datavorous/entrance-exam-dataset huggingface dataset:
The vector $\vec{a}=\alpha \hat{i}+2 \hat{j}+\beta \hat{k}$ lies in the plane of the vectors $\vec{b}=\hat{i}+\hat{j}$ and $\vec{c}=\hat{j}+\hat{k}$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Then which one of the following gives possible values of $\alpha$ and $\beta$ ?

Options:
A
$\alpha=2, \beta=2$

B
$\alpha=1, \beta=2$

C
$\alpha=2, \beta=1$

D
$\alpha=1, \beta=1$

The correct answer is:
$\alpha=1, \beta=1$

Solution:
$$
\begin{aligned}
& \overrightarrow{\mathrm{a}}=\lambda(\hat{\mathrm{b}}+\hat{\mathrm{c}}) \\
& \Rightarrow \alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}=\lambda\left(\frac{\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{2}}\right) \\
& \lambda=\sqrt{2} \alpha \text { and } \lambda=\sqrt{2} \text { and } \lambda=\sqrt{2} \beta \\
& \Rightarrow \alpha=1 \text { and } \beta=1 .
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2008
The vector equation of the plane passing through the intersection of the planes r→·i^+j^+k^=1 and r→·i^-2j^=-2, and the point 1, 0, 2 is:

Options:
A r→·i^+7j^+3k^=7
B r→·i^-7j^+3k^=73
C r→·i^+7j^+3k^=73
D r→·3i^+7j^+3k^=7

The correct answer is:
r→·i^+7j^+3k^=7

Solution:
r→·i^+j^+k^=1

r→·i^-2j^=-2

point 1, 0, 2

Eqn of plane

r→·i^+j^+k^-1+λ{r→·i^-2j^+2}=0

r→·{i^1+λ+j^1-2λ+k^1}-1+2λ=0

Point i^+0j^+2k^=r→

∴i^+2k^·{i^1+λ+j^1-2λ+k^1}-1+2λ=0

1+λ+2-1+2λ=0

λ=-23

∴r→·i^13+j^73+k^=73

r→·[i^+7j^+3k^]=7

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The vector equation of the plane through the line of intersection of the planes x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x-y+z=0 is

Options:
A r→×i^+k^+2=0
B r→⋅i^-k^-2=0
C r→×i^-k^+2=0
D r→⋅(i^-k^)+2=0

The correct answer is:
r→⋅(i^-k^)+2=0

Solution:
Planes passing through line of intersection of the planes x+y+z=1 and 2x+3y+4z=5 is
x+y+z-1+λ2x+3y+4z-5=0
⇒(1+2λ)x+(1+3λ)y+(1+4λ)z-(1+5λ)=0
Since required plane is perpendicular to x-y+z=0, hence
1⋅(1+2λ)-1⋅(1+3λ)+1⋅(1+4λ)=0⇒λ=-13
Hence, required plane is x-z+2=0 i.e. r→⋅(i^-k^)+2=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The vector $(\hat{i} \times \vec{a} \cdot \vec{b}) \hat{i}+(\hat{j} \times \vec{a} \vec{b}) \hat{j}+(\hat{k} \times \vec{a} \cdot \vec{b}) \hat{k}$ is equal to:

Options:
A
$\vec{b} \times \vec{a}$

B
$\vec{a}$

C
$\vec{a} \times \vec{b}$

D
$\vec{b}$

The correct answer is:
$\vec{a} \times \vec{b}$

Solution:
$(\hat{i} \times \vec{a} \cdot \vec{b}) \hat{i}+(\hat{j} \times \vec{a} \cdot \vec{b}) \hat{j}+(\hat{k} \times \vec{a} \cdot \vec{b}) \hat{k}$
$=(\hat{i} \cdot \vec{a} \times \vec{b}) \hat{i}+(\hat{j} \cdot \vec{a} \times \vec{b}) \hat{j}+(\hat{k} \cdot \vec{a} \times \vec{b}) \hat{k}$
$(\because \vec{a} \times \vec{b} \cdot \vec{c}=\vec{a} \cdot \vec{b} \times \vec{c})$
$=(\vec{a} \times \vec{b}) \hat{i}+(\vec{a} \times \vec{b}) \hat{j}+(\vec{a} \times \vec{b}) \hat{k}$
$==\vec{a} \times \vec{b}$

Topic: Mathematics, Vector Algebra, JEE Main
The vector $(\hat{i} \times \vec{a} \cdot \vec{b}) \hat{i}+(\hat{j} \times \vec{a} \vec{b}) \hat{j}+(\hat{k} \times \vec{a} \cdot \vec{b}) \hat{k}$ is equal to:

Options:
A
$\vec{b} \times \vec{a}$

B
$\vec{a}$

C
$\vec{a} \times \vec{b}$

D
$\vec{b}$

The correct answer is:
$\vec{a} \times \vec{b}$

Solution:
$(\hat{i} \times \vec{a} \cdot \vec{b}) \hat{i}+(\hat{j} \times \vec{a} \cdot \vec{b}) \hat{j}+(\hat{k} \times \vec{a} \cdot \vec{b}) \hat{k}$
$=(\hat{i} \cdot \vec{a} \times \vec{b}) \hat{i}+(\hat{j} \cdot \vec{a} \times \vec{b}) \hat{j}+(\hat{k} \cdot \vec{a} \times \vec{b}) \hat{k}$
$(\because \vec{a} \times \vec{b} \cdot \vec{c}=\vec{a} \cdot \vec{b} \times \vec{c})$
$=(\vec{a} \times \vec{b}) \hat{i}+(\vec{a} \times \vec{b}) \hat{j}+(\vec{a} \times \vec{b}) \hat{k}$
$==\vec{a} \times \vec{b}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2013 (09 Apr Online)
The vector sum of two forces is perpendicular to their vector differences. In that case, the force:

Options:
A are equal to each other
B are equal to each other in magnitude
C are not equal to each other in magnitude
D cannot be predicted

The correct answer is:
are equal to each other in magnitude

Solution:
$(\vec{A}+\vec{B}) \cdot(\vec{A}-\vec{B})=0$
$\begin{aligned}
& \Rightarrow A^2-\vec{A} \cdot \vec{B}+\vec{B} \cdot \vec{A}+B^2=0 \\
& \Rightarrow \mathrm{A}=\mathrm{B} \quad(\because \vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A})
\end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main
The vectors \( 2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \widehat{\mathrm{k}}, \quad 5 \hat{\mathrm{i}}+\mathrm{a} \hat{\mathrm{j}}+\widehat{\mathrm{k}} \) and \( -\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \widehat{\mathrm{k}} \) are coplanar when \( a \) is

Options:
A \( -9 \)
B \( 9 \)
C \( -18 \)
D \( 18 \)

The correct answer is:
\( 18 \)

Solution:
Three vectors are coplanar if their scalar triple product is zero.

So,
2i^+3j^-2k^×-i^+2j^+3k^.5i^+aj^+k^=0

⇒ 23-2-1235ak .5i^+aj^+5k^=0
⇒ 13i^-4j^+7k^.5i^+aj^+5k^=0
⇒ 65-4a+7=0
⇒ a=18

Topic: Mathematics, Vector Algebra, JEE Main
The vectors 3a→-5b→ and 2a→+b→ are mutually perpendicular and the vectors a→+4b→ and -a→+b→ are also mutually perpendicular. Then the angle between the vectors a→ and b→, is

Options:
A cos-1⁡19543
B π-cos-1⁡19543
C cos-1⁡9543
D π-cos-19543

The correct answer is:
cos-1⁡19543

Solution:
∵ (3a→-5b→)⋅(2a→+b→)=0
∴ a→⋅b→=6a→2-5b→27 ...(i)
and (a→+4b→)⋅(b→-a→)=0
∴ a→⋅b→=4b→2-a→23 ...(ii)
∴ From equation (i) and (ii),
a→=4325b→ and a→⋅b→=a→b→cos⁡θ
∴ cos⁡θ=19b→24325 b→2=19543

Topic: Mathematics, Vector Algebra, JEE Main
The vectors $3 \mathbf{i}+\mathbf{j}-5 \mathbf{k}$ and $a \mathbf{i}+b \mathbf{j}-15 \mathbf{k}$ are collinear, if

Options:
A $a=3, b=1$
B $a=9, b=1$
C $a=3, b=3$
D $a=9, b=3$

The correct answer is:
$a=9, b=3$

Solution:
$\frac{3}{a}=\frac{1}{b}=\frac{-5}{-15} \Rightarrow a=9, b=3$.

Topic: Mathematics, Vector Algebra, JEE Main
The vectors $\mathrm{A}$ and $\mathrm{B}$ are such that $|\mathbf{A}+\mathbf{B}|=|\mathbf{A}-\mathbf{B}| .$ The angle between the two vectors will be

Options:
A $0^{\circ}$
B $60^{\circ}$
C $90^{\circ}$
D $45^{\circ}$

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

Solution:
Given, $|A+B|=|A-B|$
$\therefore \quad(A+B)^{2}=(A-B)^{2}$
$A^{2}+B^{2}+2 A B \cos \theta=A^{2}+B^{2}-2 A B \cos \theta$
$2 A B \cos \theta+2$ AB $\cos \theta=0$
$4 A B \cos \theta=0$
Here, $\quad A \neq 0, B \neq 0$
$\therefore \quad \cos \theta=0$
$\theta=90^{\circ}$

Topic: Physics, Mathematics in Physics, JEE Main
The vectors $\overrightarrow{A B}=3 \mathbf{i}+4 \mathbf{k}$, and $\overrightarrow{A C}=5 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k}$ are the sides of a triangle $A B C$. The length of the median through $A$ is

Options:
A $\sqrt{18}$
B $\sqrt{72}$
C $\sqrt{33}$
D $\sqrt{288}$

The correct answer is:
$\sqrt{33}$

Solution:
P.V. of $\overrightarrow{A D}=\frac{(3+5) i+(0-2) j+(4+4) k}{2}$ $=4 \mathbf{i}-\mathbf{j}+4 \mathbf{k}$
$|\overrightarrow{A D}|=\sqrt{16+16+1}=\sqrt{33}$

Topic: Mathematics, Vector Algebra, JEE Main
The vectors $\mathbf{A B}=3 \mathbf{i}-2 \mathbf{j}+2 \mathbf{k}$ and $\mathbf{B C}=\mathbf{i}-2 \mathbf{k}$ are the adjacent sides of a parallelogram. The angle between its diagonals is

Options:
A $\frac{\pi}{2}$
B $\frac{\pi}{3}$ or $\frac{2 \pi}{3}$
C $\frac{3 \pi}{4}$ or $\frac{\pi}{4}$
D None of these

The correct answer is:
None of these

Solution:
Given, $\mathbf{A B}=3 \mathbf{i}-2 \mathbf{j}+2 \mathbf{k}$
and $\quad \mathbf{B C}=\mathbf{i}-2 \mathbf{k}$
$\begin{aligned} \text { Diagonals } \mathbf{A C} & =\mathbf{A B}+\mathbf{B C} \\ & =3 \mathbf{i}-2 \mathbf{j}+2 \mathbf{k}+\mathbf{i}-2 \mathbf{k} \\ & =4 \mathbf{i}-2 \mathbf{j}\end{aligned}$
and
$\begin{aligned} \mathbf{B D} & =\mathbf{B} \mathbf{C}-\mathbf{A} \mathbf{B} \\ & =\mathbf{i}-2 \mathbf{k}-(3 \mathbf{i}-2 \mathbf{j}+2 \mathbf{k}) \\ & =-2 \mathbf{i}+2 \mathbf{j}-4 \mathbf{k}\end{aligned}$
$\therefore \quad \cos \theta=\frac{\mathbf{A C} \cdot \mathbf{B D}}{|\mathbf{A C}||\mathbf{B D}|}$
$\begin{aligned} & =\frac{(4 \mathbf{i}-2 \mathbf{j}) \cdot(-2 \mathbf{i}+2 \mathbf{j}-4 \mathbf{k})}{\sqrt{4^2+(-2)^2} \sqrt{(-2)^2+(2)^2+(-4)^2}} \\ & =\frac{-8-4}{\sqrt{20} \sqrt{24}}=-\frac{12}{4 \sqrt{30}}=-\frac{3}{\sqrt{30}}=-\sqrt{\frac{3}{10}} \\ & \Rightarrow \quad \theta=\cos ^{-1}\left(-\sqrt{\frac{3}{10}}\right) \\ & \text { or } \quad \pi-\cos ^{-1}\left(-\sqrt{\frac{3}{10}}\right) \\ & \end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
The vectors $\overrightarrow{\mathrm{AB}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}} \& \overrightarrow{\mathrm{AC}}=5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ are the sides of a triangle $\mathrm{ABC}$. The length of the median through $\mathrm{A}$ is

Options:
A
$\sqrt{288}$

B
$\sqrt{18}$

C
$\sqrt{72}$

D
$\sqrt{33}$

The correct answer is:
$\sqrt{33}$

Solution:
P.V of $\overrightarrow{\mathrm{AD}}=\frac{(3+5) \mathrm{i}+(0-2) \mathrm{j}+(4+4) \mathrm{k}}{2}$
$=4 \mathrm{i}-\mathrm{j}+4 \mathrm{k}$ or $|\overrightarrow{\mathrm{AD}}|=\sqrt{16+16+1}=\sqrt{33}$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2003
The vectors are given by $\dot{\mathbf{A}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{B}=3 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+2 \mathbf{k}$. Another vector $\mathbf{C}$ has the same magnitude as $\mathbf{B}$ but has the same direction as A. Then which of the following vectors represents C?

Options:
A $\frac{7}{3}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$
B $\frac{3}{7}(i-2 \hat{j}+2 \hat{k})$
C $\frac{7}{9}(i-2 \hat{j}+2 \hat{k})$
D $\frac{9}{7}(\vec{i}+2 \hat{j}+2 \hat{k})$

The correct answer is:
$\frac{7}{3}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$

Solution:
(a) Given, $A=i+2 j+2 \hat{k}$ and $B=3 \hat{i}+6 j+2 \hat{k}$
So, $\quad C=\frac{\hat{i}+2 \hat{j}+2 \hat{k}}{\sqrt{1+4+4}} \times \sqrt{3^{2}+6^{2}+2^{2}}$
$=\frac{\hat{i}+2 \hat{j}+2 \hat{k}}{3} \times \sqrt{49}=\frac{7}{3}(\hat{i}+2 \hat{j}+2 \hat{\mathbf{k}})$

Topic: Physics, Mathematics in Physics, JEE Main
The vehicles carrying inflammable fluids usually have metallic chains touching the ground :

Options:
A To protect tyres from catching dirt from ground
B To alert other vehicles
C It is a custom
D To conduct excess charge due to air friction to ground and prevent sparking

The correct answer is:
To conduct excess charge due to air friction to ground and prevent sparking

Solution:
Static charge is developed due to air friction. This can result in combustion. So, metallic chains is used to discharge excess charge.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The velocities of three particles of masses $20 \mathrm{~g}$, $30 \mathrm{~g}$ and $50 \mathrm{~g}$ are $10 \hat{\mathbf{i}}, 10 \hat{\mathbf{j}}$ and $10 \hat{\mathbf{k}}$, respectively. The velocity of the centre of mass of the three particles is

Options:
A $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \mathbf{k}$
B $10(\mathbf{i}+\mathbf{j}+\mathbf{k})$
C $20 \hat{i}+30 \hat{\mathbf{j}}+5 \mathbf{k}$
D $2 \hat{\mathbf{i}}+30 \hat{\mathbf{j}}+50 \hat{\mathbf{k}}$

The correct answer is:
$2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \mathbf{k}$

Solution:
$\begin{aligned} & m_1=20 \mathrm{~g}, m_2=30 \mathrm{~g}, m_3=50 \mathrm{~g} \\ & \mathbf{v}_1=10 \hat{\mathbf{i}}, \mathbf{v}_2=10 \hat{\mathbf{j}}, \mathbf{v}_3=10 \hat{\mathbf{k}}\end{aligned}$ $\begin{aligned} & v_{\text {cm }}=\frac{m_1 \mathbf{v}_1+m_2 \mathbf{v}_2+m_3 \mathbf{v}_3}{m_1+m_2+m_3} \\ &=\frac{20 \times 10^{-3} \times 10 \hat{\mathbf{i}}+30 \times 10^{-3} \times 10 \hat{\mathbf{i}}+50 \times 10^{-3} \times 10 \hat{\mathbf{k}}}{(20+30+50) \times 10^{-3}} \\ &=\frac{200 \hat{\mathbf{i}}+300 \hat{\mathbf{j}}+500 \hat{\mathbf{k}}}{100} \\ &=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
The velocity and acceleration of a point-like body at the initial moment of its motion are v0 =8 m s-1 and a=2 m s-2 respectively. The acceleration vector of the body remains constant. The minimum radius of curvature of the trajectory of the body is

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

The correct answer is:
8 m

Solution:
The acceleration vector shall change the component of velocity uII along the acceleration vector.
r=v2an
The radius of curvature rmin means v is minimum and an is maximum. This is at the point P when the component of velocity parallel to the acceleration vector becomes zero, that is uII = 0.
∴ R=u⊥2a=422=8 m

Topic: Physics, Laws of Motion, JEE Main
The velocity and acceleration vectors of a particle undergoing circular motion are vâ=2i^ m s-1 and aâ=2i^+4j^ m s-2 respectively at an instant of time. What is the radius (in m) of the circle?

Options:

The correct answer is:
1

Solution:
It can be observed that the component of acceleration perpendicular to velocity is a=4 m s-2
∴ Radius=v2ac=224=1 m

Topic: Physics, Laws of Motion, JEE Main
The velocity of 4 gas molecules are given by $1 \mathrm{~km} / \mathrm{s}, 3 \mathrm{~km} / \mathrm{s}, 5 \mathrm{~km} / \mathrm{s}$ and $7 \mathrm{~km} / \mathrm{s}$. Calculate
the difference between average and RMS velocity.

Options:
A $0.338$
B $0.438$
C $0.583$
D $0.683$

The correct answer is:
$0.583$

Solution:
The average velocity
$$
\begin{aligned}
v_{\mathrm{av}} &=\frac{v_{1}+v_{2}+v_{3} \ldots v_{n}}{N} \\
&=\frac{1+3+5+7}{4}=4 \mathrm{~km} / \mathrm{s}
\end{aligned}
$$
Root mean square velocity
$$
v_{\mathrm{rms}}=\sqrt{\frac{v_{1}^{2}+v_{2}^{2}+v_{3}^{2}+v_{4}^{2}+\ldots+v_{n}^{2}}{N}}
$$
$$
\begin{array}{l}
=\sqrt{\frac{1+(3)^{2}+(5)^{2}+(7)^{2}}{4}} \\
=\sqrt{21}=4.583 \mathrm{~km} / \mathrm{s}
\end{array}
$$
Difference between average velocity and root mean square velocity $=4.583-4$
$$
=0.583 \mathrm{~km} / \mathrm{s}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
The velocity of a body moving with a uniform acceleration of $2 \mathrm{~m} \cdot / \mathrm{sec}^2$ is $10 \mathrm{~m} / \mathrm{sec}$. Its velocity after an interval of $4 \mathrm{sec}$ is

Options:
A 12 m/sec
B 14 m/sec
C 16 m/sec
D 18 m/sec

The correct answer is:
18 m/sec

Solution:
$v=u+a t=10+2 \times 4=18 \mathrm{~m} / \mathrm{sec}$

Topic: Physics, Motion In One Dimension, JEE Main
The velocity of a particle executing SHM varies with displacement (x) as 4v2=50–x2. The time period of oscillations is x7 s. The value of x is ______.

[Take π=227]

Options:

The correct answer is:
88

Solution:
Given:

4v2=50-x2

⇒v=1250-x2

Comparing with standard equation of SHM, v=ωA2-x2, we get

ω=12

Now,

T=2πω=4π=887

Hence, required x=88.

Topic: Physics, Oscillations, JEE Main, JEE Main 2023 (30 Jan Shift 2)
The velocity of a particle is v=v0+gt+Ft2 m s-1. Its position is x=0 at t=0 ; then its displacement after time (t=1 s) is :

Options:
A v0+g+F
B v0+g2+F3
C v0+g2+F
D v0+2g+3F

The correct answer is:
v0+g2+F3

Solution:
v=v0+gt+Ft2

dsdt=v0+gt+Ft2

∫ds=∫01v0+gt+Ft2dt

s=v0t+gt22+Ft3301

s=v0+g2+F3

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (17 Mar Shift 2)
The velocity of a particle is $\mathrm{v}=\mathrm{v}_0+\mathrm{gt}+\mathrm{ft}^2$. If its position is $\mathrm{x}=0$ at $\mathrm{t}=0$, then its displacement after unit time $(t=1)$ is

Options:
A
$\mathrm{v_0+2 g+3 f}$

B
$\mathrm{v_0+g / 2+f / 3}$

C
$\mathrm{v_0+g+f}$

D
$\mathrm{v_0+g / 2+f}$

The correct answer is:
$\mathrm{v_0+g / 2+f / 3}$

Solution:
$\mathrm{\int_0^x d x=\int_0^1\left(V_0+g t+\mathrm{ft}^2\right) d t}$
$\mathrm{x=v_0+g\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2007
The velocity of a particle measured from an instrument is 0.00204300 ms. The number of significant figures is

Options:
A 8
B 4
C 6
D 3

The correct answer is:
6

Solution:
In number with decimal starting zero's will be non-significant and ending zero's will be significant

Topic: Physics, Units and Dimensions, JEE Main
The velocity of a projectile at the initial point $A$ is $(2 \mathbf{i}+3 \mathbf{j}) \mathrm{m} / \mathrm{s}$. Its velocity (in $\mathrm{m} / \mathrm{s}$ ) at point $B$ is

Options:
A $-2 \mathbf{i}-3 \mathbf{j}$
B $-2 \mathbf{i}+3 \mathbf{j}$
C $2 \mathbf{i}-3 \mathbf{j}$
D $2 \mathbf{i}+3 \mathbf{j}$

The correct answer is:
$2 \mathbf{i}-3 \mathbf{j}$

Solution:
From the figure the $X$-component remain unchanged, while the $Y$-component is reverse. Then, the velocity at point $B$ is $(2 \mathrm{i}-3 \mathrm{j}) \mathrm{ms}^{-1}$.

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2013 (All India)
The velocity of a small ball of mass 0.3 g and density 8 g cc-1 when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is 1.3 g cc-1, then the value of viscous force acting on the ball will be x×10-4 N, the value of x is _____ .

[use g=10 m s-2]

Options:

The correct answer is:
25

Solution:
Let FV be the viscous force and FB be the buoyant force acting on the ball.

âââââââ

Then, when body moves with constant velocity,

FV+FB=Mg

Or, FV=Mg-FB

FV=ρBVg-ρLVg =ρB-ρLVg, where, V is the volume of small ball.

=8-1.3×103×0.3×10-38×103×10

=6.7×0.38×10-2

=67×38×10-4=25.125×10-4=25×10-4 N

Thus, the value of x=25.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2022 (29 Jul Shift 2)
The velocity of a small ball of mass M and density ρ, when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is ρ2, then the viscous force acting on the ball will be:

Options:
A 3Mg2
B Mg
C 12Mg
D 2Mg

The correct answer is:
12Mg

Solution:
When velocity becomes constant then,

Fv+Fb=Mg

Fv=Mg-Fb

=Vρbg-Vρgg

=Vgρ-ρ2

=Vgρ2

=(Vρ)g2=Mg2

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The velocity of certain ions that pass undeflected through crossed electric field $E=7.7 \mathrm{k} \mathrm{V} / \mathrm{m}$ and magnetic field $B=0.14 \mathrm{~T}$ is

Options:
A
$18 \mathrm{~km} / \mathrm{s}$

B
$77 \mathrm{~km} / \mathrm{s}$

C
$55 \mathrm{~km} / \mathrm{s}$

D
$1078 \mathrm{~km} / \mathrm{s}$

The correct answer is:
$55 \mathrm{~km} / \mathrm{s}$

Solution:
As velocity $v=\frac{E}{B}=\frac{7.7 \times 10^3}{0.14}=55 \mathrm{~km} / \mathrm{s}$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2012 (07 May Online)
The velocity of electromagnetic radiation in a medium of permittivity $\in_0$ and permeability $\mu_0$ is given by

Options:
A $\sqrt{\frac{\mu_0}{\varepsilon_0}}$
B $\sqrt{\frac{\varepsilon_0}{\mu_0}}$
C $\sqrt{\mu_0 \varepsilon_0}$
D $\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$

The correct answer is:
$\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$

Solution:
$$
v=\frac{1}{\sqrt{\mu_0 \epsilon_0}}
$$

Topic: Physics, Electromagnetic Waves, NEET, NEET 2008 (Mains)
The velocity of electromagnetic radiation in a medium of permittivity $\in_0$ and permeability $\mu_0$ is given by

Options:
A $\sqrt{\frac{\mu_0}{\varepsilon_0}}$
B $\sqrt{\frac{\varepsilon_0}{\mu_0}}$
C $\sqrt{\mu_0 \varepsilon_0}$
D $\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$

The correct answer is:
$\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$

Solution:
$$
v=\frac{1}{\sqrt{\mu_0 \epsilon_0}}
$$

Topic: Physics, Electromagnetic Waves, JEE Main
The velocity of kerosene oil in a horizontal pipe is $5 \mathrm{~m} / \mathrm{s}$. If $g=10 \mathrm{~m} / \mathrm{s}^2$ then the velocity head of oil will be

Options:
A $1.25 \mathrm{~m}$
B $12.5 \mathrm{~m}$
C $0.125 \mathrm{~m}$
D $125 \mathrm{~m}$

The correct answer is:
$1.25 \mathrm{~m}$

Solution:
Velocity head $h=\frac{v^2}{2 g}=\frac{(5)^2}{2 \times 10}=1.25 \mathrm{~m}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The velocity of sound is generally greater in solids than in gases because

Options:
A the density of solid is high, but its elasticity is low.
B the density of solid is high, but the elasticity of solid is very high.
C both the density and elasticity of solids are low.
D the density of solids is low but the elasticity is high.

The correct answer is:
the density of solid is high, but the elasticity of solid is very high.

Solution:
v=Elasticity coefficientρ

The speed of sound is higher for more elastic mediums.

As solids are more elastic compared to gases, speed is more for solids.

Topic: Physics, Waves and Sound, JEE Main
The velocity of the bullet becomes one third after it penetrates 4 cm in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at 4+xcm inside the block. The value of x is

Options:
A 2.0
B 1.0
C 0.5
D 1.5

The correct answer is:
0.5

Solution:
According to the given information for the first 4 cm, u=v0 and v=v03. Applying equation of motion for constant acceleration,

v2=u2+2as

⇒v032=v2+2a4⇒a=-8 v0298=-v029

When the body stops, v=0

⇒0=v02+2a4+x

⇒v02=2v0294+x

⇒4.5=4+x

⇒x=0.5 cm

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2022 (27 Jul Shift 2)
The velocity of the electromagnetic wave is parallel to:

Options:
A $\vec{B} \times \vec{E}$
B $\vec{E} \times \vec{B}$
C $\vec{E}$
D $\vec{B}$

The correct answer is:
$\vec{E} \times \vec{B}$

Solution:

Topic: Physics, Electromagnetic Waves, JEE Main
The velocity of the electron in the tenth Bohr's orbit of the hydrogen atom is \( \mathrm{y} \times 10^{5} \mathrm{~m} / \mathrm{s} \). The value of \( \mathrm{y} \) is
Give your answer up to two places of decimal.

Options:

The correct answer is:
2.19

Solution:
Velocity for Bohr atom for single electron system can be given as:

Vn=2.188×106×Zn m/s.

For hydrogen atom, Z=1

∴ V1=2.188×106×11V1=2.188×106 m/s

∴V10 = 2.188×106×110 V10= 2.188×105 m/s.

Topic: Chemistry, Structure of Atom, JEE Main
The velocity of the electrons liberated by electromagnetic radiation of wavelength λ=18.0 nm, from stationary He+ ions in the ground state, is

Options:
A 2.3×106 ms-1
B 1.3×106 ms-1
C 2.3×105 ms-1
D 1.3×103 ms-1

The correct answer is:
2.3×106 ms-1

Solution:
Step I : Determine binding energy of electron :
As we know Eâ¡ = - 1 3 . 6 Z 2 n 2 = - 1 3 . 6 2 2 1 2 For ground state
= â 54.4 eV
â´ The binding energy of electron is Eb = â E = 54.4 eV
Step II : Determine kinetic energy of electron.
â´ T eâ¡ = hâ¡ c Î» - Eâ¡ b
= 1 2 4 2 n meV 1 8 n mâ¡ - 5 4 . 4 eV
= (69 â 54.4) eV = 14.6 eV
= 23.36 Ã 10â19 joule
â´ Tâ¡ eâ¡ = 1 2 mv 2
vâ¡ = 2 Teâ¡ mâ¡
= 2 Ã 2 3 . 3 6 Ã 1 0 - 1 9 9 . 1 Ã 1 0 - 3 1 m/s = 2 . 3 Ã 1 0 6 m/s

Topic: Physics, Atomic Physics, JEE Main
The velocity of upper layer of water in a river is 36 km h-1. Shearing stress between horizontal layers of water is 10-3 N m-2. Depth of the river is _____ m. (Co-efficient of viscosity of water is 10-2 Pa s)

Options:

The correct answer is:
100

Solution:
The viscous force formula is given by,

F=-ηAdvdy

Therefore, shear stress FA=-ηdvdy. Here negative sign tells us that the direction of force is opposite to the velocity gradient.

⇒FA=ηdvdy

Now v=36×518=10 m s-1 and if the depth is d

dvdy=10d

⇒10-3=10-2×10d⇒d=100 m

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2022 (25 Jun Shift 1)
The velocity of water in a river is 18 km h-1 near the surface. If the river is 5 m deep, find the shearing stress between the horizontal layers of water. The coefficient of viscosity of water=10-2 poise.

Options:
A 10-4 N m-2
B 10-3 N m-2
C 10-2 N m-2
D 10-1 N m-2

The correct answer is:
10-3 N m-2

Solution:
Viscous force =Fv=-ηA·dvdh

=10-3×A×18×10003600×5 , η=10-2 poiseη=10-3 decapoise mks

∴ FvA=185×3600=1200×5

⇒FvA=10-3 N m-2

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2014 (19 Apr Online)
The velocity of water in a river is 18âkm h-1 near the surface. If the river is 5âm deep, find the shearing stress between the horizontal layers of water. The coefficient of viscosity of water =10-2 poise.

Options:
A 10-1 N m-2
B 10-2 N m-2
C 10-3 N m-2
D 10-4 N m-2

The correct answer is:
10-3 N m-2

Solution:
η= 10 −2 poise =10-3 Pa s
v=18 km h-1=180003600=5 m s-1
d=5 m
Viscous force F=ηAdvdx
FA=ηdvdx
=10-3×55
=10-3 N m-2

Topic: Physics, Mechanical Properties of Fluids, JEE Main