datavorous/entrance-exam-dataset huggingface dataset:
Impulse can be obtained from the-

Options:
A Velocity-time diagram
B Force-displacement diagram
C Velocity-displacement diagram
D Force-time diagram

The correct answer is:
Force-time diagram

Solution:
Concept:
Impulse (J): It is defined as the integral of force with respect to time. It is a vector quantity. Or it is also defined as a change in the linear moment $(P)$ with respect to time.
$J=F \times d t=\Delta P$
According to Newton's second law, the Force can be defined as a moment per time.
$\therefore \mathbf{F}=\frac{\mathbf{d}(\mathbf{m v})}{\mathbf{d t}}$
This is known as the Impulse momentum equation.
Linear impulse $=\bar{P}=$ change in momentum in that direction.
Application:
(i) A cricket player lowers his hands while catching the ball: by doing so the time of impact increases and hence the effect of force decreases.
(ii) When a person falls from a certain height on the floor, he receives more injuries as compared to falling on a heap of sand. It is because of increasing the time of impact hence decreasing the impact of force.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
In ( 0,â2Ï ), the total number of points where fx=max.sinâ¡x,cosâ¡x,1âcosâ¡x is not differentiable, are equal to

Options:
A 3
B 4
C 5
D 6

The correct answer is:
3

Solution:
The graph of fx=max.sinâ¡x,cosâ¡x,1âcosâ¡x is
â fx is not differentiable at x= Ï 4 ,â Ï 2 ,â 5Ï 3 .

â fx is not differentiable at 3 points

Topic: Mathematics, Continuity and Differentiability, JEE Main
In 1 L saturated solution of AgClKspAgCl=1.6×10-10 , 0.1 mol of CuCl KspCuCl=1.0×10-6 is added. The resultant concentration of Ag+ in the solution is 1.6×10-x . The value of " x " is

Options:

The correct answer is:
7

Solution:
Let the solubility of AgCl is x mol litre-1
AgCl⇌Ag+x+Cl-x
And that of CuCl is y mol litre-1
CuCl⇌Cu+y+Cl-y
Ksp of AgCl=Ag+Cl-1
1.6×10-10=xx+y ...(i)
Similarly Ksp of CuCl=Cu+Cl-
1.6×10-6=yx+y ...(ii)
On solving (i) and (ii)
Ag+=1.6×10-7
x=7

Topic: Chemistry, Ionic Equilibrium, JEE Main
In 3 d series, the metal having the highest M2+/M standard electrode potential is

Options:
A Cr
B Fe
C Cu
D Zn

The correct answer is:
Cu

Solution:
Cr+2/Cr→-0.90 V

Fe+2/Fe→-0.44 V

Cu+2/Cu→+0.34 V

Zn+2/Zn→-0.76 V

So Ans. Cu+2/Cu

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2022 (27 Jun Shift 2)
In 5 minutes, a body cools from 75 °C to 65 °C at room temperature of 25 °C. The temperature of body at the end of next 5 minutes is C∘.

Options:

The correct answer is:
57

Solution:
By newton's law of cooling (with approximation),
ΔTΔt=-CTag-Ts

1st -10 °C5 min=-C70 °C-25 °C

⇒C=245 min-1

2nd T-655 min=-CT+652-25=-245T+152

⇒9( T-65)=-(T+15)
⇒ 10 T=570
⇒T=57 °C

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2021 (22 Jul Shift 1)
In ψ321, the sum of angular momentum, spherical nodes and angular node is

Options:
A 6h+4π2π
B 6h2π+3
C 6h+2π2π
D 6h+8π2π

The correct answer is:
6h+4π2π

Solution:
ψ321:n=1,l=2,m=1
Angular momentum =h2πl(l+1)=6h2π
Spherical nodes =3-2-1=0; Angular node =2
Sum of all the above =6h2π+2=6h+4π2π

Topic: Chemistry, Structure of Atom, JEE Main
In a $0.2$ molal aqueous solution of a weak acid $\mathrm{HX}$ the degree of ionization is $0.3$. Taking $\mathrm{k}_{\mathrm{f}}$ for water as $1.85$, the freezing point of the solution will be nearest to

Options:
A
$-0.360^{\circ} \mathrm{C}$

B
$-0.260^{\circ} \mathrm{C}$

C
$+0.480^{\circ} \mathrm{C}$

D
$-0.480^{\circ} \mathrm{C}$

The correct answer is:
$-0.480^{\circ} \mathrm{C}$

Solution:
No Solution Available

Topic: Chemistry, Solutions, JEE Main, JEE Main 2003
In
(a) 2-Methyl pentane
(b) \( 2,2 \)-Dimethyl butane
(c) 4-Ethyl-2,3-dimethyloctane
(d) 4-(l-Methylethyl) heptane
How many of them have isopropyl group?

Options:

The correct answer is:
3

Solution:
An iso group means a methyl branching on the second carbon.

(a) 2-Methyl pentane- It contains a branching on the secondary carbon.

(b) 2, 2-Dimethyl butane: There are two substituents on the second carbon, so it is called as neo.

(c) 4-Ethyl-2, 3-dimethyloctane - It contains a branching on the second carbon, so it is an iso group.

(d) 4-(1-Methyl ethyl) heptane -1-methyl ethyl group is called as an iso group.

A, C, D contain iso group.

Topic: Chemistry, General Organic Chemistry, JEE Main
In a ΔABC, ab=2+3, and ∠C=60°. Then the ordered pair (∠A,∠B) is equal to:

Options:
A 105°, 15°
B 15°, 105°
C 45°, 75°
D (75°, 45°)

The correct answer is:
105°, 15°

Solution:
ab>1⇒∠A>∠B

A+B=120°

⇒sin(120°−B)sinB=2+3

⇒  32cotB+12=2+3

⇒ cotB=2332+3

⇒cotB=2+3

⇒   B=15°

Hence, (∠A,∠B)=(105°,15°)

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2015 (10 Apr Online)
In a ÎABC, if â A = â B = 1 2 sin - 1 6 + 1 2 3 + sin - 1 1 3 and length of the side opposite to â C is c = 6 â 3 1 4 , then the area of ÎABC is

Options:

The correct answer is:
27

Solution:
2 â B = sin - 1 6 + 1 2 3 + sin - 1 1 3
= tan - 1 6 + 1 3 - 2 + tan - 1 1 2
= tan - 1 6 + 1 3 - 2 + 1 2 1 - 6 + 1 3 - 2 Â· 1 2
= tan - 1 6 + 1 2 + 3 - 2 2 3 - 2 - 6 + 1
= tan - 1 1 2 + 2 + 3 - 2 6 - 2 - 6 - 1
= tan - 1 2 3 + 3 - 3
= tan - 1 3 3 - 3
= tan - 1 - 3
= 2 Ï 3
According to question
â A = â B = 1 2 sin - 1 6 + 1 2 3 + sin - 1 1 3
â â A = â B = 1 2 2 Ï 3
â â A = â B = Ï 3
â´ â C = Ï - â A - â B
= Ï - 2 Ï 3
= Ï 3
Since Î ABC is an equilateral Î
â´ ar Î ABC = 3 4 a 2
= 3 4 c 2
= 3 4 Â· 6â 3 1 / 4 2
= 3 4 3 â 6 Ã 6
= 3 Ã 3 Ã 3
= 2 7

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
In a ∆ABC, the sides BC, CA and AB are consecutive positive integers in increasing order. Let a→,b→ and c→ are position vectors of the vertices A,B and C respectively. If c→-a→·b→-c→=0, then the value of a→×b→+b→×c→+c→×a→ is equal to

Options:

The correct answer is:
12

Solution:
c → − a → ⋅ b → − c → =0⇒ AC → ⋅ CB → =0⇒ΔABC is right angle at C.
From diagram
k2+k+12=k+22
k2+k2+2k+1=k2+4k+4
k2-2k-3=0âk=3,-1â´kâ -1
So, sides are 3,4,5
Area of âABC=12aâÃbâ+bâÃcâ+câÃaâ=12Ã3Ã4
âaâÃbâ+bâÃcâ+câÃaâ=12

Topic: Mathematics, Vector Algebra, JEE Main
In a $\triangle A B C, a=2 \mathrm{~cm}, b=3 \mathrm{~cm}, c=4 \mathrm{~cm}$ then $\cos A$ equals to -

Options:
A $\frac{8}{7}$
B $\frac{7}{8}$
C $\frac{1}{8}$
D $\frac{1}{7}$

The correct answer is:
$\frac{7}{8}$

Solution:
By the cosine rule,
$\cos A=\frac{b^2+c^2-a^2}{2 b c}$
$\cos A=\frac{3^2+4^2-2^2}{2(3)(4)}$
$\cos A=\frac{21}{24}$
$\cos A=\frac{7}{8}$

Topic: Mathematics, Properties of Triangles, JEE Main
In a $\triangle A B C, b=\sqrt{3}+1, c=\sqrt{3}-1, \angle A=60^{\circ}$ then the value of $\tan \left(\frac{B-C}{2}\right)$ is

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

The correct answer is:
$1$

Solution:
$\tan \left(\frac{B-C}{2}\right)=\left(\frac{b-c}{b+c}\right) \cot \left(\frac{A}{2}\right)$
putting the value of $\mathrm{b}, \mathrm{c}$ and $\angle \mathrm{A}$
$\tan \left(\frac{B-C}{2}\right)=\frac{(\sqrt{3}+1)-(\sqrt{3}-1)}{(\sqrt{3}+1)+(\sqrt{3}-1)} \cot \left(30^{\circ}\right)$
$\Rightarrow \quad \tan \left(\frac{B-C}{2}\right)=1$

Topic: Mathematics, Properties of Triangles, JEE Main
In a $\triangle A B C$, if $a=4 \mathrm{~cm}, b=6 \mathrm{~cm}$ and $c=8 \mathrm{~cm}$ then $r$ equals

Options:
A $\sqrt{\frac{5}{3}}$
B $\sqrt{\frac{3}{5}}$
C $\frac{\sqrt{5}}{3}$
D None of these

The correct answer is:
$\sqrt{\frac{5}{3}}$

Solution:
If $a=4 \mathrm{~cm}, b=6 \mathrm{~cm}$ and $c=8 \mathrm{~cm}$ then, $2 s=a+b+c$
$\mathrm{s}=9 \mathrm{~cm}$
But, $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$ $\quad$ [by Heron's formula]
$=\sqrt{9 \times 5 \times 3 \times 1}$
so, $\quad r=\frac{\Delta}{s}$
or, $\quad r=\frac{3 \sqrt{15}}{9}$
or, $\quad r=\frac{\sqrt{15}}{3} \mathrm{~cm}$
or, $\quad r=\sqrt{\frac{5}{3}} \mathrm{~cm}$

Topic: Mathematics, Properties of Triangles, JEE Main
In a $\triangle A B C$, if $a=13, b=14$ and $c=15$, then the value of $\sin \left(\frac{A}{2}\right)$ is

Options:
A $\frac{3}{5}$
B $\frac{1}{\sqrt{5}}$
C $\frac{7}{\sqrt{65}}$
D $6$

The correct answer is:
$\frac{1}{\sqrt{5}}$

Solution:
we know that, $\begin{aligned} 2 s & =a+b+c \\ 2 s & =42 \\ s & =21\end{aligned}$
$s-a=8, s-b=7$, and $s-c=6$
$\sin \left(\frac{A}{2}\right)=\sqrt{\frac{(s-b)(s-c)}{b c}}$
$=\sqrt{\frac{7 \times 6}{14 \times 15}}=\frac{1}{\sqrt{5}}$

Topic: Mathematics, Properties of Triangles, JEE Main
In a $\triangle A B C$, if $a=18 \mathrm{~cm}$ and $b=24 \mathrm{~cm}$ and $c=30 \mathrm{~cm}$ then the value of $r_1, r_2$ and $r_3$ are

Options:
A $12 \mathrm{~cm}, 18 \mathrm{~cm}, 36 \mathrm{~cm}$
B $12 \mathrm{~cm}, 8 \mathrm{~cm}, 30 \mathrm{~cm}$
C $12 \mathrm{~cm}, 10 \mathrm{~cm}, 30 \mathrm{~cm}$
D $12 \mathrm{~cm}, 18 \mathrm{~cm}, 36 \mathrm{~cm}$

The correct answer is:
$12 \mathrm{~cm}, 18 \mathrm{~cm}, 36 \mathrm{~cm}$

Solution:
But, $\quad \Delta=\sqrt{s(s-a)(s-b)(s-c)}$
$\Delta=216$ sq. units
Then, $r_1=\frac{\Delta}{s-a}=\frac{216}{18}=12 \mathrm{~cm}$
or, $\quad r_2=\frac{\Delta}{s-b}=\frac{216}{12}=18 \mathrm{~cm}$
or, $\quad r_3=\frac{\Delta}{s-c}=\frac{216}{6}=36 \mathrm{~cm}$
so, $r_1, r_2, r_3$ are $12 \mathrm{~cm}, 18 \mathrm{~cm}$, and $36 \mathrm{~cm}$

Topic: Mathematics, Properties of Triangles, JEE Main
In a $\triangle A B C$, the sides $a, b$ and $c$ are in $A$. P. Then $\left(\tan \frac{A}{2}+\tan \frac{C}{2}\right): \cot \frac{B}{2}$ is equal to

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

The correct answer is:
$2: 3$

Solution:
$\left(\tan \frac{A}{2}+\tan \frac{C}{2}\right): \cot \frac{B}{2}$
$=\left[\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}+\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}\right]: \sqrt{\frac{s(s-b)}{(s-c)(s-a)}}$
$=\frac{(s-c)+(s-c)}{\sqrt{s}}: \sqrt{s}$
$=2 s-(a+c): s$
$\Rightarrow \mathbf{b}: \frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}$
$\Rightarrow 2 b: a+b+c=2 b: 3 b$ $[\because \mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P. $\therefore 2 \mathrm{~b}=\mathrm{a}+\mathrm{c}]$
$=2: 3$

Topic: Mathematics, Properties of Triangles, JEE Main
In $A^{+} B^{-}$ionic compound, radii of $A^{+}$and $B^{-}$ions are $180 \mathrm{pm}$ and $187 \mathrm{pm}$ respectively. The crystal structure of this compound will be

Options:
A $\mathrm{NaCl}$ type
B $\mathrm{CsCl}$ type
C ZnS type
D Similar to diamond

The correct answer is:
$\mathrm{CsCl}$ type

Solution:
$r_{+} / r_{-}=\frac{180}{187}=0.962$
which lies in the range of $0.732-1.000$, hence
co-ordination number $=8$ i.e. the structure is $\mathrm{CsCl}$ type.

Topic: Chemistry, Solid State, JEE Main
In a balanced equation $\mathrm{H}_2 \mathrm{SO}_4+x \mathrm{HI} \rightarrow \mathrm{H}_2 \mathrm{~S}+y \mathrm{I}_2+z \mathrm{H}_2 \mathrm{O}$, the values of $x, y, z$ are

Options:
A $x=3, y=5, z=2$
B $x=4, y=8, z=5$
C $x=8, y=4, z=4$
D $x=5, y=3, z=4$

The correct answer is:
$x=8, y=4, z=4$

Solution:
To balance the number of iodide atoms, HI is multiplied by 2. Iodine is oxidised with oxidation state changing from -1 to 0.
Net increase in electron is '2' Sulphur is reduced, oxidation state changing from +6 to \(-2 \Rightarrow\) electrons
To balance 8 electrons, multiply \(H I\) and \(I_2\) by 4
Hence,
\(\begin{array}{ll}
\mathrm{H}_2 \mathrm{SO}_4+x \mathrm{HI} \longrightarrow \mathrm{H}_2 \mathrm{~S}+y \mathrm{I}_2+2 \mathrm{H}_2 \mathrm{O} \\
\Rightarrow \mathrm{H}_2 \mathrm{SO}_4+8 \mathrm{HI} \longrightarrow \mathrm{H}_2 \mathrm{~S}+4 \mathrm{I}_2+4 \mathrm{H}_2 \mathrm{O} \\
\therefore x=8 \\
x=4 \\
z=4
\end{array}\)

Topic: Chemistry, Redox Reactions, JEE Main
In a binary compound, atoms of element A form a hcp structure and those of element M occupy 2/3 of the tetrahedral voids of the hcp structure. The formula of the binary compound is:

Options:
A M2 A3
B M4 A3
C M4 A
D MA3

The correct answer is:
M4 A3

Solution:
In HCP unit cell number of effective number of atoms is 6,

so A=6.

Number of tetrahedral voids

= 2 x number of atoms forming the lattice

=2 X 6

M12×23 A6
M8 A6
M4 A3

Topic: Chemistry, Solid State, JEE Main, JEE Main 2021 (18 Mar Shift 1)
In a binomial distribution $B\left(n, p=\frac{1}{4}\right)$, if the probability of at least one success is greater than or equal to $\frac{9}{10}$, then $n$ is greater than

Options:
A
$\frac{1}{\log _{10}{ }^4-\log _{10}{ }^3}$

B
$\frac{1}{\log _{10} 4+\log _{10} 3}$

C
$\frac{9}{\log _{10}{ }^4-\log _{10} 3}$

D
$\frac{4}{\log _{10}{ }^4-\log _{10} 3}$

The correct answer is:
$\frac{1}{\log _{10}{ }^4-\log _{10}{ }^3}$

Solution:
$$
1-q^n \geq \frac{9}{10} \Rightarrow\left(\frac{3}{4}\right)^n \leq \frac{1}{10} \Rightarrow n \geq-\log _{\frac{3}{4}} 10 \Rightarrow n \geq \frac{1}{\log _{10}{ }^4-\log _{10}{ }^3}
$$

Topic: Mathematics, Probability, JEE Main, JEE Main 2009
In a binomial distribution B(n,p), the sum and product of the mean & variance are 5 and 6 respectively, then find 6(n+p-q) is equal to :-

Options:
A 51
B 52
C 53
D 50

The correct answer is:
52

Solution:
For the binomial distribution B(n, p),

Mean =np and Variance =npq

Sum of mean and variance =np+npq=5

⇒np(1+q)=5

⇒n2p2(1+q)2=25 ....(1)

Product of mean and variance =np.npq=6

⇒n2p2q=6 ....(2)

Put the value of n2p2 from equation (2) into equation (1):

6q(1+q)2=25

⇒6q2+12q+6=25q

⇒6q2-13q+6=0

⇒(3q-2)(2q-3)=0

⇒q=23, 32 where q=23 accepted

∴ p=13⇒n·13+n·13·23=5

⇒3n+2n9=5

⇒n=9

Therefore,

6(n+p-q)=69+13-23=52

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (01 Feb Shift 1)
In a biprism experiment, the slit separation is $1 \mathrm{~mm}$. Using monochromatic light
of wavelength $5000 Å$, an interference pattern is obtained on the screen. Where
should the screen be moved, so that the change in fringe width is $12 \cdot 5 \times 10^{-5} \mathrm{~m}$ ?

Options:
A Away or towards the slit by 25 am
B Away or towards the slit by $12 \cdot 5 \mathrm{~cm}$
C Away from the slit by $5 \mathrm{~cm}$
D Towards the slit by $10 \mathrm{~cm}$

The correct answer is:
Away or towards the slit by 25 am

Solution:
$\mathrm{d}=1 \times 10^{-3} \mathrm{~m}$
$\lambda=5000 Å=5 \times 10^{-7} \mathrm{~m}$
$\beta_{2}-\beta_{1}=12.5 \times 10^{-5} \mathrm{~m}$
$\mathrm{D}_{2}-\mathrm{D}_{1}=?$
$\beta_{1}=\frac{\lambda \mathrm{D}_{1}}{\mathrm{~d}} \quad \beta_{2}=\frac{\lambda \mathrm{D}_{2}}{\mathrm{~d}}$
$\beta_{2}-\beta_{1}=\frac{\lambda}{\mathrm{d}}\left(\mathrm{D}_{2}-\mathrm{D}_{1}\right)$
$\therefore \quad\left(\mathrm{D}_{2}-\mathrm{D}_{1}\right)=\frac{\mathrm{d}}{\lambda}\left(\beta_{2}-\beta_{1}\right)$
$\quad\left(\mathrm{D}_{2}-\mathrm{D}_{1}\right)=\frac{10^{-3}}{5 \times 10^{-7}} \times 12.5 \times 10^{-5}=\frac{12.5}{50}=25 \mathrm{~cm}$

Topic: Physics, Wave Optics, JEE Main
In a biprism experiment with sodium light, bands of width 0.0195 cm are observed at 100 cm from slit. On introducing a convex lens 30 cm away from the slit between biprism and screen, two images of the slit are seen 0.7 cm apart at 100 cm distance from the slit. Calculate the wavelength of sodium light.

Options:
A λ=4450 A∘
B λ=4850 A∘
C λ=2650 A∘
D λ=6750 A∘

The correct answer is:
λ=4850 A∘

Solution:
Lens will from image of two sources at 70 cm from lens (i.e. screen)
From magnification of lens m=vu=hiho
7030=0.7/2d/2
On solving, d = 0.3 cm
Now, fringes width β=λDd
So, λ=βdD=0.0195×10-2×0.3×10-21=5850 A∘

Topic: Physics, Wave Optics, JEE Main
In a body-centred cubic lattice given below, what are the three distances \( \mathrm{AB}, \mathrm{AC} \) and \( \mathrm{AA} \) '?

Options:
A \( a, \sqrt{2} a, \frac{\sqrt{3} a}{2} \)
B \( a, \frac{\sqrt{3} a}{2}, \sqrt{2} a \)
C \( \frac{\sqrt{3} \mathrm{a}}{2}, \sqrt{2} \mathrm{a}, \mathrm{a} \)
D \( a, \frac{a}{\sqrt{2}}, \frac{\sqrt{3} a}{2} \)

The correct answer is:
\( a, \sqrt{2} a, \frac{\sqrt{3} a}{2} \)

Solution:
Edge length =AB=AD=BC=CD=a
∴ AC = AB 2 + BC 2 = 2 a
Now, AA ′ + A ′ G = 2 AA ′ = 3 a
∴ AA ′ = 3 a 2

Topic: Chemistry, Solid State, JEE Main
In a Bohr's model of an atom, when an electron jumps from n = 1 to n = 3, how much energy will be absorbed?

Options:
A 2.389×10-12 ergs
B 0.239×10-10 ergs
C 2.15×10-11 ergs
D 0.1936×10-10 ergs

The correct answer is:
0.1936×10-10 ergs

Solution:
Energy of an atom when n = 1
E1=-1312(1)2=-1312 kJ mol-1
Similarly energy when n=3,E3=-1312(3)2
=-145.7 kJ mol-1
The energy absorbed when an electron jumps from n=1 to n=3
E3-E1=-145.7--1312=1166.3 kJ mol-1
=1166.36.023×1023=193.6×10-23kJ
=193.6×10-20J1 Joule=107 ergs
193.6×10-13ergs=0.1936×10-10ergs

Topic: Chemistry, Structure of Atom, JEE Main
In a bolt factory, machines A,B and C manufacture respectively 20%, 30% and 50% of the total bolts. Of their output 3, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product. If the bolt drawn is found the defective then the probability that it is manufactured by the machine C is

Options:
A 514
B 928
C 37
D 27

The correct answer is:
514

Solution:
Let X≡Event that product is defective, then

PXA=3100PXB=4100PXC=2100

Now,

PCX=PCPXCPAPXA+PBPXB+PCPXC

⇒PCX=50100×210020100×3100+30100×4100+50100×2100

⇒PCX=10060+120+100

⇒PCX=514

Hence this is the correct option.

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (08 Apr Shift 1)
In a bombing attack, there is 50% chance that a bomb will hit the target. At least two independent hits are required to destroy the target completely. Then the minimum number of bombs, that must be dropped to ensure that there is at least 99% chance of completely destroying the target, is .....

Options:

The correct answer is:
11

Solution:
let probability of hitting the target =p⇒p=12

Let n be the minimum number of bombs

According to given condition
1-C0nP0(1-P)n+C1nP1(1-P)n-1≥99100

⇒ 2n≥(n+1)100

n=10⇒210≥1100 Reject n=11⇒211≥1200 Select

Topic: Mathematics, Probability, JEE Main, JEE Main 2020 (05 Sep Shift 2)
In a borax bead test under hot condition, a metal salt (one from the given) is heated at point B of the flame, resulted in green colour salt bead. The spin-only magnetic moment value of the salt is _______ BM (Nearest integer)
[Given atomic number of \(\mathrm{Cu}=29, \mathrm{Ni}=28, \mathrm{Mn}=25, \mathrm{Fe}=26\)]

Options:

The correct answer is:
6

Solution:
$\mathrm{Fe}^{+3}$ will give green coloured bead when heated at point B.
Number of unpaired $\mathrm{e}^{-}$in $\mathrm{Fe}^{+3}=5$
$\mu=5.92$
Nearest integer $=6$

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2024 (05 Apr Shift 1)
In a box containing 100 eggs, 10 eggs are rotten. The probability that out of a sample of 5 eggs none is rotten if the sampling is with replacement is

Options:
A ${ }^{\left(\frac{1}{10}\right)^5}$
B ${ }^{\left(\frac{1}{5}\right)^5}$
C ${ }^{\left(\frac{9}{5}\right)^5}$
D $\left(\frac{9}{10}\right)^5$

The correct answer is:
$\left(\frac{9}{10}\right)^5$

Solution:
Let $P($ freshegg $)=\frac{90}{100}=\frac{9}{10}=p$
$P($ rottenegg $)=\frac{10}{100}=\frac{1}{10}=q ; \quad n=5, \quad r=5$
So the probability that none egg is rotten
$={ }^5 C_5\left(\frac{9}{10}\right)^5 \cdot\left(\frac{1}{10}\right)^0$ $=\left(\frac{9}{10}\right)^5$

Topic: Mathematics, Probability, JEE Main
In a box of 10 electric bulbs, two are defective. Two bulbs are selected at random one after the other from the box. The first bulb after selection being put back in the box before making the second selection. The probability that both the bulbs are without defect is

Options:
A $\frac{9}{25}$
B $\frac{16}{25}$
C $\frac{4}{5}$
D $\frac{8}{25}$

The correct answer is:
$\frac{16}{25}$

Solution:
Here $P$ (without defected) $=\frac{8}{10}=\frac{4}{5}=p$
$P_{\text {(defected) }}=\frac{2}{10}=\frac{1}{5}=q$ and $n=2 \quad r=2$
Hence required probability $={ }^n C_r p^r \cdot q^{n-r}$
$={ }^2 C_2\left(\frac{4}{5}\right)^2 \cdot\left(\frac{1}{5}\right)^0=\frac{16}{25}$.

Topic: Mathematics, Probability, JEE Main
In a box, there are 20 cards, out of which 10 are labelled as A and the remaining 10 are labelled as B . Cards are drawn at random, one after the other and with replacement, till a second A card is obtained. The probability that the second A card appears before the third B card is:

Options:
A 916
B 1116
C 1316
D 1516

The correct answer is:
1116

Solution:
PA=12 PB=12

PAA+ABA+BBA+ABBA+BBAA+BABA=14+18+18+116+116+116=1116

Topic: Mathematics, Probability, JEE Main, JEE Main 2020 (09 Jan Shift 1)
In a buffer solution having equal concentrations of B- and HB the value of Kb for B- is 10-9 . The pH of buffer solution is

Options:
A 9
B 5
C 4
D 6

The correct answer is:
5

Solution:
For acidic buffer pH can be find out as follows
pH=pKa+log10[salt][acid]
As here Kb=10-9
So Ka=10-5
pKa=5
As pH=5+â¡log10[B-][HB]
As [B-]=[HB]
pH=5+log 1=5 + 0 = 5

Topic: Chemistry, Ionic Equilibrium, JEE Main
In a building there are 15 bulbs of 45W , 15 bulbs of 100W,15 small fans of 10W and 2 heaters of 1kW . The voltage of electric main supply is 220V . The minimum fuse capacity (rated value) of the building will be:

Options:
A 5A
B 25A
C 15A
D 20A

The correct answer is:
20A

Solution:
Total power is 15×45+15×100+15×10+2×1000
=4325W
So current =4325220=19.66 A≈ 20 A

so we have to use a fuse which can tolerate atleast 20 A current.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (07 Jan Shift 2)
In a Cu-voltameter, mass deposited in 30 s is 'm' g. If the time-current graph is shown in the following figure:

What is the electrochemical equivalent of Cu?

Options:
A m/2
B m/3
C m/4
D m63.5

The correct answer is:
m/3

Solution:
Total current passed in 30 sec = Area of current-time curve
=( 1 2 Ã100Ã10 )+( 100Ã10 )+( 100Ã10 )+( 1 2 Ã100Ã10 )
= 3 Ã 100 Ã 10 mA Ã s
= 3000 Ã 10â3 amp s=3 C
W=EQF or WQ= EF=Z=m3

Topic: Chemistry, Electrochemistry, JEE Main
In a capillary tube experiment, a vertical $30 \mathrm{~cm}$ long capillary tube is dipped in water. The water rises up to a height of $10 \mathrm{~cm}$ due to capillary action. If this experiment is conducted in a freely falling elevator, the length of the water column becomes

Options:
A $10 \mathrm{~cm}$
B $20 \mathrm{~cm}$
C $30 \mathrm{~cm}$
D Zero

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

Solution:
The length of the water column will be equal to full length of capillary tube.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
In a car race on straight road, car A takes a time t less than car B at the finish and passes finishing point with a speed v more than that of car B. Both the cars start from rest and travel with constant acceleration a1 and a2 respectively. Then v is equal to:

Options:
A 2a1a2a1+a2t
B a1+a22t
C a1a2t
D 2a1a2t

The correct answer is:
a1a2t

Solution:
Given,
Initially both car is at rest, so u1=u2=0
Acceleration of car A & B is a1 & a2
Let us assume,
Time of reach to destination of car A is, t1=t0
Time of reach to destination of car B is, t2=t0+t
Using second equation of motion, we have
u1t+12a1t12=u2t+12a2t22
⇒0×t+12a1t02=0×t+12a2t0+t2
⇒a1a2t0=t0+t
⇒t0=ta1a2-1
From first equation of motion, we have
v1=a1t0 &v2=a2t0+t
⇒v=v1-v2=a1-a2t0-a2t
⇒v=a1-a2ta1a2-1-a2t
⇒v=ta1a2a1-a2-a2a2a1-a2-a2
⇒v=ta1a2-a2a1a1-a2
⇒v=a1.a2 t

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2019 (09 Jan Shift 2)
In a car race, sound signals emitted by the two cars are detected by a detector on the straight track at the end of the race. The frequency observed for the two cars is 330 Hz and 360 Hz respectively, while the original frequency is 300 Hz for both the cars. The race ends with a separation of 100 m between the cars. Assume both cars move with constant velocity and velocity of sound is 330 m s-1. Find the time taken by the winning car (in second).

Options:

The correct answer is:
4.00

Solution:
330=330330âv1Ã300âââv1=30m s-1

360=330330âv2Ã300

330âv2=33012Ã10=275

v2=55m s-1 ; ÎV=25m s-1

100=ÎVÃTâââââT=4s

Topic: Physics, Waves and Sound, JEE Main
In a carnot engine, the temperature of reservoir is 527°C and that of sink is 200 K. If the workdone by the engine when it transfers heat from reservoir to sink is 12000 kJ, the quantity of heat absorbed by the engine from reservoir is _____ ×106 J.

Options:

The correct answer is:
16

Solution:
Efficiency of Carnot engine is η=1-TLTH=WQ, where, Q is heat absorbed.

Here, TH=527+273=800 K

Now, η=1-200800=12Q

∴ 34=12000Q

∴ Q=16×106 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2022 (27 Jun Shift 1)