datavorous/entrance-exam-dataset huggingface dataset:
The radius r, length l and resistance R of a metal wire was measured in the laboratory as

r=0.35±0.05 cm, R=100±10 ohm, l=15±0.2 cm

The percentage error in resistivity of the material of the wire is :

Options:
A 25.6%
B 39.9%
C 37.3%
D 35.6%

The correct answer is:
39.9%

Solution:
The resistance of the wire is given as, R=ρlA

Therefore, ρ=RAl

ρ=Rπr2l

For percentage error, we can write Δρρ=ΔRR+2Δrr+Δll

Δρρ×100=ΔRR×100+2Δrr×100+Δll×100

=10100+2×0.050.35+0.215×100

=110+27+175×100

Hence, Δρρ=39.9%

Topic: Physics, Mathematics in Physics, JEE Main
The radius r, length l and resistance R of a metal wire was measured in the laboratory as

r=0.35±0.05 cm, R=100±10 ohm, l=15±0.2 cm

The percentage error in resistivity of the material of the wire is :

Options:
A 25.6%
B 39.9%
C 37.3%
D 35.6%

The correct answer is:
39.9%

Solution:
The resistance of the wire is given as, R=ρlA

Therefore, ρ=RAl

ρ=Rπr2l

For percentage error, we can write Δρρ=ΔRR+2Δrr+Δll

Δρρ×100=ΔRR×100+2Δrr×100+Δll×100

=10100+2×0.050.35+0.215×100

=110+27+175×100

Hence, Δρρ=39.9%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2024 (01 Feb Shift 1)
The radius R of the soap bubble is doubled under isothermal condition. If T be the surface tension of the soap bubble, the work done in doing so is given by

Options:
A 32ÏR2T
B 24ÏR2T
C 8πR2T
D 4πR2T

The correct answer is:
24πR2T

Solution:
The surface energy of a bubble
E = Tension × surface area
E=T×A
For soap bubble
E=2T×A
E=2T×4πR2
Here, R = radius of the bubble
If the radius is doubled then we have
E′=2T×4πR2
E′=2T×4π2R2
E′=32πR2T

Work done = E'-E

⇒ 32πR2T-8πR2T=24πR2T

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The radius ratio of KF is 0.98. The structure of KF is the type.

Options:
A NaCl
B ZnS
C CsCl
D CaF2

The correct answer is:
CsCl

Solution:
As r+r- lies in the range 0.732 - 1.000. Hence co-ordination number is 8 and KF is of CsCl type.

Topic: Chemistry, Solid State, JEE Main
The radius ratio of KF is 0.98 . The structure of KF is of the type

Options:
A NaCl
B ZnS
C CsCl
D Graphite

The correct answer is:
CsCl

Solution:
As r+r- lies in the range 0.732-1.000. Hence the coordination number is 8 and crystal is of CsCl type.

Topic: Chemistry, Solid State, JEE Main
The random or zig-zag motion of the colloidal particles in the dispersion medium is referred to as

Options:
A Electro-osmosis
B Electrophoresis
C Brownian movement
D Tyndall effect

The correct answer is:
Brownian movement

Solution:
Correct option is 3. Brownian movement
Colloidal sol particles in a dispersion medium are continuously bombarded from all sides as a result of which it causes zig-zag motion which is known as Brownian motion

Topic: Chemistry, Surface Chemistry, JEE Main
The random variable X follows binomial distribution B(n, p), for which the difference of the mean and the variance is 1.

If 2 P(X=2)=3 P(X=1), then n2P(X>1) is equal to

Options:
A 15
B 11
C 12
D 16

The correct answer is:
11

Solution:
Given that the difference between the mean and variance is 1.

⇒np-npq=1

⇒np(1-q)=1

⇒np2=1

Also given that,

⇒2P(X=2)=3P(X=1)

⇒2·C2np2qn-2=3·C1np·qn-1

⇒2·n·(n-1)2·p=3·n·q

⇒(n-1)p=3(1-p)

⇒1p2-1p=3(1-p)

⇒(1-p)(1+p)p=3(1-p)

⇒1+p=3p

⇒ p=12

∴n=4.

Now,

n2P(x>1)=n2(1-P(x=1)-P(x=0))=161-C14·124-124=11

Hence this is the correct option.

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (13 Apr Shift 2)
The range of a for which the equation x2+ax-4=0 has its smaller root in the interval -1, 2 is

Options:
A -â, -3
B 0,Â 3
C 0,Â â
D -â,Â -3âªÂ 0,Â â

The correct answer is:
-â, -3

Solution:
Clearly, f- 1> 0, f2< 0
since, f0=- 4 < 0
â 1 - a - 4 > 0â - a - 3 > 0â - a > 3
or a < - 3 and 4+2a - 4 < 0
â a < 0
â aâ - â, - 3.

Topic: Mathematics, Quadratic Equation, JEE Main
The range of a projectile for a given initial velocity is maximum when the angle of projection is $45^{\circ}$. The range will be minimum, if the angle of projection is

Options:
A $90^{\circ}$
B $180^{\circ}$
C $60^{\circ}$
D $75^{\circ}$

The correct answer is:
$90^{\circ}$

Solution:
Range $\frac{u^2 \sin 2 \theta}{g}$; when $\theta=90^{\circ}, R=0$ i.e. the body will fall at the point of projection after completing one dimensional motion under gravity.

Topic: Physics, Motion In Two Dimensions, JEE Main
The range of fx=4sin-1x2x2+1 is

Options:
A [0,2π]
B [0,π]
C [0,2π)
D [0,π)

The correct answer is:
[0,2π)

Solution:
Given,

fx=4sin-1x2x2+1

Now taking, x21+x2=1-11+x2<1

⇒0≤x21+x2<1

Now taking sin-1 we get,

⇒0≤sin-1x21+x2<π2

⇒0≤4sin-1x21+x2<2π

Hence, the range of 4sin-1x2x2+1 is [0,2π)

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (13 Apr Shift 2)
The range of $f(x)=\cos 2 x-\sin 2 x$ contains the set

Options:
A $[2,4]$
B $[-1,1]$
C $[-2,2]$
D $[-4,4]$

The correct answer is:
$[-1,1]$

Solution:
$\begin{array}{ll} f(x) =\sqrt{2}\left[\sin \left(\frac{\pi}{4}-2 x\right)\right] \\ \therefore-\sqrt{2} \leq f(x) \leq \sqrt{2} \text { and }[-1,1] \subset[-\sqrt{2}, \sqrt{2}]\end{array}$

Topic: Mathematics, Functions, JEE Main
The range of $f(x)=\cos (x / 3)$ is

Options:
A $(-1 / 3,1 / 3)$
B $[-1,1]$
C $(1 / 3,-1 / 3)$
D $(-3,3)$

The correct answer is:
$[-1,1]$

Solution:
$f(x)=\cos (x / 3)$
We know that $-1 \leq \cos (x / 3) \leq 1$.

Topic: Mathematics, Functions, JEE Main
The range of $f(x)=\cos x-\sin x$ is

Options:
A $(-1,1)$
B $[-1,1)$
C $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
D $[-\sqrt{2}, \sqrt{2}]$

The correct answer is:
$[-\sqrt{2}, \sqrt{2}]$

Solution:
Since maximum and minimum values of $\cos x-\sin x$ are $\sqrt{2}$ and $-\sqrt{2}$ respectively, therefore range of $f(x)$ is $[-\sqrt{2}, \sqrt{2}]$.

Topic: Mathematics, Functions, JEE Main
The range of $f(x)=\sec \left(\frac{\pi}{4} \cos ^2 x\right),-\infty \lt x \lt \infty$ is

Options:
A $[1, \sqrt{2}]$
B $[1, \infty)$
C $[-\sqrt{2},-1] \cup[1, \sqrt{2}]$
D $(-\infty,-1] \cup[1, \infty)$

The correct answer is:
$[1, \sqrt{2}]$

Solution:
$f(x)=\sec \left(\frac{\pi}{4} \cos ^2 x\right)$
We know that, $0 \leq \cos ^2 x \leq 1$ at $\cos x=0, f(x)=1$ and at $\cos x=1, f(x)=\sqrt{2} ; \therefore 1 \leq x \leq \sqrt{2} \Rightarrow$ $x \in[1, \sqrt{2}]$.

Topic: Mathematics, Functions, JEE Main
The range of \(f(x)=\sqrt{\frac{a-|x|}{(a+1)-|x|}},(a>0)\) is

Options:
A \([0, a]\)
B \([0, \infty)-\left[-\sqrt{\frac{a}{a+1}}, \sqrt{\frac{a}{a+1}}\right]\)
C \(\left[0, \sqrt{\frac{a}{a+1}}\right] \cup(1, \infty)\)
D \(\left[0, \sqrt{\frac{a}{a+1}}+1\right]\)

The correct answer is:
\(\left[0, \sqrt{\frac{a}{a+1}}\right] \cup(1, \infty)\)

Solution:
Given function is
\(f(x)=\sqrt{\frac{a-|x|}{(a+1)-|x|}},(a > 0)\)
\(\because f(x) \geq 0, \forall x \in\) domain of \(f(x)\).
Now, let \(\frac{a-|x|}{(a+1)-|x|}=y\)
\(\begin{aligned}
& \Rightarrow a-|x|=y(a+1)-y|x| \quad[\because \text { assuming }|x| \neq a+1] \\
& \Rightarrow(y-1)|x|=y(a+1)-a \\
& \Rightarrow|x|=\frac{y(a+1)-a}{y-1} \geq 0, \forall x \in \text { domain of } f(x) \\
& \therefore y \in\left(-\infty, \frac{a}{a+1}\right] \cup(1, \infty) \quad(\text { as } a > 0)
\end{aligned}\)
So, range of \(f(x)=\sqrt{y} \in\left[0, \sqrt{\frac{a}{a+1}}\right] \cup(1, \infty)\)
\(\text {[as } \sqrt{y} \geq 0]\)
Hence, option (c) is correct.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2019 (23 Apr Shift 1)
The range of $\tan ^{-1} x$ is

Options:
A $\left(\pi, \frac{\pi}{2}\right)$
B $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
C $(-\pi, \pi)$
D $(0, \pi)$

The correct answer is:
$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Solution:
Given,
$\begin{aligned}
& f(x)=\tan ^{-1} x \\
& {\left[\begin{array}{cc}
\text { Solution : } & -\frac{\pi}{2} \lt f(x) \lt \frac{\pi}{2} \\
\text { Interval Notation : } & \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)
\end{array}\right]} \\
&
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The range of the function $f(x)=\frac{x}{1+|x|}, x \in R$, is

Options:
A
$R$

B
$(-1,1)$

C
$R-\{0\}$

D
$[-1,1]$

The correct answer is:
$(-1,1)$

Solution:
$f(x)=\frac{x}{1+|x|}, x \in R$
If $x>0,|x|=x \Rightarrow f(x)=\frac{x}{1+x}$ which is not defined for $x=-1$
If $x < 0,|x|=-x \Rightarrow f(x)=\frac{x}{1-x}$ which is not defined for $x=1$ Thus $f(x)$ defined for all values of $\mathrm{R}$ except 1 and $-1$
Hence, range $=(-1,1)$.

Topic: Mathematics, Functions, JEE Main, JEE Main 2012 (07 May Online)
The range of the function fx=23x2-4x+5 is

Options:
A -∞,2113
B -∞,2113
C 2113,∞
D 2113,∞

The correct answer is:
2113,∞

Solution:
fx is defined if 3x2-4x+5≥0
⇒3x2-43x+53≥0⇒x-232+119≥0
Which is true for all real x
∴ Domain f=-∞,∞
Let, y=23x2-4x+5
⇒y24=3x2-4x+5i.e. 3x2-4x+5-y24=0
For x to be real, 16-125-y24≥0

⇒y2≥113
∴ Range of y=2113,∞.

Topic: Mathematics, Functions, JEE Main
The range of the function $f(x)=\sqrt{3 x^2-4 x+5}$ is

Options:
A $\left(-\infty, \sqrt{\frac{11}{3}}\right]$
B $\left(-\infty, \sqrt{\frac{11}{5}}\right)$
C $\left[\sqrt{\frac{11}{3}}, \infty\right)$
D $\left(\sqrt{\frac{11}{5}}, \infty\right)$.

The correct answer is:
$\left[\sqrt{\frac{11}{3}}, \infty\right)$

Solution:
$f(x)$ is defined if $3 x^2-4 x+5 \geq 0$
$\Rightarrow 3\left[x^2-\frac{4}{3} x+\frac{5}{3}\right] \geq 0 \Rightarrow 3\left[\left(x-\frac{2}{3}\right)^2+\frac{11}{9}\right] \geq 0$
Which is true for all real $x \therefore$ Domain of $f(x)=$ $(-\infty, \infty)$
Let $y=\sqrt{3 x^2-4 x+5}$
$\Rightarrow y^2=3 x^2-4 x+5$ i.e. $3 x^2-4 x+\left(5-y^2\right)=0$
For $x$ to be real , $16-12\left(5-y^2\right) \geq 0 \Rightarrow y \geq \sqrt{\frac{11}{3}}$
$\therefore \quad$ Range of $y=\left[\sqrt{\frac{11}{3}}, \infty\right)$

Topic: Mathematics, Quadratic Equation, JEE Main
The range of the function fx=3-x+2+x is

Options:
A 5,10
B 22,11
C 5,13
D 2,7

The correct answer is:
5,10

Solution:
Given,

fx=3-x+2+x

Now differentiation the above function,

f'(x)=-123-x+122+x

Point of maxima and minima,

f'(x)=0

⇒3-x-2+x22+x3-x=0

⇒2+x=3-x

⇒x=12

And domain is (-2≤x≤3)

So at x=12, maxima

The maximum value of the function,

f12=52+52=10

The minimum value of function at x=-2 or x=3

f-2=5 and f3=5

Therefore, the range of the function,

Range∈5,10

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (30 Jan Shift 2)
The range of the function $f(x)={ }^{7-x} P_{x-3}$ is

Options:
A
$\{1,2,3\}$

B
$\{1,2,3,4,5\}$

C
$\{1,2,3,4\}$

D
$\{1,2,3,4,5,6\}$

The correct answer is:
$\{1,2,3\}$

Solution:
$f(x)={ }^{7-x} P_{x-3}$
$7-x \geq 0 \Rightarrow x \leq 7$
$x-3 \geq 0 \Rightarrow x \geq 3$,
and $7-x \geq x-3 \Rightarrow x \leq 5$
$\Rightarrow 3 \leq x \leq 5 \Rightarrow x=3,4,5 \Rightarrow$ Range is $\{1,2,3\}$

Topic: Mathematics, Functions, JEE Main, JEE Main 2004
The range of the function f(x)=log53+cos3π4+x+cosπ4+x+cosπ4-x-cos3π4-x is :

Options:
A 15, 5
B [0, 2]
C (0, 5)
D [-2, 2]

The correct answer is:
[0, 2]

Solution:
f(x)=log5(3-2sinx+2cosx)

-2≤(-2sinx+2cosx)≤2

⇒1≤(3-2sinx+2cosx)≤5

⇒log51≤f(x)≤log55

⇒0≤f(x)≤2

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (01 Sep Shift 2)
The range of the function fx=sin-1x2-13-cos-1x2+23 is (where, x represents the greatest integer value of x)

Options:
A -π,0
B -π,0
C 0,π
D 0,π,-π

The correct answer is:
-π,0

Solution:
x2-13=x2+23-1=x2+23-1=k let
Now, the function is sin-1k-cos-1k+1

For the above function to be defined,
-1≤k≤1 & -1≤k+1≤1
∴k=-1,0
∴ Range =sin-1-1-cos-10,sin-10-cos-11=-π,0

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The range of the function fx=sin-1x21+x2, xâR is

Options:
A -π2,π2
B 0,π2
C 0,π2
D -π2,π2

The correct answer is:
0,π2

Solution:
x21+x2=1-11+x2∈0,1
∴ Range of fx∈sin-10,sin-11
∈0,π2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The range of the function fx=sin⁡πx+1x4+1 (where, · is the greatest integer function) is

Options:
A 0,1
B -1,1
C 0
D None of these

The correct answer is:
0

Solution:
Since, x+1 is an integer
â´fx=sinâ¡Ïx+1x4+1=0
Hence, range of f=Rf=0

Topic: Mathematics, Functions, JEE Main
The range of the function fx=tan⁡πx+1x4+1 (where, · is the greatest integer function) is

Options:
A 0,1
B -1,1
C 0
D -∞,∞

The correct answer is:
0

Solution:
Since, x+1 is an integer
â´fx=tanâ¡Ïx+1x4+1=0
Hence, Range of f=Rf=0

Topic: Mathematics, Functions, JEE Main
The range of the function fx=x2ln⁡x for x∈1,e is a,b, where a+b is equal to

Options:
A e2
B e2+1
C e+1
D 2e2

The correct answer is:
e2

Solution:
f â² x =2xâlnx+ x 2 1 x
=x2lnâ¡x+1
So, fx is increasing in 1,e
â´ Minimum value of fx=f1=0
And maximum value of fx=fe=e2
â´a,b is 0,e2
âa+b=e2

Topic: Mathematics, Application of Derivatives, JEE Main
The range of the function $f(x)=\frac{x+2}{|x+2|}$

Options:
A $\{0,1\}$
B $\{-1,1\}$
C $R$
D $R-\{-2\}$

The correct answer is:
$\{-1,1\}$

Solution:
$\begin{array}{r}
f(x)=\frac{x+2}{|x+2|} \\
f(x)=\left\{\begin{aligned}
-1, & x \lt -2 \\
1, & x\gt-2
\end{aligned}\right.
\end{array}$
$\therefore$ Range of $f(x)$ is $\{-1,1\}$

Topic: Mathematics, Functions, JEE Main
The range of the function \( f(x)=\sqrt{9-x^{2}} \) is

Options:
A \( (0,3) \)
B \( [0,3] \)
C \( (0,3] \)
D \([0,3) \)

The correct answer is:
\( [0,3] \)

Solution:
Given that, $f(x)=\sqrt{9-x^{2}}$
$x^{2}=9 \Rightarrow x \pm 3$
So, domain is given by $x \in[-.3,3]$
Therefore, range of $f(x) \in[0,3]$

Topic: Mathematics, Indefinite Integration, KCET, KCET 2017
The range of the function y=2sin-1â¡x2+12+cos-1â¡x2-12 is (where, Â· denotes the greatest integer function)

Options:
A 0,π
B π,3π2
C π
D π,3π2

The correct answer is:
π,3π2

Solution:
Let, x2-12=x2+12-1=Î±âI
âx2+12-1=Î±
â´y=2sin-1â¡Î±+1+cos-1â¡Î±, Î±=-1,0
At Î±=-1
y=2sin-1â¡0+cos-1â¡-1=Ï
At Î±=0
y=2sin-1â¡1+cos-1â¡0=3Ï2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The range of the function $y=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-x^{2}}\right)$ is

Options:
A $[0, \sqrt{3 / 2}]$
B [0,1]
C $[0,3 / \sqrt{2}]$
D $[0, \infty)$

The correct answer is:
$[0,3 / \sqrt{2}]$

Solution:
Given function is $y=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-x^{2}}\right)$
Here, domain is $\left[0, \frac{\pi}{4}\right]$
$$
\begin{array}{l}
\begin{aligned}
x=& 0 \\
y &=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-0}\right) \\
&=3 \sin \left(\frac{\pi}{4}\right)=3 \times \frac{1}{\sqrt{2}}=\frac{3}{\sqrt{2}} \\
\text { When } x=\frac{\pi}{4}, y &=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-\frac{\pi^{2}}{16}}\right) \\
&=3 \sin (0)=0
\end{aligned}
\end{array}
$$
Hence, range of given function is $\left[0, \frac{3}{\sqrt{2}}\right]$

Topic: Mathematics, Functions, WBJEE, WBJEE 2014
The range of the observations 20,28,40,12,30,15,50 is _____

Options:
A 18
B 38
C 28
D 16

The correct answer is:
38

Solution:
Given range is 20,28,40,12,30,15,50

Now,

Least value =12

Highest vaue =50

Range=Highest value - Least value

Range=50-12=38.

Topic: Mathematics, Statistics, JEE Main
The range of the projectile projected at an angle of 15∘ with horizontal is 50 m. If the projectile is projected with same velocity at an angle of 45∘ with horizontal, then its range will be

Options:
A 100 m
B 1002 m
C 502 m
D 50 m

The correct answer is:
100 m

Solution:
The data given is

θ1=15°θ2=45°R1=50 m

The formula for range of a projectile is given by

R=u2 sin2θg ...(i)

Substituting the values in equation (i)

R1=u2sin2θ1g⇒50=u2sin30°g⇒u2g=100

Let the new range be R2. The magnitude of the range is

R2=u2sin2θ2g⇒R2=u2gsin90°⇒R2=100 m

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The range of values of \( m \) for which the line \( y=m x \) and the curve \( y=\frac{x}{x^{2}+1} \) enclose a region, is

Options:
A \( (-1,1) \)
B \( (0,1) \)
C 1]
D \( (1, \infty) \)

The correct answer is:
\( (0,1) \)

Solution:
Given:

y=mx and y=xx2+1

Hence points of intersection are:

x=0

and

x2+1=1m

x2=1m-1

1m-1>0

1-mm>0

m-1m<0

m∈0,1

Topic: Mathematics, Functions, JEE Main
The ranges and heights for two projectiles projected with the same initial velocity at angles 42° and 48° with the horizontal are R1, R2 and H1, H2 respectively. Choose the correct option:

Options:
A R1=R2 and H1=H2
B R1=R2 and H1<H2
C R1>R2 and H1=H2
D R1<R2 and H1<H2

The correct answer is:
R1=R2 and H1<H2

Solution:
RangeR=u2sin2θg and same for θ and 90-θ. So, same for 42° and 48° Maximum height H=u2sin2θ2g, so height is high for higher θ. So, H for 48° is higher than H for 42°.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2021 (01 Sep Shift 2)
The rank of the matrix
$\left[\begin{array}{ccc}-1 & 2 & 5 \\ 2 & -4 & a-4 \\ 1 & -2 & a+1\end{array}\right]$ is

Options:
A 1 if $a=6$
B 2 if $a=1$
C 3 If $a=2$
D 1 if $a=-6$

The correct answers are:
2 if $a=1$, 1 if $a=-6$

Solution:
$A=\left|\begin{array}{ccc}-1 & 2 & 5 \\ 2 & -4 & a-4 \\ 1 & -2 & a+1\end{array}\right|=\left|\begin{array}{ccc}0 & 0 & a+6 \\ 0 & 0 & -a-6 \\ 1 & -2 & a+1\end{array}\right|$
(Operating $R_1 \rightarrow R_1+R_3$ and $R_2 \rightarrow R_2-2 R_3$ )
$=\left|\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & -a-6 \\ 1 & -2 & a+1\end{array}\right|$ (Operating $R_1 \rightarrow R_1+R_2$ )
$a=-6, A=\left|\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & -2 & -5\end{array}\right|, \quad \therefore \rho(A)=1$
Where $\rho(A)=$ number of non-zero rows
When $a=6, A=\left|\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & -12 \\ 1 & -2 & 7\end{array}\right|$ $\therefore \quad \rho(A)=2$
When $a=1, A=\left|\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & -7 \\ 1 & -2 & 2\end{array}\right|, \therefore \quad \rho(A)=2$
When $a=2, A=\left|\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & -8 \\ 1 & -2 & 3\end{array}\right|, \therefore \rho(A)=2$.

Topic: Mathematics, Matrices, JEE Main
The rank of the matrix $A=\left[\begin{array}{cccc}2 & 3 & 1 & 4 \\ 0 & 1 & 2 & -1 \\ 0 & -2 & -4 & 2\end{array}\right]$ is

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

The correct answer is:
$2$

Solution:
Given $A=\left[\begin{array}{cccc}2 & 3 & 1 & 4 \\ 0 & 1 & 2 & -1 \\ 0 & -2 & -4 & 2\end{array}\right] \quad\left(R_2 \rightarrow 2 R_2+R_3\right)$
$A=\left[\begin{array}{cccc}2 & 3 & 1 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & -2 & -4 & 2\end{array}\right]$
Since every minor of order 3 in $A$ is 0 and there exists a minor order 3 i.e. $\left[\begin{array}{cc}2 & 3 \\ 0 & -2\end{array}\right]$ in $A$ which is non-zero. Thus, rank $=2$.

Topic: Mathematics, Matrices, JEE Main
The rate coefficient $(k)$ for a particular reactions is $1.3 \times 10^{-4} \mathrm{M}^{-1} \mathrm{~s}^{-1}$ at $100^{\circ} \mathrm{C}$, and $1.3 \times 10^{-3}$ $\mathrm{M}^{-1} \mathrm{~s}^{-1}$ at $150^{\circ} \mathrm{C}$. What is the energy of activation $\left(\mathrm{E}_{\mathrm{A}}\right)$ (in $\left.\mathrm{kJ}\right)$ for this reaction? $(\mathrm{R}=$ molar gas constant $=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )

Options:
A
16

B
60

C
99

D
132

The correct answer is:
60

Solution:
According to Arrhenius equation
$$
\begin{aligned}
&\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \\
&\log \frac{1.3 \times 10^{-3}}{1.3 \times 10^{-4}}=\frac{E_a}{2.303 \times 8.314}\left[\frac{1}{373}-\frac{1}{423}\right] \\
&1=\frac{E_a}{2.303 \times 8.314}\left[\frac{1}{373}-\frac{1}{423}\right] \\
&E_a=60 \mathrm{~kJ} / \mathrm{mole}
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2014 (12 Apr Online)
The rate constant for a first order reaction is 20 min-1. The time required for the initial concentration of the reactant to reduce to its 132 level is ______×10-2 min. (Nearest integer)

Options:

The correct answer is:
17

Solution:
For a first order reaction, we know:
t1/2=0.693k=0.69320=0.03465 min
From the relation of half lives(where 'n' is the no. of half-lives):

⇒12n=132

=125

Thus, No. of Half life =5

∴Time required =0.03465×5=0.17325 min=17.325×10-2 min

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (31 Jan Shift 2)
The rate constant for a first order reaction is given by the following equation :

lnk=33.24-2.0×104 K T

The Activation energy for the reaction is given by kJmol-1. (In Nearest integer) (Given : R=8.3 J K-1 mol-1)

Options:

The correct answer is:
166

Solution:
lnk=lnA-EART

Given: lnk=33.24-2.0×104 T

∴ on comparing EAR=2.0×104

∴ EA=2.0×104×R

⇒EA=2.0×104×8.3 J

⇒EA=16.6×104 J=166kJ

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2022 (27 Jun Shift 1)
The rate constant for the first order decomposition of a certain reaction is described by the equation:
log k s -1 = 14.34 - 1.25 Ã 10 4 K T
At what temperature will its half-life period be 256 min ?

Options:
A 239.33 kJ ; 669 K
B 259.33 kJ
C 329.33 kJ
D 539.33 kJ

The correct answer is:
239.33 kJ ; 669 K

Solution:
When half-life = 256 min,
k = ln 2 t 1 / 2 = 0.693 256 Ã 6 0 s -1
= 4.5 Ã 1 0 - 5 s -1
⇒ 1.25× 10 4 T = 14.34− [ log 4.5× 10 −5 ]
⇒ 14.34 -[ -5+0.65 ]
â 1 9 Â· 3 4 - 0 Â· 6 5 = 1 8 Â· 6 9
T = 12500 18.69 = 669K

Topic: Chemistry, Chemical Kinetics, JEE Main
The rate constant is doubled when temperature increases from $27^{\circ} \mathrm{C}$ to $37^{\circ} \mathrm{C}$. Activation energy in $k J$ is

Options:
A $34$
B $54$
C $100$
D $50$

The correct answer is:
$54$

Solution:
$\log \frac{K_2}{K_1}=\frac{E_a}{2.303 R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$
If $\frac{K_2}{K_1}=2$
$\log 2=\frac{E_a}{2.303 \times 8.314}\left[\frac{1}{300}-\frac{1}{310}\right]$
$E_a=.3010 \times 2.303 \times 8.314\left(\frac{300 \times 310}{10}\right)$
$=53598.59 \mathrm{Jmol}^{-1}=54 \mathrm{~kJ}$

Topic: Chemistry, Chemical Kinetics, JEE Main
The rate constant of a reaction increases by five times on increase in temperature from 27°C to 52°C. The value of activation energy in kJmol-1 is ___ .(Rounded-off to the nearest integer) R=8.314 J K-1 mol-1

Options:

The correct answer is:
52

Solution:
T1=300 K, T2=325 K, K2=5 K

lnK2 K1=EaR1 T1-1 T2

or, ln5=Ea8.3141300-1325

or, Ea=0.7×2.303×8.314×12×325

=52271 J=52.271 kJ

Nearest integer answer will be 52 kJ

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The rate constant of a zero order reaction is $2.0 \times$ $10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$. If the concentration of the reactant after 25 seconds is $0.5 \mathrm{M}$. What is the initial concentration?

Options:
A
$0.5 \mathrm{M}$

B
$1.25 \mathrm{M}$

C
$12.5 \mathrm{M}$

D
$1.0 \mathrm{M}$

The correct answer is:
$1.0 \mathrm{M}$

Solution:
For a zero order reaction
$$
\begin{aligned}
& \text { Rate constant }=\mathrm{k}=\frac{a-x}{t} \\
& 2 \times 10^{-2}=\frac{a-0.5}{25} \\
& a-0.5=0.5 \\
& a=1.0 \mathrm{M}
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2013 (23 Apr Online)
The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at 25°C are 3.0×10-4s-, 104.4kJmol-1 and 6×1014s-1 respectively. The value of the rate constant as T→∞ is

Options:
A 2.0×1018s-1
B 6.0×1014s-1
C infinity
D 3.6×1030s-1

The correct answer is:
6.0×1014s-1

Solution:
k=Ae-Ea/RT (Arrhenius equation)
when T→∞
k=Ae0 , or k=A
Thus k=6×1014s-1

Topic: Chemistry, Chemical Kinetics, JEE Main
The rate constants $\mathrm{k}_1$ and $\mathrm{k}_2$ for two different reactions are $10^{16} \cdot \mathrm{e}^{-2000 / T}$ and $10^{15} \cdot \mathrm{e}^{-1000 / T}$ respectively. The temperature at which $\mathrm{k}_1=\mathrm{k}_2$ is

Options:
A $\frac{1000}{2.303} \mathrm{~K}$
B $1000 \mathrm{~K}$
C $\frac{2000}{2.303} \mathrm{~K}$
D $2000 \mathrm{~K}$

The correct answer is:
$\frac{1000}{2.303} \mathrm{~K}$

Solution:
$$
K_1=10^{16} \mathrm{e}^{-\frac{2000}{T}}
$$
or $\log \mathrm{K}_1=16-\frac{2000}{2.303 \mathrm{~T}}$
$$
K_2=10^{15} e^{\frac{1000}{T}}
$$
or $\log K_2=15-\frac{1000}{2.303 \mathrm{~T}}$
If $\mathrm{T}=\frac{1000}{2.303} \mathrm{~K}$, then $\mathrm{K}_1=\mathrm{K}_2$

Topic: Chemistry, Chemical Kinetics, JEE Main