datavorous/entrance-exam-dataset huggingface dataset:
A prism is made up of material of refractive index 2. The angle of the prism is A . If the angle of minimum deviation is equal to the angle of the prism, the value of A is

Options:
A 30°
B 45°
C 60°
D 90°

The correct answer is:
90°

Solution:
∵ μ=sin⁡A+δm2sin⁡A2
Here, μ is refractive index δm is minimum deviation angle, A is angle of prism.
∴ 2=sin⁡A+A2sin⁡A2
⇒ 2sin⁡A2=sin⁡A
Hence, A=90°

Topic: Physics, Ray Optics, JEE Main
A prism is made up of material of refractive index $\sqrt{3}$. The angle of the prism is $Å$. If the angle of minimum deviation is equal to the angle of the prism, then the value of $A$ is:

Options:
A $30^{\circ}$
B $45^{\circ}$
C $60^{\circ}$
D $75^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
$\begin{aligned} & \mu=\sqrt{3}, \delta_m=A \\ & \text { For prism } \mu=\frac{\sin \frac{A+\delta_m}{2}}{\sin \frac{A}{2}} \\ & \sqrt{3}=\frac{\sin \frac{A+A}{2}}{\sin \frac{A}{2}} \\ & \sqrt{3}=\frac{\sin A}{\sin \frac{A}{2}} \Rightarrow \sqrt{3}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} \\ & \cos \frac{A}{2}=\frac{\sqrt{3}}{2} \\ & \Rightarrow \quad \cos \frac{A}{2}=\cos 30^{\circ} \\ & \therefore \quad \frac{A}{2}=30^{\circ} \Rightarrow A=60^{\circ} \\ & \end{aligned}$

Topic: Physics, Ray Optics, JEE Main
A prism of angle A=1° μ=1.5. A good estimate for the minimum angle of deviation (in degrees) is close to N10. Value of N is ………

Options:

The correct answer is:
5

Solution:
Given,

Angle of prism, A=1°

Refractive index of prism, μ=1.5

For this prism, minimum angle of deviation is given by

δ=(μ-1)A

⇒δ=(1.5-1)×1°=12=510=N10

⇒N=5

Topic: Physics, Ray Optics, JEE Main, JEE Main 2020 (05 Sep Shift 2)
A prism of refractive index 32 and refracting angle is 45° is placed in air. One of the two refracting surfaces of the prism is silvered and a ray of monochromatic light enters the prism from the other face at an angle θ. If the ray retraces its path, then what is the value of θ(in degree)?

Options:

The correct answer is:
60

Solution:
The monochromatic ray of light will retrace its path if it strikes the mirror perpendicularly.

Using Snell's law of refraction,

1×sinθ=μ×sin45°

⇒sinθ=32×sin45°

sinθ=32×12=32

⇒θ=60°

Topic: Physics, Ray Optics, JEE Main
A prism of refractive index n1 and another prism of refractive index n2 are stuck together (as shown in the figure). n1 and n2 depend on λ, the wavelength of light, according to the relation n1=1.2+10.8×10-14λ2 and n2=1.45+1.8×10-14λ2

The wavelength for which rays incident at any angle on the interface BC pass through without bending at that interface will be ____ nm.

Options:

The correct answer is:
600

Solution:
For no bending, n1=n2
1.2+10.8×10-14λ2=1.45+1.8×10-4λ2
On solving,
9×10-14=25λ2
λ=6×10-7
λ=600 nm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (27 Jul Shift 1)
A problem in mathematics is given to three students $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and their respective probability of solving the problem is $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$. Probability that the problem is solved is

Options:
A
$\frac{3}{4}$

B
$\frac{1}{2}$

C
$\frac{2}{3}$

D
$\frac{1}{3}$

The correct answer is:
$\frac{3}{4}$

Solution:

Topic: Mathematics, Probability, JEE Main, JEE Main 2002
A process has ΔH=200 J mol-1 and ΔS=40 JK-1mol-1. Out of the values given below choose the minimum temperature above which the process will be spontaneous:

Options:
A 20 K
B 5K
C 12K
D 4K

The correct answer is:
5K

Solution:
âG=âH-TâS ....... (Gibbs Helmholtz Equation)
For a reaction to be spontaneous, âG<0
So, âH-TâS<0
âH<TâS
âHâS<T
i.e. Minimum temp. should be ⇒âHâS=20040=5K

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2019 (10 Jan Shift 1)
A process will be spontaneous at all temperatures if:

Options:
A ∆H<0 and ∆S<0
B ∆H<0 and ∆S>0
C ∆H>0 and ∆S>0
D ∆H>0 and ∆S<0

The correct answer is:
∆H<0 and ∆S>0

Solution:
For a spontaneous process at all temperature ΔG should be less than zero at all temperature ∆G<0
∆G=∆H-T∆S
∆H<0
and ∆S>0 will make ΔG<0; at all temperature.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2019 (10 Apr Shift 1)
A ; Product A is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
A professor reads a greeting card received on his 50th birthday with + 2.5 D glasses keeping the card 25 cm away. Ten years letter, he reads his farewell letter with the same glasses but he has to keep the letter 50 cm away. What should be the power (in dioptre) of the lens now so that he can keep the letter at the same distance as ten years ago?

Options:

The correct answer is:
4.5

Solution:
Firstly find out distance of retina from his eye lens i.e., image distance in first case (we are assuming distance between eye lens and retina is not changing which is not true practically we are taking that liberty to make problem more mathematically convenient. However it doesn't effect base line of problem)
1 v - 1 - 2 5 = 1 fâ¡ 1
1 v + 1 2 5 = 1 fâ¡ 1
for second case image distance should be same
1 v + 1 5 0 = 1 fâ¡ 2
â 1 2 5 - 1 5 0 = 1 fâ¡ 1 - 1 fâ¡ 2
â 1 cm - 1 5 0 = 1 fâ¡ 1 - 1 fâ¡ 2 = P 1 - P 2 = 2 D
Let power of his eye lens in two cases are (Pe)1, (Pe)2 respectively
P 1 = 2.5 + P e 1
P 2 = 2.5 + P e 2
â P 1 - P 2 = P e 1 - P e 2
= 2 D
So loss in power is 2 D so he should use additional power of 2 D i.e. 4.5 D

Topic: Physics, Experimental Physics, JEE Main
A progressive wave travelling along the positive x- direction is represented by yx,t=A sin(kx-ωt+ϕ) . Its snapshot at t=0 is given in the figure.
For this wave, the phase ϕ is:

Options:
A π2
B 0
C π
D -π2

The correct answer is:
π

Solution:
y=Asin(kx-ωt+ϕ)

at x=0, t=0, y=0 and slope is negative

⇒ϕ=π

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2019 (12 Apr Shift 1)
A progressive wave travelling along the positive \( x- \) direction is represented by \( y(x, \mathrm{t})=\mathrm{A} \sin (\mathrm{k} x-\omega \mathrm{t}+\phi) \). Its snapshot
at \( \mathrm{t}=0 \) is given in the figure.
For this wave, the phase \( \phi \) is:

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

The correct answer is:
\( \pi \)

Solution:
y=Asin(kx-ωt+ϕ)

at x=0, t=0, y=0 and slope is negative

⇒ϕ=π

Topic: Physics, Waves and Sound, JEE Main
A projectile can have the same range $R$ for two angles of projection. If $T_1$ and $T_2$ be the time of flights in the two cases, then the product of the two time of flights is directly proportional to

Options:
A
$1 / R^2$

B
$1 / \mathrm{R}$

C
$\mathrm{R}$

D
$\mathrm{R}^2$

The correct answer is:
$\mathrm{R}$

Solution:
Range is same for complimentary angles. $\mathrm{T}_1=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}$ and $\mathrm{T}_2=\frac{2 \mathrm{u} \sin (90-\theta)}{\mathrm{g}}$ and $R=\frac{u^2 \sin 2 \theta}{\mathrm{g}}$
$$
\therefore \mathrm{T}_1 \mathrm{~T}_2=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}} \times \frac{2 \mathrm{u} \cos \theta}{\mathrm{g}}=\frac{2 \mathrm{R}}{\mathrm{g}} \text {. }
$$

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2004
A projectile can have the same range $\mathrm{R}$ for two angles of projection. If $\mathrm{t}_1$ and $\mathrm{t}_2$ be the times of flights in the two cases, then the product of the two time of flights is proportional to

Options:
A
$\mathrm{R}^2$

B
$1 / R^2$

C
$1 / \mathrm{R}$

D
$\mathrm{R}$

The correct answer is:
$\mathrm{R}$

Solution:
$\mathrm{t}_1 \mathrm{t}_2=\frac{2 \mathrm{u}^2 \sin 2 \theta}{\mathrm{g}^2}=\frac{2 \mathrm{R}}{\mathrm{g}}$

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2005
A projectile fired at 30° to the ground is observed to be at same height at time 3 s and 5 s after projection, during its flight. The speed of projection of the projectile is m s-1.
(Given g=10 m s-2)

Options:

The correct answer is:
80

Solution:
The time taken to reach maximum height is

T'=usinθg.

The total time of flight is T=(3+5) s=8 s.

Hence,

T'=T2=3+52=usinθg

⇒4=(u)×1210

⇒u=80 m s-1

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (11 Apr Shift 1)
A projectile fired with initial velocity $u$ at some angle ${ }^\theta$ has a range ${ }^R$. If the initial velocity be doubled at the same angle of projection, then the range will be

Options:
A 2R
B R/2
C R
D 4R

The correct answer is:
4R

Solution:
$R=\frac{u^2 \sin 2 \theta}{g} \therefore R \propto u^2$. If initial velocity be doubled then range will become four times.

Topic: Physics, Motion In Two Dimensions, JEE Main
A projectile is fired at an angle of $45^{\circ}$ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is

Options:
A $60^{\circ}$
B $\tan ^{-1}\left(\frac{1}{2}\right)$
C $\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
D $45^{\circ}$

The correct answer is:
$\tan ^{-1}\left(\frac{1}{2}\right)$

Solution:
Height of projectile
$$
\begin{aligned}
& H=\frac{u^2 \sin ^2 \theta}{2 g} \\
& H=\frac{u^2 \sin ^2 45^{\circ}}{2 g} \\
& H=\frac{u^2}{4 g}
\end{aligned}
$$
Range of projectile
$$
\begin{aligned}
R & =\frac{u^2 \sin 2 \theta}{g} \\
& =\frac{u^2 \sin 90^{\circ}}{g} \\
R & =\frac{u^2}{g} \\
\therefore \quad \frac{R}{2} & =\frac{u^2}{2 g} \\
\therefore \quad \tan \alpha & =\frac{H}{R / 2}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{u^2 / 4 g}{u^2 / 2 g} \\
\tan \alpha & =\frac{1}{2} \\
\alpha & =\tan ^{-1}\left(\frac{1}{2}\right)
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2011 (Mains)
A projectile is fired from the surface of the Earth of radius R with a velocity ηve, where ve is the escape velocity and η<1. Neglecting the air resistance, the orbital velocity of the projectile is

Options:
A ve1-Î·2
B veη25
C 25veη
D 2η5ve

The correct answer is:
ve1-η2

Solution:
Velocity at the surface=vs=Î·ve
Velocity in orbit=v0
By conversation of mechanical energy,
12mvs2+-GMmR=12mv02-GMmrâ¦1
By Newton's law, GMmr2=mv02râââââ¦2
Solving equations (1) and (2)
v0=ve1-Î·2

Topic: Physics, Gravitation, JEE Main
A projectile is fired from the surface of the earth with a velocity of 5 m s-1 and angle θ with the horizontal. Another projectile fired from another planet with a velocity of 3 m s-1 at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is: (given =9.8 m s-2)

Options:
A 3.5
B 5.9
C 16.3
D 110.8

The correct answer is:
3.5

Solution:
The Equation of the Trajectory of a Projectile

y=xtanθ-gx22u2cosθ

y ,x are Variable and θ is constant

so gu2=constant

g52=a32
a=9.8×925
a=3.5

Topic: Physics, Motion In Two Dimensions, JEE Main
A projectile is fired with the velocity v0= 2 m s-1 at an angle 60o with horizontal. At top of its trajectory, it explodes into three fragments of equal mass. The first fragment retraces the path, second moves vertically upward with speed 3v02. The speed of the third fragment (in m s-1) is

Options:

The correct answer is:
5

Solution:
Let velocity of third fragment be →V
Pi→=Pf→
⇒3mv02i^=−mv02i^+m3v02j^+mv→
v → =4 v 0 2 i ^ − 3 v 0 2 j ^
⇒v=5v02

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A projectile is given an initial velocity of \( \hat{i}+2 \hat{j} \). The Cartesian equation of its path is: \( \left(g=10 m / s^{2}\right) \)
(Here, \( \hat{i} \) is unit vector along horizontal and \( \hat{j} \) is unit vector vertically upwards)

Options:
A \( y=2 x-5 x^{2} \)
B \( y=x-5 x^{2} \)
C \( 4 y=2 x-5 x^{2} \)
D \( y=2 x-25 x^{2} \)

The correct answer is:
\( y=2 x-5 x^{2} \)

Solution:
tanâ¡θ=usinâ¡θucosâ¡θ=21

The desired equation is :

y=xtanâ¡θ-gx22u2cos2â¡θ

or y=x2-10 x22 22+122 152

or y=2x-5x2

Topic: Physics, Motion In Two Dimensions, JEE Main
A projectile is given an initial velocity of i^+2j^ m s-1, where i ^ is along the ground and j ^ is along the vertical upward. If g=10 m s-2, the equation of its trajectory is :

Options:
A 4y=2x-5x2
B 4y=2x-25x2
C y=x-5x2
D y=2x-5x2

The correct answer is:
y=2x-5x2

Solution:
v → = i ^ + 2 j ^
⇒ x = t ..... (i)
y = 2 t - 1 2 1 0 t 2 ..... (ii)
From (i) and (ii)
y = 2 x - 5 x 2

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2013 (07 Apr)
A projectile is launched from the origin with speed \( v \) at an angle \( \theta \) from the horizon. At the highest point in the trajectory, the projectile breaks into two pieces, \( A \) and \( B \) of masses \( m \) and \( 2 m \) respectively. Immediately after the breakup, piece \( A \) is at rest relative to the ground. Neglect air resistance. Which of the following sentences most accurately describes what happens next?

Options:
A Piece \( B \) will hit the gorund first, since it is more massive.
B Both pieces have zero vertical velocity immedialtely after th breakup, and therefore they hit the ground at the same time.
C Piece \( A \) will hit the ground first, because it will have a downward velocity immediatly after the breakup.
D There is no way of knowing which piece will hit the ground first, because not enough information is given about the breakup.

The correct answer is:
Both pieces have zero vertical velocity immedialtely after th breakup, and therefore they hit the ground at the same time.

Solution:
At the highest point before explosion the vertical component of velocity is zero. So, as per law of conservation of momentum, after explosion also the pieces will have the same vertical component of velocity. So, in the vertical direction they will be under free fall with zero initial velocity. Hence, they will hit the ground at the same time.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A projectile is launched with an initial speed of 40 m s-1 at an angle 30° above the ground. The projectile lands on a hillside 2.0 s later. The net displacement from where the projectile was launched to where it hits the target is (Take g=10 m s-2)

Options:
A 203 m
B 302 m
C 40 m
D 2013 m

The correct answer is:
2013 m

Solution:
Horizontal displacement covered,

x=40cos30°×2=403 m

Vertical displacement covered,

y=40sin30°×2+12-10×22=40-20=20 m

Hence, required displacement will be,

x2+y2=2013 m

Topic: Physics, Motion In Two Dimensions, JEE Main
A projectile is projected at 30° from horizontal with initial velocity 40 m s-1. The velocity of the projectile at t=2 s from the start will be:

Options:
A 403 m s-1
B Zero
C 20 m s-1
D 203 m s-1

The correct answer is:
203 m s-1

Solution:
It is given that v=40 m s-1.

The components of the velocity will be vy=vsin30°=40 m s-12=20 m s-1vx=vcos30°=4032 m s-1=203 m s-1

The time taken to reach maximum height is T=usin30°g=20 m s-110 m s-2=2 s

Hence, at T=2 s, as vy=0 m s-1 therefore vnet=vx=203 m s-1.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (11 Apr Shift 2)
A projectile is thrown in the upward direction making an angle of \(60^{\circ}\) with the horizontal with a velocity of \(140 \mathrm{~ms}^{-1}\). Then the time after which its velocity makes an angle \(45^{\circ}\) with the horizontal is (Acceleration due to gravity, \(g=10 \mathrm{~ms}^{-2}\) )

Options:
A \(0.5124 \mathrm{~s}\)
B \(51.24 \mathrm{~s}\)
C \(5.124 \mathrm{~s}\)
D \(512.4 \mathrm{~s}\)

The correct answer is:
\(5.124 \mathrm{~s}\)

Solution:
Given, angle of projection, \(\theta=60^{\circ}\) and velocity,
\(u=140 \mathrm{~ms}^{-1}\)
This velocity have divided into two components, horizontal component, \(u_x=u \cos 60^{\circ}=140 \cos 60^{\circ}\) and vertical component, \(u_y=u \sin 60^{\circ}=140 \sin 60^{\circ}\) Let after time \(t\), the inclination of particle with horizontal be \(45^{\circ}\) and at time \(t\) velocity along \(x=v_x\) and along \(y=v_y\).
Now,
\(\begin{aligned}
\tan 45^{\circ} & =\frac{v_y}{v_x} \\
& =v_x=v_y
\end{aligned}\)
Since, the horizontal component of velocity remains constant, i.e.,
\(\begin{aligned}
v_x & =u_x=140 \cos 60^{\circ} \\
\text{and } v_y & =u_y-g t \\
140 \cos 60^{\circ} & =140 \sin 60^{\circ}-10 t \quad\left[\because v_y=v_x\right] \\
140 \times \frac{1}{2} & =140 \times \frac{\sqrt{3}}{2}-10 t \\
70 & =70 \sqrt{3}-10 t \\
10 t & =70 \sqrt{3}-70 \\
t & =\frac{70(\sqrt{3}-1)}{10} \\
& =7 \times(0.7320)=5.124 \mathrm{~s}
\end{aligned}\)

Topic: Physics, Motion In Two Dimensions, AP EAMCET, AP EAMCET 2019 (20 Apr Shift 1)
A projectile is thrown into air with velocity $15 \mathrm{~m} / \mathrm{s}$ at an angle $30^{\circ}$ with the horizontal. After what time its direction of motion is perpendicular to its initial direction? (Assume, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Options:
A $3 \mathrm{~s}$
B $2 \mathrm{~s}$
C $0.5 \mathrm{~s}$
D $1.5 \mathrm{~s}$

The correct answer is:
$3 \mathrm{~s}$

Solution:
The given projectile motion is as shown in figure,
Let $t$ be the time after which the initial direction is perpendicular to its direction of motion.
As, velocity of a projectile has two components. So, its initial velocity is given by
$\mathbf{v}_1=u \cos \theta \hat{\mathbf{i}}+u \sin \theta \hat{\mathbf{j}}$
Since, in projectile motion the horizontal component of velocity remains same, while the vertical component varies, so its velocity after time $t$,
$\begin{aligned} & \mathbf{v}_2=u_y \hat{\mathbf{i}}+v_y \hat{\mathbf{j}} \\ & \mathbf{v}_2=u \cos \theta \hat{\mathbf{i}}+(u \sin \theta-g t) \hat{\mathbf{j}}\end{aligned}$
$\left[\because v_y=u_y-g t=u \sin \theta-g t\right] \ldots$
As, velocities are perpendicular, so $\mathbf{v}_1 \cdot \mathbf{v}_2=0$
From Eqs (i) and (ii), we get
$\begin{array}{ll}\Rightarrow & (u \cos \theta \dot{\mathbf{i}}+u \sin \theta \mathbf{j}) \cdot(u \cos \theta \mathbf{i}+(u \sin \theta-g t) \mathbf{j})=0 \\ \Rightarrow & u^2 \cos ^2 \theta+u \sin \theta(u \sin \theta-g t)=0 \\ \Rightarrow & u^2 \cos ^2 \theta+u^2 \sin ^2 \theta-u \sin \theta \times g t=0 \\ \Rightarrow & u^2\left(\cos ^2 \theta+\sin ^2 \theta\right)-u g t \sin \theta=0 \\ \Rightarrow & u^2=u g t \sin \theta \quad\left(\because \sin ^2 \theta+\cos ^2 \theta=1\right) \\ \Rightarrow & t=\frac{u}{g \sin \theta}\end{array}$
Here, $u=15 \mathrm{~m} / \mathrm{s}, g=10 \mathrm{~m} / \mathrm{s}^2$ and $\theta=30^{\circ}$
$\therefore \quad t=\frac{15}{10 \times \sin 30^{\circ}}=\frac{15 \times 2}{10}=3 \mathrm{~s}$

Topic: Physics, Motion In Two Dimensions, TS EAMCET, TS EAMCET 2020 (09 Sep Shift 2)
A projectile moving vertically upwards with a velocity of $200 \mathrm{~ms}^{-1}$ breaks into two equal parts at a height of $490 \mathrm{~m}$. One part starts moving vertically upwards with a velocity of $400 \mathrm{~ms}^{-1}$. How much time it will take, after the break up with the other part to hit the ground?

Options:
A
$2 \sqrt{10} \mathrm{~s}$

B
$5 \mathrm{~s}$

C
$10 \mathrm{~s}$

D
$\sqrt{10} \mathrm{~s}$

The correct answer is:
$10 \mathrm{~s}$

Solution:
Momentum before explosion
= Momentum after explosion
$$
\begin{gathered}
m \times 200 \hat{j}=\frac{m}{2} \times 400 \hat{j}+\frac{m}{2} v \\
=\frac{m}{2}(400 \hat{j}+v) \\
\Rightarrow \quad 400 \hat{j}-400 \hat{j}=v \\
\therefore \quad v=0
\end{gathered}
$$
i.e., the velocity of the other part of the mass, $v=0$
Let time taken to reach the earth by this part be $t$
Applying formula, $h=u t+\frac{1}{2} g t^2$
$$
\begin{aligned}
& 490=0+\frac{1}{2} \times 9.8 \times t^2 \\
& \Rightarrow \quad t^2=\frac{980}{9.8}=100 \\
& \therefore \quad t=\sqrt{100}=10 \mathrm{sec}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A projectile moving vertically upwards with a velocity of $200 \mathrm{~ms}^{-1}$ breaks into two equal parts at a height of $490 \mathrm{~m}$. One part starts moving vertically upwards with a velocity of $400 \mathrm{~ms}^{-1}$. How much time it will take, after the break up with the other part to hit the ground?

Options:
A
$2 \sqrt{10} \mathrm{~s}$

B
$5 \mathrm{~s}$

C
$10 \mathrm{~s}$

D
$\sqrt{10} \mathrm{~s}$

The correct answer is:
$10 \mathrm{~s}$

Solution:
Momentum before explosion
= Momentum after explosion
$$
\begin{gathered}
m \times 200 \hat{j}=\frac{m}{2} \times 400 \hat{j}+\frac{m}{2} v \\
=\frac{m}{2}(400 \hat{j}+v) \\
\Rightarrow \quad 400 \hat{j}-400 \hat{j}=v \\
\therefore \quad v=0
\end{gathered}
$$
i.e., the velocity of the other part of the mass, $v=0$
Let time taken to reach the earth by this part be $t$
Applying formula, $h=u t+\frac{1}{2} g t^2$
$$
\begin{aligned}
& 490=0+\frac{1}{2} \times 9.8 \times t^2 \\
& \Rightarrow \quad t^2=\frac{980}{9.8}=100 \\
& \therefore \quad t=\sqrt{100}=10 \mathrm{sec}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2012 (12 May Online)
A projectile of mass 1.2 kg undergoes a perfectly inelastic collision with a trolley of mass 3.6 kg as shown in the figure. At the time of the collision, the speed of the projectile is 5 m s-1 at an angle of 37Â° with the horizontal. The trolley is free to move, only along a horizontal rail which coincides with the direction of the projectile's horizontal motion. Assuming that the trolley doesn't topple, calculate the amount of heat energy (in J) released in the collision. [Take sin37Â°=35 and cos37Â°=45]

Options:

The correct answer is:
12.6

Solution:
Using the conservation of linear momentum in the horizontal direction we get,

1.2×4=4.8×v

⇒v=1 m s-1

The loss in kinetic energy is equal to the amount of heat energy released.

∆K=12×1.2×52-12×4.8×12

∆K=12.6 J

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A projectile of mass m is fired from ground with speed u at an angle Î¸ to horizontal. What is the power of gravitational force, when it makes an angle Î¸2 with the horizontal?

Options:
A -mgu cosθ tanθ2
B -mgu tanθ cosθ2
C -2mgu tanθ sinθ2
D -2mgu sinθ tanθ2

The correct answer is:
-mgu cosθ tanθ2

Solution:
vx=ucos⁡θ=vcos⁡θ2 , where v is the speed at this instant
vy=vsinθ2=ucosθ sinθ2cosθ2=ucosθ tanθ2

P=F→.v→=-mgu cosθ tanθ2

Topic: Physics, Motion In Two Dimensions, JEE Main
A projectile of mass m is fired from the surface of the earth at an angle Î± = 6 0 â from the vertical. The initial speed v0 is equal to GMeRe. The maximum height that projectile can rise from the surface of the earth is Rex. Find the value of x?

G = Universal Gravitational constant
Me = Mass of earth
Re = Radius of earth

Options:

The correct answer is:
2

Solution:
By conservation of angular momentum about C at position P and F
mv 0 3 2 R e = mv r max
â v = 3 R e v 0 2 r max ...(i)
By conservation of mechanical energy between P and F
1 2 mv 0 2 - GM e m R e = 1 2 mv 2 - GM e m r max ...(ii)
Put (i) in (ii)
1 2 mv 0 2 - GM e m R e = 1 2 m 3 R 2 v 0 2 4 r max 2 - GM e m r max
v 0 = GM e R
â 1 2 m GM e R e - GM e m R e = 3 8 mR e 2 r max 2 GM e R - GM e m r max
â 1 2 R e - 1 R e = 3 R 8 r max 2 - 1 r max
â - 1 2 R e = 3 R e - 8 r max 8 r max 2
â 4 r max 2 - 8 R e r max + 3 R e 2 = 0
Solving the quadratic in rmax
we get
r max = 8 R e Â± 1 6 R e 8 = 8 R e Â± 4 R e 8
â r max = 8 R e + 4 R e 8 = 1 2 R e 8 = 3 R e 2
â´ h max = r max - R e = 3 R e 2 - R e = R e 2
â´ x = 2

Topic: Physics, Gravitation, JEE Main
A projectile of mass m is fired from the surface of the earth at an angle α = 6 0 ∘ from the vertical. The initial speed v0 is equal to GMeRe. The maximum height that projectile can rise from surface of earth is Rex. Find the value of x ?

G = Universal Gravitational constant
Me = Mass of earth
Re = Radius of earth

Options:
A 2
B

4

C

6

D

8

The correct answer is:
2

Solution:
By conservation of angular momentum about C at position P and F
mv 0 3 2 R e = mv r max
â v = 3 R e v 0 2 r max ...(i)
By conservation of mechanical energy between P and F
1 2 mv 0 2 - GM e m R e = 1 2 mv 2 - GM e m r max ...(ii)
Put (i) in (ii)
1 2 mv 0 2 - GM e m R e = 1 2 m 3 R 2 v 0 2 4 r max 2 - GM e m r max
v 0 = GM e R
â 1 2 m GM e R e - GM e m R e = 3 8 mR e 2 r max 2 GM e R - GM e m r max
â 1 2 R e - 1 R e = 3 R 8 r max 2 - 1 r max
â - 1 2 R e = 3 R e - 8 r max 8 r max 2
â 4 r max 2 - 8 R e r max + 3 R e 2 = 0
Solving the quadratic in rmax
we get
r max = 8 R e Â± 1 6 R e 8 = 8 R e Â± 4 R e 8
â r max = 8 R e + 4 R e 8 = 1 2 R e 8 = 3 R e 2
â´ h max = r max - R e = 3 R e 2 - R e = R e 2
â´ x = 2

Topic: Physics, Gravitation, JEE Main