \mathrm{~cm}$. The wheel is free to rotate about its axis. Initially the wheel is at rest. The string is now pulled by a force of $40 \mathrm{~N}$. The angular velocity of the wheel after $10 \mathrm{~s}$ is $x \mathrm{rad} / \mathrm{s}$, where $x$ is _______

Options:

The correct answer is:
100

Solution:
$\begin{aligned} & \tau=\mathrm{FR}=\mathrm{I} \alpha \Rightarrow 40 \times 0.1=0.4 \alpha \\ & \alpha=10 \mathrm{rad} / \mathrm{s}^2 \\ & \mathrm{~W}_{\mathrm{f}}=10 \times 10=100 \mathrm{rad} / \mathrm{s}\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2024 (09 Apr Shift 1)
A string of $7 \mathrm{~m}$ length has a mass of $0.035 \mathrm{~kg}$. If tension in the string is $60.5 \mathrm{~N}$, then speed of a wave on the string is

Options:
A $77 \mathrm{~m} / \mathrm{s}$
B $102 \mathrm{~m} / \mathrm{s}$
C $110 \mathrm{~m} / \mathrm{s}$
D $165 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$110 \mathrm{~m} / \mathrm{s}$

Solution:
Given : Length $(l)=7 \mathrm{~m}$
$\operatorname{Mass}(\mathrm{M})=0.035 \mathrm{~kg}$ and tension $(\mathrm{T})=60.5$
$\mathrm{N}$. We know that mass of string per unit length $(\mathrm{m})$
$$
=\frac{0.035}{7}=0.005 \mathrm{~kg} / \mathrm{m}
$$
and speed of
$$
\text { wave }=\sqrt{\frac{T}{m}}=\sqrt{\frac{60.5}{0.005}}=110 \mathrm{~m} / \mathrm{s} .
$$

Topic: Physics, Waves and Sound, JEE Main
A string of area of cross-section 4 mm2 and length 0.5 is connected with a rigid body of mass 2 kg. The body is rotated in a vertical circular path of radius 0.5 m. The body acquires a speed of 5 m s-1 at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is _____ ×10-5. (Use Young's modulus 1011 N m-2 and g=10 m s-2)

Options:

The correct answer is:
30

Solution:
Given here, A=4 mm2, l=0.5 m, m=2 kg and vb=5 m s-1.

Tension at the bottom of the circular motion is given by T=mg+mvb2R

Using Young's modulus, we have

Strain produced in the string when the body is at the bottom of the circle is ∆ll=TAY

=mg+mvb2lAY =20+2520.53×10-6×1011=30×10-5

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2022 (28 Jul Shift 2)
A string of length 1.5 m with its two ends clamped is vibrating in the fundamental mode. The amplitude at the centre of the string is 4 mm. The minimum distance between the two points having amplitude of 2 mm is:

Options:
A 1 m

B 75 m

C 60 m

D 50 m

The correct answer is:
1 m

Solution:
λ=2l=3 m
Equation of standing wave
y=2Asin⁡kxcos⁡ωt
y=A as amplitude is 2A.
A=2Asin⁡kx
2πλx1=π6⇒x1=14 m
And 2πλ.x2=5π6⇒x2=1.25 m⇒x2-x1=1 m

Topic: Physics, Waves and Sound, JEE Main
A string of length 1.5 m with its two ends clamped is vibrating in the fundamental mode. The amplitude at the centre of the string is 4 mm. The minimum distance between the two points having amplitude of 2 mm is:

Options:

The correct answer is:
1

Solution:
λ=2L=3 m
Equation of standing wave
y=2Asinkxcosωt
y=A as amplitude is 2A.
A=2Asinkx
2πλx1=π6⇒x1=14 m
And 2πλx2=5π6⇒x2=1.25 m⇒x2-x1=1 m

Topic: Physics, Waves and Sound, JEE Main
A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to

Options:
A 20.0 cm
B 10.0 cm
C 16.6 cm
D 33.3 cm

The correct answer is:
20.0 cm

Solution:
V=fλ
Tμ=fλ
85×10-3=100λ
λ=40 cm
λ2=20 cm

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2019 (10 Jan Shift 1)
A string of length 1 m is fixed at one end with a bob of mass 100 g and the string makes 2Ï rev s-1 around a vertical axis through a fixed point. The angle of inclination of the string with vertical is

Options:
A tan-1⁡5/8
B tan-1⁡3/5
C cos-1⁡8/5
D cos-1⁡5/8

The correct answer is:
cos-1⁡5/8

Solution:
T cosâ¡Î¸=mg .....(i)
T sinâ¡Î¸=mÏ2r .....(ii)
So tanâ¡Î¸=Ï2rg ;
âsinâ¡Î¸cosâ¡Î¸=Ï2l sinâ¡Î¸gâcosâ¡Î¸=gÏ2l
âcosâ¡Î¸=1016Ã1=58

â Î¸=cos-158

Topic: Physics, Laws of Motion, JEE Main
A string of length $2 \mathrm{~m}$ is fixed at both ends. If this string vibrates in its fourth normal mode with a frequency of $500 \mathrm{~Hz}$ then the waves would travel on it's with a velocity of

Options:
A $125 \mathrm{~m} / \mathrm{s}$
B $250 \mathrm{~m} / \mathrm{s}$
C $500 \mathrm{~m} / \mathrm{s}$
D $1000 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$500 \mathrm{~m} / \mathrm{s}$

Solution:
$\begin{aligned} & \qquad \lambda=\frac{2 l}{p} \\ & \text { For string } \\ & \text { where } p=\text { No. of loops = Order of vibration } \\ & \text { Hence for forth mode } p=4 \Rightarrow \lambda=\frac{1}{2} \\ & \text { Hence } v=n \lambda \\ & \qquad 500 \times \frac{2}{2}=500 \mathrm{~Hz}\end{aligned}$

Topic: Physics, Waves and Sound, JEE Main
A string of length ' $L$ ' fixed at one end carries a mass ' $m$ ' at the other end. The string makes $\frac{3}{\pi}$ r.p.s. around the vertical axis through fixed end. The tension in the string is

Options:
A 72 mL
B 18 mL
C 9 mL
D 36 mL

The correct answer is:
36 mL

Solution:
Consider the free body diagram
On considering horizontal force balance:
$\mathrm{T} \sin \theta=\mathrm{m} \omega^2 \mathrm{R}$
Using the geometry, $\mathrm{R}=\mathrm{L} \sin \theta$
$\therefore \mathrm{T}=\left(\mathrm{m} \omega^2 \mathrm{~L}\right)$
Given, $\omega=\left(\frac{3}{\pi}\right)$ r.p.s. $=\left(\frac{3}{\pi}\right)(2 \pi) \mathrm{rad} / \mathrm{s}=6 \mathrm{rad} / \mathrm{s}$
$\therefore \mathrm{T}=36 \mathrm{~mL}$

Topic: Physics, Work Power Energy, JEE Main
A string of mass per unit length Î¼ is clamped at both ends such that one end of the string is at x = 0 and the other is at x = L. When the string vibrates in fundamental mode, the amplitude of the mid-point O of the string is a, and tension in the string is T, Find the total oscillation energy stored in the string.

Options:
A Ï 2 a 2 T 4 L
B π a T 2 L
C - π 2 a 2 T 3 L
D π 2 a T 6 L

The correct answer is:
π 2 a 2 T 4 L

Solution:
Î»2=L
Î»=2L
Ï=Î¼V2
The amplitude at a distance x from the origin is given by
A = asinkx
consider an element of mass dm and length dx of string at a distance x from the origin
The total energy of this element = its maximum KE
= 1 2 d m Ï 2 A 2
= 1 2 Î¼ d xâ¡ 4 Ï 2 f 2 a 2 sin 2 kâ¡ x
(Because Î¼= dm dx and Ï=2Ïf and it is given that A = a sinkx )
total energy of string = â« 0 L 2 Ï 2 Î¼ f 2 a 2 sin 2 kâ¡ x d x
= Ï 2 Î¼ f 2 a 2 [ x - sin 2 kâ¡ x 2 ] 0 L
= Ï 2 Î¼ f 2 a 2 L
= Ï 2 Î¼ V 2 Î» 2 a 2 L = Ï 2 4 L 2 T a 2 L = Ï 2 a 2 T 4 L

Topic: Physics, Waves and Sound, JEE Main
A string vibrates with a frequency of $200 \mathrm{~Hz}$. When its length is doubled and tension is altered, it begins to vibrate with a frequency of $300 \mathrm{~Hz}$. The ratio of the new tension to the original tension is

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

The correct answer is:
9:1

Solution:
Given, $L_{2}=2 L_{1}, f_{1}=200 \mathrm{~Hz}, f_{2}=300 \mathrm{~Hz}$
We know that, frequency of vibrating string is given as
$$
f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}
$$
where, $\mu=$ mass per unit length of string
and $L=$ length of the string.
$$
\begin{array}{ll}
\Rightarrow & f \propto \frac{\sqrt{T}}{L} \\
\Rightarrow \quad & \frac{f_{2}}{f_{1}}=\sqrt{\frac{T_{2}}{T_{1}}}, \frac{L_{1}}{L_{2}} \\
\Rightarrow \quad \sqrt{\frac{T_{2}}{T_{1}}}=\frac{f_{2} L_{2}}{f_{1} L_{1}}=\frac{300}{200} \times \frac{2 L_{1}}{L_{1}} \\
\Rightarrow \quad & \sqrt{\frac{T_{2}}{T_{1}}}=3 \\
& \frac{T_{2}}{T_{1}}=9 \\
\therefore \quad & T_{2}: T_{1}=9: 1
\end{array}
$$

Topic: Physics, Waves and Sound, JEE Main
A strip of copper and another germanium are cooled from room temperature to $80 \mathrm{~K}$. The resistance of

Options:
A each of these decreases
B copper strip increases and that of germanium decreases
C copper strip decreases and that of germanium increases
D each of these increases.

The correct answer is:
copper strip decreases and that of germanium increases

Solution:
Copper is conductor and germanium is a semiconductor. When cooled, the resistance of copper strip decreases and that of germanium increases.

Topic: Physics, Semiconductors, JEE Main
A strip of copper and another of germanium are cooled from room temperature to $80 \mathrm{~K}$. The resistance of

Options:
A
each of these decreases

B
copper strip increases and that of germanium decreases

C
copper strip decreases and that of germanium increases

D
each of these increases

The correct answer is:
copper strip decreases and that of germanium increases

Solution:
No Solution Available

Topic: Physics, Current Electricity, JEE Main, JEE Main 2003
A strong base an abstract an $\alpha$-hydrogen from

Options:
A Alkane
B Alkene
C Amine
D Ketone

The correct answer is:
Ketone

Solution:
No Solution Available

Topic: Chemistry, Aldehydes and Ketones, NEET, NEET 2008 (Mains)
A strong magnetic field is applied on a stationary electron. Then, the electron

Options:
A moves in the direction of the field
B moves in an opposite direction of the field
C remains stationary
D starts spinning

The correct answer is:
remains stationary

Solution:
Force on an electron in an uniform magnetic field is given as
$F=q v B$
Since, the particle is stationary, i.e., $v=0$
$\Rightarrow \quad F=0$
So, the electron remains stationary.

Topic: Physics, Magnetic Effects of Current, KCET, KCET 2021
A structural steel rod has a radius of $10 \mathrm{~mm}$ and length of $1.0 \mathrm{~m}$. A $100 \mathrm{kN}$ force stretches it along its length. Young's modulus of structural steel is $2 \times 10^{11} \mathrm{Nm}^{-2}$. The percentage strain is about

Options:
A
$0.16 \%$

B
$0.32 \%$

C
$0.08 \%$

D
$0.24 \%$

The correct answer is:
$0.16 \%$

Solution:
Given: $F=100 \mathrm{kN}=10^5 \mathrm{~N}$
$$
\begin{aligned}
& \mathrm{Y}=2 \times 10^{11} \mathrm{Nm}^{-2} \\
& \ell_0=1.0 \mathrm{~m}
\end{aligned}
$$
radius $\mathrm{r}=10 \mathrm{~mm}=10^{-2} \mathrm{~m}$
From formula, $Y=\frac{\text { Stress }}{\text { Strain }}$
$$
\Rightarrow \quad \text { Strain }=\frac{\text { Stress }}{Y}=\frac{F}{A Y}
$$
$$
\begin{aligned}
& =\frac{10^5}{\pi r^2 Y}=\frac{10^5}{3.14 \times 10^{-4} \times 2 \times 10^{11}} \\
& =\frac{1}{628}
\end{aligned}
$$
Therefore $\%$ strain $=\frac{1}{628} \times 100=0.16 \%$

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2012 (07 May Online)
A structure that is traversing the middle lamella and connecting the cytoplasm of neighbouring cells is called

Options:
A Primary wall junction
B Plasmodesmata
C Desmosomes
D Secondary wall

The correct answer is:
Plasmodesmata

Solution:
The middle lamella is the layer mainly of calcium pectate which holds or glues the different neighbouring cells together. The cell wall and Middle lamellae may be traversed by cytoplasmic channel called plasmodesmata which connects the cytoplasm of neighbouring cells.

Topic: Biology, Cell - The Unit of Life, NEET
A student appeared in an examination consisting of 8 true-false type questions. The student guesses the answers with equal probability. The smallest value of n, so that the probability of guessing at least n correct answers is less than 12, is :

Options:
A 5
B 6
C 3
D 4

The correct answer is:
5

Solution:
PE<12

⇒∑r=n8Cr8128-r12r<12

⇒∑r=n8Cr8128<12

⇒Cn8+Cn+18+…+C88<128

⇒256-C08+C18+…+Cn-18<128

⇒C08+C18+…+Cn-18>128

⇒1+8+28+56+70+C58+…+Cn-18>128

⇒163+C58+…+Cn-18>128

⇒n-1≥4

⇒n≥5

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (27 Jul Shift 2)
A student has studied the decomposition of a gas AB3 at 25°C. He obtained the following data

p (mm Hg)	50	100	200	400

Relative t12s	4	2	1	0.5

The order of the reaction is

Options:
A 0.5
B 2
C 1
D 0 (zero)

The correct answer is:
2

Solution:
By observing the data given, by doubling the pressure, half-life is reduced to half. This result is following second order kinetics. Mathematically it can be proved as follows,

The half-life and initial pressure of nth order reaction can be related as follows,

t12∝Po1-n

t121t122=P011-nP021-n

⇒42=501001-n

⇒2=121-n

⇒2=2n-1

⇒n-1=1

⇒n=2

Hence, the order of the reaction is second order.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (24 Jan Shift 2)
A student has to answer 10 out of 13 questions in an examination. The number of ways in which he can answer, if he must answer atleast three of the first five questions, is

Options:
A 276
B 600
C 840
D 640

The correct answer is:
276

Solution:
The possible cases are:
Required number of ways =C53×C87+C54×C86+C55×C85
=5×42×8+5×8×72+1×8×7×63×2
=80+140+56=276

Topic: Mathematics, Permutation Combination, JEE Main
A student in a town in India, where the price per unit (1 unit =1kWh ) of electricity is ₹ 5.00, purchases a 1kVA UPS (uninterrupted power supply) battery. A day before the exam, 10 friends arrive to the student's home with their laptops and all connect their laptops to the UPS. Assume that each laptop has a constant power requirement of 90 W. Consider the following statements.
I. All the 10 laptops can be powered by the UPS, if connected directly.
II. All the 10 laptops can be powered, if connected using an extension box with a 3 A fuse.
III. If all the 10 friends use the laptop for 5 h, then the cost of the consumed electricity is about ₹ 22.50 .
Select the correct option with the true statements.

Options:
A I only

B I and II only

C I and III only

D II and III only .

The correct answer is:
I and III only

Solution:
Power requirement for 1 laptop, P1=90 W
So, power r quirement for 10 laptops,
P=10×P1=10×90-900 W=0.9 kW
In 5 h, electrical energy used by all laptops,

E=P×t=0.9×5

=4.5kWh

cost of electrical energy used is
cost =E× Unit cost

=4.5×5
=₹ 22.50

So, statement III is correct.
For laptop charger, input voltage is 220 V
So, current when all 10 laptops are connected through an extension,

I=PV=900220≈4.1 A

As, line current exceeds current rating of fuse, therefore 3 A fuse cannot be used. So, statement II is incorrect.

Topic: Physics, Current Electricity, JEE Main
A student in the laboratory measures thickness of a wire using screw gauge. The readings are 1.22 mm,1.23 mm,1.19 mm and 1.20 mm. The percentage error is x121%. The value of x is _____ .

Options:

The correct answer is:
150

Solution:
The average value of the readings is xavg=1.19+1.20+1.22+1.234=1.21

Now, absolute error in measurement Δxavg=0.02+0.01+0.01+0.024=0.064=0.015

Percentage error in measurement is =0.0151.21×100%=150121%

So, the value of x=150.

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2022 (28 Jun Shift 2)