datavorous/entrance-exam-dataset huggingface dataset:
The work of $146 \mathrm{~kJ}$ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by $7^{\circ} \mathrm{C}$. The gas is $\left(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$

Options:
A
monoatomic

B
diatomic

C
triatomic

D
a mixture of monoatomic and diatomic

The correct answer is:
diatomic

Solution:
$146=C_v \Delta T$
$\Rightarrow C_v=21 \mathrm{~J} / \mathrm{mol} \mathrm{K}$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2006
The Wurtz-Fittig reaction involves condensation of:

Options:
A
two molecules of aryl halides

B
one molecule of each of aryl-halide and alkyl-halide.

C
one molecule of each of aryl-halide and phenol.

D
two molecules of aralkyl-halides.

The correct answer is:
one molecule of each of aryl-halide and alkyl-halide.

Solution:
Reaction between alkyl halide, aryl halide and sodium in presence of ether is known as Wurtz fitting reaction
$$
\begin{aligned}
& \mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}+2 \mathrm{Na}+\mathrm{ClCH}_3 \longrightarrow \\
& \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3+2 \mathrm{NaCl}
\end{aligned}
$$
Toluene

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2013 (22 Apr Online)
The x and y coordinates of a particles at any time t are given by x=7t+4t2 and y=5t. where x and y are in metre and t in second. The acceleration of particle at t=5 s is

Options:
A zero
B 8 m s-2
C 20 m s-2
D 40 m s-2

The correct answer is:
8 m s-2

Solution:
a=ax2+ay2
= d 2 x dt 2 2 + d 2 y dt 2 2 1 2
Hence, a=d2xdt2=8 m s-2

Topic: Physics, Motion In One Dimension, JEE Main
The x-coordinate of the incentre of the triangle that has the coordinates of midpoints of its sides as 0,1, 1,1 and 1,0 is

Options:
A 1+2
B 1 - 2
C 2 + 2
D 2 - 2

The correct answer is:
2 - 2

Solution:
Let x1, y1, x2, y2 and x3, y3 be the vertices of a triangle,

So,
x1+x2=0, x2+x3=2, x3+x1=2 [mid point formula x1+x22 ]
Similarly,

y1+y2=2, y2+y3=2, y3+y1=0
So, the vertices are 0,0, 0,2, 2,0.
We know that x-coordinate of incentre x¯=ax1+bx2+cx3a+b+c

a=2,b=2,c=22,x1=0,x2=2,x3=0

∴ x¯=0+4+02+2+22
⇒ x¯=2-2

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2013 (07 Apr)
The x-intercept of the common tangent to the parabolas y2=32x and x2=108y is

Options:
A -18
B -12
C -9
D -6

The correct answer is:
-18

Solution:
Tangent to y2=32x is y=mx+8m and

tangent to x2=108y is y=mx-27m2
∴8m=-27m2⇒m3=-827
⇒m=-23
∴x-intercept is 27m=27-23=-18

Topic: Mathematics, Parabola, JEE Main
The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t=43 s is:

Options:
A 332π2 cm s-2
B -π232 cm s-2
C π232 cm s-2
D -332π2 cm s-2

The correct answer is:
-332π2 cm s-2

Solution:
x=sin⁡2π8t
a=-ω2x=-2π8×2π8×sin⁡2π8×43
⇒a=π216×32=-3π232 cm s-2

Topic: Physics, Oscillations, JEE Main
The $x-t$ graph shown in figure represents

Options:
A Velocity of the body is continuously changing
B Instantaneous velocity
C The body travels with constant speed up to time $t_1$ and then stops
D Constant velocity

The correct answer is:
The body travels with constant speed up to time $t_1$ and then stops

Solution:
The body travels with constant speed up to time $t_1$ and then stops
Up to time $t_1$, slope of the graph is constant and after $t_1$ slope is zero i.e. the body travels with constant velocity upto time $t_1$ and then stops moving. Constant velocity definitely means the speed is also constant. From $t_1$ onwards, the displacement is not changing. It is constant, which means velocity is also zero. That means the body stops.

Topic: Physics, Motion In One Dimension, JEE Main
The X-Y plane be taken as the boundary between two transparent media M1 and M2. M1 in Z≥0 has a refractive index of 2 and M2 with Z<0 has a refractive index of 3. A ray of light travelling in M1 along the direction given by the vector A→=43i-33j-5k^, is incident on the plane of separation. The value of difference between the angle of incident in M1 and the angle of refraction in M2 will be _____ degree.

Options:

The correct answer is:
15

Solution:
Given: A→=43i-33j-5k^

As incident vector A makes i angle with normal Z-axis and refracted vector R makes r angle with normal Z- axis.

The angle made by the incident ray with the normal is given as

i=cos-1AzA=cos-15(43)2+(33)2+52

=cos-1510⇒i=60°

Applying Snell's law of refraction at the interface of two media

μ1sini=μ2sinr

⇒2sin60°=3×sinr

⇒r=45°

Difference between i and r =60°-45°=15°

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (29 Jul Shift 1)
The x-z plane separates two media A and B of refractive indices n1=1.5 and n2=2.A ray of light travels from A to B. Its directions in the two media are given by unit vectors Âµâ1=ai^+bj^, Âµâ2,=ci^+dj^, then -

Options:
A ac=43
B ac=34
C bd=43
D bd=34

The correct answer is:
ac=43

Solution:
n1sin i=n2sin r
1.5 μ→1×j^=2 μ→2×j^
1.5 [(ai^+bj^×j^]=2 [ (ci^+dj^) × j^]
1.5 ak^=2ck^
ac=21.5=43

Topic: Physics, Ray Optics, JEE Main
The $\mathrm{Y}$ axis of centre of gravity of semicircular plate $1.32 \mathrm{~m}$ diameter from its base as shown in figure.

Options:
A 0.14 m
B 0.21 m
C 0.28 m
D None of these

The correct answer is:
0.28 m

Solution:
Considering the strip parallel to $\mathrm{X}$-axis
Area of strip, $d A=2 x . d y$
The distance of the C.G. of this area from the $x$-axis is $y$
$\therefore$ Moment of this area about the $\mathrm{x}$-axis
$\begin{aligned} & =y \cdot d A \\ & =y \cdot 2 x d y \\ & =2 x y d y.....(i)\end{aligned}$
But, we know $x^2+y^2=R^2$
or, $\mathrm{x}=\sqrt{R^2-y^2}$
Substituting the above value of $x$ in equation (i), we get
Moment of area $\mathrm{dA}$ about the $x$-axis,
$=2 \sqrt{R^2-y^2} \cdot \mathrm{y} \cdot \mathrm{dy}$
The moment of total area $A$ about the $x$-axis will be obtained by integrating the above equation from 0 to $R$.
$\therefore$ Moment of area $\mathrm{A}$ about $\mathrm{x}$-axis
$=\int_0^{\mathrm{R}} 2 \sqrt{\mathrm{R}^2-\mathrm{y}^2} \cdot \mathrm{y} \mathrm{dy}$
( $\therefore$ y varies from 0 to $R$ )
$=-\int_0^{\mathrm{R}} 2 \sqrt{\mathrm{R}^2-\mathrm{y}^2} \cdot(-2 \mathrm{y}) \mathrm{dy}=-\left[\frac{\left(\mathrm{R}^2-\mathrm{y}^2\right)^{3 / 2}}{3 / 2}\right]_0^{\mathrm{R}}$
$=-\frac{2}{3}\left[0-R^2\right]=\frac{2 R^3}{3}......(ii)$
Also the moment of total area $A$ about $x$-axis $=A \times \bar{y}$
Where
$A=$ Total area of semi-circle $=\frac{\pi \mathrm{R}^2}{2}$
$\bar{y}=$ Distance of C.G of area $A$ from $x$-axis
$\therefore$ Moment of total $\mathrm{A}$ about $\mathrm{x}$-axis $=\frac{\pi \mathrm{R}^2}{2} \times \overline{\mathrm{y}}....(iii)$
Equating the two values given by equations (ii) and (iii),
$\begin{aligned} & \frac{\pi \mathrm{R}^2}{2} \times \overline{\mathrm{y}}=\frac{2 \mathrm{R}^3}{3} \\ & \text { or } \overline{\mathrm{y}}=\frac{2 R^3}{3} \times \frac{2}{\pi R^2} \\ & \text { or } \overline{\mathrm{y}}=\frac{4 R}{3 \pi}\end{aligned}$
Calculation:
Given that,
$\begin{aligned} & R=\frac{D}{2}=\frac{1.32}{2}=0.66 \\ & \bar{y}=\frac{4 R}{3 \pi} \\ & \bar{y}=\frac{4 X 0.66}{3 X \frac{22}{7}} \\ & \bar{y}=0.28 \mathrm{~m}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
The yellow precipitate formed during the chromyl chloride test is chemically

Options:
A chromic acid
B lead chromate
C lead acetate
D sodium chromate

The correct answer is:
lead chromate

Solution:
The yellow precipitate formed during the chromyl chloride test is chemically lead chromate $\left(\mathrm{PbCrO}_{4}\right)$.

Topic: Chemistry, d and f Block Elements, JEE Main
The yield of the product in the reaction
$$
\mathrm{A}_{2}(\mathrm{~g})+2 \mathrm{~B}(\mathrm{~g}) \rightleftharpoons \mathrm{C}(\mathrm{g})+\mathrm{Q} \mathrm{kJ}
$$
would be higher at

Options:
A high temperature and high pressure
B high temperature and low pressure
C low temperature and high pressure
D lowtemperature and low pressure

The correct answer is:
low temperature and high pressure

Solution:
$A_{2}(g)+2 B(g) \rightleftharpoons C(g)+Q k J$ Since, the reaction is exothermic, So, it is favoured by low temperature.
In addition, the number of moles of products is lesser than the number of moles of reactants, thus high pressure favours the forward reaction.

Topic: Chemistry, Chemical Equilibrium, JEE Main
The Young's modulus of a rubber string of length 12 cm and density 1.5 kg m-3 is 5×108 N m-2. When this string is suspended vertically, the increase in its length due to its own weight is (take g=10 m s-2)

Options:
A 2.16×10−10 m
B 9.6×10−11 m
C 9.6×10−3 m
D 2.16×10−3 m

The correct answer is:
2.16×10−10 m

Solution:
Hooke's law for elasticity, σ=γε

γ=σε=FAl∆l here, l is the length of string undergone through tension.

Here, the string is suspended vertically and the string is elongated due to its own weight therefore, weight is acting at the centre of mass which in the given case is located at the half-length of string, therefore, l=12 cm2=6 cm or 0.06 m

γ=σε=FAl∆lγ=FAll2∆l=mgV0.062∆l ⋯⋯A area×l length=V volumeγ=ρg0.062∆l ⋯⋯ρ=mV⇒∆l=ρg0.062γ=1.5 kg m-310 m s-20.06 m25×108 N m-2⇒∆l=2.16×10-10 m

Topic: Physics, Mechanical Properties of Solids, JEE Main
The Young's modulus of a steel wire of length 6 m and cross-sectional area 3 mm2, is 2 × 1111 N/m2 . The wire is suspended from its support on a given planet. A block of mass 4 kg is attached to the free end of the wire. The acceleration due to gravity on the planet is 14 of its value on the earth. The elongation of wire is (Take g on the earth = 10 m/s2 ):

Options:
A 1 cm
B 1 mm
C 0.1 mm
D 0.1 cm

The correct answer is:
0.1 mm

Solution:
The external force or tension F acting on the wire is its own weight. It is given by

F= mg=4 kg×104 m s-2=10 N

The formula to calculate the Young's modulus of the material of the wire is given by

Y= FA∆LL= FL∆LA..........................(1)

where, A is the cross-sectional area, L is the original length and ∆L is the elongation of the wire.

Substitute the given values of the known parameters into equation (1) and solve to calculate the required elongation of the wire.

2×1011 N/m2= 10 N×6 m∆L×3 mm2×1 m2106 mm2∆L= 10 N×6 m2×1011 N/m2×3 mm2×1 m2106 mm2= 0.0001 m×103 mm1 m= 0.1 mm

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2023 (01 Feb Shift 2)
The Young's modulus of a wire of length $L$ and radius $r$ is $Y N / m^2$. If the length and radius are reduced to $L / 2$ and $r / 2$, then its Young's modulus will be

Options:
A $Y / 2$
B $Y$
C $2 Y$
D $4 Y$

The correct answer is:
$Y$

Solution:
Young's modulus of wire does not varies with dimension of wire. It is the property of given material.

Topic: Physics, Mechanical Properties of Solids, JEE Main
The Young's modulus of the material of a wire is $6 \times 10^{12} \mathrm{~N} / \mathrm{m}^2$ and there is no transverse strain in it, then its modulus of rigidity will be

Options:
A $3 \times 10^{12} \mathrm{~N} / \mathrm{m}^2$
B $2 \times 10^{12} \mathrm{~N} / \mathrm{m}^2$
C $10^{12} \mathrm{~N} / \mathrm{m}^2$
D None of the above

The correct answer is:
$3 \times 10^{12} \mathrm{~N} / \mathrm{m}^2$

Solution:
$\quad Y=2 n(1+\sigma)$
For no transverse strain $(\sigma=0)$
$Y=2 \eta \Rightarrow \eta=\frac{Y}{2}=3 \times 10^{12} \mathrm{~N} / \mathrm{m}^2$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The zener diode has a Vz=30 V. The current passing through the diode for the following circuit is___mA.

Options:

The correct answer is:
9

Solution:
i=604000 A

i1=305000 A

i-i1=604000-305000=91000A

current from zener diode

i2=i-i1=9 mA

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (26 Feb Shift 2)
The zener diode normally operates under reverse bias condition, the major use of this fact is in the applications where we require

Options:
A large volume of current
B a constant voltage
C a current that is increasing without any change in applied voltage
D all of the above

The correct answer is:
a constant voltage

Solution:
In reverse bias, once Zener diode attains the breakdown voltage, there is no change in voltage across the diode even if we change the current in the circuit by changing the position of rheostat and that's why the voltage across Zener diode is constant

Topic: Physics, Semiconductors, JEE Main
The Zeta potential is related to which property of colloids?

Options:
A Colour
B Brownian movement
C Charge on surface of colloidal particle
D Tyndall effect

The correct answer is:
Charge on surface of colloidal particle

Solution:
The combination of the two layers of opposite charges around the colloidal particle is called Helmholtz electrical double layer. According to modern views, the first layer of ions is firmly held and is termed fixed layer while the second layer is mobile which is termed diffused layer. Since separation of charge is a seat of potential, the charges of opposite signs on the fixed and diffused parts of the double layer results in a difference in potential between these layers in the same manner as potential difference is developed in a capacitor. This potential difference between the fixed layer and the diffused layer of opposite charges is called the electrokinetic potential or zeta potential.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2022 (28 Jun Shift 1)
Themaximum value of $4 \sin ^{2} x-12 \sin x+7$ is

Options:
A 25
B 4
C does not exist
D None of these

The correct answer is:
None of these

Solution:
$$
\begin{array}{l}
\begin{array}{l}
4 \sin ^{2} x-12 \sin x+7 \\
=4\left(\sin ^{2} x-3 \sin x\right)+7 \\
=4\left[\left(\sin x-\frac{3}{2}\right)^{2}-\frac{9}{4}\right]+7 \\
=4\left(\sin x-\frac{3}{2}\right)^{2}-9+7 \\
=4\left(\sin x-\frac{3}{2}\right)^{2}-2
\end{array} \\
\qquad \begin{array}{l}
=4 \\
=4
\end{array}
\end{array}
$$

Topic: Mathematics, Trigonometric Equations, JEE Main
Themaximum value of the finction $f(x)=3 x^{3}-18 x^{2}+27 x-40$

on the set $\mathrm{S}=\left\{x \in R: x^{2}+30 \leq 11 x\right\}$ is :

Options:
A -122
B -222
C 122
D 222

The correct answer is:
122

Solution:
Consider the function,

$f(x)=3 x(x-3)^{2}-40$

$\operatorname{Now} S=\left\{x \in \infty k: x^{2}+30 \leq 11 x\right\}$

So $x^{2}-11 x+30 \leq 0 \quad \Rightarrow \quad x \circ \in[5,6]$

$\therefore f(x)$ will have maximum value for $x=6$ The maximum value of function is,

$f(6)=3 \times 6 \times 3 \times 3-40=122$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (11 Jan Shift 1)
There are 3 bags which are known to contain 2 white and 3 black balls; 4 white and 1 black balls and 3 white and 7 black balls respectively. A ball is drawn at random from one of the bags and found to be a black ball. Then the probability that it was drawn from the bag containing most black balls is

Options:
A $\frac{3}{5}$
B $\frac{5}{19}$
C $\frac{3}{4}$
D None of these

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

Solution:
Consider the following events :
$A \rightarrow$ Ball drawn is black; $E_1 \rightarrow$ Bag I is chosen;
$E_2 \rightarrow$ Bag II is chosen and $E_3 \rightarrow$ Bag III is chosen.
Then, $P\left(E_1\right)=\left(E_2\right)=P\left(E_3\right)=\frac{1}{3}, P\left(\frac{A}{E_1}\right)=\frac{3}{5}$.

Topic: Mathematics, Probability, JEE Main
There are 3 oranges, 5 apples and 6 mangoes in a fruit basket. Number of ways in which at least one fruit can be selected from the basket is

Options:
A 168
B 167
C 125
D 124

The correct answer is:
167

Solution:
Number of ways =(3 + 1) (5 + 1) (6 + 1) -1 = 167

Topic: Mathematics, Permutation Combination, JEE Main
There are 3 sections in a question paper and each section contains 5 questions. A candidate has to answer a total of 5 questions, choosing at least one question from each section. Then the number of ways, in which the candidate can choose the questions, is:

Options:
A 3000
B 1500
C 2255
D 2250

The correct answer is:
2250

Solution:
A→5Q B→5Q C→5Q

A1,A2,A3,A4,A5 B1,B2,B3,B4,B5
C1,C2,C3,C4,C5

A1A2A3B1C1⇒C13×C35×C15×C15=750, where three questions are picked from any of the subject and one each from other two subjects.

A1A2B1B2C1⇒C23×C25×C25×C15=1500, where one question is picked from one of the subject and two each from rest of the two subjects.

∴Total number of ways=2250

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2020 (05 Sep Shift 2)
There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is _____

Options:

The correct answer is:
5626

Solution:
$\begin{array}{|l|l|l|}
\hline {\begin{array}{c}
\text{From} \\
\text{Group A}
\end{array}} & {\begin{array}{c}
\text{From} \\
\text{Group B}
\end{array}} & \text{Ways of selection} \\
\hline 4 \mathrm{M} & 4 \mathrm{~W} & { }^4 \mathrm{C}_4{ }^4 \mathrm{C}_4=1 \\
\hline 3 \mathrm{M} 1 \mathrm{~W} & 1 \mathrm{M} 3 \mathrm{~W} & { }^4 \mathrm{C}_3{ }^5 \mathrm{C}_1{ }^5 \mathrm{C}_1{ }^4 \mathrm{C}_3=400 \\
\hline 2 \mathrm{M} 2 \mathrm{~W} & 2 \mathrm{M} 2 \mathrm{~W} & { }^4 \mathrm{C}_2{ }^5 \mathrm{C}_2{ }^5 \mathrm{C}_2{ }^4 \mathrm{C}_2=3600 \\
\hline 1 \mathrm{M} 3 \mathrm{~W} & 3 \mathrm{M} 1 \mathrm{~W} & { }^4 \mathrm{C}_1{ }^5 \mathrm{C}_3{ }^5 \mathrm{C}_3{ }^4 \mathrm{C}_1=1600 \\
\hline 4 \mathrm{~W} & 4 \mathrm{M} & { }^5 \mathrm{C}_4{ }^5 \mathrm{C}_4=25 \\
\hline \text{ Total } & \text{5626} \\
\hline
\end{array}$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2024 (04 Apr Shift 2)
There are 5 letters and 5 different envelopes. The number of ways in which all the letters can be put in wrong envelope, is

Options:
A 119
B 44
C 59
D 40

The correct answer is:
44

Solution:
Required numbers
$$
=5 !\left[1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}\right]=44
$$
Note that if $r(0 \leq r \leq n)$ objects occupy the original places and none of the remaining $(\mathrm{n}-\mathrm{r})$ objects occupies its original places then the number of such arrangements
$$
\begin{array}{l}
={ }^{n} C_{r} \cdot(n-r) ! \\
{\left[1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\ldots .+(-1)^{n-2} \frac{1}{(n-r) !}\right]}
\end{array}
$$

Topic: Mathematics, Permutation Combination, VITEEE, VITEEE 2012
There are 5 students in class 10,6 students in class 11 and 8 students in class 12. If the number of ways, in which 10 students can be selected from them so as to include at least 2 students from each class and at most 5 students from the total 11 students of class 10 and 11 is 100k, then k is equal to

Options:

The correct answer is:
238

Solution:
Class 10th 11th 12th Total student 568

From these students, we can choose two students from class 10, three students from class 11 & 5 students from class 12.

So, total number of ways is C25×C36×C58

Again, From these students, we can choose two students from class 10, two students from class 11 & 6 students from class 12.

Number of selection =C25×C26×C68

Again, from these students, we can choose three students from class 10, two students from class 11 & 5 students from class 12. So, total number of ways is C35×C26×C58

⇒ Total number of ways =C25×C36×C58+C25×C26×C68+C35×C26×C58 =C25×C36×C38+C25×C26×C28+C25×C26×C38

=5×42×1×6×5×43×2×1×8×7×63×2×1+5×42×1×6×52×1×8×72×1+5×42×1×6×52×1×8×7×63×2×1

=23800

According to question 100k=23800

⇒k=238

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (25 Jul Shift 1)
There are 6 positive numbers and 8 negative numbers. Three numbers are chosen from them at random and multiplied. The probability that the product is a negative number is

Options:
A 1134
B 1733
C 1635
D 1135

The correct answer is:
1135

Solution:
Total number of ways =14C3
If three numbers are multiplied and the product is a negative number then it should be either 3 negative numbers or 2 positive and 1 negative number
Total favourbale number of ways =8C3+6C28C1
Hence, required probability =8C3+6C28C114C3
=56+120364
=4491

Topic: Mathematics, Probability, JEE Main
There are 7 distinguishable rings. The number of possible five-rings arrangements on the four fingers (except the thumb) of one hand (the order of the rings on each finger is to be counted and it is not required that each finger has a ring) is equal to

Options:
A 214110
B 211410
C 124110
D 141120

The correct answer is:
141120

Solution:
There are 7C5 ways of selecting the rings to be worn.
If a,b,c,d are the number of the rings on the fingers, we need to find the non-negative integers such that a+b+c+d=5.
The numbers of such quadruples is 5+4-1C4-1=8C3.
For each set of 5 rings, there are 5! arrangements.
So, the total number of required arrangements is 7C5Ã8C3Ã5!=141120

Topic: Mathematics, Permutation Combination, JEE Main
There are 8 equal resistances $R$. Two are connected in parallel, such four groups are connected in series, the total resistance of the system will be

Options:
A $R / 2$
B $2 R$
C $4 R$
D $8 R$

The correct answer is:
$2 R$

Solution:
Resistance of parallel group=R/2
Total equivalent resistance $=4 . R / 2=2 R$

Topic: Physics, Current Electricity, JEE Main
There are 1010 radioactive nuclei in a given radioactive element. Its half-life time is 1 min. How many nuclei will remain after 30 s?(2=1.414)

Options:
A 7×109
B 105
C 2×1010
D 4×1010

The correct answer is:
7×109

Solution:
Recall the relationship between the present amount of element and the initial amount of element in terms of the decay constant, N=N0e-λt, now the relation between decay constant and half-life, t12=0.693λ, ⇒N=N0e-0.693tt12

⇒N=1010e-0.35=7×109.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2021 (27 Aug Shift 1)
There are 10 coins in a box out of which 8 are normal and the remaining are with heads on both sides. A coin is chosen at random from the box and tossed 6 times. If it shows heads each time, then the probability that the selected coin has head on both sides is

Options:
A $\frac{16}{17}$
B $\frac{32}{41}$
C $\frac{8}{9}$
D $\frac{12}{13}$

The correct answer is:
$\frac{16}{17}$

Solution:
$P($ both side head $)=\frac{2}{10}$
$P($ Fair coin $)=\frac{8}{10}$
$\begin{aligned} & P\left(\frac{\text { Head }}{\text { Fair }}\right)=\left(\frac{1}{2}\right)^6 \\ & P\left(\frac{\text { Head }}{\text { Unfair }}\right)=1\end{aligned}$
Probability $=\frac{\frac{2}{10} \times 1}{\frac{2}{10} \times 1+\left(\frac{1}{2}\right)^6 \times \frac{8}{10}}=\frac{16}{17}$

Topic: Mathematics, Probability, TS EAMCET, TS EAMCET 2023 (14 May Shift 2)
There are 10 points in a plane, of which no three points are collinear except 4 . Then, the number of distinct triangles that can be formed by joining any three points of these ten points, such that at least one of the vertices of every triangle formed is from the given 4 collinear points is

Options:
A $80$
B $100$
C $96$
D $116$

The correct answer is:
$96$

Solution:
Number of required triangles:
(Select 1 point from 4 collinear and 2 from remaining 6 ) OR (select 2 from collinear points \& 1 from remaining 6 )
$$
\begin{aligned}
& =\left({ }^4 C_1 \times{ }^6 C_2\right)+\left({ }^4 C_2 \times{ }^6 C_1\right) \\
& =(4 \times 15)+(6 \times 6)=96
\end{aligned}
$$

Topic: Mathematics, Permutation Combination, TS EAMCET, TS EAMCET 2023 (14 May Shift 1)
There are 12 volleyball players in all in a college, out of which a team of 9 players is to be formed. If the captain always remains the same, then in how many ways can the team be formed

Options:
A $36$
B $108$
C $99$
D $165$

The correct answer is:
$165$

Solution:
Required number of ways $={ }^{11} \mathrm{C}_2 = 165$.
{Since, captain already be chosen, so now from 11 players 8 are to be chosen}.

Topic: Mathematics, Permutation Combination, JEE Main
There are 15 players in a cricket team, out of which 6 are bowlers, 7 are batsmen and 2 are wicketkeepers. The number of ways, a team of 11 players be selected from them so as to include at least 4 bowlers, 5 batsmen and 1 wicketkeeper, is

Options:

The correct answer is:
777

Solution:
We have,

15 : Players

6 : Bowlers

7 : Batsman

2 : Wicket keepers

Total number of ways for at least 4 bowlers, 5 batsman and 1 wicket keeper

=C46×C67×C12+C46×C57×C22+C56×C57×C12

=777

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (20 Jul Shift 1)