datavorous/entrance-exam-dataset huggingface dataset:
An angle between the lines whose direction cosines are given by the equations, $l+3 m+5 n=$ 0 and $5 l m-2 m n+6 n l=0$, is

Options:
A
$\cos ^{-1}\left(\frac{1}{8}\right)$

B
$\cos ^{-1}\left(\frac{1}{6}\right)$

C
$\cos ^{-1}\left(\frac{1}{3}\right)$

D
$\cos ^{-1}\left(\frac{1}{4}\right)$

The correct answer is:
$\cos ^{-1}\left(\frac{1}{6}\right)$

Solution:
Given
$$
\begin{aligned}
&l+3 m+5 n=0 \\
&\text { and } 5 l m-2 m n+6 n l=0
\end{aligned}
$$
From eq. (1) we have
$$
l=-3 m-5 n
$$
Put the value of $l$ in eq. (2), we get;
$$
\begin{aligned}
&5(-3 m-5 n) m-2 m n+6 n(-3 m-5 n)=0 \\
&\Rightarrow 15 m^2+45 m n+30 n^2=0 \\
&\Rightarrow m^2+3 m n+2 n^2=0 \\
&\Rightarrow m^2+2 m n+m n+2 n^2=0 \\
&\Rightarrow(m+n)(m+2 n)=0 \\
&\therefore m=-n \text { or } m=-2 n \\
&\text { For } m=-n, l=-2 n \\
&\text { And for } m=-2 n, l=n
\end{aligned}
$$
$$
\begin{aligned}
&\therefore(l, m, n)=(-2 n,-n, n) \text { Or }(l, m, n) \\
&=(n,-2 n, n) \\
&\Rightarrow(l, m, n)=(-2,-1,1) \text { Or }(l, m, n) \\
&=(1,-2,1)
\end{aligned}
$$
Therefore, angle between the lines is given as:
$$
\begin{aligned}
&\cos (\theta)=\frac{(-2)(1)+(-1) \cdot(-2)+(1)(1)}{\sqrt{6} \cdot \sqrt{6}} \\
&\Rightarrow \cos (\theta)=\frac{1}{6} \Rightarrow \theta=\cos ^{-1}\left(\frac{1}{6}\right)
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
An angle between the plane x+y+z=5 and the line of intersection of the planes, 3x+4y+z-1=0 and 5x+8y+2z+14=0 is

Options:
A cos-1⁡317
B cos-1⁡317
C sin-1⁡317
D sin-1⁡317

The correct answer is:
sin-1⁡317

Solution:
Let n→ be the vector along the line of intersection of given planes.

⇒n→=i^j^k^341582=i^(0)−j^(6−5)+k^(24−20)=−j^+4k^

Now, angle between the plane x+y+z=5 and the line of intersection of the planes 3x+4y+z-1=0 and 5x+8y+2z+14=0 is equal to π2-cos-1⁡-1+43 17

=π2-cos-1⁡317

=sin-1⁡317 ∵sin-1x+cos-1x=π2

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2018 (15 Apr)
An angle between the plane, $x+y+z=5$ and the line of intersection of the planes, $3 x+4 y+z-1$ $=0$ and
$5 x+8 y+2 z+14=0$, is

Options:
A
$\cos ^{-1}\left(\frac{3}{\sqrt{17}}\right)$

B
$\cos ^{-1}\left(\sqrt{\frac{3}{17}}\right)$

C
$\sin ^{-1}\left(\frac{3}{\sqrt{17}}\right)$

D
$\sin ^{-1}\left(\sqrt{\frac{3}{17}}\right)$

The correct answer is:
$\sin ^{-1}\left(\sqrt{\frac{3}{17}}\right)$

Solution:
Normal to $3 x+4 y+z=1$ is $3 \hat{i}+4 \hat{j}+k$.
Normal to $5 x+8 y+2 z=-14$ is $5 \hat{i}+8 \hat{j}+2 \hat{k}$
The line of intersection of the planes is perpendicular to both normals, so, direction ratios of the intersection line are directly proportional to the cross product of the normal vectors.
Therefore the direction ratios of the line is $-\hat{j}+4 \hat{k}$
Hence the angle between the plane $x+y+z+5=0$ and the intersection line is $\sin ^{-1}\left(\frac{-1+4}{\sqrt{17} \sqrt{3}}\right)=\sin ^{-1}\left(\sqrt{\frac{3}{17}}\right)$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
An angle of intersection of the curves, x2a2+y2 b2=1 and x2+y2=ab, a>b, is :

Options:
A tan-12ab
B tan-1a+bab
C tan-1a-bab
D tan-1a-b2ab

The correct answer is:
tan-1a-bab

Solution:
Given equations are x2a2+y2 b2=1 ...1 and x2+y2=ab.....2 where a>b.

From 1,

b2x2+a2y2=a2b2

Substituting value of y2 from 2 in the above equation

b2x2+a2ab-x2=a2b2

⇒b2-a2x2=a2b2-a3b
⇒x2=ba2b-ab2-a2=a2ba+b

∴x=a2ba+b

Substituting value of x2 in 2, we get

y2=ab2a+b

∴y=ab2a+b
Point of intersection is a2ba+b, ab2a+b

Now, from 1,
x2a2+y2b2=1

⇒2xa2+2yb2dydx=0
⇒dydx=m1=-b2xa2y

From 2,
x2+y2=ab

⇒2x+2y·dydx=0

⇒dydx=m2=-xy

Let θ be the angle between curve 1 & 2.
∴tanθ=m1-m21+m1m2=-b2xa2y+xy1+b2x2a2y2=xya2-b2b2x2+a2y2=xya2-b2a2b2

∴tanθa2ba+b, ab2a+b=a3b3(a+b)2·a2-b2a2b2=(a-b)ab

Hence, θ=tan-1a-bab

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (31 Aug Shift 2)
An annular ring with inner and outer radii $R_1$ and $R_2$ is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, $F_1 / F_2$ is

Options:
A
$\frac{\mathrm{R}_2}{\mathrm{R}_1}$

B
$\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^2$

C
1

D
$\frac{\mathrm{R}_1}{\mathrm{R}_2}$

The correct answer is:
$\frac{\mathrm{R}_1}{\mathrm{R}_2}$

Solution:
$\frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{\mathrm{R}_1 \omega^2}{\mathrm{R}_2 \omega^2}=\frac{\mathrm{R}_1}{\mathrm{R}_2}$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2005
An ant starts from the origin and crawls 10 cm along the x-axis and then 20 cm along the y-axis. The dot product of the ant's displacement vector with the position vector of a point that makes 45° with the x-axis and has a magnitude of 2 cm is

Options:
A 30cm
B 302cm
C 302cm
D 15cm

The correct answer is:
30cm

Solution:
Ant's displacement vector =10i^+20j^=a→

Position vector of the given point

=2cos45° i^+2sin45° j^=i^+j^=b→

Therefore, required dot product will be,

a→·b→=10i^+20j^·i^+j^=10+20=30 cm

Topic: Physics, Mathematics in Physics, JEE Main
An antenna is mounted on a 400 m tall building. What will be the wavelength of signal that can be radiated effectively by the transmission tower upto a range of 44 km ?

Options:
A 605 m
B 302 m
C 37.8 m
D 75.6 m

The correct answer is:
605 m

Solution:
d=2RhT

⇒44×103=2×6400×103×hT

⇒hT=151.25 m

∵ lantenna≥λ4

⇒λ≤4l

considering, hT=1= length of antenna

λ=4×151.25=605 m

Topic: Physics, Communication System, JEE Main, JEE Main 2021 (27 Aug Shift 2)
An antenna is placed in a dielectric medium of dielectric constant 6.25. If the maximum size of that antenna is 5.0 mm, it can radiate a signal of minimum frequency of _____ GHz.

(Given μr=1 for dielectric medium)

Options:

The correct answer is:
6

Solution:
We know that, refractive index is related to velocity as n=cv.

Here, n=εrμr=6.251=2.5

Since, effective length of antenna l>λ4⇒l>v4f

So, f>v4l=c4l2.5=3×1082.5×4×5×10-3=6×109 Hz=6 GHz.

Topic: Physics, Communication System, JEE Main, JEE Main 2022 (24 Jun Shift 2)
An anteriorly convex curve is seen in which region of the vertebral column?

Options:
A Thoracic
B Lumbar
C Sacral
D Both (A) and (C)

The correct answer is:
Lumbar

Solution:
The vertebral column plays the most important role in maintaining the posture and contour of the body. It is formed by 26 units called vertebrae. The first vertebra is the atlas and it articulates with the occipital condyles. The vertebral column is differentiated into Cervical (7) vertebrae, thoracic vertebrae (12), lumbar vertebrae (5), sacral vertebrae (5 fused to form 1 bone) and coccygeal vertebrae (4 fused to form 1 bone). In the thoracic and sacral regions, the vertebral column forms a curvature which is convex posteriorly (concave anteriorly) while in the cervical and lumbar region it forms a curvature which is concave posteriorly (convex anteriorly)

Topic: Biology, Locomotion and Movement, NEET
An antiseptic dettol is a mixture of two compounds 'A' and 'B' where A has 6π electrons and B has 2π electrons. What is B?

Options:
A Bithional
B

Terpineol

C Chlorophenol
D Chloroxyenol

The correct answer is:

Terpineol

Solution:
dettol is a mixture of chloroxylenol and terpineol. chloroxylenol has 6π electrons and terpineol has 2π electrons.

Hence,B is terpineol.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2022 (25 Jun Shift 2)
An aqueous KCl solution of density 1.20 g mL-1 has a molality of 3.30 mol kg-1. The molarity of the solution in molL-1 is____.(The Nearest integer)
[Molar mass of KCl=74.5 g]

Options:

The correct answer is:
3

Solution:
Molality m=M×10001000×d-M×Msolute

3.3=M×10001000×1.20-M×74.5

3960-245.85M=1000M

1245.85M=3960

M=39601245.85=3.17≈3M

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (26 Aug Shift 1)
An aqueous solution containing 1 M each of Au3+,Cu2+,Ag+,Li+ is being electrolysed by using inert electrodes. The value of standard potentials are
EAg+/Ago=0.80V,ECu+/Cuo=0.34V and EAu3+/Auo=1.50V,ELi+/Lio=-3.03V
with increasing voltage, the sequence of deposition of metals on the cathode will be

Options:
A Li, Cu, Ag, Au
B Cu, Ag, Au
C Au, Ag, Cu
D Au, Ag, Cu, Li

The correct answer is:
Au, Ag, Cu

Solution:
Only Au3+,Ag+ and Cu2+ will deposit at cathode. Li will not deposit at cathode because SRP of H2 is 0.00V, so after Cu2+ , H2 will evolve at cathode.

Topic: Chemistry, Electrochemistry, JEE Main
An aqueous solution contains 0.10 M H2S and 0.20 M HCl. If the equilibrium constant for the formation of HS- from H2S is 1.0×10-7 and that of S2- from HS- ions is 1.2×10-13, then, the concentration of S2- ions in the aqueous solution is:

Options:
A 5×10-19
B 5×10-8
C 3×10-20
D 6×10-21

The correct answer is:
3×10-20

Solution:
HCl→H++Cl-0.2+x+y0.2

H2S⇌H++HS-0.1-x0.2+x+yx-y; Ka1=10-7

HS-→H++S2-x-y0.2+x+yy; Ka2=1.2×10-13

H+=0.2+x+yM=0.2M

Ka1=H+HS-H2S=0.2×HS-0.1-x=0.2HS-0.1 (x=negligible)

HS-=Ka1×0.10.2=12×10-7

Ka2=H+S2-HS-=0.2 MS2-12×10-7

S2-=Ka2×12×10-70.2=1.2×10-13×12×10-70.2=3×10-20 M

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2018 (08 Apr)
An aqueous solution contains an unknown concentration of Ba2+ . When 50 mL of a 1 M solution of Na2SO4 is added BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is 1×10-10 . What is the original concentration of Ba2+ ?

Options:
A 1.0×10-10M
B 5×10-9M
C 2×10-9M
D 1.1×10-9M

The correct answer is:
1.1×10-9M

Solution:
Ba2+=10-9 M (in 500 mL solution)
SO42- in 500 mL solution will be
50Ã1=MÃ500
M =0.1
Ba2+450 mL+Na2SO450 mL, 1MâBaSO4+2Na+
KspBaSO4=Ba2+SO42-
10-10=Ba2+Ã0.1
Now, we have to calculate Ba2+ in original solution (450 mL)
10-9Ã500=450ÃM
M=10-9Ã500450=109Ã10-9=1.11Ã10-9M

Topic: Chemistry, Ionic Equilibrium, JEE Main
An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1 M solution of Na2SO4 is added, BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is 1×10-10. What is the original concentration of Ba2+?

Options:
A 1.0×10-10M
B 5×10-9M
C 2×10-9M
D 1.1×10-9M

The correct answer is:
1.1×10-9M

Solution:
Ba+2+SO42−⇌BaSO4(s)ppt
Final conc. of [SO42−]=MV1V1+V2=1×50500=0.1M
Final conc.of [Ba+2] When BaSO4 start precipitating
KSP=QSP=[Ba+2][SO42−]
10−10=[Ba+2](0.1M)
Initial conc. [Ba+2]; initial volume was =500-50=450 ml
M1V1=M2V2
M1=M2V2V1=10−9×500450
M1=1.1×10−9M

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2018 (08 Apr)
An aqueous solution of a metal bromide MBr20.05 M is saturated with H2S. What is the minimum pH at which MS will precipitate? Ksp for MS=6.0Ã10-21. Concentration of saturated H2S=0.1 M; K1=10-7 and K2=1.3Ã10-13 for H2S.
[Report your answer by rounding it upto nearset whole number]

Options:

The correct answer is:
1.00

Solution:
For the precipitation of MS form a solution of MBr2 by passing H2S
K sp MS = M 2+ S 2â
or 6Ã10-21=0.05S2-
or S2-=1.2Ã10-19M
i.e. precipitation of MS will start when H2S provides
1.2Ã10-19 M ion of S2-
For H 2 Sâ2 H + + S 2â
K=K1ÃK2=[H+]2[S2-][H2S]
or 10-7Ã1.3Ã10-13=[H+]2[1.2Ã10-19]0.1
or H+=1.04Ã10-1
â´pH=0.9826

Topic: Chemistry, Ionic Equilibrium, JEE Main
An aqueous solution of a salt MX2 at certain temperature has a Van't Hoff factor of 2. What is the degree of dissociation for this solution of the salt?

Options:
A 0.50
B 0.33
C 0.67
D 0.80

The correct answer is:
0.50

Solution:
α=i-1k-1
=2-13-1=12=0.50
Here,
i= van't hoff factor
α =degree of dissociation
k= total number of ions(dissociated)

Topic: Chemistry, Solutions, JEE Main, JEE Main 2016 (10 Apr Online)
An aqueous solution of a salt X turns blood red on treatment with SCN- and blue on treatment with K4FeCN6 , X also gives a positive chromyl chloride test. The salt X is:

Options:
A FeCl3
B FeNO33
C CuCl2
D CuNO32

The correct answer is:
FeCl3

Solution:
FeCl 3 + 3 SCN aq - ⇌ Fe SCN 3 + 3 Cl - Blood red
4 FeCl 3 + 3 K 4 Fe CN 6 → 1 2 KCl + Fe 4 Fe CN 6 Prussian blue
2 FeCl 3 + 3 H 2 SO 4 → Fe 2 SO 4 3 + 6 HCl
K 2 Cr 2 O 7 + 2 H 2 SO 4 → 2 KHSO 4 + 2 CrO 3 + H 2 O
CrO 3 + 2 HCl → CrO 2 Cl 2 + H 2 O Chromyl chloride
CrO 2 Cl 2 +4NaOH→ Na 2 CrO 4 Yellow +2NaCl+2H 2 O
Na 2 CrO 4 + Pb CH 3 COO 2 → PbCrO 4 + 2 CH 3 COONa Yellow ppt.

Topic: Chemistry, d and f Block Elements, JEE Main
An aqueous solution of a salt X turns blood red on treatment with SCN- and blue on treatment with K4FeCN6 , X also gives a positive chromyl chloride test. The salt X is:

Options:
A FeCl3
B FeNO33
C CuCl2
D CuNO32

The correct answer is:
FeCl3

Solution:
FeCl 3 + 3 SCN aq - ⇌ Fe SCN 3 + 3 Cl - Blood red
4 FeCl 3 + 3 K 4 Fe CN 6 → 1 2 KCl + Fe 4 Fe CN 6 Prussian blue
2 FeCl 3 + 3 H 2 SO 4 → Fe 2 SO 4 3 + 6 HCl
K 2 Cr 2 O 7 + 2 H 2 SO 4 → 2 KHSO 4 + 2 CrO 3 + H 2 O
CrO 3 + 2 HCl → CrO 2 Cl 2 + H 2 O Chromyl chloride
CrO 2 Cl 2 +4NaOH→ Na 2 CrO 4 Yellow +2NaCl+2H 2 O
Na 2 CrO 4 + Pb CH 3 COO 2 → PbCrO 4 + 2 CH 3 COONa Yellow ppt.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2015 (10 Apr Online)
An aqueous solution of an inorganic salt on treatment with $\mathrm{HCl}$ gives a white precipitate. This solution contains

Options:
A $\mathrm{Hg}_2^{2+}$
B $\mathrm{Hg}^{2+}$
C $\mathrm{Zn}^{2+}$
D $C d^{2+}$

The correct answer is:
$\mathrm{Hg}_2^{2+}$

Solution:
$\mathrm{Hg}_2^{2+}$ will produce white pot with dill $\mathrm{HCl}$.
$\mathrm{H}_{\mathrm{g}_2}^{+2}+2 \mathrm{HCl} \rightarrow \underset{\text { (White ppt) }}{\mathrm{H}_{g_2} \mathrm{Cl} 2 \downarrow+2 \mathrm{H}^{+}}$
This precipitation is possible due to very low value of the solubility products,

Topic: Chemistry, Practical Chemistry, JEE Main
An aqueous solution of \( \mathrm{CuSO}_{4} \) is subjected to electrolysis using inert electrodes. The \( \mathrm{pH} \) of
the solution will

Options:
A decrease
B increase or decrease depending on the strength of the current
C increase
D remains unchanged

The correct answer is:
decrease

Solution:
(A)
At cathode $: \mathrm{Cu}^{+2}(a q)+2 e^{-} \rightarrow \mathrm{Cu}(\mathrm{s})$
At anode $: 2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{O}_{2}+4 \mathrm{H}^{+}+4 e^{-}$
Production of $\mathrm{H}^{+}$ions will decrease $\mathrm{pH}$

Topic: Chemistry, Electrochemistry, KCET, KCET 2019
An aqueous solution of methanol in water has vapour pressure

Options:
A Equal to that of water
B Equal to that of methanol
C More than that of water
D Less than that of water

The correct answer is:
More than that of water

Solution:
Methanol has low boiling point than $\mathrm{H}_2 \mathrm{O}$ $\mathrm{H}_2 \mathrm{O}$ Lower is boiling point of solvent more is vapour pressure.

Topic: Chemistry, Solutions, JEE Main
An aqueous solution of NiCl2 was heated with excess sodium cyanide in presence of strong oxidizing agent to form Ni(CN)62-. The total change in number of unpaired electrons on metal centre is ___________.

Options:

The correct answer is:
2

Solution:
Ni(CN)62-

Ni+4→d6 strong field ligand

Pairing will be there zero unpaired electron

NiCl2→Ni2+→d8

Change=2

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (20 Jul Shift 2)
An aqueous solution of oxalic acid dihydrate contains its $6.3 \mathrm{~g}$ in $250 \mathrm{ml}$. The volume of $0.1 \mathrm{~N}$ $\mathrm{NaOH}$ required to completely neutralize $10 \mathrm{ml}$ of this solution

Options:
A
$4 \mathrm{ml}$

B
$20 \mathrm{ml}$

C
$2 \mathrm{ml}$

D
$40 \mathrm{ml}$

The correct answer is:
$40 \mathrm{ml}$

Solution:
Normality of oxalic acid solution
$$
=\frac{6.3 \times 1000}{63 \times 250}=0.4 \mathrm{~N}
$$
Now from
$$
\begin{aligned}
& N_1 V_1=N_2 V_2 \\
& 0.4 \times 10=0.1 \times V_2 \\
& V_2=40 \mathrm{~mL}
\end{aligned}
$$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2012 (12 May Online)
An aqueous solution of volume 300 cm3 contains 0.63 g of protein. The osmotic pressure of the solution at 300 K is 1.29 mbar. The molar mass of the protein is gmol-1.

Given : R=0.083 LbarK-1 mol-1

Options:

The correct answer is:
40535

Solution:
The osmotic pressure of a non-electrolytic solution can be calculated as follows,

π=CRT

π = osmotic pressure

C = molarity=Number of molesVolume of solution in L

1.29×10-3=0.63(MW)(0.3)×(0.083)×300

MW=0.63×0.083×300(0.3)×1.29×10-3

= 40534.88

≈40535 (Nearest integer)

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (10 Apr Shift 2)
An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to:

Options:
A 0.044 H
B 0.065 H
C 80 H
D 0.08 H

The correct answer is:
0.065 H

Solution:
Given that the lamp draws a current of 10 A upon a 80 V D.C. supply, the resistance is,

R = 8010 = 8 Ω

When connected to an AC supply with an inductor in series, as shown in the figure above, the current,

I = 10 A

and the voltage drop over the lamp is 80 V and VL over the inductance.

Applying the rules of L-R circuit,

VL2+802=2202
⇒VL=48400-6400=210 V
The Inductive reactance of the given inductor of inductance L, for the frequency of supply f,

XL=2πfL ...(1)

The voltage drop across the inductance,
VL= I XL=210
⇒XL=2πfL=VLI=21010=21 Ω

Substituting values in equation (1)
L=212πf=212π×50=21100 π=0.065 H

Topic: Physics, Alternating Current, JEE Main, JEE Main 2016 (03 Apr)
An arc of a circle of radius $R$ subtends an angle ${ }^{\frac{\pi}{2}}$ at the centre. It carries a current $i$. The magnetic field at the centre will be

Options:
A $\frac{\mu_0 i}{2 R}$
B $\frac{\mu_0 i}{8 R}$
C $\frac{\mu_0 i}{4 R}$
D $\frac{2 \mu_0 i}{5 R}$

The correct answer is:
$\frac{\mu_0 i}{8 R}$

Solution:
$B=\frac{\mu_0}{4 \pi} \frac{\theta i}{r}=\frac{\mu_0}{4 \pi} \times \frac{\pi}{2} \times \frac{i}{R}=\frac{\mu_0 i}{8 R}$

Topic: Physics, Magnetic Effects of Current, JEE Main
An arc of a circular loop of radius R is kept in the horizontal plane and a constant magnetic field B is applied in the vertical direction as shown in the figure. If the arc carries current I then find the force on the arc.

Options:
A IRB

B IRB4
C 2IRB
D IRB2

The correct answer is:
2IRB

Solution:
If this loop is close net force on it will be zero
so F + Fstraight = 0
â F = - F straight
F = I 2 RB
F = 2 IRB

Topic: Physics, Magnetic Effects of Current, JEE Main
An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If OP→=u→, OR→=v→ and OQ→=αu→+βv→, then α, β2, are the roots of the equation

Options:
A x2+x-2=0
B x2-x-2=0
C 3x2-2x-1=0
D 3x2+2x-1=0

The correct answer is:
x2-x-2=0

Solution:
Given,

An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R,

And OP→=u→, OR→=v→ and OQ→=αu→+βv→,

So, plotting the diagram of the given value we get,

Now given, OQ→=αu→+βv→

Where, OR →=v→ and OP →=u→

Now OP →=OQ →=radius, so R will lie on angle bisector of OQ→ and OP→

Now using the perpendicular condition we get,

OQ→·OP→=0

⇒OQ→·OP→=αu→2+β·v→·u→=0

⇒αu→2+β·v→u→cos45°=0

⇒α+β·cos45°=0 as u→=v→=radius

⇒α=-β2

Now solving, OQ→·OR→=OQ→OR→cos45°=12r2

⇒αv→+βv→·v→=r22

⇒α·r22+β·r2=r22

⇒α2+β=12

⇒β=2 as α=-β2

So, α=-1

Now finding the equation with roots α=-1, β2=2 we get, x2-x-2=0

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (10 Apr Shift 1)
An arithmetic progression is written in the following way
The sum of all the terms of the $10^{\text {th }}$ row is_______

Options:

The correct answer is:
1505

Solution:
$\begin{aligned} & 2,5,11,20, \ldots . . \\ & \text { General term }=\frac{3 n^2-3 n+4}{2} \\ & \begin{aligned} & \mathrm{T}_{10}=\frac{3(100)-3(10)+4}{2} \\ & \quad= 137 \\ & 10 \text { terms with c.d. }=3 \\ & \text { sum }=\frac{10}{2}(2(137)+9(3)) \\ &=1505\end{aligned}\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (08 Apr Shift 2)
An army vehicle of mass $1000 \mathrm{~kg}$ is moving with a velocity of $10 \mathrm{~m} / \mathrm{s}$ and is acted upon by a forward force of $1000 \mathrm{~N}$ due to the engine and a retarding force of $500 \mathrm{~N}$ due to friction. What will be its velocity after $10 \mathrm{~s}$

Options:
A $5 \mathrm{~m} / \mathrm{s}$
B $10 \mathrm{~m} / \mathrm{s}$
C $15 \mathrm{~m} / \mathrm{s}$
D $20 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$15 \mathrm{~m} / \mathrm{s}$

Solution:
$v=u+\frac{F}{m} t=10+\left(\frac{1000-500}{1000}\right) \times 10=15 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Laws of Motion, JEE Main
An aromatic compound 'A' having molecular formula C7H6O2, on treating with aqueous ammonia and heating forms compound ′B′. The compound ′B′ on reaction with molecular bromine and potassium hydroxide provides compound ′C′ having molecular formula C6H7N. The structure of ′A′ is:

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2019 (10 Jan Shift 2)
An aromatic compound ' \( \mathrm{A} \) ' \( \left(\mathrm{C}_{7} \mathrm{H}_{9} \mathrm{~N}\right) \) on reacting with \( \mathrm{NaNO}_{2} / \mathrm{HCl} \) at \( 0^{\circ} \mathrm{C} \) forms benzyl
alcohol and nitrogen gas. The number of isomers possible for the compound ' \( A \) ' is

Options:
A \( 7 \)
B \( 06 \)
C \( 55 \)
D \( 13 \)

The correct answer is:
\( 55 \)

Solution:
The reaction involved is
No. of isomeric forms of \( C_{7} H_{9} N \)
o-Toluidine
\( m \)-Toluidine
\( \mathrm{CH}_{3} \)
\( p \)-Toluidine
\( \mathrm{CH}_{2} \mathrm{NH}_{2} \)

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
An aromatic compounds among other things should have a $\pi$-electron cloud containing electrons where $n$ can't be

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

The correct answer is:
1/2

Solution:
According to Huckel Ruel, all aromatic compounds must have $(4 n+2) \pi$ electrons where $n$ is an integer, i.e., $n=0,1,2,3, \ldots$ and possesses unusual stability due to the complete delocalisation of $\pi$-electrons.

Topic: Chemistry, General Organic Chemistry, JEE Main
An arrangement of spring, strings, pulley and masses is shown in the figure below.
The pulley and the strings are massless and M>m. The spring is light with spring constant k. If the string connecting m to the ground is detached, then immediately after detachment

Options:
A the magnitude of the acceleration of m is zero and that of M is g
B the magnitude of the acceleration of m is M-mg/m and that of M is zero
C the accelerations of both masses are same
D the elongation in the spring is M-mg/k

The correct answer is:
the magnitude of the acceleration of m is M-mg/m and that of M is zero

Solution:
At t=0, T=mg

When string is cut, mass M remains static and spring snaps. For mass m at the instant string is cut, forces on m are

Acceleration of m is upwards as spring snaps,

am=Fnetm=Mg-mgm

=M-mmg

Topic: Physics, Laws of Motion, KVPY, KVPY 2018 (SB/SX)
An artificial cell is made by encapsulating \(0.2 \mathrm{M}\) glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a \(0.05 \mathrm{M}\) solution of \(\mathrm{NaCl}\) at \(300 \mathrm{~K}\) is _______ \(\times 10^{-1}\) bar. (nearest integer).
[Given : \(\mathrm{R}=0.083 \mathrm{~L} \mathrm{bar} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\)]
Assume complete dissociation of \(\mathrm{NaCl}\)

Options:

The correct answer is:
25

Solution:
$\begin{aligned} & \mathrm{NaCl} \longrightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-} \\ & 0.05 \mathrm{M} \quad 0.05 \mathrm{M} \quad 0.05 \mathrm{M} \\ & \text { Total } \mathrm{C}_1=0.05+0.05=0.1 \mathrm{M}(\mathrm{NaCl}) \\ & \mathrm{C}_2=0.2 \mathrm{M}(\text { glucose }) \\ & \pi=\left(\mathrm{C}_2-\mathrm{C}_1\right) \mathrm{RT} \\ & =(0.2-0.1) \times 0.083 \times 300 \\ & =2.49 \mathrm{bar} \\ & =24.9 \times 10^{-1} \text { bar }\end{aligned}$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2024 (05 Apr Shift 1)