datavorous/entrance-exam-dataset huggingface dataset:
The least value of the quadratic polynomial, f x = 2 p 2 + 1 x 2 + 2 4 p 2 - 1 x + 4 2 p 2 + 1 for real values of p and x is

Options:

The correct answer is:
3

Solution:
y = 2 p 2 x 2 + 4x + 4 + x 2 - 2 x + 1 + 3
y = 2 p 2 x + 2 2 + x - 1 2 + 3
p = 0 , x = 1
y min = 3

Topic: Mathematics, Circle, JEE Main
The least value of z where z is complex number which satisfies the inequality ez+3z-1z+1loge2≥log257+9i, i=-1, is equal to :

Options:
A 3
B 5
C 2
D 8

The correct answer is:
3

Solution:
ez+3z-1z+1loge2≥log257+9i

⇒2z+3z-1z+1≥log216

⇒2z+3z-1z+1≥23

z+3z-1z+1≥3

z+3z-1≥3z+1

z2+2z-3≥3z+3

⇒z2+z-6≥0

⇒z-3z+2≥0⇒z-3≥0

⇒z≥3 ⇒zmin=3

Topic: Mathematics, Complex Number, JEE Main
The least value of z where z is complex number which satisfies the inequality ez+3z-1z+1loge2≥log257+9i, i=-1, is equal to :

Options:
A 3
B 5
C 2
D 8

The correct answer is:
3

Solution:
ez+3z-1z+1loge2≥log257+9i

⇒2z+3z-1z+1≥log216

⇒2z+3z-1z+1≥23

z+3z-1z+1≥3

z+3z-1≥3z+1

z2+2z-3≥3z+3

⇒z2+z-6≥0

⇒z-3z+2≥0⇒z-3≥0

⇒z≥3 ⇒zmin=3

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (16 Mar Shift 2)
The left-hand derivative of $\mathrm{f}(x)=[x] \sin (\pi x)$, at $x=\mathrm{k}, \mathrm{k}$ is an integer and $[\cdot]$ is the greatest integer function, is

Options:
A $(-1)^{\mathrm{k}}(\mathrm{k}-1) \pi$
B $(-1)^{\mathrm{k}-1}(\mathrm{k}-1) \pi$
C $(-1)^{\mathrm{k}} \mathrm{k} \pi$
D $(-1)^{\mathrm{k}-1} \mathrm{k} \pi$

The correct answer is:
$(-1)^{\mathrm{k}}(\mathrm{k}-1) \pi$

Solution:
$\begin{aligned} f(x)= & {[x] \sin (\pi x) } \\ \text { LHD } & =\lim _{h \rightarrow 0} \frac{f(k-h)-f(k)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{[k-h] \sin \pi(k-h)-[k] \sin k \pi}{-h} \\ & =\lim _{h \rightarrow 0} \frac{(k-1) \sin (k \pi-\pi h)-k \sin k \pi}{-h} \\ & =\lim _{h \rightarrow 0} \frac{(-1)^{k+1}(k-1) \sinh \pi-0}{-h} \quad \ldots[\because k \in I] \\ & =(-1)^k(k-1) \pi\end{aligned}$

Topic: Mathematics, Limits, JEE Main
The length and breadth of a metal sheet are 3.124 m and 3.002 m respectively. The area of this sheet up to the correct number of significant figures is

Options:
A 9.378 m2
B 9.37 m2
C 9.378248 m2
D 9.3782 m2

The correct answer is:
9.378 m2

Solution:
Area of metal sheet =3.124Ã3.002ââm2

=9.378248ââm2

Now, the result must have significant figures equal to the least of figures being multiplied, so, area of metal sheet =9.378ââm2 (four significant digits)

Topic: Physics, Units and Dimensions, JEE Main
The length and width of a rectangular room are measured to be $3.95 \pm 0.05 \mathrm{~m}$ and $3.05 \pm 0.05 \mathrm{~m}$, respectively, the area of the floor is

Options:
A $12.05 \pm 0.01 \mathrm{~m}^{2}$
B $12.05 \pm 0.005 \mathrm{~m}^{2}$
C $12.05 \pm 0.34 \mathrm{~m}^{2}$
D $12.05 \pm 0.40 \mathrm{~m}^{2}$

The correct answer is:
$12.05 \pm 0.34 \mathrm{~m}^{2}$

Solution:
$\mathrm{A}=\ell \mathrm{B}$
$\mathrm{d} \mathrm{A}=\ell \mathrm{dB}+\mathrm{B} \mathrm{d} \ell$
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{A}}=\frac{\ell \mathrm{dB}}{\ell \mathrm{B}}+\frac{\mathrm{Bd} \ell}{\ell \mathrm{B}}$
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{A}}=\frac{\mathrm{dB}}{\mathrm{B}}+\frac{\mathrm{d} \ell}{\ell}$
$\quad=\frac{0.05}{3.05}+\frac{0.05}{3.95}$
$\quad=0.016+0.012$
$\quad=0.028 \times 12.05$
$\mathrm{dA}=0.33$
$12.05 \pm 0.34$ Ans. (C)

Topic: Physics, Mathematics in Physics, JEE Main
The length of a dipole antenna for a carrier wave of frequency 6Ã 10 8 âHz is ( take speed of EM wave 3Ã108 m s-1 )

Options:
A 0.25ââm
B 1 m
C 2 m
D 0.1 m

The correct answer is:
0.25  m

Solution:
Length of the antenna
=λ2=c2υ

( Where c = speed of light = 3×108 m s-1, υ = frequency = 6×108 Hz )

putting these values in formula above

Length = 3×1082×6×108=0.25 m

Topic: Physics, Communication System, JEE Main
The length of a given cylindrical wire is increased by 100%. Due to the consequent decrease in diameter, the change in the resistance of the wire will be

Options:
A 200%
B 100%
C 50%
D 300%

The correct answer is:
300%

Solution:
Let the length of the wire be l, radius of the wire be r
∴ Resistance Râ¡ = ρ lâ¡ π râ¡ 2 ρ = resistivity of the wire
Now lâ¡ is increased by 100%
∴ lâ¡ 'â¡ = lâ¡ + 1 0 0 1 0 0 lâ¡ = 2 lâ¡
As length is increased, its radius is going to be decreased in such a way that the volume of the cylinder remains constant.
π râ¡ 2 × lâ¡ = π râ¡ 'â¡ 2 × lâ¡ 'â¡ ⇒ râ¡ 'â¡ 2 = râ¡ 2 × lâ¡ lâ¡ 'â¡ = râ¡ 2 × lâ¡ 2 lâ¡ = râ¡ 2 2
∴ The new resistance Râ¡ 'â¡ 2 = ρ lâ¡ 'â¡ π râ¡ 'â¡ 2 = ρ 2 lâ¡ π × râ¡ 2 2 = 4 Râ¡
∴ Change in resistance = Râ¡ 'â¡ - Râ¡ = 3 Râ¡
∴ % change = 3 Râ¡ Râ¡ × 1 0 0 %â¡ = 3 0 0 %.â¡

Topic: Physics, Current Electricity, NEET
The length of a given cylindrical wire is increased by $100 \%$. Due to the consequent decrease in diameter the change in the resistance of the wire will be

Options:
A
$200 \%$

B
$100 \%$

C
$50 \%$

D
$300 \%$

The correct answer is:
$300 \%$

Solution:
$\mathrm{L}_1=2 \mathrm{l}$ or $\left(\pi \mathrm{r}^2 \mathrm{l}\right)=\left(\pi \mathrm{r}_2^2\right)(2 \mathrm{l})$
$\Rightarrow \mathrm{r}_2=\frac{\mathrm{r}}{\sqrt{2}}$$;$ $\mathrm{R}=\rho \frac{1}{\pi^2}$
$\mathrm{R}_{\text {new }}=(\rho) \frac{21}{(\pi)(\mathrm{r} / \sqrt{2})^2}=\frac{(\rho) 4 \mathrm{l}}{(\pi) \mathrm{r}^2}=4 \times \mathrm{R}$
$\therefore \Delta \mathrm{R}=4 \mathrm{R}-\mathrm{R}=3 \mathrm{R}$
$\frac{\Delta \mathrm{R}}{\mathrm{R}} \%=\frac{3 \mathrm{R}}{\mathrm{R}} \times 100=300 \%$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2003
The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be _____ %

Options:

The correct answer is:
300

Solution:
As volume is constant, so, length doubled and area reduces to half.

Resistance of a wire is given by R=ρlA, where, ρ is resistivity of a material.

Here, R1=ρlA and R2=ρ2lA2=4R

So percentage increase in the resistance will be =4R-RR×100%=300%

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (25 Jun Shift 2)
The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is 2 s. The magnent is cut into three equal parts of length L 3 each,and three parts are then placed on each other with their like poles together. The time period of this combination will be

Options:
A 2 s
B 23 s
C 23 s
D 2/3 s

The correct answer is:
23 s

Solution:
The time period of oscillations of magnet
T=2ÏIMBH â¦i
Where I= moment of inertia of magnet
=mL212 (m, being the mass of magnet)
Now magnet is cut into three equal parts.
When the three equal parts of magnet are placed on one another with their like poles together, then
Iâ²=112m3ÃL32Ã3
=112mL29=I9
and Mâ²= pole strength ÃL3Ã3=M
Hence,
â Tâ²=13ÃT
Tâ²=23 s

Topic: Physics, Experimental Physics, JEE Main
The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2 \mathrm{~s}$. The magnet is cut along its length into three equal parts and three parts are then placed on each other with their like poles together. The time period of this combination will be

Options:
A $2 \mathrm{~s}$
B $2 / 3 \mathrm{~s}$
C $2 \sqrt{3} \mathrm{~s}$
D $2 / \sqrt{3} \mathrm{~s}$

The correct answer is:
$2 / 3 \mathrm{~s}$

Solution:
The time period of oscillations of magnet
$$
T=2 \pi \sqrt{\left(\frac{I}{M H}\right)}
$$
where $I=$ moment of inertia of magnet $=\frac{m L^{2}}{12}(m$, being the mass of magnet $)$ $M=$ pole strength $\times L$ and $H=$ horizontal component of earth's magnetic field. When the three equal parts of magnet are placed on one another with their like poles together, then
$$
\begin{aligned}
I^{\prime} &=\frac{1}{12}\left(\frac{m}{3}\right) \times\left(\frac{L}{3}\right)^{2} \times 3 \\
&=\frac{1}{12} \frac{m L^{2}}{9}=\frac{I}{9}
\end{aligned}
$$
and $\quad M^{\prime}=$ pole strength $\times \frac{L}{3} \times 3=M$
Hence, $T^{\prime}=2 \pi \sqrt{\left(\frac{I / 9}{M H}\right)}$
$$
\begin{aligned}
\Rightarrow \quad T^{\prime} &=\frac{1}{3} \times T \\
T^{\prime} &=\frac{2}{3} \mathrm{~s}
\end{aligned}
$$

Topic: Physics, Magnetic Properties of Matter, JEE Main
The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is 150 ms. The magnet is cut along its length into three equal parts and three parts are then placed on each other with their like poles together. The time period of this combination (in ms) will be

Options:

The correct answer is:
50

Solution:
The time period of oscillations of magnet
T=2πIMBH …i
Where I= moment of inertia of magnet
=mL212 (m, being the mass of magnet)
When the three equal parts of magnet are placed on one another with their like poles together, then
I′=112m3×L32×3
=112mL29=I9
and M′= pole strength ×L3×3=M
Hence,
⇒ T′=13×T
⇒ T′=1503=50 ms

Topic: Physics, Experimental Physics, JEE Main
The length of a magnet is large compared to its width and breadth. The time period of its width and breadth. The time period of its oscillation in a vibration magnetometer is $2 \mathrm{~s}$. The magnet is cut along its length into three equal parts and three parts are then placed on each other with their like poles together. The time period of this combination will be

Options:
A
$2 \mathrm{~s}$

B
$2 / 3 \mathrm{~s}$

C
$2 \sqrt{3} \mathrm{~s}$

D
$2 / \sqrt{ } 3 \mathrm{~s}$

The correct answer is:
$2 / 3 \mathrm{~s}$

Solution:
Time period of vibration, $T=2 \pi \sqrt{\frac{1}{M B}}$
Where $\ell=$ moment of inertia of magnet, $\mathrm{M}=$ magnetic moment $\mathrm{I}=\frac{\mathrm{m} \ell^2}{12}$ and $\mathrm{M}=$ pole strength $\times \ell$
$$
I^{\prime}=\frac{1}{12}\left(\frac{m}{3}\right)\left(\frac{\ell}{3}\right)^2 \times 3=\frac{\mathrm{l}}{9}
$$
and $M^{\prime}=$ pole strength (will remain the same) $\times(\ell / 3) \times 3=M$.
$$
\mathrm{T}^{\prime}=\frac{\mathrm{T}}{\sqrt{9}}=\frac{2}{9} \mathrm{~s} .
$$

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2004
The length of a metal wire is ℓ1, when the tension in it is T1 and is ℓ2 when the tension is T2. The natural length of the wire is:

Options:
A ℓ1ℓ2
B ℓ1 T2-ℓ2 T1 T2-T1
C ℓ1 T2+ℓ2 T1 T2+T1
D ℓ1+ℓ22

The correct answer is:
ℓ1 T2-ℓ2 T1 T2-T1

Solution:
T1=kℓ1-ℓ0

T2=kℓ2-ℓ0

T1 T2=ℓ1-ℓ0ℓ2-ℓ0

T1ℓ2-T1ℓ0= T2ℓ1-T2ℓ0

T2ℓ1-T1ℓ2 T2-T2=ℓ0

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2021 (20 Jul Shift 2)
The length of a metal wire is l1 when the tension in it is T1 and is l2 when the tension is T2. The natural length of wire is

Options:
A l1+l22
B l1l2
C l1T2-l2T1T2-T1
D l1T2+l2T1T2+T1

The correct answer is:
l1T2-l2T1T2-T1

Solution:
Let the original length be l.
In first case,
l1=l+dl1
Using Hooke's law, x=TK
Where, x is the displacement.
We have, l1=l+T1K ... (i)
Similarly, l2=l+T2K ... (ii)
Solving Equations. (i) and (ii), we get
l=l1T2-l2T1T2-T1

Topic: Physics, Mechanical Properties of Solids, JEE Main
The length of a metal wire is $L_{1}$ when the tension is $T_{1}$ and $L_{2}$ when the tension is $T_{2}$ The unstretched length of wire is

Options:
A $\frac{L_{1}+L_{2}}{2}$
B $\sqrt{L_{1} L_{2}}$
C $\frac{T_{2} L_{1}-T_{1} L_{2}}{T_{2}-T_{1}}$
D $\frac{T_{2} L_{1}+T_{1} L_{2}}{T_{2}+T_{1}}$

The correct answer is:
$\frac{T_{2} L_{1}-T_{1} L_{2}}{T_{2}-T_{1}}$

Solution:
Let the initial length of the metal wire is $L$.
The strain at tension $T_{1}$ is $\Delta L_{1}=L_{1}-L$
The strain at tension $T_{2}$ is $\Delta L_{2}=L_{2}-L$
Suppose, the Young's modulus of the wire is $Y$, then
$\frac{\frac{T_{1}}{A}}{\frac{\Delta L_{1}}{L}}=\rac{\frac{T_{2}}{A}}{\frac{\Delta L}{L}}$
where, $A$ is an cross-section of the wire assumed to be same at all the situations. $\begin{aligned} \Rightarrow \frac{T_{1}}{A} \times \frac{L}{\Delta L_{1}} &=\frac{T_{2}}{A} \times \frac{L}{\Delta L_{2}} \\ \Rightarrow \quad \frac{T_{1}}{\left(L_{1}-L\right)} &=\frac{T_{2}}{\left(L_{2}-L\right)} \\ & T_{1}\left(L_{2}-L\right)=T_{2}\left(L_{1}-L\right) \cdot L=\frac{T_{2} L_{1}-T_{1} L_{2}}{T_{2}-T_{1}} \end{aligned}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The length of a metallic rod is $5 \mathrm{~m}$ at $0^{\circ} \mathrm{C}$ and becomes $5.01 \mathrm{~m}$, on heating upto $100^{\circ} \mathrm{C}$. The linear expansion of the metal will be

Options:
A $2.33 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
B $6.0 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
C $4.0 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
D $2.0 \times 10^{-5} /{ }^{\circ} \mathrm{C}$

The correct answer is:
$2.0 \times 10^{-5} /{ }^{\circ} \mathrm{C}$

Solution:
$\alpha=\frac{\Delta L}{L_0 \times \Delta \theta}=\frac{0.01}{5 \times 100}=2 \times 10^{-5} /{ }^{\circ} \mathrm{C}$

Topic: Physics, Thermal Properties of Matter, JEE Main
The length of a potentiometer wire is \( 400 \mathrm{~cm} \). Leclanche cell gets balanced against \( 150 \mathrm{~cm} \) length of the wire. If the length of the wire is altered to \( 600 \mathrm{~cm} \), find the balancing length.

Options:
A \( 225 \mathrm{~cm} \)
B \( 300 \mathrm{~cm} \)
C \( 250 \mathrm{~cm} \)
D \( 200 \mathrm{~cm} \)

The correct answer is:
\( 225 \mathrm{~cm} \)

Solution:
Since the resistance remains the same in both cases

xℓ=x'ℓ'

E400150=E400ℓ'

or ℓ'=225 cm

Topic: Physics, Current Electricity, JEE Main
The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA For a cell of emf 5 V and internal resistance of 20 Ω the null point on it is found to be at 1000 cm The resistance of whole wire is:

Options:
A 80Ω
B 120Ω
C 60Ω
D 100Ω

The correct answer is:
100Ω

Solution:
Potential gradient =51000=VP1200
VP=6 V
And RP=VPI=660×10-3=100 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The length of a rectangular plate is measured by a meter scale and is found to be \( 10.0 \mathrm{~cm} \). Its width is measured by vernier calipers as \( 1.00 \mathrm{~cm} \). The least count of the meter scale and vernier calipers are \( 0.1 \mathrm{~cm} \) and \( 0.01 \mathrm{~cm} \) respectively. In the previous question, what is the minimum possible error in area measurement?

Options:
A \( 02 \mathrm{~cm}^{2} \)
B \( 01 \mathrm{~cm}^{2} \)
C \( 03 \mathrm{~cm}^{2} \)
D 0

The correct answer is:
0

Solution:
minimum possible error is always zero

Topic: Physics, Mathematics in Physics, JEE Main
The length of a seconds pendulum at a height h=2R from earth surface will be:

(Given: R= Radius of earth and acceleration due to gravity at the surface of earth g=π2 m s-2)

Options:
A 29m
B 49m
C 89m
D 19m

The correct answer is:
19m

Solution:
The acceleration due to gravity at height 2R from the surface of the Earth will be,

g'=gR2R+h2⇒g'=gR2R+2R2⇒g'=g9

Length of the second's pendulum at the surface of the Earth is 1 m.

Since, time period of a pendulum is given by,

T=2πlg⇒T'T=l'l×gg'⇒1=l'l×gg'⇒l'=l9=19 m

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (25 Jul Shift 2)
The length of a second's pendulum on the surface of earth is $1 \mathrm{~m}$. What will be the length of a second's pendulum on the moon?

Options:

None

Solution:
A simple pendulum having time period $T=2 \mathrm{~s}$ is called second pendulum.
For a simple pendulum , $T=2 \pi \sqrt{\frac{l}{g}}$
where, $l=$ length of the pendulum and $g=$ acceleration due to gravity on surface of the earth.
$$
\begin{aligned}
&T_e=2 \pi \sqrt{\frac{l_e}{g_e}} \\
&T_e^2=4 \pi^2\left(\frac{l_e}{g_e}\right)
\end{aligned}
$$
On the surface of the moon, the time period
$$
\begin{aligned}
&\left(T_m\right)=2 \pi \sqrt{\frac{l_m}{g_m}} \Rightarrow T_m^2=4 \pi^2\left(\frac{l_m}{g_m}\right) \\
&\left(\frac{T_e}{T_m}\right)^2=\frac{4 \pi^2}{4 \pi^2}\left(\frac{l_e}{g_e}\right) \times\left(\frac{g_m}{l_m}\right)
\end{aligned}
$$
Dividing eq. (i) from (ii),
$$
\left(\frac{T_e}{T_m}\right)=\sqrt{\frac{l_e}{l_m} \times \frac{g_m}{g_e}}
$$
$T_e=T_m=2$ sec to maintain the second's pendulum time period.
$$
1=\sqrt{\frac{l_e}{l_m} \times \frac{g_m}{g_e}}
$$
But the acceleration due to gravity at moon is $1 / 6$ of the acceleration due to gravity at earth,
i.e, $\left(\because g_m=\frac{g_c}{6}\right)$
Squaring Eq. (iii) and putting this value,
$$
\begin{aligned}
&1=\frac{l_e}{l_m} \times \frac{g_e / 6}{g_e}=\frac{l_e}{l_m} \times \frac{1}{6} \\
&\frac{l_e}{6 l_m}=1 \Rightarrow 6 l_m=l_e \\
&l_m=\frac{1}{6} l_e=\frac{1}{6} \times 1=\frac{1}{6} m \\
&l_m=\frac{1}{6} m
\end{aligned}
$$
On the surface of the moon, the time period
$$
\begin{aligned}
&\left(T_m\right)=2 \pi \sqrt{\frac{l_m}{g_m}} \Rightarrow T_m^2=4 \pi^2\left(\frac{l_m}{g_m}\right) \\
&\left(\frac{T_e}{T_m}\right)^2=\frac{4 \pi^2}{4 \pi^2}\left(\frac{l_e}{g_e}\right) \times\left(\frac{g_m}{l_m}\right)
\end{aligned}
$$
Dividing eq. (i) from (ii),
$$
\left(\frac{T_e}{T_m}\right)=\sqrt{\frac{l_e}{l_m} \times \frac{g_m}{g_e}}
$$
$T_e=T_m=2 \mathrm{sec}$ to maintain the second's pendulum time period.
$$
1=\sqrt{\frac{l_e}{l_m} \times \frac{g_m}{g_e}}
$$
But the acceleration due to gravity at moon is $1 / 6$ of the acceleration due to gravity at earth,
i.e, $\left(\because g_m=\frac{g_e}{6}\right)$
Squaring Eq. (iii) and putting this value,
$$
\begin{aligned}
&1=\frac{l_e}{l_m} \times \frac{g_e / 6}{g_e}=\frac{l_e}{l_m} \times \frac{1}{6} \\
&\frac{l_e}{6 l_m}=1 \Rightarrow 6 l_m=l_e \\
&l_m=\frac{1}{6} l_e=\frac{1}{6} \times 1=\frac{1}{6} m \\
&l_m=\frac{1}{6} m
\end{aligned}
$$

Topic: Physics, Oscillations
The length of a simple pendulum executing simple harmonic motion is increased by $21 \%$. The percentage increase in the time period of the pendulum of increased length is

Options:
A
$11 \%$

B
$21 \%$

C
$42 \%$

D
$10 \%$

The correct answer is:
$10 \%$

Solution:
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{l}}{\mathrm{g}}} ; \log \mathrm{T}=\log (2 \pi)+\frac{1}{2} \log \left(\frac{\mathrm{l}}{\mathrm{g}}\right) \Rightarrow \log \mathrm{T}=\log (2 \pi)+\frac{1}{2} \log (1)-\frac{1}{2} \log (\mathrm{g})$
Differentiating
$\frac{\Delta \mathrm{T}}{\mathrm{T}}=0+\frac{1}{2} \times \frac{\Delta \mathrm{l}}{\mathrm{l}}-0 \Rightarrow \frac{\Delta \mathrm{T}}{\mathrm{T}} \times 100=\frac{1}{2} \times \frac{\Delta \mathrm{l}}{\mathrm{l}} \times 100$ $=\frac{1}{2} \times 21=10.5 \approx 10 \%$
Note: In this method, the $\%$ error obtained is an approximate value on the higher side. Exact value is less than the obtained one.

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2003
The length of a simple pendulum executing simple harmonic motion is increased by $21 \%$. The percentage increase in the time period of the pendulum of increased length is

Options:
A
$11 \%$

B
$21 \%$

C
$42 \%$

D
$10 \%$

The correct answer is:
$10 \%$

Solution:
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{l}}{\mathrm{g}}} ; \log \mathrm{T}=\log (2 \pi)+\frac{1}{2} \log \left(\frac{\mathrm{l}}{\mathrm{g}}\right) \Rightarrow \log \mathrm{T}=\log (2 \pi)+\frac{1}{2} \log (1)-\frac{1}{2} \log (\mathrm{g})$
Differentiating
$\frac{\Delta \mathrm{T}}{\mathrm{T}}=0+\frac{1}{2} \times \frac{\Delta \mathrm{l}}{\mathrm{l}}-0 \Rightarrow \frac{\Delta \mathrm{T}}{\mathrm{T}} \times 100=\frac{1}{2} \times \frac{\Delta \mathrm{l}}{\mathrm{l}} \times 100$ $=\frac{1}{2} \times 21=10.5 \approx 10 \%$
Note: In this method, the $\%$ error obtained is an approximate value on the higher side. Exact value is less than the obtained one.

Topic: Physics, Mathematics in Physics, JEE Main
The length of a simple pendulum executing simple harmonic motion is increased by 21%. The percentage increase in the time period of the pendulum of increased length is

Options:
A 11%
B 21%
C 42%
D 10%

The correct answer is:
10%

Solution:
Let the lengths of the pendulum be (100l) and (121l)

Since we know the time period of a simple pendulum is given by T=2πlg
∴ T'T=121100=1110
∴ Fractional change = T'-TT=11-1010=110
∴ Percentage change = 10%

Topic: Physics, Oscillations, JEE Main
The length of a sonometer wire tuned to a frequency of 250 Hz is 0.60 m. The frequency of tuning fork with which the vibrating wire will be in tune when the length is made 0.40 m is

Options:
A 250 Hz
B 375 Hz
C 56 Hz
D 384 Hz

The correct answer is:
375 Hz

Solution:
n1L1=n2L2⇒250×0.6=n2×0.4⇒n2=375
⇒ n2=375 Hz

Topic: Physics, Waves and Sound, JEE Main
The length of a telescope, for which it is known that it has an objective lens of focal length 144 cm and an eye piece of focal length 6 cm, is

Options:
A 1.5 m
B 1.44 m
C 1.38 m
D 1.56 m

The correct answer is:
1.5 m

Solution:
The length of a telescope is

l=f0+fe

Given,

f0=144 cm=1.44 m, fe=6.0 cm=0.06 m
l=1.44+0.06=1.5 m

Topic: Physics, Ray Optics, JEE Main
The length of a telescope is $36 \mathrm{~cm}$. The focal lengths of its lenses can be

Options:
A $30 \mathrm{~cm}, 6 \mathrm{~cm}$
B $-30 \mathrm{~cm},-6 \mathrm{~cm}$
C $30 \mathrm{~cm},-6 \mathrm{~cm}$
D $-30 \mathrm{~cm}, 6 \mathrm{~cm}$

The correct answer is:
$30 \mathrm{~cm}, 6 \mathrm{~cm}$

Solution:
Total length $L=f_o+f_e$ and both lenses are convex.

Topic: Physics, Ray Optics, JEE Main
The length of a wire becomes l1 and l2 when 100 N and 120 N tension are applied respectively. If 10l2=11l1, then the natural length of wire will be 1xl1. Here the value of x is

Options:

The correct answer is:
2

Solution:
Young's modulus is given by Y=FA∆ll

⇒FA=YΔll⇒Δl=FlAY

⇒l+Δl=l1⇒F1LAY+L=LF1AY+1=l1 ...(i)

Similarly,

l2=F2AY+1L

⇒10l2=11l1

It is given that 10l1=11l2

⇒10×LF2AY+1=11LF1AY+1

⇒10F2AY+10=11F1AY+11

Substituting F1=100 N, F2=120 N

⇒1200AY−1100AY=1⇒100AY=1

⇒1AY=1100

Using equation (i),

L=l1F1AY+1=l1100100+1=l12

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2023 (11 Apr Shift 1)
The length of a wire is $1.0 \mathrm{~m}$ and the area of cross-section is $1.0 \times 10^{-2} \mathrm{~cm}^2$. If the work done for increase in length by 0.2 $\mathrm{cm}$ is 0.4 joule, then Young's modulus of the material of the wire is

Options:
A $2.0 \times 10^{10} \mathrm{~N} / \mathrm{m}^2$
B $4 \times 10^{10} \mathrm{~N} / \mathrm{m}^2$
C $2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
D $2 \times 10^{10} \mathrm{~N} / \mathrm{m}^2$

The correct answer is:
$2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

Solution:
$W=\frac{1}{2} \frac{Y A l^2}{L}=0.4=\frac{1}{2} \times \frac{Y \times 10^{-6} \times\left(0.2 \times 10^{-2}\right)^2}{1}$
$\therefore \gamma=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The length of a wire of a potentiometer is $100 \mathrm{~cm}$, and the e.m.f. of its standard cell is $\mathrm{E}$ volt. It is employed to measure the e.m.f of a battery whose internal resistance is $0.5 \Omega$. If the balance point is obtained at $\mathrm{l}=30$ $\mathrm{cm}$ from the positive end, the e.m.f. of the battery is

Options:
A
$\frac{30 \mathrm{E}}{100.5}$

B
$\frac{30 \mathrm{E}}{(100-0.5)}$

C
$\frac{30(\mathrm{E}-0.5 \mathrm{i})}{100}$, where $\mathrm{i}$ is the current in the potentiometer wire

D
$\frac{30 \mathrm{E}}{100}$

The correct answer is:
$\frac{30 \mathrm{E}}{100}$

Solution:
Potential $\propto \mathrm{R}$
$\mathrm{R} \propto$ length $\Rightarrow$ Potential difference $\propto l$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2003
The length of an elastic string is a m when the tension is 4 N, and 'b m when the tension is 5 N. The length in metre when the tension is 9 N, is

Options:
A 4a-5b
B 5b-4a
C 9b-9a
D a+b

The correct answer is:
5b-4a

Solution:
T1=Kl1-l
T2=K(l2-l)
So, T 1 T 2 = l 1 −l l 2 −l
∴ T1l2-T1l=T2l1-T2l
T2-T1l=T2l1-T1l2
l= T2l1-T1l2(T2-T1)

l=5a-4b .......(i)

5-4=k(b-a)

k=1b-a ...(ii)
9=kl′ (l′= change in length)
9 =1b-a ×l'⇒l'= 9b-9a
Hence, final length

lfinal=l+l'

lfinal=5a-4b+9b-9a

lfinal=5b-4a

Topic: Physics, Mechanical Properties of Solids, JEE Main
The length of an object is measured using a vernier system whose main scale is 30 cm long with 600 divisions. If 19 divisions of the main scale coincide with 20 divisions of the vernier scale, then its least count is

Options:
A 0.25 cm
B 0.025Â cm
C 0.35 cm
D 0.025 mm

The correct answer is:
0.025 mm

Solution:
âµ Least count =MSD-VSD
âµ MSD=30100600=5Ã10-4 m
and VSD=1920ÃMSD
=1920Ã5Ã10-4=4.75Ã10-4
â´ Least count =5-4.75Ã10-4=0.025 mm

Topic: Physics, Experimental Physics, JEE Main