datavorous/entrance-exam-dataset huggingface dataset:
A variable plane remains at constant distance $\mathrm{p}$, from the origin.If it meets coordinate axes at points $A, B, C$ then the locus of the centroid of $\Delta$ $\mathrm{ABC}$ is

Options:
A $\mathrm{x}^{-2}+\mathrm{y}^{-2}+\mathrm{z}^{-2}=9 \mathrm{p}^{-2}$
B $x^{-3}+y^{-3}+z^{-3}=9 p^{-3}$
C $x^{2}+y^{2}+z^{2}=9 p^{2}$
D $x^{3}+y^{3}+z^{3}=9 p^{3}$

The correct answer is:
$\mathrm{x}^{-2}+\mathrm{y}^{-2}+\mathrm{z}^{-2}=9 \mathrm{p}^{-2}$

Solution:
Let $\mathrm{A} \equiv(\mathrm{a}, 0,0), \mathrm{B} \equiv(0, \mathrm{~b}, 0), \mathrm{C} \equiv(0,0, \mathrm{c})$, then equation of the plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Its distance from the origin, $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{p^{2}} \ldots$ (i)
If $(x, y, z)$ be centroid of $\Delta A B C$, then
$$
\mathrm{x}=\frac{\mathrm{a}}{3}, \mathrm{y}=\frac{\mathrm{b}}{3}, \mathrm{z}=\frac{\mathrm{c}}{3}...(ii)
$$
Eliminating a,b,c from (i) and (ii) required locus is
$x^{-2}+y^{-2}+z^{-2}=9 p^{-2}$

Topic: Mathematics, Three Dimensional Geometry, VITEEE, VITEEE 2016
A variable, the opposite external potential (Eext) is applied to the cell Zn | Zn2+ 1M∥Cu2+ 1 M| Cu, of potential 1.1 V. When Eext<1.1 V and Eext>1.1 V, respectively electrons flow from

Options:
A anode to cathode in both the cases.
B anode to cathode and cathode to anode.
C cathode to anode and anode to cathode.
D cathode to anode in both cases.

The correct answer is:
anode to cathode and cathode to anode.

Solution:
For a Daniel cell,
Ecell=0.34-(-0.76)=1.10 V.
When Eext<1.10 V, electrons flow from anode to cathode in the external circuit.
When Eext>1.10 V, electrons flow from cathode to anode in the external circuit (reverse reaction).

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2015 (10 Apr Online)
A vector $\overline{\mathrm{a}}$ has components 1 and $2 \mathrm{p}$ with respect to a rectangular Cartesian system. This system is rotated through a certain angle about origin in the counter clock wise sense. If, with respect to the new system, $\bar{a}$ has components 1 and $(p+1)$, then

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

The correct answer is:
$\mathrm{p}=\frac{-1}{3}$ or $\mathrm{p}=1$

Solution:
$$
\begin{aligned}
\bar{a} & =1 \cdot \hat{i}+2 p \hat{j} \\
& =\hat{i}+2 p \hat{j}
\end{aligned}
$$
Let $\bar{b}$ be the vector obtained on rotation with components 1 and $(p+1)$. Then,
$$
\begin{aligned}
& \overline{\mathrm{b}}=\hat{\mathrm{i}}+(\mathrm{p}+1) \hat{\mathrm{j}} \\
& |\overline{\mathrm{a}}|=|\overline{\mathrm{b}}|
\end{aligned}
$$
...[Magnitude remains unchanged after rotation]
$$
\begin{aligned}
& \Rightarrow|\vec{a}|^2=|\bar{b}|^2 \\
& \Rightarrow 1+(2 p)^2=1+(p+1)^2 \\
& \Rightarrow 4 p^2=p^2+2 p+1 \\
& \Rightarrow 3 p^2-2 p-1=0 \\
& \Rightarrow(3 p+1)(p-1)=0 \\
& \Rightarrow p=-\frac{1}{3} \text { or } p=1
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, MHT CET, MHT CET 2023 (13 May Shift 1)
A vector a→ has components 3p and 1 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to new system, a→ has components p+1 and 10, then a value of p is equal to:

Options:
A 1
B -54
C 45
D -1

The correct answer is:
-1

Solution:
a→Old=3pi^+j^

a→New=(p+1)i^+10j^

⇒a→Old=a→New

⇒9p2+1=p2+2p+1+10

8p2-2p-10=0

4p2-p-5=0

(4p-5)(p+1)=0⇒p=54,-1

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (18 Mar Shift 1)
A vector a has components \( 3 p \) and \( 1 \) with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to new system, a has components \( p+1 \) and \( \sqrt{10} \), then a value of \( p \) is equal to:

Options:
A \( 1 \)
B \( -\frac{5}{4} \)
C \( \frac{4}{5} \)
D \( -1 \)

The correct answer is:
\( -1 \)

Solution:
a→Old=3pi^+j^

a→New=(p+1)i^+10j^

⇒a→Old=a→New

⇒ap2+1=p2+2p+1+10

8p2-2p-10=0

4p2-p-5=0

(4p-5)(p+1)=0→p=54,-1

Topic: Mathematics, Vector Algebra, JEE Main
A vector a→=αi^+2j^+βk^α,β∈R lies in the plane of the vectors, b→=i^+j^ and c→=i^-j^+4k^. If a→ bisects the angle between b→ and c→, then

Options:
A a→⋅i^+3=0
B a→⋅i^+1=0
C a→⋅k^+2=0
D a→⋅k^+4=0

The correct answer is:
a→⋅k^+2=0

Solution:
As, a→ lies in the plane of the vectors b→ and c→, So, a→=λi^+j^2+i^+j^+4k^32=λ323i^+3j^+i^-j^+4k^=λ324i^+2j^+4k^
Compare with a→=ai^+2j^+βk^
2λ32=2⇒λ=32
a→=4i^+2j^+4k^
Not in option
So, now consider a→=μi^+j^2-i^-j^+4k^32
a→=μ323i^+3j^-i^+j^-4k^
=μ322i^+4j^-4k^
Compare with a→=αi^+2j^+βk^
4μ32=2⇒μ=322
a→=i^+2j^-2k^
a→.k^+2=0

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2020 (07 Jan Shift 1)
A vector a→ is parallel to the line of intersection of the plane determined by the vectors i^,i^+j^ and the plane determined by the vectors i^-j^,i^+k^. The obtuse angle between a→ and the vector b→=i^-2j^+2k^ is

Options:
A 3π4
B 2π3
C 4π5
D 5π6

The correct answer is:
3π4

Solution:
The normal to the plane determined by the vectors i^, i^+j^ will be

n→1=i^×i^+j^=k^

The normal to the plane determined by the vectors i^-j^, i^+k^ will be

n→2=i^+k^×i^-j^

=i^+j^-k^

Now the vector along the line of intersection will be n→1×n→2

=k^×i^+j^-k^=-i^+j^

i.e. a→=-i^+j^

Given b→=i^-2j^+2k^

a→·b→=-3

Let a→∧b→=θ

Now cosθ=a→·b→a→b→=-32×3

θ=3π4

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (26 Jul Shift 2)
A vector A→ is rotated by a small angle ∆θ radians ∆θ≪1 to get a new vector B→ . In that case B→-A→ is :

Options:
A A→1-∆θ22
B 0
C A→∆θ
D B→∆θ-A→

The correct answer is:
A→∆θ

Solution:
B→=A→

B→-A→=A→2+A→2+2A→A→cos180°-∆θ

=A→2+A→2-2A→2cos∆θ

=2A→2-2A→2cos∆θ

=2A→21-cos∆θ

∵1-cos∆θ=2sin2∆θ2

=2A→2×2sin2∆θ2

=2A→sin∆θ2

For small ∆θ, sin∆θ2≃∆θ2

B→-A→=2A→∆θ2

=A→ ∆θ

Topic: Physics, Mathematics in Physics, JEE Main
A vector A→ is rotated by a small angle ∆θ radians ∆θ≪1 to get a new vector B→ . In that case B→-A→ is :

Options:
A A→1-∆θ22
B 0
C A→∆θ
D B→∆θ-A→

The correct answer is:
A→∆θ

Solution:
B→=A→

B→-A→=A→2+A→2+2A→A→cos180°-∆θ

=A→2+A→2-2A→2cos∆θ

=2A→2-2A→2cos∆θ

=2A→21-cos∆θ

∵1-cos∆θ=2sin2∆θ2

=2A→2×2sin2∆θ2

=2A→sin∆θ2

For small ∆θ, sin∆θ2≃∆θ2

B→-A→=2A→∆θ2

=A→ ∆θ

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2015 (11 Apr Online)
A vector $\vec{F}_1$ is along the positive $X$-axis. If its vector product with another vector $\vec{F}_2$ is zero then $\vec{F}_2$ could be

Options:
A $4 \hat{j}$
B $(-\hat{i}+\hat{j})$
C $(\hat{j}+\hat{k})$
D $(-4 \hat{i})$

The correct answer is:
$(-4 \hat{i})$

Solution:
The vector product of two vectors $\overrightarrow{\mathrm{A}}$ and $\vec{B}$ is defined by $\vec{A} \times \vec{B}=\hat{n}|A||B| \sin x$, where $\mathrm{A}$ is the unit vector perpendicular to both $\mathrm{A}$ and $\mathrm{B}$ vectors and $\mathrm{x}$ is the angle between them.
Here in this question the vector product of $\mathrm{P} \overrightarrow{1}$ and $\mathrm{P} \overrightarrow{2}$ vectors is zero. This is only possible in two cases: 1) Any of the vectors is zero itself or 2 ) the sin of the angle between them is zero.
From the given options, the vector parallel to the given vector $\hat{1}$ is $-4 \hat{i}$.

Topic: Physics, Mathematics in Physics, JEE Main
A vector has magnitude same as that of A→=3j^+4j^ and is parallel to B→=4i^+3j^. The x and y components of this vector in first quadrant are x and 3 respectively where x=____.

Options:

The correct answer is:
4

Solution:
Magnitude of vector A→=32+42=5

Now, unit vector along B→=(4i^+3j^)42+32=45i^+35j^

Now, N→=A→B^=5×45i^+35j^=4i^+3j^

∴x=4

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2024 (30 Jan Shift 2)
A vector in x-y plane makes an angle of 30o with y-axis. The magnitude of y-component of vector is 23. The magnitude of x-component of the vector will be :

Options:
A 13
B 6
C 2
D 3

The correct answer is:
2

Solution:
The angle made by the vector on x axis is

θ=90°-30°=60°

Let the vector be a→.

It is given that

ay=23⇒asin60°=23⇒a32=23⇒a=4

x component is ax=acos60°⇒ax=4×12=2

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2023 (15 Apr Shift 1)
A vector $\vec{n}$ is inclined to $x$-axis at $45^{\circ}$, to $y$-axis at $60^{\circ}$ and at an acute angle to $z$-axis. If $\vec{n}$ is a normal to a plane passing through the point $(\sqrt{2},-1,1)$ then the equation of the plane is :

Options:
A
$4 \sqrt{2} x+7 y+z-2$

B
$2 x+y+2 z=2 \sqrt{2}+1$

C
$3 \sqrt{2} x-4 y-3 z=7$

D
$\sqrt{2} x-y-z=2$

The correct answer is:
$2 x+y+2 z=2 \sqrt{2}+1$

Solution:
Direction cosines of $\vec{n}$ are $\frac{1}{2}, \frac{1}{4}, \frac{1}{2}$.
Equation of the plane,
$$
\begin{aligned}
& \frac{1}{2}(x-\sqrt{2})+\frac{1}{4}(y+1)+\frac{1}{2}(z-1)=0 \\
\Rightarrow & 2(x-\sqrt{2})+(y+1)+2(z-1)=0 \\
\Rightarrow & 2 x+y+2 z=2 \sqrt{2}-1+2 \\
\Rightarrow & 2 x+y+2 z=2 \sqrt{2}+1
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main
A vector $\vec{n}$ is inclined to $x$-axis at $45^{\circ}$, to $y$-axis at $60^{\circ}$ and at an acute angle to $z$-axis. If $\vec{n}$ is a normal to a plane passing through the point $(\sqrt{2},-1,1)$ then the equation of the plane is :

Options:
A
$4 \sqrt{2} x+7 y+z-2$

B
$2 x+y+2 z=2 \sqrt{2}+1$

C
$3 \sqrt{2} x-4 y-3 z=7$

D
$\sqrt{2} x-y-z=2$

The correct answer is:
$2 x+y+2 z=2 \sqrt{2}+1$

Solution:
Direction cosines of $\vec{n}$ are $\frac{1}{2}, \frac{1}{4}, \frac{1}{2}$.
Equation of the plane,
$$
\begin{aligned}
& \frac{1}{2}(x-\sqrt{2})+\frac{1}{4}(y+1)+\frac{1}{2}(z-1)=0 \\
\Rightarrow & 2(x-\sqrt{2})+(y+1)+2(z-1)=0 \\
\Rightarrow & 2 x+y+2 z=2 \sqrt{2}-1+2 \\
\Rightarrow & 2 x+y+2 z=2 \sqrt{2}+1
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2013 (09 Apr Online)
A vector of length 3 perpendicular to each of the vectors $3 \mathbf{i}+\mathbf{j}-4 \mathbf{k}$ and $6 \mathbf{i}+5 \mathbf{j}-2 \mathbf{k}$ is

Options:
A $2 \mathbf{i}-2 \mathbf{j}+\mathbf{k}$
B $-2 \mathbf{i}+2 \mathbf{j}+\mathbf{k}$
C $2 \mathbf{i}+2 \mathbf{j}-\mathbf{k}$
D None of these

The correct answer is:
$2 \mathbf{i}-2 \mathbf{j}+\mathbf{k}$

Solution:
Let the vector is $x \mathbf{i}+y \mathbf{j}+\mathbf{z} \mathbf{k}$. Now according to the conditions, $\sqrt{x^2+y^2+z^2}=3 \Rightarrow x^2+y^2+z^2=9$
$6 x+5 y-2 z=0$ and $3 x+y-4 z=0$
[ it is perpendicular to both vectors, hence by $a_1 b_1+a_2 b_2+a_3 b_3=0$ ]
On solving the equation (i), (ii) and (iii), we get $x=2, \quad y=-2$ and $z=1$.
Therefore, the required vector is $2 \mathbf{i}-2 \mathbf{j}+\mathbf{k}$.
Trick : By inspection, the vector $2 \mathbf{i}-2 \mathbf{j}+\mathbf{k}$ is of length 3 and also perpendicular to the given vectors.

Topic: Mathematics, Vector Algebra, JEE Main
A vector of magnitude \( 3 \), bisecting the angle between the vectors \( \vec{a}=2 \hat{\mathrm{i}}+\widehat{\mathrm{j}}-\widehat{\mathrm{k}} \) and \( \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\widehat{\mathrm{k}} \) and making an obtuse angle with \( \vec{b} \) is -

Options:
A \( \frac{3 i-j}{\sqrt{6}} \)
B \( \frac{\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}}{\sqrt{14}} \)
C \( \frac{3(\hat{i}+3 \hat{j}-2 \hat{k})}{\sqrt{14}} \)
D \( \frac{3 \hat{\mathbf{i}}-\hat{\mathbf{j}}}{\sqrt{10}} \)

The correct answer is:
\( \frac{3(\hat{i}+3 \hat{j}-2 \hat{k})}{\sqrt{14}} \)

Solution:
We have a→= 2i^ + j ^- k^

b→= i^ -2 j ^+ k^

Internal and External angle bisecting Vectors between a→ and b→ is given by a→a→ + b→b→ and a→a→ - b→b→ respectively.

Internal angle bisector= a→a→ + b→b→=p→say

⇒p→=a→ + b→6

⇒p→=3i^ - j ^6

External Angle bisector=

=a→a→ - b→b→=q→say

⇒q→=a→ - b→6

⇒q→=i^ +3 j ^-2 k^6

Since p→.b→>0 and q→.b→<0

hence required vector will along q→.

Let required vector be

r→=kq→

Since r→=3

⇒k=3614

Hence required vector

⇒r→=3i^ +3 j ^-2 k^14

Topic: Mathematics, Vector Algebra, JEE Main
A vector of magnitude $\sqrt{51}$ which makes equal angles with the vectors $\bar{a}=\frac{1}{3}(\bar{i}-2 \bar{j}+2 \bar{k})$, $\bar{b}=\frac{1}{5}(-4 \bar{i}-3 \bar{k})$ and $\bar{c}=\bar{j}$, is

Options:
A $5 \bar{i}-\bar{j}+5 \bar{k}$
B $-5 \bar{i}+\bar{j}-5 \bar{k}$
C $-5 \bar{i}+\bar{j}+5 \bar{k}$
D $-\bar{i}-\bar{j}+7 \bar{k}$

The correct answer is:
$-5 \bar{i}+\bar{j}+5 \bar{k}$

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Vector Algebra, JEE Main
A vector $\mathbf{P}$ directed along the $X$-axis is added to vector $\mathbf{Q}$ which has a magnitude of $10 \mathrm{~m}$. The resultant vector is directed along the $Y$-axis, with a magnitude that is 2 times that of $\mathbf{P}$. The magnitude of $\mathbf{P}$ is

Options:
A $\sqrt{10} \mathrm{~m}$
B $5 \sqrt{2} \mathrm{~m}$
C $6 \mathrm{~m}$
D $2 \sqrt{5} \mathrm{~m}$

The correct answer is:
$2 \sqrt{5} \mathrm{~m}$

Solution:
By parallelogram law of vector addition, angle made by resultant with $\mathbf{P}$ is given by;
$\tan \alpha=\frac{Q \sin \theta}{P+Q \cos \theta}$
Here $\alpha=90^{\circ}$ so ; $\tan 90^{\circ}=\infty=\frac{Q \sin \theta}{P+Q \cos \theta}$
$\Rightarrow \quad P+Q \cos \theta=0$ or $Q \cos \theta=-P$
Now, magnitude of resultant,
$R=\sqrt{P^2+Q^2+2 P Q \cos \theta}$
Here, we have $R=2 P$ and $Q \cos \theta=-P$ and $Q=10 \mathrm{~m}$
So, $\quad 2 P=\sqrt{P^2+100-2 P^2}$
$\Rightarrow \quad 4 P^2=100-P^2 \Rightarrow 5 P^2=100$
or $\quad P=\sqrt{20}=2 \sqrt{5} \mathrm{~m}$
Magnitude of $P$ is $2 \sqrt{5} \mathrm{~m}$

Topic: Physics, Mathematics in Physics, JEE Main
A vector r→ is equally inclined with the vectors a→=cos⁡θi^+sin⁡θj^, b→=-sin⁡θi^+cos⁡θj^ and c→=k^, then the angle between r→ and a→ is

Options:
A cos-112
B cos-113
C cos-113
D cos-112

The correct answer is:
cos-113

Solution:
a→,b→,c→ are unit vectors and mutually perpendicular vectors and r→ is equally inclined to all 3 vectors
⇒r→=λa→+b→+c→
⇒r→=λcos⁡θ-sin⁡θi^+sin⁡θ+cos⁡θj^+k^
Angle between r→ and a→ is
cos⁡α=r→⋅a→r→a→=λcos2⁡θ-sin⁡θcos⁡θ+sin2⁡θ+sin⁡θcos⁡θ1×3λ
cos⁡α=13⇒α=cos-113

Topic: Mathematics, Vector Algebra, JEE Main
A vector v→ in the first octant is inclined to the x axis at 60°, to the y-axis at 45° and to the z-axis at an acute angle. If a plane passing through the points 2,-1,1 and a,b,c, is normal to v→, then

Options:
A 2a+b+c=1
B a+b+2c=1
C a+2b+c=1
D 2a-b+c=1

The correct answer is:
a+2b+c=1

Solution:
Direction cosines l,m and n of given vector v→,

l=cos60∘=12

m=cos45∘=12

n=cosγ

v→=cos60°i^+cos45°j^+cosγk^....(1)

We know,

l2+m2+n2=1

⇒14+12+cos2γ=1 (γ→Acute)

⇒cosγ=12

⇒ γ=60°

Equation of plane is 12x-2+12y+1+12z-1=0

⇒x2+y2+z2=12

⇒ x+2y+z=1

a,b,c lies on it,

⇒ a+2b+c=1

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (30 Jan Shift 2)
A vehicle of mass 200 kg is moving along a levelled curved road of radius 70 m with angular velocity of 0.2 rad s-1. The centripetal force acting on the vehicle is:

Options:
A 560 N
B 2800 N
C 2240 N
D 14 N

The correct answer is:
560 N

Solution:
The centripetal force is given by

F=Mv2R=MRω2

The given data is

M=200 kgR=70 mω=0.2 rad s-1

The magnitude of the force is

F=200×70×0.22 N

⇒F=560 N

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (13 Apr Shift 2)
A vehicle starts moving in a straight line with an acceleration, $a=4 \mathrm{~m} / \mathrm{s}^2$, with initial velocity equal to zero. After accelerating for time $t_1$, the vehicle moves uniformly and for time $t_2$, the vehicle finally decelerates for time $t_1$ eventually coming to a stop. The total time taken during the motion is $10 \mathrm{~s}$ and the average velocity during the motion is $5.1 \mathrm{~m} / \mathrm{s}$. The time taken by the vehicle during acceleration is

Options:
A $2 \mathrm{~s}$
B $2.5 \mathrm{~s}$
C $1.5 \mathrm{~s}$
D $1.8 \mathrm{~s}$

The correct answer is:
$1.5 \mathrm{~s}$

Solution:
For first part of journey, $x=0, a=4 \mathrm{~ms}^{-2}$, time $=t_1$. Velocity attained by particle at end of time $t_1$ is $\quad v_1=u+a t_1=4 t_1$
$\therefore$ Displacement in first $t_1$ seconds is
$$
s_1=\frac{1}{2} a t_1^2=\frac{1}{2} \times 4 \times t_1^2=2 t_1^2
$$
For second part of journey, initial velocity, $v_1=4 t_1$, time $=t_2$ and acceleration $=0$.
$\therefore$ Displacement in next $t_2$ seconds is $s_2=v_1 t_2=4 t_1 t_2$ For third part of journey,
Initial velocity, $v_1=4 t_1$
Final velocity $=0$
Time interval $=t_1$
So, acceleration $=\frac{0-4 t_1}{t_1}=-4 \mathrm{~ms}^{-2}$
$\therefore$ Displacement in third part is
$$
s_3=\frac{-16 t_1^2}{-8}=2 t_1^2 \quad\left(\because v_2^2-v_1^2=2 a s_3\right)
$$
Now, $\quad v_{\text {avg }}=\frac{\text { Total displacement }}{\text { Total time }}$
$$
\begin{aligned}
& 5.1=\frac{s_1+s_2+s_3}{t_1+t_2+t_1} \\
& {\left[\therefore \text { third time interva] }=t_1\right]} \\
& 5.1=\frac{2 t_1^2+4 t_1 t_2+2 t_1^2}{10}
\end{aligned}
$$
$$
\begin{array}{lc}
\Rightarrow & 5 \mathrm{l}=4 t_1^2+4 t_1\left(10-2 t_1\right) \\
\Rightarrow & 5 \mathrm{l}=4 t_1^2+40 t_1-8 t_1^2 \\
\Rightarrow & 4 t_1^2-40 t_1+5 \mathrm{l}=0 \\
\Rightarrow & 4 t_1^2-34 t_1-6 t_1+5 \mathrm{l}=0
\end{array}
$$
$$
\begin{aligned}
& \Rightarrow \quad\left(2 t_1-3\right)\left(2 t_1-17\right)=0 \Rightarrow t_1=\frac{3}{2} \\
& \text { or } \\
& t_1=\frac{17}{2} \Rightarrow t_1=1.5 \mathrm{~s} \\
&
\end{aligned}
$$
(As $t_1=8.5 \mathrm{~s}$ is not possible. It exceeds total time.)

Topic: Physics, Motion In One Dimension, JEE Main
A vehicle travels 4 km with speed of 3 km h-1 and another 4 km with speed of 5 km h-1, then its average speed is :

Options:
A 4.25 km h-1
B 3.50 km h-1
C 4.00 km h-1
D 3.75 km h-1

The correct answer is:
3.75 km h-1

Solution:
Average speed in this case is given by, 2vav=1v1+1v2

⇒2vav=13+15

⇒2vav=815

⇒vav=154=3.75 km h-1

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2023 (30 Jan Shift 2)
A vehicle travels half the distance with speed v and the remaining distance with speed 2v. Its average speed is:

Options:
A 3v4
B v3
C 2v3
D 4v3

The correct answer is:
4v3

Solution:
Let the total distance travelled be d.

Time taken in both part is,

t1=d2v and t2=d22v.

Now, average speed can be written as

vavg=dt1+t2=dd2v+d4v=4v3

Topic: Physics, Motion In One Dimension, NEET, NEET 2023 (All India)
A vehicle without passengers is moving on a frictionless horizontal road with velocity ' $u$ ' can be stopped in a distance ' $d$ '. Now $40 \%$ of it's weight is added. If the retardation remains same the stopping distance at velocity ' $u$ ' is

Options:
A (1.6)d
B (1.4)d
C d
D (1.2)d

The correct answer is:
(1.4)d

Solution:
Given, $\mathrm{m}_1=\mathrm{m}$ and $\mathrm{m}_2=1.4 \mathrm{~m}$
$\because \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{aS}$
$\therefore 0=\mathrm{u}^2-2 \mathrm{ad}$
$\therefore \mathrm{a}=\frac{\mathrm{u}^2}{2 \mathrm{~d}}$
$\therefore$ Retarding force $\mathrm{f}=\mathrm{ma}=\frac{\mathrm{mu}^2}{2 \mathrm{~d}}$
$\therefore \frac{\mathrm{d}^{\prime}}{\mathrm{d}}=1.4$ or $\mathrm{d}^{\prime}=1.4 \mathrm{~d}$

Topic: Physics, Motion In One Dimension, MHT CET, MHT CET 2022 (07 Aug Shift 1)
A velocity $\frac{1}{4} \mathrm{~m} / \mathrm{s}$ is resolved into two components along $\mathrm{OA}$ and $\mathrm{OB}$ making angles $30^{\circ}$ and $45^{\circ}$ respectively with the given velocity. Then the component along $\mathrm{OB}$ is

Options:
A
$\frac{1}{8} \mathrm{~m} / \mathrm{s}$

B
$\frac{1}{4}(\sqrt{3}-1) \mathrm{m} / \mathrm{s}$

C
$\frac{1}{4} \mathrm{~m} / \mathrm{s}$

D
$\frac{1}{8}(\sqrt{6}-\sqrt{2}) \mathrm{m} / \mathrm{s}$

The correct answer is:
$\frac{1}{8}(\sqrt{6}-\sqrt{2}) \mathrm{m} / \mathrm{s}$

Solution:
Component along $\mathrm{OB}=\frac{\frac{1}{4} \sin 30^{\circ}}{\sin \left(45^{\circ}+30^{\circ}\right)}=\frac{1}{8}(\sqrt{6}-\sqrt{2}) \mathrm{m} / \mathrm{s}$.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2004
A velocity selector consists of electric field E→=E k^ and magnetic field B→=B j^ with B=12 mT. The value E required for an electron of energy 728eV moving along the positive x-axis to pass undeflected is

(Given, mass of electron =9.1×10-31 kg)

Options:
A 192 kV m-1
B 192 mV m-1
C 9600 kV m-1
D 16 kV m-1

The correct answer is:
192 kV m-1

Solution:
Fiven that E→=Ek^ and B→=12 j^ mT

Kinetic energy=728 eV

Kinetic energy =12mv2

⇒728 eV=12×9.1×10-31×v2

⇒728×1.6×10-19=12×9.1×10-31×v2

⇒v=16×106 m s-1

For electron to move undeflected net force on it should be zero.

⇒eE=evB

⇒E=vB=16×106×12×10-3

⇒E=192×103 V m-1=192 kV m-1

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (26 Jul Shift 2)
A (+)vely charged particle is placed near an infinitely long straight conductor where there is zero gravity. Then

Options:
A the charged particle will not move
B it will move parallel to the straight conductor
C it will move perpendicular to the straight conductor
D it will move with constant acceleration

The correct answer is:
the charged particle will not move

Solution:
As the charged particle is at rest, no force will act on it due to the magnetic field produced by the conductor at the site of the charge. Hence, it will remain at rest.

Topic: Physics, Electrostatics, VITEEE, VITEEE 2016
A vernier calliper has 20 divisions on the vernier scale which coincide with 19 divisions on the main scale. The least count of the instrument is 0.1 mm. The length of one main scale division is

Options:
A 0.5 mm
B 1 mm
C 2 mm
D 0.25 mm

The correct answer is:
2 mm

Solution:
20 division on vernier scale = 19 division on the main scale
20 VRS = 19 MSR
L.C = MSR - VSR = 0.1 mm
âMSR-1920MSR=0.1 mm
âMSR=2 mm

Topic: Physics, Experimental Physics, JEE Main
A vernier callipers has 20 divisions on the vernier scale, which coincides with $19^{\text {th }}$ division on the main scale. The least count of the instrument is $0.1 \mathrm{~mm}$. One main scale division is equal to _____$\mathrm{mm}$.

Options:
A $0.5$
B $2$
C $5$
D $1$

The correct answer is:
$2$

Solution:
$\begin{aligned} & 20 \mathrm{VSD}=19 \mathrm{MSD} \\ & 1 \mathrm{VSD}=\frac{19}{20} \mathrm{MSD} \\ & \text { L.C. }=1 \mathrm{MSD}-1 \mathrm{VSD} \\ & 0.1 \mathrm{~mm}=1 \mathrm{MSD}-\frac{19}{20} \mathrm{MSD} \\ & 0.1=\frac{1}{20} \mathrm{MSD} \\ & 1 \mathrm{MSD}=2 \mathrm{~mm}\end{aligned}$

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2024 (05 Apr Shift 2)
A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of negligible thickness. The piston is free to move along the length of the cylinder. The length of the cylinder above piston is l1, and that below the piston is l2, such that l1>l2. Each part of the cylinder contains n moles of an ideal gas at equal temperature T. If the piston is stationary, its mass m will be given by:
(R is universal gas constant and g is the acceleration due to gravity)

Options:
A RTngl1-3l2l1l2
B nRTgl1-l2l1l2
C nRTg1l2+1l1
D RTgl2l1+l2l1l2

The correct answer is:
nRTgl1-l2l1l2

Solution:
P1+mgA=P2
⇒nRTl1A+mgA=nRTl2A
⇒m=nRTg1l2-1l1
=nRTgl1-l2l1l2

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2019 (12 Jan Shift 2)
A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of negligible thickness. The piston is free to move along the length of the cylinder. The length of the cylinder above piston is l1, and that below the piston is l2, such that l1>l2. Each part of the cylinder contains n moles of an ideal gas at equal temperature T. If the piston is stationary, its mass m will be given by:
(R is universal gas constant and g is the acceleration due to gravity)

Options:
A RTngl1-3l2l1l2
B nRTgl1-l2l1l2
C nRTg1l2+1l1
D RTgl2l1+l2l1l2

The correct answer is:
nRTgl1-l2l1l2

Solution:
P1+mgA=P2
⇒nRTl1A+mgA=nRTl2A
⇒m=nRTg1l2-1l1
=nRTgl1-l2l1l2

Topic: Physics, Thermodynamics, JEE Main
A vertical electric field of magnitude 4.9×105 N C-1 just prevents a water droplet of a mass 0.1 g from falling. The value of charge on the droplet will be : (Given g=9.8 m s-2)

Options:
A 1.6×10-9C
B 2.0×10-9C
C 3.2×10-9C
D 0.5×10-9C

The correct answer is:
2.0×10-9C

Solution:
To balance gravitational force, electric force on the liquid drop should act in upward direction.

For translational equilibrium along vertical direction,

qE=mg

⇒q=mgE=0.1×10-3×9.84.9×105=2.0×10-9C

Topic: Physics, Electrostatics, JEE Main, JEE Main 2022 (24 Jun Shift 1)