Jee Main 2024 01 02 2024 Shift 1
Question 1
The dimensions of angular impulse is equal to
(1) $\left[M^{1} L^{2} T^{-1}\right]$
(2) $\left[M^{1} L^{2} T^{1}\right]$
(3) $\left[M^{1} L^{2} T^{2}\right]$
(4) $\left[M^{1} L^{1} T^{-1}\right]$
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Answer: (1)
Solution:
Angular impulse $=$ Change in angular momentum
$[J]=[m v r]$
$[J]=\left[M^{1} L^{2} T^{-1}\right]$
Question 2
A vernier caliper has 10 main scale divisions coinciding with 11 vernier scale divisions. 1 main scale division equals $5 mm$. The least count of the device is
(1) $\frac{1}{2} mm$
(2) $\frac{5}{12} mm$
(3) $\frac{5}{11} mm$
(4) $0.3 mm$
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Answer: (3)
Solution:
$10 M=11 V$
$ \begin{aligned} \Rightarrow 1 V & =\frac{10}{11} \times 5 mm \\ \Rightarrow LC & =|M-V| \\ & =\frac{5}{11} mm \end{aligned} $
Question 3
On increasing temperature, the elasticity of a material
(1) Increases
(2) Decreases
(3) Remains constant
(4) May increase or decrease
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Answer: (2)
Solution:
$E=\frac{\text { Stress }}{\text { Strain }}$
As temperature increases, strain increases
$\therefore \quad$ Elasticity decreases
Question 4
Determine the lowest energy of photon emitted in Balmer series of hydrogen atom.
(1) $10.02 eV$
(2) $1.88 eV$
(3) $1.65 eV$
(4) $2.02 eV$
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Answer: (2)
Solution:
For $3 \rightarrow 2$ transitions
$\Delta E=13.6\left(\frac{1}{4}-\frac{1}{9}\right)$
$=13.6 \times \frac{5}{36}$
$=1.88 eV$
Question 5
de Broglie wavelength of proton $=\lambda$ and that of an $\alpha$ particle is $2 \lambda$. The ratio of velocity of proton to that of $\alpha$ particle is :
(1) 8
(2) $\frac{1}{8}$
(3) 4
(4) $\frac{1}{4}$
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Answer: (1)
Solution:
$\lambda=\frac{h}{p}$
$\Rightarrow \lambda=\frac{h}{m v _p}$
and $2 \lambda=\frac{h}{4 m v _{\alpha}}$
$\Rightarrow \frac{1}{2}=\frac{4 v _{\alpha}}{v _p}$
$\Rightarrow \frac{v _p}{v _{\alpha}}=8$
Question 6
2 moles of monoatomic gas and 6 moles of diatomic gas are mixed. Molar specific heat, for constant volume, of mixture shall be $(R$ is universal gas constant)
(1) $1.75 R$
(2) $2.25 R$
(3) $2.75 R$
(4) $2.50 R$
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Answer: (2)
Solution:
$\left(C _V\right) _{\text {mix }}=\left(\frac{2 \times \frac{3}{2}+6 \times \frac{5}{2}}{2+6}\right) R$
$=\frac{(3+15) R}{8}=\frac{9}{4} R$
Question 7
A gas undergoes a thermodynamic process from state $\left(P _1 V _1 T _1\right)$ to state $\left(P _2, V _2, T _2\right)$. For the given process if $P V^{\frac{3}{2}}=$ constant, find the work done by the gas.
(1) $\frac{\left(P _2 V _2-P _1 V _1\right)}{2}$
(2) $\frac{\left(P _1 V _1-P _2 V _2\right)}{2}$
(3) $\frac{3}{2}\left(P _1 V _1-P _2 V _2\right)$
(4) $2\left(P _1 V _1-P _2 V _2\right)$
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Answer: (4)
Solution:
$W=\frac{P _1 V _1-P _2 V _2}{\alpha-1}$
$ \begin{aligned} & =\frac{P _1 V _1-P _2 V _2}{\left(\frac{3}{2}-1\right)} \\ & =2\left(P _1 V _1-P _2 V _2\right) \end{aligned} $
Question 8
For measuring resistivity, the relation $R=\rho \frac{l}{A}=\frac{\rho l}{\pi r^{2}}$ is used. Percentage error in resistance $(R)$, in length $(I)$ and in radius $(r)$ are given $x, y$ and $z$ respectively. Find percentage error in resistivity $\rho$.
(1) $x+y+2 z$
(2) $x+2 y+z$
(3) $\frac{x}{2}+y+z$
(4) $x+2 z-y$
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Answer: (1)
Solution:
$\frac{\Delta \rho}{\rho}=\frac{\Delta R}{R}+\frac{2 \Delta r}{r}+\frac{\Delta l}{l}$
$=x+2 z+y$
Question 9
Two capacitors are charged as shown. When both the positive terminals and negative terminals of capacitors are connected the energy loss will be
(1) $\frac{1}{2} C V^{2}$
(2) $\frac{3}{4} C V^{2}$
(3) $\frac{1}{4} C V^{2}$
(4) $2 CV^{2}$
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Answer: (3)
Solution:
$V _c=\frac{C V+2 C V}{2 C}=\frac{3 V}{2}$
$\therefore \quad$ Energy loss $=\frac{1}{2} C V^{2}+\frac{1}{2} C(2 V)^{2}-\frac{1}{2} 2 C\left(\frac{3 V}{2}\right)^{2}$
$$ =\frac{1}{4} C V^{2} $$
Question 10
A moving coil galvanometer has resistance $50 \Omega$ and full deflection current is $5 mA$. The resistance needed to convert this galvanometer into voltmeter of range 100 volt is
(1) $19550 \Omega$
(2) $18500 \Omega$
(3) $19850 \Omega$
(4) $18760 \Omega$
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Answer: (1)
Solution:
$\lg (G+R)=100 V$
$5 \times 10^{-3}(50+R)=100^{20}$
$50+R=20000$
$R=19550 \Omega$
Question 11
In the voltage regulator circuit shown below, the reverse breakdown voltage of zener diode is $5 V$ and power dissipated across it is $100 mW$. Find $R _S$
(1) $120 \Omega$
(2) $250 \Omega$
(3) $1000 \Omega$
(4) $1500 \Omega$
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Answer: (1)
Solution:
$i _{1000 \Omega}=5 mA$
$ \begin{aligned} & i _z=\frac{P}{V _z}=20 mA \\ & \therefore \quad i _R=25 mA \\ & V _R=3 V \\ & \therefore \quad R=\frac{3}{25} \times 10^{3}=120 \Omega \end{aligned} $
Question 12
Two strings are identical and fixed at both ends with tension $6 N$ each. If the tension in one string fixed at both end is changed from $6 N$ to $52 N$, then find beats frequency.
Linear mass density $=1 kg / m$
(1) $2.38 Hz$
(2) $3.25 Hz$
(3) $2.75 Hz$
(4) $5.25 Hz$
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Answer: (1)
Solution:
$f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}$
$f _1=\frac{1}{2 L} \sqrt{\frac{T _1}{\mu}}$
$f _2=\frac{1}{2 L} \sqrt{\frac{T _2}{\mu}}$
Beats frequency $=\Delta f=f _2-f _1=\frac{1}{2 L}\left(\sqrt{\frac{52}{\mu}}-\sqrt{\frac{6}{\mu}}\right)$
$=\frac{1}{2}(\sqrt{52}-\sqrt{6})$
$=\frac{1}{2}(7.21-2.45)$
$=2.38 Hz$
Question 13
A particle is moving in a circle of radius $R$ in time period of $T$. This moving particle is projected at angle $\theta$ with horizontal \& attains a maximum height of $4 R$. Angle $\theta$ can be given as ( $g$ is acceleration due to gravity)
(1) $\sin ^{-1}\left(\frac{T}{2 \pi} \sqrt{\frac{2 g}{R}}\right)$
(2) $\sin ^{-1}\left(\frac{T}{\pi} \sqrt{\frac{g}{R}}\right)$
(3) $\sin ^{-1}\left(\frac{T}{\pi} \sqrt{\frac{2 g}{R}}\right)$
(4) $\sin ^{-1}\left(T \sqrt{\frac{2 g}{R}}\right)$
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Answer: (3)
Solution:
$\frac{2 \pi R}{T}=u$
$\frac{u^{2} \sin ^{2} \theta}{2 g}=4 R$
$\frac{4 \pi^{2} R^{2}}{T^{2} 2 g} \sin ^{2} \theta=4 R$
$\sin ^{2} \theta=\frac{2 g T^{2}}{\pi^{2} R}=\left(\frac{T}{\pi} \sqrt{\frac{2 g}{R}}\right)^{2}$
Question 14
A block of mass $20 kg$ is placed on rough surface having co-efficient of friction 0.04 as shown in figure.
Find acceleration of system when it released.
(1) $3 m / s$
(2) $2 m / s$
(3) $1 m / s$
(4) $4 m / s$
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Answer: (2)
Solution:
Maximum friction $\left(F _{\max }\right)=0.04 \times 20 \times 10=8 N$
Pulley force $(F)=60 N$
Acceleration $(a)=\frac{60-8}{26}=2 m / s$
Question 15
In single slit diffraction with slit width $0.1 mm$, light of wavelength $6000 \mathring{A}$ is used. A convex lens of focal length $20 cm$ is used to focus the diffracted ray. Find width of central maxima.
(1) $24 mm$
(2) $2.4 mm$
(3) $12 mm$
(4) $1.2 mm$
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Answer: (2)
Solution:
Angular width $=\frac{2 \lambda}{a}$
$$ \begin{aligned} \text { Linear width } & =\frac{2 \lambda}{a} f \\ & =\frac{2 \times 6000 \times 10^{-10} \times 20 \times 10^{-2}}{0.1 \times 10^{-3}} \\ & =2 \times 6 \times 2 \times 10^{-4} \\ & =24 \times 10^{-4} \\ & =2.4 mm \end{aligned} $$
Question 16
Two particles each of mass $2 kg$ are placed as shown in $x y$ plane. If the distance of centre of mass from origin is $\frac{4 \sqrt{2}}{x}$, find $x$
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Answer: (2)
Solution:
$\vec{r} _{cm}=-2 \hat{i}+2 \hat{j}$
$ \therefore \quad r=2 \sqrt{2} $
$x=2$
Question 17
Eight identical batteries $(5 \vee, 1 \Omega)$ are connected as shown :
The reading of the ideal voltmeter is volts.
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Answer: (0)
Solution:
$\varepsilon=8 \times 5=40 V$
$r=8 \times 1=8 \Omega$
$\Rightarrow i=5 A$
$\Rightarrow$ Voltmeter reads
$=5-i r=0$ volts
Question 18
A bullet, of mass $10^{-2} kg$ and velocity $200 m / s$ gets embedded inside the bob (mass $1 kg$ ) of a simple pendulum as shown. The maximum height the system rises by is $cm$.
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Answer: (20)
Solution:
Momentum conservation :
$10^{-2} \times 200 \simeq 1 \times v$
Energy conservation :
$v=\sqrt{2 g h}$
$\Rightarrow h=\frac{v^{2}}{2 g}=\frac{4}{20} m=20 cm$
Question 19
The length of a seconds pendulum if it is placed at height $2 R$ ( $R$ : radius of earth) is $\frac{10}{x \pi^{2}}$ metres. Find $x$.
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Answer: (9)
Solution:
$T=2 \pi \sqrt{\frac{l}{g}}$
$ \begin{aligned} & \Rightarrow \quad 2=2 \pi \sqrt{\frac{l}{g _0 / 9}} \\ & \Rightarrow \quad 2=2 \pi \times 3 \sqrt{\frac{l}{10}} \\ & \Rightarrow \quad \frac{I}{10}=\frac{1}{9 \pi^{2}} \\ & \Rightarrow \quad I=\frac{10}{9 \pi^{2}} m \end{aligned} $
Question 20
Nuclear mass and size of nucleus of an element $A$ are 64 and 4.8 femtometer. If size of nucleus of element $B$ is 4 femtometer then its nuclear mass will be $\frac{1000}{x}$ then
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Answer: (27)
Solution:
$R^{3}=\alpha A$
$ \begin{aligned} & \frac{\left(4.8^{3}\right)}{4^{3}}=\frac{64}{M} \\ & M=\frac{16 \times 4 \times 16 \times 4}{48 \times 48 \times 48} \times 10^{3} \end{aligned} $
Question 21
In a series $LCR$ circuit connected to an $AC$ source, value of the elements are $L _0, C _0 \& R _0$ such that circuit is in resonance mode. If now capacity of capacitor is made $4 C _0$, the new value of inductance, for circuit to still remain in resonance, is $\frac{L _0}{n}$. Find $n$.
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Answer: (4)
Solution:
$\frac{1}{\sqrt{L C}}=$ fixed
$\Rightarrow L C=$ fixed
$\Rightarrow L=\frac{L _0}{4}$
Question 22
The current through a conductor varying with time as $i=3 t^{2}+4 t^{3}$.
Find amount of charge (in C) passes through cross section of conductor in internal $t=1 sec$ to $t=2 sec$.
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Answer: (22)
Solution:
$Q=\int i \cdot d t$
$ \begin{aligned} & =\int _1^{2}\left(3 t^{2}+4 t^{3}\right) \cdot d t=\left(t^{3}+t^{4}\right) _1^{2} \\ & =(8+16)-(2) \\ & =22 C \end{aligned} $
Question 23
Distance between virtual magnified image, (size three times of object) of an object placed in front of convex lens and object is $20 cm$. The focal length of lens is $x cm$, then $x$ is
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Answer: (15)
Solution:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \quad \frac{v}{u}=3$
$3 x-x=20$
$x=20$
$\frac{1}{-30}-\frac{1}{-10}=\frac{1}{f}$
$\frac{1}{10}-\frac{1}{30}=\frac{1}{f}$
$\frac{2}{30}=\frac{1}{f} \Rightarrow f=15$
