datavorous/entrance-exam-dataset huggingface dataset:
A nuclear reactor starts producing a radionuclide of half-life T at a constant rate R starting at time t=0 The activity of the radionuclide at t=T is found to the A. Then RA is

Options:
A 2:1
B 3:2
C 4:1
D 2:3

The correct answer is:
2:1

Solution:
dNdt=R-λN
∫ONdNR-λN=∫0tdt
N=Rλ(1-e-λt)
At t=T=ln2λ; A=R2

⇒RA=2

Topic: Physics, Nuclear Physics, JEE Main
A nuclear transformation is denoted by $X(n, \alpha){ }_3^7 \mathrm{Li}$. Which of the following is the nucleus of element $X$ ?

Options:
A
${ }^{12} \mathrm{C}_6$

B
${ }_5^{10} \mathrm{~B}$

C
${ }_5^{9} \mathrm{~B}$

D
${ }_4^{11} \mathrm{~Be}$

The correct answer is:
${ }_5^{10} \mathrm{~B}$

Solution:
$\mathrm{X}+{ }_0^1 \mathrm{n} \rightarrow{ }_2^4 \mathrm{He}+{ }_3^7 \mathrm{Li}$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2005
A nucleoside on hydrolysis gives

Options:
A an aldopentose and a nitrogenous base
B an aldopentose and phosphoric acid
C an aldopentose, a nitrogenous base and phosphoric acid
D a nitrogenous base and phosphoric acid

The correct answer is:
an aldopentose and a nitrogenous base

Solution:
A nucleoside on hydrolysis gives an aldopentose and a nitrogenous base.
Nucleoside is a unit formed by the attachment of a base to an aldopentose sugar.

Topic: Chemistry, Biomolecules, JEE Main
A nucleus A, with a finite de-broglie wavelength λA, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to half of that of B. The de-Broglie wavelengths λB and λC of B and C are respectively:

Options:
A λA,λA2
B λA2,λA
C λA,2λA
D 2λA,λA

The correct answer is:
λA2,λA

Solution:
hλA=hλB-hλC …(i)
Also λC=2λB ...(ii)
From equation (i) & (ii),
1λA=1λB-12λB
λB=λA2; λC=λA

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2019 (08 Apr Shift 2)
A nucleus at rest disintegrates into two smaller nuclei with their masses in the ratio of $2: 1$. After disintegration they will move :

Options:
A in the same direction with same speed.
B in opposite directions with the same speed.
C in opposite directions with speed in the ratio of $2: 1$ respectively.
D in opposite directions with speed in the ratio of $1: 2$ respectively.

The correct answer is:
in opposite directions with speed in the ratio of $1: 2$ respectively.

Solution:
By conservation of momentum
$\begin{aligned}& \mathrm{p}_{\mathrm{i}}=\mathrm{p}_{\mathrm{f}} \\
& \mathrm{O}=\mathrm{m}_1 \mathrm{u}_{1+} \mathrm{m}_2 \mathrm{u}_2 \\& \frac{\mathrm{u}_1}{\mathrm{u}_2}=-\left[\frac{1}{2}\right] \text { as } \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{2}{1}
\end{aligned}$
move in opposite direction with speed ratio $1: 2$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2024 (09 Apr Shift 2)
A nucleus breaks into two nuclear parts, which have their velocity ratio $2: 1$. The ratio of their nuclear radii will be

Options:
A $\sqrt{2}$
B $\frac{1}{2}$
C $\frac{1}{2^{1 / 3}}$
D $\frac{1}{\sqrt{2}}$

The correct answer is:
$\frac{1}{2^{1 / 3}}$

Solution:
By law of conservation of momentum, the two parts will have equal and opposite momentum, assuming the nucleus was initially at rest.
$$
\begin{aligned}
& \therefore \mathrm{m}_1 \mathrm{v}_1=\mathrm{m}_2 \mathrm{v}_2 \text { or } \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{\mathrm{v}_2}{\mathrm{v}_1}=\frac{1}{2} \\
& \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^3=\frac{1}{2} \\
& \therefore \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{1}{2^{1 / 3}}
\end{aligned}
$$

Topic: Physics, Nuclear Physics, JEE Main
A nucleus disintegrates into two nuclear parts which have their velocities in the ratio $2: 1$. The ratio of their nuclear sizes will be

Options:
A
$2^{1 / 3}: 1$

B
$1: 3^{1 / 2}$

C
$3^{1 / 2}: 1$

D
$1: 2^{1 / 3}$

The correct answer is:
$1: 3^{1 / 2}$

Solution:
$\frac{R_1}{R_2}=\left(\frac{m_2}{2 m_2}\right)^{1 / 3}$
$\Rightarrow \frac{R_1}{R_2}=1: 2^{1 / 3}$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2004
A nucleus disintegrates into two smaller parts, which have their velocities in the ratio 3:2. The ratio of their nuclear sizes will be x313. The value of ‘x’ is:

Options:

The correct answer is:
2

Solution:
Given here, v1v2=32

Using conservation of momentum during disintegration, we have

m1v1=m2v2⇒m1 m2=23

Since, m=43πR3ρ

We have m∝R3 ∵nuclear mass density, ρ=constant

m1m2=R1R23

⇒R1R2=2313

So, x=2.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2023 (25 Jan Shift 2)
A nucleus has a half-life of 30 minutes. At $3 \mathrm{PM}$ its decay rate was measured as 120,000 counts/sec. What will be the decay rate at $5 \mathrm{PM} ?$

Options:
A 120,000 counts/sec
B 60,000 counts/sec
C 30,000 counts/sec
D 7,500 counts/sec

The correct answer is:
7,500 counts/sec

Solution:
Given $\mathrm{T}=30$ minutes. $\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=120 \mathrm{~K} \frac{\text { counts }}{\mathrm{sec}}$
After each half life, activity is reduced to half therefore after $\mathrm{n}$ half lives activity reduces to $\left(\frac{1}{2}\right)^{\mathrm{n}}$.
Also $\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{N}$
$\frac{\mathrm{dN}}{\mathrm{dt}}$ at $5 \mathrm{P} . \mathrm{M}$. will be equal to activity remaining after four half lives.
i.e. $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{16} \mathrm{th}\right)$ of the initial activity
$\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)_{\text {at } 5 \mathrm{PM}}=\left(\frac{1}{16}\right.$ th $)$ of the initial activity
$$
\begin{array}{l}
\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)_{5 \mathrm{P}_{. \mathrm{M}}}=\left(\frac{1}{16}\right)\left(\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}\right)_{3 \mathrm{PM}}
\end{array}
$$

Topic: Physics, Nuclear Physics, KVPY, KVPY 2010 (SB/SX)
A nucleus has mass number A1 and volume V1. Another nucleus has mass number A2 and volume V2. If relation between mass number is A2=4A1, then V2V1= _______.

Options:

The correct answer is:
4

Solution:
For a nucleus we can write volume: V=43πR3

Now, R=R0A13

Therefore, V=43πR03A

⇒V2V1=A2A1=4

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2024 (31 Jan Shift 2)
A nucleus moving with velocity v→ emits an α- particle. If the velocities of α- particle and the remaining nucleus be v1→ and v2→ and their masses be m1 and m2 respectively, then

Options:
A all velocity vectors v→ , v1→ and v2→ must be parallel
B v→ must be parallel to v1→+v2→
C v→ must be parallel to m1v1→+m2v2→
D none of above

The correct answer is:
v→ must be parallel to m1v1→+m2v2→

Solution:
From conservation of linear momentum,
pâinitial =pâfinal
m1v1â+m2v2â=m1+m2vâ
vâ must be parallel to m1v1â+m2v2â

Topic: Physics, Laws of Motion, NEET
A nucleus n mâ¡ Xâ¡ emits one Î± particle and two Î²- particles. The resulting nucleus is

Options:
A n - 4 m&ApplyFunction; - 6 Z&ApplyFunction;
B n m&ApplyFunction; - 6 Z&ApplyFunction;
C n m⁡ - 4 X⁡
D n - 2 m&ApplyFunction; - 4 Y&ApplyFunction;

The correct answer is:
n m⁡ - 4 X⁡

Solution:
When an alpha particle 2 4 He is emitted, the mass number and the atomic number of the daughter nucleus decreases by four and two respectively. When a beta particle (β–) is emitted, the atomic number of the daughter nucleus increases by one but the mass number remains the same.
∴ n mâ¡ Xâ¡ â¶ α n - 2 mâ¡ - 4 Yâ¡ â¶ 2 β - n mâ¡ - 4 Xâ¡

Topic: Physics, Nuclear Physics, NEET
A nucleus of mass 20 u emits a Î³ -photon of 6 MeV . If the emission assume to occur when nucleus is free and rest, then the nucleus will have kinetic energy nearest to Take,1 u=1.6Ã10-27kg .

Options:
A 10 keV
B 1 keV
C 0.1Â keV
D 100Â keV

The correct answer is:
1 keV

Solution:
Due to momentum of photon, nucleus possesses recoil momentum,
âµ E=mc2
=mcc p=mc
=pc
â´ Momentum of Î³ -photon
p1=Ec=p ......(i)
According to the law of conservation of linear momentum
pi=pf
O=p1+p2
Where, p2 is linear momentum of nucleus
â p2=-p1
=Ec [in magnitude] .....(ii)
âµ Kinetic energy =12mv2 [From equation (i)]
=12mv2m
â´p=mv
=12Ec2m [from equation (ii)]
â´ KE=126Ã106Ã1.6Ã10-19/3Ã108220Ã1.6Ã10-27
=0.16Ã10-15J
=1.6Ã10-161.6Ã10-16 keV
=1 keV

Topic: Physics, Nuclear Physics, JEE Main
A nucleus of mass M at rest splits into two parts having masses M'3 and 2M'3M'<M. The ratio of de Broglie wavelength of two parts will be

Options:
A 1:2
B 2:1
C 1:1
D 2:3

The correct answer is:
1:1

Solution:
Since the initial momentum of the system is zero. To keep the final momentum zero, the momentum of both parts should be equal and opposite in direction.

P→1=P→2

Here P→ is momentum

De-Broglie wavelength is given by λ=hP

Hence, both will have same de Broglie wavelength.

Therefore, λ1:λ2=1:1.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2022 (26 Jul Shift 2)
A nucleus of mass M emits γ-ray photon of frequency v. The loss of internal energy by the nucleus is: [Take c as the speed of electromagnetic wave]

Options:
A hv
B 0
C hv1-hv2Mc2
D hv1+hv2Mc2

The correct answer is:
hv1+hv2Mc2

Solution:
Energy of γ ray Eγ=hv

Momentum of γ ray Pγ=hλ=hvc
Total momentum is conserved.

P→γ+P→Nu=0
where P→Nu= momentum of decayed nuclei

⇒Pγ=PNu
⇒hvc=PNu
⇒KE of nuclei

=12Mv2=PNu22M=12Mhvc2
Loss in internal energy =Eγ+KENu
=hv+12Mhvc2
=hv1+hv2Mc2

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2021 (20 Jul Shift 1)
A nucleus of mass $M+\Delta m$ is at rest and decays into two daughter nuclei of equal mass $\frac{M}{2}$ each. Speed of light is $\mathrm{c}$.
40. The binding energy per nucleon for the parent nucleus is $E_1$ and that for the daughter nuclei is $E_2$. Then

Options:
A
$\mathrm{E}_2=2 \mathrm{E}_1$

B
$E_1>E_2$

C
$E_2>E_1$

D
$\mathrm{E}_1=2 \mathrm{E}_2$

The correct answer is:
$E_2>E_1$

Solution:
After decay, the daughter nuclei will be more stable hence binding energy per nucleon will be more than that of their parent nucleus.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2010
A nucleus of mass $M+\Delta m$ is at rest and decays into two daughter nuclei of equal mass $\frac{M}{2}$ each. Speed of light is $\mathrm{C}$.
The speed of daughter nuclei is

Options:
A
$c \frac{\Delta m}{M+\Delta m}$

B
$c \sqrt{\frac{2 \Delta m}{M}}$

C
$c \sqrt{\frac{\Delta \mathrm{m}}{\mathrm{M}}}$

D
$c \sqrt{\frac{\Delta m}{M+\Delta m}}$

The correct answer is:
$c \sqrt{\frac{2 \Delta m}{M}}$

Solution:
Conserving the momentum
$$
\begin{gathered}
0=\frac{\mathrm{M}}{2} \mathrm{~V}_1-\frac{\mathrm{M}}{2} \mathrm{~V}_2 \\
\mathrm{~V}_1=\mathrm{V}_2 \quad \quad \ldots \ldots (1) \\
\Delta \mathrm{mc}^2=\frac{1}{2} \cdot \frac{\mathrm{M}}{2} \mathrm{~V}_1^2+\frac{1}{2} \cdot \frac{\mathrm{M}}{2} \cdot \mathrm{V}_2^2 \quad \quad \ldots \ldots (2)\\
\Delta \mathrm{mc}^2=\frac{\mathrm{M}}{2} \mathrm{~V}_1^2 \\
\frac{2 \Delta \mathrm{mc}^2}{\mathrm{M}}=\mathrm{V}_1^2 \\
\mathrm{~V}_1=\mathrm{c} \sqrt{\frac{2 \Delta \mathrm{m}}{\mathrm{M}}}
\end{gathered}
$$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2010
A nucleus of mass M+Δm is at rest and decays into two daughter nuclei of equal mass M2 each. Speed of light is c.

The speed of daughter nuclei is

Options:
A cΔmM+Δm
B cΔmM+Δm
C c2ΔmM
D cΔmM

The correct answer is:
c2ΔmM

Solution:
The decrease in mass Δm will appear as the kinetic energy of the two daughter nuclei. Since the two nuclei formed are of equal masses, they will share the available energy equally among themselves. If v is the speed of a daughter nucleus, then 12M2v2=12Δmc2
or v= c2ΔmM

Topic: Physics, Nuclear Physics, JEE Main
A nucleus of mass M+âm is at rest and decays into two daughter nuclei of mass M2 each. If the speed of light is c, then the speed of any one of the daughter nuclei is

Options:
A c∆mM+∆m
B c2∆mM
C c∆mM
D c∆mM+∆m

The correct answer is:
c2∆mM

Solution:
According to the conservation of momentum
0=M2V1-M2V2
V1=V2=v
∆mc2=12.M2V12+12.M2V22

⇒v=c2∆mM

Topic: Physics, Nuclear Physics, JEE Main
A nucleus of uranium decays at rest into nuclei of thorium and helium. Then :

Options:
A the helium nucleus has less momentum than the thorium nucleus
B the helium nucleus has more momentum than the thorium nucleus
C the helium nucleus has less kinetic energy than the thorium nucleus
D the helium nucleus has more kinetic energy than the thorium nucleus

The correct answer is:
the helium nucleus has more kinetic energy than the thorium nucleus

Solution:
In an explosion a body breaks up into two pieces of unequal masses both part will have numerically equal momentum and lighter part will have more velocity. $\mathrm{U} \rightarrow \mathrm{Th}+\mathrm{He}$

$$

\mathrm{KE}_{\mathrm{Th}}=\frac{\mathrm{P}^{2}}{2 \mathrm{~m}_{\mathrm{Th}}}, \mathrm{KE}_{\mathrm{He}}=\frac{\mathrm{P}^{2}}{2 \mathrm{~m}_{\mathrm{He}}}

$$

sinc $\mathrm{m}_{\mathrm{He}}$ is less so $\mathrm{KE}_{\mathrm{He}}$ will be more.

Topic: Physics, Nuclear Physics, JEE Main
A nucleus with mass number 184 initially at rest emits an α-particle. If the Q value of the reaction is 5.5 MeV, calculate the kinetic energy of the α- particle.

Options:
A 5.0 MeV
B 5.5 MeV
C 0.12 MeV
D 5.38 MeV

The correct answer is:
5.38 MeV

Solution:
124 mv2+12180 m4v1802=5.5 MeV

⇒ 124 mv21+4541802=5.5 MeV
⇒ K.E.α=5.51+45·41802 MeV=5.38 MeV

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2021 (22 Jul Shift 1)
A nucleus with mass number 220, initially at rest, emits an Î±â particle. If the Q- value of the reaction is 5.5 MeV, the kinetic energy of the Î±- particle is

Options:
A 4.4 MeV
B 5.4 MeV
C 5.6 MeV
D 6.5 MeV

The correct answer is:
5.4 MeV

Solution:
Let K1 & K2 and P1 & P2 are the K.E. and momentum of the α- particle and remaining nucleus, then
K1+K2=5.5 MeV ......(i)
From the conservation of linear momentum
P1=P2⇒2K1×4m ⇒K1=54K2 ......(ii)
From (i) and (ii)
K1=5.5×5455=5.4 MeV

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A nucleus with mass number 220 initially at rest emits an Î±-particle. If the Q value of the reaction is 5.5 MeV, then calculate the kinetic energy of the Î±-particle.

Options:
A 4.4 MeV
B 5.4 MeV
C 5.6 MeV
D 6.5 MeV

The correct answer is:
5.4 MeV

Solution:
Given that K1+K2=5.5 MeV ... (i)
From conservation of linear momentum,
p1=p2
2K1(216m)=2K2(4m) (As p=2Km)
K2=54K1 ... (ii)
Solving Eqs. (i) and (ii), we get
K2=K.E. of Î±-particle 5.4 MeV.

Topic: Physics, Nuclear Physics, JEE Main
A nucleus with mass number 242 and binding energy per nucleon as 7.6 MeV breaks into two fragment each with mass number 121. If each fragment nucleus has binding energy per nucleon as 8.1 MeV, the total gain in binding energy is _______ MeV.

Options:

The correct answer is:
121

Solution:
Binding energy is given by E=∆mc2

where ∆m is the mass defect.

The energy per nucleon of the nucleus having mass number 242 is 7.6 MeV.

The initial binding energy is,

BE=242×7.6 MeV

The energy per nucleon of the nucleus with mass number 121 is 8.1 MeV.

Therefore, binding energy is

BE'=2121×8.1 MeV

The gain in the binding energy is

BE'-BE = (8.1 – 7.6) × 242 MeV=121 MeV

= 121 MeV

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2023 (08 Apr Shift 1)
A nucleus with $\mathrm{Z}=92$ emits the following in a sequence:
$\alpha, \beta^{-}, \beta^{-} \alpha, \alpha, \alpha, \alpha, \alpha, \beta^{-}, \beta^{-}, \alpha, \beta^{+}, \beta^{+}, \alpha$. Then $\mathrm{Z}$ of the resulting nucleus is

Options:
A
$76$

B
$78$

C
$82$

D
$74$

The correct answer is:
$78$

Solution:
No. of $\alpha$ particles emitted $=8, \quad$ No. of $\beta^{-}$particles emitted $=4, \quad$ No.of $\beta^{+}$particles emitted $=2$
$z=92-2 \times 8+4-2=78$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2003
A nucleus with Z = 92 emits the following in a sequence: Î± , Î± , Î² - , Î² - , Î± , Î± , Î± , Î± , Î² - , Î² - , Î± , Î² + , Î² + , Î±. What is the atomic number of the resulting nucleus?

Options:

The correct answer is:
78

Solution:
The nucleus emits 8 α particles i.e., 8 2 He 4
∴ Decrease in Z = 8 × 2 = 16
Four β - particles are emitted i.e., 4 - 1 β 0
∴ Increase in Z = 4 × 1 = 4
2 positrons are emitted i.e., 2 1 β 0
∴ Decrease in Z = 2 × 1 = 2
∴ Z of resultant nucleus = 92 - 16 + 4 - 2 = 78.

Topic: Physics, Atomic Physics, JEE Main
A nucleus ${ }_{Z}^{A} X$ emits an $\alpha$-particle. The resultant nucleus emits a $\beta^{+}-$particle. The respective atomic and mass number of final nucleus will be

Options:
A Z-3, A-4
B Z-1, A-4
C Z-2, A-4
D Z, A-2

The correct answer is:
Z-3, A-4

Solution:
\begin{array}{l}
{ }_{Z}^{A} X \rightarrow_{2}^{4} H e+_{Z-2}^{A-2} Y \\
{ }_{Z-2}^{A-4} Y \rightarrow e^{+}+{ }_{Z-3}^{A-4} Y^{\prime}
\end{array}
During $\beta^{+}$ emission.
$$
{ }_{1} \mathrm{p}^{1} \rightarrow{ }_{0}^{n}{ }^{1}+\beta^{+}
$$
The proton changes into neutron. So, charge number decreases by 1 but mass number remains unchanged.

Topic: Physics, Nuclear Physics, VITEEE, VITEEE 2011
A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5 Ω. When a resistance of 15 Ω is used for shunting null point moves to 300 cm. The internal resistance of the cell is ______ Ω.

Options:

The correct answer is:
5

Solution:
Let potential gradient be x.

For the first case, we can write

εr+5×5=200x ......(1)

and after adding resistance, we can write

εr+15×15=300x .....(2)

From both equations, we get

⇒r=5 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (29 Jan Shift 2)
A number equal to 2 times the mean and with a frequency equal to k is inserted in a data having n observations. If the new mean is 43 times the old mean, then the value of kn is

Options:

The correct answer is:
0.5

Solution:
x¯new=43x¯
From given information
x¯new=nx¯+2kx¯n+k=43x¯
⇒3n+6k=4n+4k
⇒2k=n
⇒kn=0.5

Topic: Mathematics, Statistics, JEE Main
A number equal to \( 2 \) times the mean and with a frequency equal to \( k \) is inserted in a data having \( n \) observations. If the new mean is \( \frac{4}{3} \) times the old mean, then the value of \( \frac{k}{n} \) is

Options:

The correct answer is:
0.5

Solution:
x¯new=43x¯

From given information

x¯new=nx¯+2kx¯n+k=43x¯

⇒3n+6k=4n+4k

⇒2k=n

⇒kn=0.5

Topic: Mathematics, Statistics, JEE Main
A number equal to 4 times of the mean and a frequency equal to k is inserted in the data of n observations. If the new mean is 75 times the old mean, then nk is equal to

Options:

The correct answer is:
6.5

Solution:
x-new=75x-
From the given information,
x-new=nx-+4kx-n+k=75x-
⇒5n+20k=7n+7k
⇒13k=2n
⇒nk=132

Topic: Mathematics, Statistics, JEE Main
A number is called a palindrome if it reads the same backward as well as forward. For example 285582 is a six digit palindrome. The number of six digit palindromes, which are divisible by 55, is ________.

Options:

The correct answer is:
100

Solution:
Palindrome Numbers only in this form will be divisible by 55.

Numbers can be between 0-9

Number of ways of selecting value of a=P110

Similarly number of ways of choosing value of b=P110 (As a and b can also be same)

Therefore, Total number of ways of getting palindrome numbers divisible by 55 is =1×10×10×1=100

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (27 Aug Shift 1)
A number is the reciprocal of the other. If the arithmetic mean of the two numbers be $\frac{13}{12}$, then the numbers are

Options:
A $\frac{1}{4}, \frac{4}{1}$
B $\frac{3}{4}, \frac{4}{3}$
C $\frac{2}{5}, \frac{5}{2}$
D $\frac{3}{2}, \frac{2}{3}$

The correct answer is:
$\frac{3}{2}, \frac{2}{3}$

Solution:
Suppose that required numbers $a$ and $b$ Therefore according to the conditions $a=\frac{1}{b}$
and $\frac{a+b}{2}=\frac{13}{12} \Rightarrow$ $a+b=\frac{13}{6}$
$\Rightarrow \quad a+\frac{1}{a}=\frac{13}{6}$ $\Rightarrow 6 a^2-13 a+6=0$
$\Rightarrow\left(a-\frac{3}{2}\right)\left(a-\frac{2}{3}\right)=0$
$\Rightarrow \quad a=\frac{3}{2}$ and $b=\frac{2}{3}$
or $a=\frac{2}{3}$ and $b=\frac{3}{2}$

Topic: Mathematics, Sequences and Series, JEE Main
A number $n$ is chosen at random from $\{1,2,3,4, \ldots, 1000\}$. The probability that $n$ is a number that leaves remainder 1 when divided by 7 , is :

Options:
A $\frac{71}{500}$
B $\frac{143}{1000}$
C $\frac{72}{500}$
D $\frac{71}{1000}$

The correct answer is:
$\frac{71}{500}$

Solution:
Multiples of 7 in $\{1,2, \ldots, 1000\}$ are $7,14,21, \ldots, 994$.
Let the number of terms be $N$
$\therefore \quad 994=7+(N-1) \cdot 7$
$\Rightarrow \quad \frac{987}{7}=(N-1)$
$\Rightarrow \quad N-1=141$
$\Rightarrow \quad N=142$
$\therefore$ Number of terms which leaves remainder 1 when divided by $7=142$ and $n(S)=1000$
$\therefore$ Required probability $=\frac{n(E)}{n(S)}$
$=\frac{142}{1000}=\frac{71}{500}$

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2006
A number $n$ is chosen at random from $\{1,2,3,4, \ldots, 1000\}$. The probability that $n$ is a number that leaves remainder 1 when divided by 7 , is :

Options:
A $\frac{71}{500}$
B $\frac{143}{1000}$
C $\frac{72}{500}$
D $\frac{71}{1000}$

The correct answer is:
$\frac{71}{500}$

Solution:
Multiples of 7 in $\{1,2, \ldots, 1000\}$ are $7,14,21, \ldots, 994$.
Let the number of terms be $N$
$\therefore \quad 994=7+(N-1) \cdot 7$
$\Rightarrow \quad \frac{987}{7}=(N-1)$
$\Rightarrow \quad N-1=141$
$\Rightarrow \quad N=142$
$\therefore$ Number of terms which leaves remainder 1 when divided by $7=142$ and $n(S)=1000$
$\therefore$ Required probability $=\frac{n(E)}{n(S)}$
$=\frac{142}{1000}=\frac{71}{500}$

Topic: Mathematics, Probability, TS EAMCET, TS EAMCET 2006
A number $n$ is chosen at random from the natural numbers 2 to 1001 . The probability that $n$ is a number that leaves remainder 1 when divided by 7 , is

Options:
A $\frac{73}{500}$
B $\frac{71}{1000}$
C $\frac{143}{1000}$
D $\frac{71}{500}$

The correct answer is:
$\frac{71}{500}$

Solution:
Total number of numbers from 2 to $1001=1000$
Now, the numbers, which leaves remainder 1 when divided by 7 , are $8,15,22, \ldots, 995$.
Let these are $m$ in counting, then we have
$$
\begin{aligned}
\quad 995=8+(m-1) 7 & \Rightarrow 987=(m-1) 7 \\
\Rightarrow \quad m-1=141 & \Rightarrow m=142 \\
\therefore & \text { Required probability }=\frac{142}{1000}=\frac{71}{500}
\end{aligned}
$$

Topic: Mathematics, Probability, TS EAMCET, TS EAMCET 2018 (05 May Shift 2)
A number $n$ is randomly selected from the set $\{1,2,3, \ldots ., 1000\}$. The probability that $\frac{\sum_{i=1}^n i^2}{\sum_{i=1}^n i}$ is an integer is

Options:
A
$0.331$

B
$0.333$

C
$0.334$

D
$0.332$

The correct answer is:
$0.334$

Solution:
$\frac{\sum_{i=1}^n i^2}{\sum_{i=1}^n i}=\frac{\frac{n(n+1)(2 n+1)}{6}}{\frac{n(n+1)}{2}}=\frac{2 n+1}{3}$ For $n=1,2,3, \ldots . ., 1000$ Value of $\frac{2 n+1}{3}=\frac{3}{3}, \frac{5}{3}, \frac{7}{3}, \ldots \ldots \ldots, \frac{2001}{3}$ respectively. Out of $\frac{3}{3}, \frac{5}{3}, \frac{7}{3}, \ldots \ldots \ldots, \frac{2001}{3}$ only first term $\left(\frac{3}{3}=1\right)$, fourth term $\left(\frac{9}{3}=3\right), 667^{\text {th }}$ term $\left(\frac{2001}{3}=667\right)$ are integers.
Hence, out of 1000 values of $\frac{2 n+1}{3}$, total number of integral values of $\frac{2 n+1}{3}$ $=333+1=334$
$\therefore \quad$ Required probability $=\frac{334}{1000}=0.334$

Topic: Mathematics, Probability, JEE Main, JEE Main 2012 (12 May Online)
A number of Carnot engines are operated at identical cold reservoir temperatures $\left(T_{L}\right)$. However, their hot reservoir temperatures are kept different. A graph of the efficiency of the engines versus hot reservoir temperature $\left(T_{H}\right)$ is plotted. The correct graphical representation is

Options:
A

B

C

D

The correct answer is:

Solution:
Efficiency of a Carnot's engine is given as
$\eta=1-\frac{T_{L}}{T_{H}}$
where, $T_{L}=$ temperature of cold reservoir and
$T_{H}=$ temperature of hot reservoir.
At constant $T_{L}$, we can say that, with the increase in the valuc of $T_{H}, \eta$ will also incrcasc.
So, the graph between efficiency and $T_{H}$ is correctly depicted in option (b).

Topic: Physics, Thermodynamics, JEE Main
A number x is chosen at random from the set 1,2,3,4,â¦â¦,100. Define the event: A= the chosen number x satisfies x-10x-50x-30â¥0, then PA is

Options:
A 0.20
B 0.51
C 0.71
D 0.70

The correct answer is:
0.71

Solution:
S=1,2,3……100
A: Chosen no. x satisfies
x-10x-50x-30≥0
∴x∈10,11,12……29∪50,51……100
PA=71100=0.71

Topic: Mathematics, Probability, JEE Main