datavorous/entrance-exam-dataset huggingface dataset:
If the Thomson model of the atom was correct, then the result of Rutherford's gold foil experiment would have been:

Options:
A All of the α-particles pass through the gold foil without decrease in speed.
B α-particles are deflected over a wide range of angles.
C All α-particles get bounced back by 180°.
D α-particles pass through the gold foil deflected by small angles and with reduced speed.

The correct answer is:
α-particles pass through the gold foil deflected by small angles and with reduced speed.

Solution:
According to Thomson's plum pudding model, protons are diffused in a positively charged cloud (charge is not centred) and the negatively charged particles are embedded in it. So, the α- particles would deviate by small angles and due to repulsion from protons, their speed would decrease.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2021 (27 Jul Shift 2)
If the three distinct lines $x+2 a y+a=0, x+3 b y$ $+b=0$ and $x+4 a y+a=0$ are concurrent, then the point $(a, b)$ lies on $a$ :

Options:
A
circle

B
hyperbola

C
straight line

D
parabola

The correct answer is:
straight line

Solution:
$$
\begin{aligned}
&x+2 a y+a=0 \\
&x+3 b y+b=0 \\
&x+4 a y+a=0
\end{aligned}
$$
Subtracting equation (3) from (1)
$$
\begin{aligned}
&-2 a y=0 \\
&a y=0 \Rightarrow y=0
\end{aligned}
$$
Putting value of $y$ in equation (1), we get
$$
\begin{aligned}
&x+0+a=0 \\
&x=-a
\end{aligned}
$$
Putting value of $x$ and $y$ in equation (2), we get
$$
\begin{aligned}
&-a+b=0 \\
&a=b
\end{aligned}
$$
Thus, $(a, b)$ lies on a straight line

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2014 (12 Apr Online)
If the three distinct lines x+2ay+a=0,x+3by+b=0 and x+4ay+a=0 are concurrent, then the point a,b lies on a

Options:
A circle
B straight line
C parabola
D hyperbola

The correct answer is:
straight line

Solution:
x+a2y+1=0
x+b3y+1=0
x+a4y+1=0
12aa13bb14aa=0
R2âR2-R1
R3âR3-R1
12aa03b-2ab-a02a0=0
â2ab-a=0
2a=0 or b=a
Hence, locus of a,b is y=x or x=0

Topic: Mathematics, Straight Lines, JEE Main
If the three function $f(x), g(x)$ and $h(x)$ are such that
$h(x)=f(x) \cdot g(x)$ and $f^{\prime}(x) \cdot g^{\prime}(x)=c$
where $c$ is constant, then
$\frac{f^{\prime \prime}(x)}{f(x)}+\frac{g^{\prime \prime}(x)}{g(x)}+\frac{2 c}{f(x) \cdot g(x)}$
is equal to

Options:
A $h^{\prime}(x) \cdot h^{\prime \prime}(x)$
B $\frac{\mathrm{h}(\mathrm{x})}{\mathrm{h}^{\prime \prime}(\mathrm{x})}$
C $\frac{h^{\prime \prime}(x)}{h(x)}$
D $\frac{h(x)}{h^{\prime}(x)}$

The correct answer is:
$\frac{h^{\prime \prime}(x)}{h(x)}$

Solution:
Given, $h(x)=f(x) \cdot g(x)$ and $f^{\prime}(x) \cdot g^{\prime}(x)=c$
Now, $h^{\prime}(x)=f^{\prime}(x) \cdot g(x)+f(x) \cdot g^{\prime}(x)$
$h^{\prime \prime}(x)=f^{\prime \prime}(x) \cdot g(x)+f^{\prime}(x) \cdot g^{\prime}(x)$
$+f^{\prime}(x) \cdot g^{\prime}(x)+f(x) \cdot g^{\prime \prime}(x)$
$h^{\prime \prime}(x)=f^{\prime \prime}(x) \cdot g(x)+f(x) \cdot g^{\prime \prime}(x)$
$+2 f^{\prime}(x) \cdot g^{\prime}(x)$
$h^{\prime \prime}(x)=f^{\prime \prime}(x) \cdot g(x)+f(x) \cdot g^{\prime \prime}(x)+2 c \quad \ldots$ (i)
Now, we find
$$
\frac{f^{\prime \prime}(x)}{f(x)}+\frac{g^{\prime \prime}(x)}{g(x)}+\frac{2 c}{f(x) \cdot g(x)}
$$
$$
\begin{aligned}
&=\frac{f^{\prime \prime}(x) \cdot g(x)+g^{\prime \prime}(x) \cdot f(x)+2 c}{f(x) \cdot g(x)} \\
&=\frac{h^{\prime \prime}(x)}{h(x)} \quad[\text { from Eq. (i)] }
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
If the three lines $x-3 y=p, a x+2 y=q$ and $a x+y=r$ form a right-angled triangle then :

Options:
A
$a^2-9 a+18=0$

B
$a^2-6 a-12=0$

C
$a^2-6 a-18=0$

D
$a^2-9 a+12=0$

The correct answer is:
$a^2-9 a+18=0$

Solution:
Since three lines $x-3 y=p$, $a x+2 y=q$ and $a x+y=r$ form a right angled triangle $\therefore$ product of slopes of any two lines $=-1$ Suppose $a x+2 y=q$ and $x-3 y=p$ are $\perp$ to each other.
$$
\therefore \frac{-a}{2} \times \frac{1}{3}=-1 \Rightarrow a=6
$$
Now, consider option one by one $a=6$ satisfies only option (a)
$\therefore$ Required answer is $a^2-9 a+18=0$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2013 (09 Apr Online)
If the three normals drawn to the parabola, y2=2x pass through the point a,0, a≠0, then a must be greater than :

Options:
A 12
B -12
C -1
D 1

The correct answer is:
1

Solution:
Let the equation of the normal be y=mx-2am-am3

Here 4a=2

a=12

y=mx-m-12m3

It passes through A(a, 0) then

0=am-m-12m3

m=0,m2-2(a-1)=0

For real values of m

2(a-1)>0

⇒a>1

Topic: Mathematics, Parabola, JEE Main, JEE Main 2021 (16 Mar Shift 1)
If the three planes $x=5,2 x-5 a y+3 z-2=0$ and $3 b x+y-3 z=0$ contain a common line, then $(a, b)$ is equal to

Options:
A
$\left(\frac{8}{15},-\frac{1}{5}\right)$

B
$\left(\frac{1}{5},-\frac{8}{15}\right)$

C
$\left(-\frac{8}{15}, \frac{1}{5}\right)$

D
$\left(-\frac{1}{5}, \frac{8}{15}\right)$

The correct answer is:
$\left(\frac{1}{5},-\frac{8}{15}\right)$

Solution:
Let the direction ratios of the common line be $\ell, m$ and $n$.
$$
\begin{aligned}
& \therefore \ell \times 1+\mathrm{m} \times 0+\mathrm{n} \times 0=0 \Rightarrow \ell=0 \\
& 2 \ell-5 m a+3 n=0 \Rightarrow 5 m a-3 n=0 \\
& 3 \ell b+m-3 n=0 \Rightarrow m-3 n=0
\end{aligned}
$$
Subtracting (3) from (1), we get $m(5 a-1)=0$
Now, value of $m$ can not be zero because if $m=0$ then $n=0$
$\Rightarrow \ell=m=n=0$ which is not possible.
Hence, $5 a-1=0 \Rightarrow a=\frac{1}{5}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2012 (19 May Online)
If the three points $(3 q, 0),(0,3 p)$ and $(1,1)$ are collinear then which one is true ?

Options:
A $\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=1$
B $\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=1$
C $\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=3$
D $\frac{1}{\mathrm{p}}+\frac{3}{\mathrm{q}}=1$

The correct answer is:
$\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=3$

Solution:
Hints: $\mathrm{A}(3 \mathrm{q}, 0) \mathrm{B}(0,3 \mathrm{p}) \mathrm{C}(11)$
Slope $=1 \mathrm{AC}=5 \log \mathrm{BC}$
$$
\begin{aligned}
& \frac{1-0}{1-3 q}=\frac{1-3 p}{1-0}=3, \frac{1}{1-3 q}=\frac{1-3 p}{1} \\
& 1=(1-3 p)(1-3 q), 1=1-3 q-3 p+9 p q \\
& \Rightarrow 3 p+3 q=9 p q, \frac{1}{q}+\frac{1}{p}=3
\end{aligned}
$$

Topic: Mathematics, Straight Lines, JEE Main
If the time period is doubled, then the angular momentum of the body will (provided the moment of inertia of the body is constant)

Options:
A remain constant
B quadruple
C become half
D double

The correct answer is:
become half

Solution:
As we know.

L=Iω

⇒L=I×2πT

L∝1T

Given, T2=2T

Hence L1L2=2TT

⇒ L2=L12

Topic: Physics, Rotational Motion, JEE Main
If the time period of a two meter long simple pendulum is 2 s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:

Options:
A 2π2 m s-2
B 16 m s-2
C π2 m s-2
D 9.8 m s-2

The correct answer is:
2π2 m s-2

Solution:
T=2πl g

2=2π2g

⇒g=2π2

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (25 Feb Shift 1)
If the time period $t$ of the oscillation of a drop of liquid of density $d$, radius $r$, vibrating under surface tension $s$ is given by the formula $t=\sqrt{r^{2 b} s^c d^{a / 2}}$. It is observed that the time period is directly proportional to $\sqrt{\frac{d}{s}}$. The value of $b$ should therefore be :

Options:
A
$\frac{3}{4}$

B
$\sqrt{3}$

C
$\frac{3}{2}$

D
$\frac{2}{3}$

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

Solution:
No Solution Available

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2013 (23 Apr Online)
If the time period (T) of vibration of a liquid drop depends on surface tension (S) , radius (r) of the drop and density (ρ) of the liquid, then the expression of T is

Options:
A T=k ρ r 3 /S
B T=k ρ 1/2 r 3 /S
C T=k ρ r 3 / S 1/2
D None of these

The correct answer is:
T=k ρ r 3 /S

Solution:
Let T∝ S x r y ρ z by substituting the dimension of [T]=[T] [S]=[M T −2 ], [r]=[L], [ρ]=[M L −3 ] and by comparing the power of both the sides x=−1/2, y=3/2, z=1/2 so T∝ ρ r 3 /S ⇒T=k ρ r 3 S

Topic: Physics, Units and Dimensions, JEE Main
If the total energy transferred to a surface in time t is 6.48×105 J, then the magnitude of the total momentum delivered to this surface for complete absorption will be :

Options:
A 2.46×10-3 kg m s-1
B 2.16×10-3 kg m s-1
C 1.58×10-3 kg m s-1
D 4.32×10-3 kg m s-1

The correct answer is:
2.16×10-3 kg m s-1

Solution:
As we know,

p=Ec, where c is velocity of light.

Therefore, p=6.48×1053×108=2.16×10-3 kg m s-1

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2024 (30 Jan Shift 2)
If the total maximum value of the function fx=3e2sinxsin2x, x∈0, π2, is ke, then ke8+k8e5+k8 is equal to

Options:
A e3+e6+e11
B e5+e6+e11
C e3+e6+e10
D e3+e5+e11

The correct answer is:
e3+e6+e11

Solution:
Given function is fx=3e2sinxsin2x

For maxima or minima f'(x)=0

f'x=fx2sinxcosx×ln3e2sinx+sin2x1×2sinx3e×3e2-1sin2x×cosx

⇒fxsin2xln3e2sinx-sinxcosx=0

Now on equation we get, sin2x=0 (not possible)

So, ln3e2sinx=+12

⇒3×e2sinx=e12

⇒sinx=32

⇒fmax=e38=e118e⇒k=e118

⇒ke8+k8e5+k8=e3+e6+e11

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2023 (12 Apr Shift 1)
If the total number of positive integral solutions of 15<x1+x2+x3â¤20 is k, then the value of k100 is equal to

Options:

The correct answer is:
6.85

Solution:
15<x1+x2+x3â¤20
âx1+x2+x3=16+r, where r=0,1,2,3,4.
Now, the number of positive integral solutions of x1+x2+x3=16+r is
16+r-1C3-1=15+rC2
Thus, the required number of solutions is
âr=0415+rC2=15C2+16C2+17C2+18C2+19C2
=15C2-16C3+16C3+16C2+17C2+18C2+19C2
=15C2-16C3+20C3=685

Topic: Mathematics, Permutation Combination, JEE Main
If the total number of ways of selecting two numbers from the set 1,2,3,………..,89,90 such that their sum is divisible by 3 is k, then k500 is

Options:

The correct answer is:
2.67

Solution:
Divide the numbers in 3 sets of the form 3k,3k+1 and 3k+2
S1=1,4,7,……,88
S2=2,5,8,…….,89
S3=3,6,9,……,90
For the sum of two numbers to be divisible by three, either we will choose the two numbers from S3 or one number from S1 and one from S2

i.e. k=C130×C130+C230=30×30+30×292=1335

Hence, 1335500=2.67

Topic: Mathematics, Permutation Combination, JEE Main
If the trigonometric equation tan-1⁡x=2sin-1⁡a has a solution, then the complete set of values of a is

Options:
A 12<a<12
B all real values of a
C a<12
D a<12

The correct answer is:
a<12

Solution:
∵tan-1⁡x∈-π2,π2
∴-π2<2sin-1⁡a<π2
-π4<sin-1⁡a<π4⇒a∈-12,12

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If the truth value of the Boolean expression ((p∨q)∧(q→r)∧(~r))→(p∧q) is false, then the truth values of the statements p, q, r respectively can be:

Options:
A FTF
B TFF
C TFT
D FFT

The correct answer is:
TFF

Solution:
pqrp∨qq→r~r(p∨q)∧(q→r)∧~rp∧q((p∨q)∧(q→r)∧~r)→p∧q T TTTTFFTTTTFTFTFTTTFTTTFFFTTFFTTTTFFFTTTTFFFTFTFTFTFFTFFTFTFFFTFFFFTTFFT

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (26 Aug Shift 1)
If the truth value of the statement p→~q∨r is false F, then the truth values of the statements p,q,r are respectively

Options:
A T,F,T
B T,F,F
C T,T,F
D F,T,T

The correct answer is:
T,T,F

Solution:
Recall, p→q is false if and only if p is true and q is false

Alternatively,

Truth table of statement p→~q∨r

p	q	r	~q	~q∨r	p→~q∨r

T	T	T	F	T	T

T	T	F	F	F	F

T	F	T	T	T	T

T	F	F	T	T	T

F	T	T	F	T	T

F	T	F	F	F	T

F	F	T	T	T	T

F	F	F	T	T	T

So, the truth values are p: true, q: true, r: false

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2019 (12 Apr Shift 1)
If the two circles $(x-1)^2+(y-3)^2=r^2$ and $x^2+y^2-8 x+2 y+8=0$ intersect in two distinct point, then

Options:
A
$r>2$

B
$2 < \mathrm{r} < 8$
C
$r < 2$
D
$\mathrm{r}=2$

The correct answer is:
$2 < \mathrm{r} < 8$

Solution:
$\left|r_1-r_2\right|=C_1 C_2$ for intersection
$\Rightarrow \mathrm{r}-3 < 5 \Rightarrow \mathrm{r} < 8$
and $\mathrm{r}_1+\mathrm{r}_2>\mathrm{C}_1 \mathrm{C}_2, \mathrm{r}+3>5 \Rightarrow \mathrm{r}=2$
From (1) and (2), $2 < \mathrm{r} < 8$.

Topic: Mathematics, Circle, JEE Main, JEE Main 2003
If the two circles, $x^2+y^2+2 g_1 x+2 f_1 y=0$ and $x^2+y^2+2 g_2 x+2 f_2 y=0$ touches each other, then

Options:
A $f_1 g_1=f_2 g_2$
B $\frac{f_1}{g_1}=\frac{f_2}{g_2}$
C $f_1 f_2=g_1 g_2$
D None of these

The correct answer is:
$\frac{f_1}{g_1}=\frac{f_2}{g_2}$

Solution:
Since the 2 circles touch each other
$\mathrm{C}_1 \mathrm{C}_2=\mathrm{r}_1 \pm \mathrm{r}_2$
$\begin{aligned} & \Rightarrow \sqrt{\left(g_1-g_2\right)^2+\left(f_1-f_2\right)^2}=\sqrt{g_1^2+f_1^2} \pm \sqrt{g_2^2+f_2^2} \\ & -2 g_1 g_2-2 f_1 f_2= \pm 2\left(\sqrt{g_1^2+f_1^2}\right)\left(\sqrt{g_2^2+f_2^2}\right) \\ & \left(g_1 g_2\right)^2+\left(f_1 f_2\right)^2+2 g_1 g_2 f_1 f_2=\left(g_1 g_2\right)^2+\left(f_1 f_2\right)^2+\left(g_1 f_2\right. \\ & 2=\frac{g_1 f_2}{g_2 f_1}+\frac{g_2 f_1}{g_1 f_2} \\ & \text { Let } \frac{g_1 f_2}{g_2 f_1}=a \\ & \Rightarrow 2=a+\frac{1}{a} \Rightarrow a^2+-2 a+1=0 \\ & \rightarrow a=1 \rightarrow \frac{g_1 f_2}{g_2 f_1}=1\end{aligned}$
$\frac{f_1}{g_1}=\frac{f_2}{g_2}$

Topic: Mathematics, Circle, JEE Main
If the two equations $x^2-c x+d=0$ and $x^2-a x+b=0$ have one common root and the second has equal roots, then $2(b+d)=$

Options:
A $0$
B $a+c$
C $a c$
D $-a c$

The correct answer is:
$a c$

Solution:
Let roots of $x^2-c x+d=0$ be $\alpha, \beta$ then roots of $x^2-a x+b=0$ be $\alpha, \alpha$
$\therefore \alpha+\beta=c, \alpha \beta=d, \alpha+\alpha=a, \alpha^2=b$
Hence $2(b+d)=2\left(\alpha^2+\alpha \beta\right)=2 \alpha(\alpha+\beta)=a c$

Topic: Mathematics, Quadratic Equation, JEE Main
If the two lines l1:x-23=y+1-2,z=2 and l2:x-11=2y+3α=z+52 are perpendicular, then an angle between the lines l2 and l3:1-x3=2y-1-4=z4 is

Options:
A cos-1294
B sec-1294
C cos-1229
D cos-1229

The correct answer is:
sec-1294

Solution:
Given l1:x-23=y+1-2=z-20

l2:x-11=y+32α2=z+52

l3:x-1-3=y-12-2=z-04

Since l1 is perpendicular to l2, so 3-α+0131+α24+4=0⇒α=3

Now for angle between l2 & l3,

cosθ=1×-3+(-2)α2+2×41+α24+49+4+16

cosθ=-3-α+85+α2429

Putting α=3

cosθ=229429=429

θ=cos-1429⇒θ=sec-1294

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (26 Jun Shift 1)
If the two lines x+a-1y=1 and 2x+a2y=1,a∈R-0,1 are perpendicular, then the distance of their point of intersection from the origin is

Options:
A 25
B 25
C 25
D 25

The correct answer is:
25

Solution:
If two lines a1x+b1y=c1 and a2x+b2y=c2 are perpendicular, then a1a2+b1b2=0.

Since, given lines x+a-1y=1 and 2x+a2y=1 are perpendicular, hence, we get

1⋅2+a-1⋅a2=0

⇒a3-a2+2=0

⇒a+1⋅a2-2a+2=0

⇒a=-1 or a2-2a+2=0

But, the discriminant of the quadratic equation is -22-4×1×2=4-8=-4, hence, no real roots exist.

Thus, the only possible value of a is -1.

And, hence the lines are x-2y=1 and 2x+y=1.

Point of intersection of these two lines is 35, -15

We know that, the distance of a point x, y from origin is x2+y2
Hence, distance of the point 35, -15 from origin =925+125=25.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2019 (09 Apr Shift 2)
If the two metals A and B are exposed to radiation of wavelength 350 nm. The work functions of metals A and B are 4.8 eV and 2.2 eV. Then choose the correct option

Options:
A Metal B will not emit photo-electrons
B Both metals A and B will emit photo-electrons
C Both metals A and B will not emit photoelectrons
D Metal A will not emit photo-electrons

The correct answer is:
Metal A will not emit photo-electrons

Solution:
As we know, for photoelectric emission, hν>ϕ.

Now,

hν=hcλ=1240350 eV=3.54 eV

Therefore, only metal B will emit photoelectrons.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (31 Jan Shift 2)
If the two roots of the equation, a-1 x4+x2+1+a+1x2+x+12=0 are real and distinct, then the set of all values of a is equal to

Options:
A 0,12
B -12,0 âª0,12
C -â,-2Â âª2,Â â
D -12,Â 0

The correct answer is:
-12,0 âª0,12

Solution:
a-1x4+x2+1+a+1x2+x+12=0
x2+x+1 a-1x2-x+1+a+1(x2+ x+1)=0
x2+x+12ax2+2x+2a=0
⇒ x2+x+1 ax2+x+a=0
If two Roots are real
Then, roots of ax2+x+a=0 should be real & distinct.
∴1-4a2>0
1-2a1+2a>0

As a≠0 therefore
a∈-12,0 ∪0,12

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2015 (11 Apr Online)
If the uncertainty in the position of a particle is equal to its de-Broglie wavelength, the minimum uncertainty in its velocity should be

Options:
A 14π
B v4π
C v4πm
D mv4π

The correct answer is:
v4π

Solution:
âx.m.âvâ¥h4Ï or Î».m. âvâ¥h4Ï
or hmvm. âvâ¥h4Ï
or âvâ¥v4Ï

Topic: Chemistry, Structure of Atom, JEE Main
If the uncertainty in the position of proton is \( 6 \times 10^{-8} \mathrm{~m} \), then the minimum uncertainty in its speed will be

Options:
A \( 1 \mathrm{cms}^{-1} \)
B \( 1 \mathrm{~ms}^{-1} \)
C \( 1 \mathrm{mms}^{-1} \)
D \( 100 \mathrm{~ms}^{-1} \)

The correct answer is:
\( 1 \mathrm{~ms}^{-1} \)

Solution:
Δp=mΔv=ℏΔx
or Δv=ℏmΔx=1.034×10-341.67×10-27×6×10-8
≈1ms-1

Topic: Physics, Dual Nature of Matter, JEE Main
If the uncertainty in velocity and position of a minute particle in space are, 2.4×10-26ms-1 and 10-7m respectively. The mass of the particle in g is___(Nearest integer)

(Given : h=6.626×10-34Js)

Options:

The correct answer is:
22

Solution:
ΔV=2.4×10-26 ms-1

Δx=10-7 m

∵ Δp·Δx=h4π

∴ mΔV.Δx=h4π

⇒m×2.4×10-26×10-7=6.626×10-344×π

m=6.6269.6×π×10-1

m=0.02198kg

m=21.98gm

nearest integer =22

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2022 (27 Jun Shift 1)
If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with m fraction of octahedral holes occupied by aluminium ions and n fraction of tetrahedral holes occupied by magnesium ions, m and n, respectively, are

Options:
A 12,18
B 1,14
C 12,12
D 14,18

The correct answer is:
12,18

Solution:
No. of oxygen atoms per unit cell in ccp =4O2-
No. of octahedral voids per unit cell =4Al3+
No. of Tetrahedral voids per unit cell =8Mg3+
Total negative charge due to oxygen atoms =8
Net charge must be zero.
m43+2n8+4-2=0
3m+4n=2
(A) 32+48=2 is correct.
(B) 3×1+4×14=4≠2 is incorrect.
(C) 3×12+4×12=4≠2 is incorrect
(D) 3×14+4×18=34+24=54≠2 is incorrect.

Topic: Chemistry, Solid State, JEE Main
If the unit of force and length be each increased by four times, then the unit of energy is increased by

Options:
A 16 times
B 8 times
C 2 times
D 4 times

The correct answer is:
16 times

Solution:
Since unit of energy $=$ (unit of force). (unit of length) so if we increase unit of length and force, each by four times, then unit of energy will increase by sixteen times.

Topic: Physics, Work Power Energy, JEE Main
If the unit of length and force be increased four times, then the unit of energy is

Options:
A Increased 4 times
B Increased 8 times
C Increased 16 times
D Decreased 16 times

The correct answer is:
Increased 16 times

Solution:
Energy $=$ force $\times$ distance, so if both are increased by 4 times then energy will increase by 1.6 times.

Topic: Physics, Units and Dimensions, JEE Main
If the units of mass, length and time are doubled then unit of angular momentum will be

Options:
A Doubled
B Tripled
C Quadrupled
D Eight times the original value

The correct answer is:
Quadrupled

Solution:
The dimensions of angular momentum are
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â J=ML2T-1
When units of mass, length and time are doubled, than
Â Â Â Â Â Â Â Â Â Â Â Â Â J′=2M2L22T-1
âÂ Â Â Â Â Â Â Â J′=4ML2T-1=4Â J
Unit of angular momentum is quadrupled.

Topic: Physics, Units and Dimensions, JEE Main
If the value of 1+23+632+1033+….. upto ∞log0.2513+132+133+…. upto ∞ is l, then l2 is equal to

.

Options:

The correct answer is:
3

Solution:
l=1+23+632+1033+….. upto ∞log0.2513+132+133+…. upto ∞

Assume, S=1+23+632+1033+….. upto ∞

S=1+23+632+1033+…-S3=13+232+633+…2S3=1+13+432+433+…

2S3=43+432+433+….

S=32431-13=3

Now l=3log0.25131-13

l=3log1412=312=3

⇒l2=3

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (25 Jul Shift 1)
If the value of 1+tan 1o1+tan 2o1+tan 3oâ¦â¦â¦.(1+tan 44o)(1+tan 45o) is 2Î» , then the sum of the digits of the number Î» is

Options:
A 3
B 6
C 5
D 4

The correct answer is:
5

Solution:
If, A+B=45o
tanâ¡A+B=1
âµtan(A+B)= tanA+tanB 1âtanAtanB
âtanâ¡A+tanâ¡B=1-tanâ¡Atanâ¡B
â 1+tanA (1+tanB) = 2
LHS = 1+tan 1 o ) (1+tan 44 o ) (1+tan 2 o ) 1+tan 43 o ) â¦[ 1+tan 45 o ) for each 1+tanÎ¸ 1+tan Ï 4 âÎ¸ =2
= 2 22 1+1
=223
= 2 Î»
then, Î»=23.

Hence the sum of digits of Î» is 2+3
=5

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If the value of a third-order determinant is \( 11 \), then the value of the square of the determinant formed by the cofactors will be

Options:
A \( 11 \)
B \( 121 \)
C \( 1331 \)
D \( 14641 \)

The correct answer is:
\( 14641 \)

Solution:
Given, Determinant of order 3 is 11

then, the value of determinant formed by cofactors is

C=An-1

Where n=Order of matrix.

Here n=3

∴C=113-1

⇒C=112

⇒C=121

Hence, the value of C2 is 1212=14641.

Topic: Mathematics, Determinants, JEE Main
If the value of integral ∫dxx+x2-12=ax3-x+bx2-11b+C, (where, C is the constant of integration), then a×b is equal to

Options:
A 1
B 49
C 2
D 94

The correct answer is:
49

Solution:
∫x-x2-12dx=∫x2+x2-1-2xx2-1dx
=2x33-x+23x2-132+C ⇒a=b=23

Topic: Mathematics, Indefinite Integration, JEE Main
If the value of limx→π6cosx+π31-3tan⁡x is equal to λ, then the value of 120λ2 is equal to

Options:

The correct answer is:
22.5

Solution:
Let, x=π6-t, then
limt→0cosπ2-t1-3tanπ6-t
=limt→0sin⁡t1-31-3tan⁡t3+tan⁡t

=limt→0sin⁡t3+tant4tan⁡t=34=λ

Hence, 120λ2=120×316=22.5

Topic: Mathematics, Limits, JEE Main
If the value of limx→02-cosxcos2xx+2x2 is equal to ea, then a is equal to_____.

Options:

The correct answer is:
3

Solution:
We have,

limx→02-cosxcos2xx+2x2→1∞

Hence,
=elimx→01-cosxcos2xx2×x+2
Now,

limx→01-cosxcos2xx2→00

Using L'Hospital Rule

limx→0sinxcos2x-cosx×12cos2x×-2sin2x2x

=limx→0sinxcos2x+sin2x·cosx2xcos2x

=limx→0sin3x2xcos2x

=32limx→0sin3x3x×limx→01cos2x

=32limx→0sin3x3x×1

=32

So,

elimx→01-cosxcos2xx2x+2

=e32×2=e3

⇒a=3

Topic: Mathematics, Limits, JEE Main, JEE Main 2021 (20 Jul Shift 1)
If the value of mode and mean is 60 and 66 respectively, then the value of median is

Options:
A 70
B 64
C 90
D 50

The correct answer is:
64

Solution:
$\quad$ Mode $=3$ Median -2 Mean

$\therefore \quad$ Median $=\frac{1}{3}($ mode +2 mean $)=\frac{1}{3}(60+2 \times 66)=64$

Topic: Mathematics, Statistics, JEE Main
If the value of R is increased in the potentiometer wire AB (Fig 3.4), in which direction will the balance point J shift?

Options:
A J will shift towards A.
B J will remain at the same position.
C J will shift towards B.
D The direction of J's shift cannot be determined.

The correct answer is:
J will shift towards B.

Solution:
Correct Option is : (C)
J will shift towards B.

Topic: Chemistry, Current Electricity, NEET
If the value of real number α>0 for which x2-5αx+1=0 and x2-αx-5=0 have a common real roots is 32β then β is equal to ________

Options:

The correct answer is:
13

Solution:
Given,

Quadratic equation x2-5αx+1=0 and x2-αx-5=0 have a common real root,

So, taking k to be common root then equation becomes,

k2-5αk+1=0 ......1

k2-αk-5=0 .....2

Now subtraction above equations we get,

k=64α

Now putting the value of k in equation 1 we get,

64α2-5α64α+1=0

⇒64α2-264=0

⇒3616α2=264

⇒α2=926

⇒α=32×13

Now comparing with given value 32β we get, β=13

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2023 (30 Jan Shift 2)
If the value of the definite integral A=∫010πsin⁡xdx is equal to kπ, then the absolute value of k is equal to (where, ⋅ is the greatest integer function)

Options:

The correct answer is:
5

Solution:
As period of sin⁡x is 2π,
I=5∫02πsin⁡xdx
I=5∫0π0dx+∫π2π-1dx
=5-xπ2π
=-5π
∴k=-5⇒k=5

Topic: Mathematics, Definite Integration, JEE Main
If the value of the expression tanâ¡12cos-1â¡25 is in the form of a+ b where a, bâZ, then the value of a+b b is

Options:

The correct answer is:
0.6

Solution:
We know that,
tanâ¡x2=1-cosâ¡xsinâ¡x
â´ tanâ¡12cos-1â¡25 =1-cosâ¡cos-1â¡25sinâ¡cos-1â¡25
=1-2/5sinâ¡sin-1â¡1/5
=1-2/51/5=5-2
Now, a+ b =â2+ 5
âa=â2 and b=5
Hence, a+b b = 3 5 =0.6

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main