datavorous/entrance-exam-dataset huggingface dataset:
If the range of the function $f(x)=-3 x-3$ is $\{3,-6,-9,-18\}$, then which of the following elements is not in the domain of $f$ ?

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

The correct answer is:
-1

Solution:
We have, $f(x)=-3 x-3$
(i) Let $f(x)=3$, then we get
$$
3=-3 x-3 \Rightarrow 6=-3 x \Rightarrow x=-2
$$
(ii) Let $f(x)=-6$, then we get
$$
-6=-3 x-3 \Rightarrow-3=-3 x \Rightarrow x=1
$$
(iii) Let $f(x)=-9$, then we get
$$
-9=-3 x-3 \Rightarrow-6=-3 x \Rightarrow x=2
$$
(iv) Let $f(x)=-18$, then we get
$$
-18=-3 x-3 \Rightarrow-15=-3 x \Rightarrow x=5
$$
Thus, domain of $f=\{-2,1,2,5\}$
Hence, -1 cannot be in the domain of $f$.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2017
If the range of the function fx=sin-1x+2âtan-1âx+x2+4x+1 is [p, q], then the value of p+q is

Options:

The correct answer is:
4

Solution:
Domain of fx is -1,ââ1
fx=sin-1x+2tan-1x+x2+4x+1
fâ²x=11-x2+21+x2+2x+4
=11-x2+21+x2+2x+2
âââxâ-1,ââ1, fâ²x>0
â´ââââfx is an increasing function
â´ââââp=minimum value of fx
=sin-1-1+2tan-1-1+-12+4-1+1
=-Ï2+2-Ï4-2=-Ï2-Ï2-2=-Ï-2
And q= maximum value of fx
=sin-11+2tan-11+1+4+1
=-Ï2+2Ï4+6=Ï+6
â´ So, the range of fx is -Ï-2,ââÏ+6
â´âp+q=Ï+6-Ï-2=4

Topic: Mathematics, Functions, JEE Main
If the rates of cooling of two different liquids put in exactly similar calorimeters and kept in identical surroundings are the same, then which among the following is most suitable option?

Options:
A Equal masses of the liquids at the same temperature are taken
B Equal volumes of the liquids at the same temperature are taken
C Both 1 and 2
D Insufficient information

The correct answer is:
Equal volumes of the liquids at the same temperature are taken

Solution:
d θ dt = σ A mc T 4 - T 0 4 . If the liquids put in exactly similar calorimeters and identical surrounding then we can consider T0 and A constant then d θ dt ∝ T 4 - T 0 4 mc
If we consider that equal masses of liquid (m) are taken at the same temperature then d θ dt ∝ 1 c
So for same rate of cooling c should be equal which is not possible because liquids are of different nature. Again from equation (i)
d θ dt ∝ T 4 - T 0 4 mc ⇒ d θ dt ∝ T 4 - T 0 4 V ρ c
Now if we consider that equal volume of liquid (V) are taken at the same temperature then d θ dt ∝ 1 ρ c
So for same rate of cooling multiplication of ρ × c for two liquid of different nature can be possible

Topic: Physics, Thermal Properties of Matter, JEE Main
If the ratio of H.M. and G.M. of two quantities is $12: 13$, then the ratio of the numbers is

Options:
A $1: 2$
B $2: 3$
C $3: 4$
D None of these

The correct answer is:
None of these

Solution:
Given that $\frac{H \cdot M .}{G \cdot M .}=\frac{12}{13} \Rightarrow \frac{\frac{2 a b}{a+b}}{\sqrt{a b}}=\frac{12}{13}$
or $\frac{a+b}{2 \sqrt{a b}}=\frac{13}{12}$
$\Rightarrow \quad \frac{(a+b)+2 \sqrt{a b}}{(a+b)-2 \sqrt{a b}}$ $=\frac{13+12}{13-12}=\frac{25}{1}$
$\Rightarrow \quad \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=\frac{5^2}{1}$ $\Rightarrow \quad \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{5}{1}$
$\Rightarrow \quad \frac{(\sqrt{a}+\sqrt{b})+(\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})-(\sqrt{a}-\sqrt{b})}$ $=\frac{5+1}{5-1}$
$\Rightarrow \frac{2 \sqrt{a}}{2 \sqrt{b}}=\frac{6}{4} \Rightarrow\left(\frac{a}{b}\right)^{1 / 2}$$-\frac{6}{4} \Rightarrow a: b=9: 4$

Topic: Mathematics, Sequences and Series, JEE Main
If the ratio of lengths, radii and Young's moduli of steel and brass wires in the figure are $a, b$ and $c$ respectively, then the corresponding ratio of increase in their lengths is :

Options:
A
$\frac{3 c}{2 a b^2}$

B
$\frac{2 a^2 c}{b}$

C
$\frac{3 a}{2 b^2 c}$

D
$\frac{2 a c}{b^2}$

The correct answer is:
$\frac{3 a}{2 b^2 c}$

Solution:
According to questions,
$\begin{aligned}
& \frac{\ell_{\mathrm{s}}}{\ell_{\mathrm{b}}}=\mathrm{a}, \frac{\mathrm{r}_{\mathrm{s}}}{\mathrm{r}_{\mathrm{b}}}=\mathrm{b}, \frac{\mathrm{y}_{\mathrm{s}}}{\mathrm{y}_{\mathrm{b}}}=\mathrm{c}, \frac{\Delta \ell \mathrm{s}}{\Delta \ell_{\mathrm{b}}}=? \\
& \text { As, } \mathrm{y}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell} \Rightarrow \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{Ay}} \\
& \Delta \ell_{\mathrm{s}}=\frac{3 \mathrm{mg} \ell_{\mathrm{s}}}{\pi \mathrm{r}_{\mathrm{s}}^2 \cdot \mathrm{y}_{\mathrm{s}}}\left[\because \mathrm{F}_{\mathrm{s}}=(\mathrm{M}+2 \mathrm{M}) \mathrm{g}\right] \\
& \Delta \ell_{\mathrm{b}}=\frac{2 \mathrm{Mg}\ell_{\mathrm{b}}}{\pi \mathrm{r}_{\mathrm{b}}^2 \cdot \mathrm{y}_{\mathrm{b}}}\left[\because \mathrm{F}_{\mathrm{b}}=2 \mathrm{Mg}\right]
\end{aligned}$
$\therefore \frac{\Delta \ell_{\mathrm{s}}}{\Delta \ell_{\mathrm{b}}}=\frac{\frac{3 \mathrm{Mg} \ell_{\mathrm{s}}}{\pi \mathrm{r}_{\mathrm{s}}^2 \cdot \mathrm{y}_{\mathrm{s}}}}{\frac{2 \mathrm{Mg} \cdot \ell_{\mathrm{b}}}{\pi \mathrm{r}_{\mathrm{b}}^2 \cdot \mathrm{y}_{\mathrm{b}}}}=\frac{3 \mathrm{a}}{2 \mathrm{b}^2 \mathrm{C}}$

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2013 (09 Apr Online)
If the ratio of lengths, radii and Young's modulus of steel and brass wires shown in the figure are a, b and c respectively, the ratio between the increase in lengths of brass and steel wires would be

Options:
A $\frac{b^{2} a}{2 c}$
B $\frac{b c}{2 a^{2}}$
C $\frac{b a^{2}}{2 c}$
D $\frac{a}{2 b^{2} c}$

The correct answer is:
$\frac{a}{2 b^{2} c}$

Solution:
Given, $\frac{l_{1}}{l_{2}}=a, \frac{r_{1}}{r_{2}}=b, \frac{Y_{1}}{Y_{2}}=c$
Free body diagram of the two blocks brass and steel are
Let Young's modulus of steel is $Y_{1}$ and of brass is $Y_{2}$.
$\therefore Y_{1}=\frac{F_{1} \cdot l_{1}}{A_{1} \cdot \Delta l_{1}}...(i)$
and $Y_{2}=\frac{F_{2} \cdot l_{2}}{A_{2} \cdot \Delta l_{2}}...(ii)$
Dividing Eq. (i) by(ii),
$$
\begin{array}{l}
\frac{Y_{1}}{Y_{2}}=\frac{\frac{F_{1} \cdot l_{1}}{A_{1} \cdot \Delta l_{1}}}{\frac{F_{2} \cdot l_{2}}{A_{2} \cdot \Delta l_{2}}} \\
\text { or } \frac{Y_{1}}{Y_{2}}=\frac{F_{1} \cdot A_{2} \cdot l_{1} \cdot \Delta l_{2}}{F_{2} \cdot A_{1} \cdot l_{2} \cdot \Delta l_{1}}...(ii)
\end{array}
$$
Force on steel wire from free body diagram $T=F_{1}=(2 g)$ newton
Force on brass wire from free body diagram $F_{2}=T^{\prime}=T+2 g=(4 g)$ newton
Now, putting the value of $F_{1}, F_{2}$, in Eq. (iii), we get
$$
\frac{Y_{1}}{Y_{2}}=\left(\frac{2 g}{4 g}\right) \cdot\left(\frac{\pi r_{2}^{2}}{\pi r_{1}^{2}}\right) \cdot\left[\frac{l_{1}}{l_{2}}\right] \cdot\left(\frac{\Delta l_{2}}{\Delta l_{1}}\right)
$$
or $c=\frac{1}{2}\left(\frac{1}{b^{2}}\right) \cdot a\left(\frac{\Delta l_{2}}{\Delta l_{1}}\right)$
or $\frac{\Delta l_{1}}{\Delta l_{2}}=\left(\frac{a}{2 b^{2} c}\right)$

Topic: Physics, Mechanical Properties of Solids, JEE Main
If the ratio of lengths, radii and Young's modulus of steel and brass wires shown in the figure are $a, b$ and $c$ respectively, the ratio between the increase in lengths of brass and steel wires would be

Options:
A $\frac{b^2 a}{2 c}$
B $\frac{b c}{2 a^2}$
C $\frac{b a^2}{2 c}$
D $\frac{a}{2 b^2 c}$

The correct answer is:
$\frac{b^2 a}{2 c}$

Solution:
Free body diagram of the two blocks are
Given, $\quad \frac{l_1}{l_2}=a, \frac{r_1}{r_2}=b, \frac{Y_1}{Y_2}=c$
Let Young's modulus of steel is $Y_1$ and of brass is $Y_2$.
Dividing Eq. (i) by Eq. (ii), we get
$\frac{Y_1}{Y_2}=\frac{\frac{F_1 \cdot l_1}{A_1 \cdot \Delta l_1}}{\frac{F_2 \cdot l_2}{A_2 \cdot \Delta l_2}}$
Force on steel wire from free body diagram
$T=F_1=(2 g) \text { newton }$
Force on brass wire from free body diagram
$F_2=T^{\prime}=T+2 g=(4 g) \text { newton }$
Now, putting the value of $F_1, F_2$, in Eq. (iii), we get
$\frac{Y_1}{Y_2}=\left(\frac{2 g}{4 g}\right) \cdot\left(\frac{\pi r_2^2}{\pi r_1^2}\right) \cdot\left[\frac{l_1}{l_2}\right] \cdot\left(\frac{\Delta l_2}{\Delta l_1}\right)$
or $\quad c=\frac{1}{2}\left(\frac{1}{b^2}\right) \cdot a\left(\frac{\Delta l_2}{\Delta l_1}\right)$
or $\quad \frac{\Delta l_1}{\Delta l_2}=\left(\frac{a}{2 b^2 c}\right)$

Topic: Physics, Mechanical Properties of Solids, JEE Main
If the ratio of the concentration of electrons that of holes in a semiconductor is $\frac{7}{5}$ and the ratio of currents is $\frac{7}{4}$, then what is the ratio of their drift velocities?

Options:
A
$\frac{4}{7}$

B
$\frac{5}{8}$

C
$\frac{4}{5}$

D
$\frac{5}{4}$

The correct answer is:
$\frac{5}{4}$

Solution:
$\frac{\mathrm{n}_{\mathrm{e}}}{\mathrm{n}_{\mathrm{n}}}=\frac{7}{5} \frac{\mathrm{l}_{\mathrm{e}}}{\mathrm{l}_{\mathrm{n}}}=\frac{7}{4}$
$\frac{\left(\mathrm{V}_{\mathrm{d}}\right)_{\mathrm{e}}}{\left(\mathrm{V}_{\mathrm{d}}\right)_{\mathrm{n}}} \Rightarrow \frac{\mathrm{l}_{\mathrm{e}}}{\mathrm{l}_{\mathrm{n}}} \times \frac{\mathrm{n}_{\mathrm{n}}}{\mathrm{n}_{\mathrm{e}}}=\frac{5}{4}$

Topic: Physics, Semiconductors, JEE Main, JEE Main 2006
If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of 24+134n is 6:1, then the third term from the beginning is:

Options:
A 302
B 303
C 602
D 603

The correct answer is:
603

Solution:
Given expansion is24+134n

We know that general term of an expansion x+an is Tr+1=Crnxn-rar

(5th term from beginning)

T4+1=C4n(24)n-4·1344

(5th term from end)

We know that rth term from the end in a binomial expansion is Tn-r+2 from the beginning.

T5'=Tn-4+1=C4n134n-4(24)4

It is also given that the ratio of the fifth term from the beginning to the fifth term from the end is 6:1.

Now, T5T5'=61

⇒2n-84·3n-84=6

∴(6)n-82=(6)

∴n=10

∴T3=C2102481342=45×43

=603

Hence, this is the correct option.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (06 Apr Shift 1)
If the ratio of the radius of a nucleus with 61 neutrons to that of helium nucleus is 3, then the atomic number of this nucleus is

Options:
A 27
B 47
C 51
D 61

The correct answer is:
47

Solution:
The radius is directly proportional to 13 power of mass number, i.e., R∝A13
Or A1A2=R1R23=133=127
⇒ A2A1=27
⇒ A2=A1×27
=4×27=108
⇒ Z2=A2-61
=108-61=47

Topic: Physics, Nuclear Physics, JEE Main
If the ratio of the roots of $x^2+b x+c=0$ and $x^2+q x+r=0$ be the same, then

Options:
A $r^2 c=b^2 q$
B $r^2 b=c^2 q$
C $r b^2=c q^2$
D $r c^2=b q^2$

The correct answer is:
$r b^2=c q^2$

Solution:
Let $\alpha, \beta$ be the roots of $x^2+b x+c=0$ and $\alpha^{\prime}, \beta^{\prime}$ be the roots of $x^2+q x+r=0$
Then $\alpha+\beta=-b, \alpha \beta=c, \alpha^{\prime}+\beta^{\prime}=-q, \alpha^{\prime} \beta^{\prime}=r$
It is given that $\frac{\alpha}{\beta}=\frac{\alpha^{\prime}}{\beta^{\prime}} \Rightarrow \frac{\alpha+\beta}{\alpha-\beta}=\frac{\alpha^{\prime}+\beta^{\prime}}{\alpha^{\prime}-\beta^{\prime}}$
$\Rightarrow \frac{(\alpha+\beta)^2}{(\alpha-\beta)^2}=\frac{\left(\alpha^{\prime}+\beta^{\prime}\right)^2}{\left(\alpha^{\prime}-\beta^{\prime}\right)^2} \Rightarrow \frac{b^2}{b^2-4 c}=\frac{q^2}{q^2-4 r}$
$\Rightarrow b^2 r=q^2 c$

Topic: Mathematics, Quadratic Equation, JEE Main
If the reaction $\mathrm{N}_{2}+3 \mathrm{H}_{2}=2 \mathrm{NH}_{3}$ occurs at $200^{\circ} \mathrm{C}$ and $1000 \mathrm{~atm}$ then the graph showing the correct equilibrium yield at $400^{\circ} \mathrm{C}$ is

Options:
A
B
C
D

The correct answer is:

Solution:
Since the formation of amnonia is an exothermic reaction hence on increasing temperature, reaction will proceed in backward direction i.e. formation of $\mathrm{NH}_{3}$ decreases.

Topic: Chemistry, Chemical Equilibrium, JEE Main
If the real part of the complex number 1-cosθ+2isinθ-1 is 15 for θ∈(0,π), then the value of the integral ∫0θsinx dx is equal to:

Options:
A 1
B 2
C -1
D 0

The correct answer is:
1

Solution:
Let

z=11-cosθ+2isinθ

⇒z=1-cosθ-2isinθ1-cosθ+2isinθ1-cosθ-2isinθ

⇒z=2sin2θ2-2isinθ(1-cosθ)2+4sin2θ

Put,sinθ=2sinθ2cosθ2, 1-cosθ=2sin2θ2

⇒z=2sin2θ2-4isinθ2cosθ24sin4θ2+16sin2θ2cos2θ2

Hence,

Rez=2sin2θ24sin4θ2+16sin2θ2cos2θ2

⇒12sin2θ2+8cos2θ2=15

⇒12sin2θ2+8cos2θ2=15

⇒sin2θ2+4cos2θ2=52

⇒1-cos2θ2+4cos2θ2=52

⇒3cos2θ2=32

⇒cos2θ2=12

⇒θ2=nπ±π4, n∈Z

⇒θ=2nπ±π2

For θ∈(0,π), θ=π2.

Hence,

∴∫0θsinθdθ=∫0π2sinθdθ=-cosθ0π2

=-0-1

=1

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (20 Jul Shift 2)
If the real part of the complex number z=3+2icosθ1-3icosθ,θ∈0,π2 is zero, then the value of sin23θ+cos2θ is equal to ______.

Options:

The correct answer is:
1

Solution:
We have, z=3+2icosθ1-3icosθ,θ∈0,π2

⇒z=3+2icosθ1-3icosθ×1+3icosθ1+3icosθ

⇒z=3+2icosθ1+3icosθ1+9cos2θ

⇒z=3-6cos2θ+11icosθ1+9cos2θ

Now, Rez=3-6cos2θ1+9cos2θ=0

⇒3-6cos2θ=0

⇒cos2θ=36

⇒cos2θ=12

⇒cosθ=±12

⇒θ=π4, ∵θ∈0,π2

Hence, sin23θ+cos2θ

=sin23π4+cos2π4

=sinπ-π42+cosπ42

=sinπ42+cosπ42=1

Since, sin2θ+cos2θ=1

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (27 Jul Shift 2)
If the reciprocals of 2, log3x-4â¡4 and log3x+72â¡4 are in arithmetic progression, then x is equal to

Options:
A 1
B 2
C 4
D 0

The correct answer is:
2

Solution:
Reciprocals of 2,logâ¡4logâ¡3x-4,logâ¡4logâ¡3x+72 are in A.P.
â12,logâ¡3x-4logâ¡4,logâ¡3x+72logâ¡4 are in A.P.
â12logâ¡4,logâ¡3x-4,logâ¡3x+72 are in A.P.
âlogâ¡2,logâ¡3x-4,logâ¡3x+72 are in A.P.
â2,3x-4,3x+72 are in G.P.
â3x-42=23x+72
Let 3x=y
ây-42=2y+72
ây2+16-8y=2y+7
ây2-10y+9=0
ây-1y-9=0ây=1,9
â3x=1,9âx=0,2
But, for x=0,3x-4=-3<0
So, x=2 will only form an A.P.

Topic: Mathematics, Sequences and Series, JEE Main
If the rectangular hyperbola is $x^{2}-y^{2}=64$. Then, which of the following is not correct?

Options:
A The length of latus rectum is 16
B The eccentricity is $\sqrt{2}$
C The asymptotes are parallel to each other
D The directrices are $x=\pm 4 \sqrt{2}$

The correct answer is:
The directrices are $x=\pm 4 \sqrt{2}$

Solution:
Given equation of rectangular hyperbola is $x^{2}-y^{2}=8^{2}$
Length of latusrectum $=2 \times(8)=16$ and eccentricity $=\sqrt{2}$
The asymptotes are perpendicular lines.
So, $x \pm y=0$
Now, directrices are
$x=\pm \frac{8}{\sqrt{2}}=\pm 4 \sqrt{2}$

Topic: Mathematics, Hyperbola, JEE Main
If the reflection of the ellipse \( \frac{(x-4)^{2}}{16}+\frac{(y-3)^{2}}{9}=1 \) in the line mirror \( x-y-2=0 \) is
\( k_{1} x^{2}+k_{2} y^{2}-160 x-36 y+292=0 \), then \( k_{1}+k_{2} \) is equal to

Options:

The correct answer is:
25.00

Solution:
For (x-4)216+(y-3)29=1 any point on the ellipse let be (4+4cosθ, 3+3sinθ) where θ is the eccentric angle.

Take the image of (4+4cosθ, 3+3sinθ) with respect to the line x-y-2=0 is h,k

⇒h-4cosθ-41=k-3sinθ-3-1=-24cosθ+4-3sinθ-3-22

i.e 3sinθ+5,4cosθ+2

For eliminating θ, we know,

sin2θ+cos2θ=1

⇒h-532+k-242=1

⇒16h2+9k2-160h-36k+292=0

Hence, the locus of the reflection will be 16x2+9y2-160x-36y+292=0

On comparison we will get

k1+k2=25.

Topic: Mathematics, Ellipse, JEE Main
If the refractive indices of crown glass for red, yellow and violet colours are $1.5140,1.5170$ and 1.5318 respectively and for flint glass these are $1.6434,1.6499$ and 1.6852 respectively, then the dispersive powers for crown and flint glass are respectively

Options:
A 0.034 and 0.064
B 0.064 and 0.034
C 1.00 and 0.064
D 0.034 and 1.0

The correct answer is:
0.034 and 0.064

Solution:
The dispersive power for crown glass $\omega=\frac{n_v-n_r}{n_y-1}$
$=\frac{1.5318-1.5140}{(1.5170-1)}=\frac{0.0178}{0.5170}=0.034$
and for flint glass $\omega^{\prime}=\frac{1.6852-1.6434}{(1.6499-1)}=0.064$

Topic: Physics, Ray Optics, JEE Main
If the relation $p$ (subnormal length) $=q$ (subtangent length) ${ }^2$ holds true for the curve $b y^2=(x+a)^3$, then the value of $\frac{p}{q}$ is equal to

Options:
A $\frac{8}{27}$
B $\frac{8 b}{27}$
C $\frac{8}{27 b}$
D $\frac{27}{8 b}$

The correct answer is:
$\frac{8 b}{27}$

Solution:
Given curve,
$b y^2=(x+a)^3$
On differentiating w.r.t. $x$, we get
$2 b y \frac{d y}{d x}=3(x+a)^2$
$\begin{aligned} & \text { Length of subnormal }=y \frac{d y}{d x}=y\left(\frac{3(x+a)^2}{2 b y}\right) \\ & \text { Length of subtanget }=y \frac{d x}{d y}=y\left(\frac{2 b y}{3(x+a)^2}\right)\end{aligned}$
Given, $P($ Subnormal length $)=q($ subtangent length) ${ }^2$
$\Rightarrow \quad P\left(y \frac{d y}{d x}\right)=q\left(y \frac{d x}{d y}\right)^2 \Rightarrow \frac{P}{q}=y\left(\frac{d x}{d y}\right)^3$
$\Rightarrow \quad \frac{P}{q}=y\left(\frac{2 b y}{3(x+a)^2}\right)^3=\frac{8 y^4}{27} \frac{b^3}{\left((x+a)^3\right]^2}$
$\Rightarrow \quad \frac{P}{q}=\frac{8 b}{27}\left(\frac{b y^2}{(x+a)^3}\right)^2$
$\Rightarrow \quad \frac{P}{q}=\frac{8 b}{27}$ $\left[\because b y^2=(x+a)^3\right]$

Topic: Mathematics, Application of Derivatives, JEE Main
If the remainder when x is divided by 4 is 3, then the remainder when 2020+x2022 is divided by 8 is ___ .

Options:

The correct answer is:
1

Solution:
x=4k+3

∴2020+x2022=2020+4k+32022

=4505+k+32022

=4λ+32022=16λ2+24λ+91011

=82λ2+3λ+1+11011

=8p+11011

∴ Remainder when divided by 8=1.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (25 Feb Shift 2)
If the resistance of each branch of the circuit shown below is 1 Ω, then the equivalent resistance of the circuit between the points A and B is

Options:
A 16 24 Ω
B 22 35 Ω
C 12 30 Ω
D 30 12 Î©

The correct answer is:
22 35 Ω

Solution:
⇒ R eq = ( 1 1 6 â¥ 2 â¥ 1 1 6 ) = 22 35 Ω

Topic: Physics, Current Electricity, JEE Main
If the resistivity of pure silicon is 3000 Î©m and the mobilities of electrons and holes are 0.12 m2V-1s-1 and 0.045 m2V-1s-1 respectively, then what will be the resistivity of a specimen of silicon when 1019 atoms of phosphorous are added per cubic metre?

Options:
A 2.21Ωm
B 3.21Ωm
C 4.21Ωm
D 5.21Ωm

The correct answer is:
5.21Ωm

Solution:
The resistivity of pure S is given by

ρ=1σ=1eneμe+nhμh=1en1μe+μh

or n1=1eρμe+μn=11.6×10-19×3000(0.12+0.045)

=1.26×1016m-3

When 1019 atoms of phosphorous (donor atoms of valence five) are added per m3, the semiconductor becomes n - type semiconductor.

∴ne-nh≈ne=Nd=1019 ∵nn=1.26×1016

Resitivity ρ=1neμee=11.6×10-19×1019×0.12=5.21Ωm

Topic: Physics, Semiconductors, JEE Main
If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that

Options:
A
linear momentum of the system does not change in time

B
kinetic energy of the system does not change in time

C
angular momentum of the system does not change in time

D
potential energy of the system does not change in time

The correct answers are:
linear momentum of the system does not change in time

Solution:
On a system of particles if,
$$
\Sigma \overrightarrow{\mathbf{F}}_{\mathrm{ext} .}=0
$$
then $\quad \overrightarrow{\mathbf{P}}_{\text {system }}=$ constant
No other conclusions can be drawn.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
If the r.m.s speed of chlorine molecule is 490 m s-1 at 27oC, the r.m.s speed of argon molecules at the same temperature will be (Atomic mass of argon =39.9 u, molecular mass of chlorine =70.9 u)

Options:
A 551.7 m s-1
B 651.7 m s-1
C 451.7 m s-1
D 751.7 m s-1

The correct answer is:
651.7 m s-1

Solution:
The formula to calculate the rms speed of the argon molecules is given by

vAr=3RTMAr ...1

The formula to calculate the rms speed of chlorine molecule is given by

vCl=3RTMCl ...2

Divide equation (1) by equation (2) and simplify to obtain the required ratio.

vArvCl=MClMAr ...3

Substitute the values of the known parameters into equation (3) and solve to calculate the required value.

vAr490 m s-1=7140⇒vAr= 7140×490 m s-1≈651.7 m s-1

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2023 (12 Apr Shift 1)
If the R.M.S. speed of oxygen molecules at 0°C is 160 m s-1. Find the R.M.S. speed of hydrogen molecules at 0°C.

Options:
A 40 m s-1
B 80 m s-1
C 640 m s-1
D 332 m s-1

The correct answer is:
640 m s-1

Solution:
Recall the formula of root mean square speed, vRMS=3RTM,

⇒vRMSO2 vRMSH2=MH2 MO2=232=14

⇒vRMSH2=4 vRMSO2=4×160=640 m s-1

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (27 Aug Shift 2)
If the rms value of the voltage of the waveform shown below is pq V, then what is the value of p+q? (p and q are the smallest positive integers.)

Options:

The correct answer is:
7

Solution:
V rms 2 = 1 T ∫ 0 T V 2 dt = 2 T ∫ 0 T 2 V 2 dt
T = 2 sec,
V = 2t 0 ≤ t ≤ 1
V rms 2 = 2 T ∫ 0 1 4 t 2 dt
= ∫ 0 1 4 t 2 dt
= 4 t 3 3 ] 0 1 = 4 3
Vrms=43 V

Topic: Physics, Alternating Current, JEE Main
If the Rolle's theorem holds for the function fx=2x3+ax2+bx in the interval −1,1 for the point c=12, then the value of 2a+b is:

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

The correct answer is:
-1

Solution:
Given that the function holds Rolle's theorem in the interval −1,1

⇒ f-1=f1

⇒−2+a−b=2+a+b

⇒b=-2

Now, f'x=6x2+2ax+b

∴ f'12=0⇒6×14+2a×12+b=0

⇒ a=12

⇒ 2a+b=-1

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2014 (09 Apr Online)
If the root mean square velocity of hydrogen molecule at a given temperature and pressure is 2 km s-1, the root mean square velocity of oxygen at the same condition in km s-1 is:

Options:
A 2.0
B 0.5
C 1.5
D 1.0

The correct answer is:
0.5

Solution:
Root mean square velocity is given by, Vrms=3RTM

Therefore, V1 V2=M2M1⇒2 V2=322

⇒V2=0.5 km s-1

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2024 (01 Feb Shift 2)
If the roots of the equation 10x3-cx2-54x-27=0 are in harmonic progression, then the value of c must be equal to

Options:

The correct answer is:
9

Solution:
The roots of 10x3-cx2-54x-27=0 are in H.P.
Replacing x by 1x we will get an equation whose roots will be in A.P.
⇒10x3-cx2-54x-27=0⇒27x3+54x2+cx-10=0
Hence, roots of this equation are in A.P. i.e. a-d, a, a+d
a-d+a+a+d=-5427=-2⇒a=-23
Since a is the root of the given equation
⇒27-233+54-232+c-23-10=0
⇒-8+24-10=2c3⇒c=6×32=9
c=9

Topic: Mathematics, Sequences and Series, JEE Main
If the roots of the equation ax2+bx+c=0 are in the ratio m : n , then

Options:
A mn b2=ac m+n2
B b2m+n=mn
C m+n=b2mn
D mnc2=abÂ m+n2

The correct answer is:
mn b2=ac m+n2

Solution:
Let the roots are mr and nr
mr+nr=-baârm+n=-baâ¦ 1
and mnr2=caââââââ¦ 2
Eliminating r , we get
b2mn=cam+n2

Topic: Mathematics, Quadratic Equation, JEE Main
If the roots of the equation $b x^2+c x+a=0$ be imaginary, then for all real values of $x$, the expression $3 b^2 x^2+6 b c x+2 c^2$ is

Options:
A
greater than $4 a b$

B
less than $4 a b$

C
greater than $-4 a b$

D
less than - 4ab

The correct answer is:
greater than $-4 a b$

Solution:
$$
b x^2+c x+a=0
$$
Roots are imaginary $\Rightarrow c^2-4 a b < 0 \Rightarrow c^2 < 4 a b \Rightarrow-c^2>-4 a b$ $3 b^2 x^2+6 b c x+2 c^2$
since $3 b^2>0$
Given expression has minimum value
Minimum value $=\frac{4\left(3 b^2\right)\left(2 c^2\right)-36 b^2 c^2}{4\left(3 b^2\right)}=-\frac{12 b^2 c^2}{12 b^2}=-c^2>-4 a b$.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2009
If the roots of the equation \( b x^{2}+c x+a=0 \) be imaginary, then for all real values of \( x \), the expression
\( 3 b^{2} x^{2}+6 b c x+2 c^{2} \) is

Options:
A Greater than \( 4 a b \)
B Less than \( 4 a b \)
C Greater than \( -4 a b \)
D Less than \( -4 a b \)

The correct answer is:
Greater than \( -4 a b \)

Solution:
Given bâ¡ xâ¡ 2 +cx+a=0 has imaginary roots
⇒c2-4ab<0
⇒c2<4ab
⇒ -c2>-4ab ...(i)
Let fx=3b2x2+6bcx+2c2
Here, 3b2>0
So, the given expression has a minimum value
∴ Minimum value=-D4a
=4ac-b24a
=43b22c2-36b2c24(3b2)
=-12b2c212b2=-c2>-4ab
[from Eq. (i)]

Topic: Mathematics, Quadratic Equation, JEE Main
If the roots of the equation $\left(p^2+q^2\right) x^2-2 q(p+r) x+\left(q^2+r^2\right)=0$ be real and equal, then $p, q, r$ will be in

Options:
A $\mathrm{A} . \mathrm{P}$
B $G.P$
C $H.P$.
D None of these

The correct answer is:
$G.P$

Solution:
Given equation is
$\left(p^2+q^2\right) x^2-2 q(p+r) x+\left(q^2+r^2\right)=0$
Roots are real and equal, then
$4 q^2(p+r)^2-4\left(p^2+q^2\right)\left(q^2+r^2\right)=0$
$\Rightarrow q^2\left(p^2+r^2+2 p r\right)-\left(p^2 q^2+p^2 r^2+q^4+q^2 r^2\right)=0$
$\Rightarrow q^2 p^2+q^2 r^2+2 p q^2 r-p^2 q^2-p^2 r^2-q^4-q^2 r^2=0$
$\Rightarrow 2 p q^2 r-p^2 r^2-q^4=0 \Rightarrow\left(q^2-p r\right)^2=0$
Hence $q^2=p r$. Thus $p, q . r$ in G.P.

Topic: Mathematics, Quadratic Equation, JEE Main