Jee Main 2024 31 01 2024 Shift 1
Question 1
The logical circuit shown below is equivalent to
(1) NAND
(2) NOR
(3) AND
(4) OR
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Answer: (4)
Solution:
Output, $Y=\overline{\bar{A} \cdot \bar{B}}=A+B$
Question 2
The block $M$ of mass $10 kg$ is having acceleration $2 m / s^{2}$ in the direction shown. Find mass $(m)$ of the other block.
(1) $2.5 kg$
(2) $7.5 kg$
(3) $12.5 kg$
(4) $5 kg$
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Answer: (3)
Solution:
$a=\frac{(M \sin 53-m \sin 37) g}{(M+m)}$
$m=\frac{15}{2} kg$
Question 3
If the percentage error in measuring length and diameter of a wire is $0.1 \%$ each, then the percentage error of the resistance of the wire is
(1) $0.3 \%$
(2) $0.2 \%$
(3) $0.1 \%$
(4) $0.4 \%$
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Answer: (1)
Solution:
$R=\frac{\rho l}{A}$
$R=\frac{4 \rho l}{\pi d^{2}}$
$ \begin{aligned} \frac{\Delta R}{R} & =\frac{\Delta l}{l}+2 \frac{\Delta d}{d} \\ & =0.3 \% \end{aligned} $
Question 4
4 identical particles of mass $m$ each are placed at 4 corners of a square. The gravitational force exerted on one of the mass by other masses is $\left[\frac{2 \sqrt{2}+1}{32}\right] \frac{G m^{2}}{l^{2}}$. The distance of side of square is :
(1) $2 /$
(2) $4 /$
(3) $\frac{1}{2}$
(4) $I$
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Answer: (2)
Solution:
$ \begin{aligned} & \frac{G m^{2}}{a^{2}} \\ \Rightarrow F _{\text {Net }} & =\frac{G m^{2}}{a^{2}}\left[\frac{1}{2}+\sqrt{2}\right] \\ & =\frac{2 \sqrt{2}+1}{2} \frac{G m^{2}}{a^{2}} \\ \Rightarrow 2 a^{2} & =32 l^{2} \\ \Rightarrow a & =4 l \end{aligned} $
Question 5
$\quad T-V$ graph is given for two different pressures $P _1 \&$ $P _2$. Then
(1) $P _2>P _1$
(2) $P _1=P _2$
(3) $P _2<P _1$
(4) $P _2 \leq P _1$
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Answer: (1)
Solution:
$T-V$ graph: Straight line
$\Rightarrow$ Isobaric
Also, slope $\propto P$
$\Rightarrow P _2>P _1$
Question 6
For a $1-D$ motion, relation between position $x$ and time $t$ is $t=\alpha x^{2}+\beta x$. Find the relation between velocity $v$ and acceleration $a$.
(1) $a=\alpha v$
(2) $a=-2 \alpha v$
(3) $a=-2 \alpha v^{3}$
(4) $a=2 \alpha v^{2}$
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Answer: (3)
Solution:
$t=\alpha x^{2}+\beta x$
$ \begin{aligned} & \Rightarrow \frac{d t}{d x}=2 \alpha x+\beta=\frac{1}{v} \\ & \Rightarrow \quad v=\frac{1}{2 \alpha x+\beta} \end{aligned} $
Also, $a=\frac{v d v}{d x}=\frac{-1}{2 \alpha x+\beta}\left[\frac{1}{2 \alpha x+\beta}\right]^{2} .2 \alpha$
$ \Rightarrow a=\frac{-2 \alpha}{(2 \alpha x+\beta)^{3}} $
$ \begin{aligned} & =-2 \alpha v^{3} \\ \Rightarrow \quad a & =-2 \alpha v^{3} \end{aligned} $
Question 7
Two resistances having coefficient of variation of resistivity $\alpha _1$ and $\alpha _2$ are having equal resistance. Equivalent temperature coefficient of resistivity in series and parallel carburation are.
(1) $\frac{\alpha _1+\alpha _2}{2}, \quad \alpha _1+\alpha _2$
(2) $\alpha _1+\alpha _2, \alpha _1+\alpha _2$
(3) $\alpha _1+\alpha _2, \frac{\alpha _1+\alpha _2}{2}$
(4) $\frac{\alpha _1+\alpha _2}{2}, \frac{\alpha _1+\alpha _2}{2}$
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Answer: (4)
Solution:
$R _1=R _0\left(1+\alpha _1 T\right)$
$R _2=R _0\left(1+\alpha _2 T\right)$
$\alpha _s=\frac{R _0\left(1+\alpha _1 T\right)+R _0\left(1+\alpha _2 T\right)-2 R _0}{2 R _0 T}$
$\alpha _s=\frac{\alpha _1+\alpha _2}{2}$
$ \begin{aligned} & \alpha _{|}=\frac{\frac{R _0\left(1+\alpha _1 T\right)+R _0\left(1+\alpha _2 T\right)}{R _0\left(1+\alpha _1 T\right)+R _0\left(1+\alpha _2 T\right)}-\frac{R _0}{2}}{\frac{R _0}{2} T} \\ & \alpha _{|}=\frac{\left(\alpha _1+\alpha _2\right)}{2} \frac{T}{T}=\frac{\alpha _1+\alpha _2}{2} \end{aligned} $
Question 8
An artillery of mass $M _1$, fires a shell of mass $M _2$. At the time of firing the ratio of kinetic energy is
(1) $\frac{M _2}{M _1}$
(2) $\frac{M _1+M _2}{M _1}$
(3) $\frac{M _1+M _2}{M _2}$
(4) $\frac{M _1}{M _1+M _2}$
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Answer: (1)
Solution:
By conservation of momentum
$P _1=-P _2$
$\frac{K _1}{K _2}=\frac{P _1^{2}}{2 M _1} \times \frac{2 M _2}{P _2^{2}}=\frac{M _2}{M _1}$
Question 9
The fundamental frequency of closed organ pipe is equal to the frequency of first overtone of open organ pipe of length $60 cm$. The length of closed organ pipe is
(1) $45 cm$
(2) $30 cm$
(3) $15 cm$
(4) $60 cm$
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Answer: (3)
Solution:
$\frac{V}{4 L _1}=2\left(\frac{V}{2 L _2}\right)$
$L _1=$ Length of closed organ pipe
$L _2=$ Length of open organ pipe
$L _2=4 L _1$
$L _1=\frac{L _2}{4}=15 cm$
Question 10
When a small spherical ball is dropped into a long cylindrical pipe filled with glycerine, then what will be the $v$ versus $t$ graph?
(1)
(2)
(3)
(4)
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Answer: (3)
Solution:
$F=m g-B-6 \pi \eta r v$
Where $B$ : Buoyancy
$ \begin{aligned} & \Rightarrow \quad m \frac{d v}{d t}=\underbrace{(m g-B)} _{\text {constant }}-\underbrace{(6 \pi \eta r) v} _{\text {constant }} \\ & \Rightarrow \quad v=v _0\left[1-e^{-c t}\right], c: \text { constant } \end{aligned} $
Question 11
Force $\mathbf{F}$ depends on distance $(x)$ and time $(t)$ as $F=a x^{2}+b t^{\frac{1}{2}}$, find dimension of $\frac{b^{2}}{a}$
(1) $M^{1} L^{2} T^{-3}$
(2) $M^{1} L^{-3} T^{3}$
(3) $M^{1} L^{3} T^{-3}$
(4) $M^{2} L^{2} T^{1}$
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Answer: (3)
Solution:
From dimensional analysis
$ \begin{aligned} & {[a]=\frac{[F]}{\left[L^{2}\right]}=\frac{M L T^{-2}}{L^{2}}=\left[M L^{-1} T^{-2}\right]} \\ & {[b]=\frac{[F]}{\left[T^{\frac{1}{2}}\right]}=\frac{M L T^{-2}}{T^{\frac{1}{2}}}=M L T^{\frac{-5}{2}}} \end{aligned} $
Then dimension of $\frac{b^{2}}{a}=\frac{M^{2} L^{2} T^{-5}}{M L^{-1} T^{-2}}$
$\left[\frac{b^{2}}{a}\right]=\left[M L^{3} T^{-3}\right]$
Question 12
Two charges $q \& 3 q$ are placed at a distance $r$ from each other. Find the distance from $q$ where electric field is zero.
(1) $\frac{r}{\sqrt{3}+1}$
(2) $\frac{r}{2}$
(3) $\frac{r}{\sqrt{3}-1}$
(4) $\frac{2 r}{3}$
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Answer: (1)
Solution:
$\frac{k q}{x^{2}}=\frac{k 3 q}{(r-x)^{2}}$
$x=\frac{r-x}{\sqrt{3}}$
$x=\frac{r}{(\sqrt{3}+1)}$
Question 13
The refractive index of thin prism of an apex angle $A$ is $\cot \left(\frac{A}{2}\right)$. Then the minimum angle of deviation is:
(1) $180^{\circ}-3 A$
(2) $180^{\circ}-2 A$
(3) $180^{\circ}-A$
(4) $180^{\circ}-4 A$
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Answer: (2)
Solution:
$\delta _{\min }=2 \sin ^{-1}\left[\mu \sin \frac{A}{2}\right]-A$
$ \begin{aligned} & =2 \sin ^{-1}\left[\cos \frac{A}{2}\right]-A \\ & =\pi-2 A \end{aligned} $
Question 14
In single electron atom/ion, first member of Lyman series is $\lambda$, then wavelength of second member of this series shall be
(1) $\frac{5}{27} \lambda$
(2) $\frac{5}{32} \lambda$
(3) $\frac{27}{32} \lambda$
(4) $\frac{15}{23} \lambda$
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Answer: (3)
Solution:
$\frac{1}{\lambda}=c{1-\frac{1}{4} }$ for $2 \rightarrow 1$
$ \begin{aligned} & \frac{1}{\lambda^{\prime}}=c\left(1-\frac{1}{9}\right) \text { for } 3 \rightarrow 1 \\ & \frac{\lambda^{\prime}}{\lambda}=\frac{3}{4} \times \frac{9}{8}=\frac{27}{32} \end{aligned} $
Question 15
When light of wavelength $\lambda$ is incident on a metal, the stopping potential is $8 V$. If the wavelength is made $3 \lambda$ the stopping potential becomes $2 V$. Find the threshold wavelength for the photoelectric effect.
(1) $2.6 \times 10^{-6} m$
(2) $2.8 \times 10^{-7} m$
(3) $1.24 \times 10^{-6} m$
(4) $1.24 \times 10^{-7} m$
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Answer: (3)
Solution:
$(8 e)=\frac{h c}{\lambda}-\phi _0$
(2e) $\frac{h c}{3 \lambda}-\phi _0$
$\phi _0=1 eV$
$\therefore \quad \lambda _{\text {th }}=12400 \mathring{A}$
$\lambda _{\text {th }}=1.24 \times 10^{-6} m$
Question 16
In YDSE, intensity at two sources are in ratio of $1: 9$. If sources are coherent, then intensity at central point is $l _1$ and if sources are coherent (and phase differs by $60^{\circ}$ ), then intensity at central point is $I _2$, then $\frac{I _1}{I _2}$ is
(1) $\frac{10}{13}$
(2) $\frac{5}{13}$
(3) $\frac{8}{13}$
(4) $\frac{7}{11}$
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Answer: (1)
Solution:
$l _1=1+9=10$
$I _2=1+9+2 \sqrt{9} \cos 60^{\circ}=13$
$\frac{l _1}{l _3}=\frac{10}{13}$
Question 17
Calculate the average energy density of an electromagnetic wave whose electric field is oscillating with amplitude $50 v / m$ and frequency $5 \times 10^{10} Hz$.
(1) $2 \times 10^{-6} J / m^{3}$
(2) $1.1 \times 10^{-8} J / m^{3}$
(3) $3 \times 10^{-7} J / m^{3}$
(4) $1.6 \times 10^{-7} J / m^{3}$
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Answer: (2)
Solution:
Average energy density $=\frac{1}{2} \in _0 E _0^{2}$
$ \begin{aligned} & =\frac{1}{2} \times 8.85 \times 10^{-12} \times 2500 \\ & =1.106 \times 10^{-8} J / m^{3} \end{aligned} $
Question 18
A ball dropped from height $H$ rebounds upto height $h$ after colliding with horizontal surface. If coefficient of restitution for collision is $e=\frac{1}{2}$ then $\frac{H}{h}$ shall be
Show Answer
Answer: (4)
Solution:
Defn of $e=\frac{1}{2}=\frac{\sqrt{2 g h}}{\sqrt{2 g H}}$
$ \begin{aligned} & \frac{1}{4}=\frac{h}{H} \\ \Rightarrow & 4 \end{aligned} $
Question 19
Find equivalent resistance between $A$ and $B$ for given circuit in ohms.
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Answer: (1)
Solution:
$\Rightarrow 5$ is short circuited
$ \begin{aligned} & \Rightarrow 2,2 \text { are parallel } \Rightarrow 1 \\ & \Rightarrow 1 \& 2 \text { in series }=3 \\ & R _{eq}=\frac{1}{3}(3)=1 \Omega \end{aligned} $
Question 20
A uniform disk of mass $50 kg$ is rolling with speed of $0.4 m / s$. Find minimum energy (in J) required to bring the disk to rest.
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Answer: (6)
Solution:
$KE=\frac{1}{2} m v _{\omega}^{2}+\frac{1}{2}\left(\frac{m r^{2}}{2}\right) \omega^{2}$
$K=\frac{3}{4} m v^{2}$
$=\frac{3}{4} \times 50 \times(4) \times(0.4)$
$=6 J$
Question 21
Mass defect in a nuclear reaction is $0.4 U$. The $Q$ value of the reaction is $\frac{x}{10} MeV$, find $x$. Take $1 U=$ 930.5 MeV/c ${ }^{2}$
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Answer: (372.2)
Solution:
$\phi=\left[930.5 \times 0.4 MeV / c^{2}\right] \times c^{2}$
$=372.2 MeV$
Question 22
At any instant, magnetic field inside a coil is $3000 T$ and it changes $2000 T$ in next 2 second. If average induced emf through coil is 22 Volt, then find number of turns of coil. (Area of turn is $2 \times 10^{-3} m^{2}$ )
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Answer: (22)
Solution:
From Faraday’s law
Induced emf $(e)=-N \frac{d \Phi}{d t}$
$|22|=N\left|\frac{1000}{2}\right| \times 2 \times 10^{-3}$
$N=22$
Question 23
A parallel plate capacitor with plates separated by $5 mm$ then it draws current of $I _0$ from $A C$ source. Now a dielectric of thickness $2 mm$ is inserted between plates then current drawn increases by $25 \%$. Find dielectric constant.
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Answer: (2)
Solution:
$ \begin{aligned} & i=\varepsilon _0 \omega c \\ & i^{\prime}=\varepsilon \omega c^{\prime} \\ & \Rightarrow \frac{1}{1.25}=\frac{\varepsilon _0 A}{5} \\ & \Rightarrow 5=\frac{2.5}{k}+3.75 \\ & \Rightarrow 1.25=\frac{2.5}{k} \\ & \Rightarrow k=2 \end{aligned} $
