Work, Power & Energy
769 Questions
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JEE (Main) April 2019
Q126
Errorless
Critical Thinking
MCQ
A wedge of mass M = 4m lies on a frictionless plane. A particle of mass m approaches the wedge with speed v. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by
A.
$\frac{2v^{2}}{7g}$
B.
$\frac{v^{2}}{g}$
C.
$\frac{2v^{2}}{5g}$
D.
$\frac{v^{2}}{2g}$
JEE (Main) April 2019
Q127
Errorless
Critical Thinking
MCQ
Two particles, of masses M and 2M, moving, as shown, with speeds of 10 m/s and 5m/s, collide elastically at the origin. After the collision, they move along the indicated directions with speeds $\nu_{1}$ and $\nu_{2}$ , respectively. The values of $\nu_{1}$ and $\nu_{2}$ are nearly
A.
3.2 m/s and 6.3 m/s
B.
3.2 m/s and 12.6 m/s
C.
6.5 m/s and 6.3 m/s
D.
6.5 m/s and 3.2 m/s
JEE (Main) April 2019
Q128
Errorless
Critical Thinking
MCQ
A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. What is the mass of the second body
A.
1.8 kg
B.
1.2 kg
C.
1.5 kg
D.
1.0 kg
JEE (Main) April 2019
Q129
Errorless
Graphical Questions
MCQ
A particle moves in the one dimension from rest under the influence of a force that varies with the distance travelled by the particle as shown in the figure. The kinetic energy of the particle after it has travelled 3m is
A.
6.5 J
B.
2.5 J
C.
4 J
D.
5 J
JEE (Advanced) 2019
Q130
Errorless
JEE Section
MSQ
A small particle of mass m moving inside a heavy, hollow and straight tube along the tube axis undergoes elastic collision at two ends. The tube has no friction and it is closed at one end by a flat surface while the other end is fitted with a heavy movable flat piston as shown in figure. When the distance of the piston from closed end is $L = L_{0}$ the particle speed is $v = v_{0}$ . The piston is moved inward at a very low speed V such that $V \ll \frac{dL}{L} v_{0}$ , where dL is the infinitesimal displacement of the piston. Which of the following statement(s) is/are correct
A.
The rate of which the particle strikes the piston is v/L
B.
After each collision with the piston, the particle speed increases by 2V
C.
The particle's kinetic energy increases by a factor of 4 when the piston is moved inward from $L_0$ to $\frac{1}{2} L_0$
D.
If the piston moves inward by dL, the particle speed increases by $2v\frac{dL}{L}$
JEE (Advanced) 2019
Q131
Errorless
JEE Section
Numerical
A particle is moved along a path $AB - BC - CD - DE - EF - FA$ , as shown in figure, in presence of a force $\vec{F} = (\alpha y\hat{i} + 2\alpha x\hat{j})N$ , where $x$ and $y$ are in meter and $\alpha = -1N/m$ . The work done on the particle by this force $\vec{F}$ will be ____ Joule
Correct Answer: 0.75
Explanation:
$W = F.dr$ where $d\vec{r} = dx\hat{i} + dy\hat{j}$
$W = \alpha y dx + 2\alpha x dy$ and $\alpha = -1 N / m$
$A\to B,y = 1,dy = 0$
then $W_{A \to B} = \int \alpha y dx = \alpha 1 \int_{0}^{1} dx = \alpha = -1$
$B\to C,x = 1,dx = 0$
then $W_{B \to C} = 2\alpha 1 \int_{1}^{0.5} dy = -2\alpha (0.5) = -\alpha = 1$
$C\to D,y = 0.5,dy = 0$
then $W_{C \to D} = \int_{1}^{0.5} \alpha y dx = \alpha \cdot \frac{1}{2} \int_{1}^{0.5} dx = -\frac{\alpha}{4} = \frac{1}{4}$
$D\to E,x = 0.5,dx = 0$
then $W_{D \to E} = 2\alpha \int x dy = 2\alpha \cdot \frac{1}{2} \int_{0.5}^{0} dy = -\frac{\alpha}{2} = \frac{1}{2}$
$E\to F,y = 0,dy = 0$ then $W_{EF} = 0$
$F\to A,x = 0,dx = 0$ then $W_{F\to A} = 0$
Net work done in cycle $W = \frac{3}{4} J = 0.75J$ .
$W = \alpha y dx + 2\alpha x dy$ and $\alpha = -1 N / m$
$A\to B,y = 1,dy = 0$
then $W_{A \to B} = \int \alpha y dx = \alpha 1 \int_{0}^{1} dx = \alpha = -1$
$B\to C,x = 1,dx = 0$
then $W_{B \to C} = 2\alpha 1 \int_{1}^{0.5} dy = -2\alpha (0.5) = -\alpha = 1$
$C\to D,y = 0.5,dy = 0$
then $W_{C \to D} = \int_{1}^{0.5} \alpha y dx = \alpha \cdot \frac{1}{2} \int_{1}^{0.5} dx = -\frac{\alpha}{4} = \frac{1}{4}$
$D\to E,x = 0.5,dx = 0$
then $W_{D \to E} = 2\alpha \int x dy = 2\alpha \cdot \frac{1}{2} \int_{0.5}^{0} dy = -\frac{\alpha}{2} = \frac{1}{2}$
$E\to F,y = 0,dy = 0$ then $W_{EF} = 0$
$F\to A,x = 0,dx = 0$ then $W_{F\to A} = 0$
Net work done in cycle $W = \frac{3}{4} J = 0.75J$ .
Online April 2019
Q132
Advance
ARCHIVE: JEE MAIN
MCQ
A particle moves in one dimension from rest under the influence of a force that varies with the distance travelled by the particle as shown in the Figure. The kinetic energy of the particle after it has travelled 3 m is
A.
4 J
B.
2.5 J
C.
5 J
D.
6.5 J
Online April 2019
Q133
Advance
ARCHIVE: JEE MAIN
MCQ
A uniform cable of mass M and length L is placed on a horizontal surface such that its $\left(\frac{1}{n}\right)^{\text{th}}$ part is hanging below the edge of the surface. To lift the hanging part of the cable up to the surface, the work done should be
A.
$\frac{MgL}{n^{2}}$
B.
nMgL
C.
$\frac{MgL}{2n^{2}}$
D.
$\frac{2MgL}{n^{2}}$
Online January 2019
Q134
Advance
ARCHIVE: JEE MAIN
MCQ
A block of mass $m$ , lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant $k$ . The other end of the spring is fixed, as shown in the figure. The block is initially at rest in its equilibrium position. If now the block is pulled with a constant force $F$ , the maximum speed of the block is
A.
$\frac{F}{\pi\sqrt{mk}}$
B.
$\frac{\pi F}{\sqrt{mk}}$
C.
$\frac{2F}{\sqrt{mk}}$
D.
$\frac{F}{\sqrt{mk}}$
Online January 2019
Q135
Advance
ARCHIVE: JEE MAIN
MCQ
A force acts on a 2 kg object so that its position is given as a function of time as $x = 3t^{2} + 5$ . What is the work done by this force in first 5 seconds?
A.
950 J
B.
900 J
C.
850 J
D.
875 J
Online January 2019
Q136
Advance
ARCHIVE: JEE MAIN
MCQ
A block of mass $m$ is kept on a platform which starts from rest with constant acceleration $\frac{g}{2}$ upward, as shown in figure. Work done by normal reaction on block in time $t$ is
A.
$\frac{3mg^2t^2}{8}$
B.
$-\frac{mg^2t^2}{8}$
C.
0
D.
$\frac{mg^2t^2}{8}$
Online January 2019
Q137
Advance
ARCHIVE: JEE MAIN
MCQ
A particle which is experiencing a force, given by $\vec{F} = 3\hat{i} - 12\hat{j}$ , undergoes a displacement of $\vec{d} = 4\hat{i}$ . If the particle had a kinetic energy of $3\mathrm{J}$ at the beginning of the displacement, what is its kinetic energy of the end of the displacement?
A.
15 J
B.
9 J
C.
12 J
D.
10 J
Online January 2019
Q138
Advance
ARCHIVE: JEE MAIN
MCQ
A pendulum is executing simple harmonic motion and its maximum kinetic energy is $K_{1}$ . If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is $K_{2}$ . Then
A.
$K_{2} = 2K_{1}$
B.
$K_{2} = \frac{K_{1}}{4}$
C.
$K_{2} = K_{1}$
D.
$K_{2} = \frac{K_{1}}{2}$
JEE (Advanced) 2019
Q139
Advance
ARCHIVE: JEE ADVANCED
Numerical
Aparticle is moved along a path $AB-BC-CD-DE-EF-FA$ , as shown in figure, in presence of a force $\vec{F} = (\alpha y\hat{i} + 2\alpha x\hat{j})$ N, where $x$ and $y$ are in meter and $\alpha = -1$ Nm $^{-1}$ . The work done on the particle by this force $\vec{F}$ will be____ joule.
Correct Answer: 0.75
Explanation:
Given that, $\vec{F} = \alpha y \hat{i} + 2 \alpha x \hat{j}$ where $\alpha = -1$
Since Work done is given by
$\begin{array}{r l} & W = \int \vec {F} \cdot d \vec {\ell} \\ \Rightarrow & d W = \alpha (y d x + 2 x d y) \end{array}$
For $A\to B$ , $y = 1$ , so $dy = 0$
$\Rightarrow W _ {A \rightarrow B} = \alpha \int y d x = \alpha \int_ {0} ^ {1} d x = \alpha$
For $B\to C$ , $x = 0$ , so $dx = 0$
$\Rightarrow W _ {B \rightarrow C} = 2 \alpha \int_ {1} ^ {0. 5} d y = 2 \alpha (- 0. 5) = - \alpha$
For $C\to D$ , $y = 0.5$ , so $dy = 0$
$\Rightarrow W _ {C \rightarrow D} = 0. 5 \alpha \int_ {1} ^ {0.5} d x = - \frac {\alpha}{4}$
For $D\to E$ , $x = 0.5$ , so $dx = 0$
$\Rightarrow W _ {D \rightarrow E} = 2 \alpha (0.5) \int_ {0.5} ^ {0} d y = - \frac {\alpha}{2}$
$\Rightarrow W _ {E \to F} = 0$
$\Rightarrow W _ {F \to A} = 0$
Hence total work done is
$\begin{array}{r l} & W _ {t o t a l} = \alpha - \alpha - \frac {\alpha}{4} - \frac {\alpha}{2} \\ \Rightarrow & W _ {t o t a l} = - \frac {3 \alpha}{4} = \frac {- 3 (- 1)}{4} = \frac {3}{4} \mathrm{J} = 0. 7 5 \mathrm{J} \end{array}$
Since Work done is given by
$\begin{array}{r l} & W = \int \vec {F} \cdot d \vec {\ell} \\ \Rightarrow & d W = \alpha (y d x + 2 x d y) \end{array}$
For $A\to B$ , $y = 1$ , so $dy = 0$
$\Rightarrow W _ {A \rightarrow B} = \alpha \int y d x = \alpha \int_ {0} ^ {1} d x = \alpha$
For $B\to C$ , $x = 0$ , so $dx = 0$
$\Rightarrow W _ {B \rightarrow C} = 2 \alpha \int_ {1} ^ {0. 5} d y = 2 \alpha (- 0. 5) = - \alpha$
For $C\to D$ , $y = 0.5$ , so $dy = 0$
$\Rightarrow W _ {C \rightarrow D} = 0. 5 \alpha \int_ {1} ^ {0.5} d x = - \frac {\alpha}{4}$
For $D\to E$ , $x = 0.5$ , so $dx = 0$
$\Rightarrow W _ {D \rightarrow E} = 2 \alpha (0.5) \int_ {0.5} ^ {0} d y = - \frac {\alpha}{2}$
$\Rightarrow W _ {E \to F} = 0$
$\Rightarrow W _ {F \to A} = 0$
Hence total work done is
$\begin{array}{r l} & W _ {t o t a l} = \alpha - \alpha - \frac {\alpha}{4} - \frac {\alpha}{2} \\ \Rightarrow & W _ {t o t a l} = - \frac {3 \alpha}{4} = \frac {- 3 (- 1)}{4} = \frac {3}{4} \mathrm{J} = 0. 7 5 \mathrm{J} \end{array}$
JEE (Advanced) 2018
Q140
Allen
JEE Advanced PYQ
MSQ
A particle of mass m is initially at rest at the origin. It is subjected to a force and starts moving along the x-axis. Its kinetic energy K changes with time as $dK/dt = \gamma t$ , where $\gamma$ is a positive constant of appropriate dimensions. Which of the following statements is (are) true?
A.
The force applied on the particle is constant
B.
The speed of the particle is proportional to time
C.
The distance of the particle from the origin increases linearly with time
D.
The force is conservative
NEET 2018
Q141
Errorless
Conservation of Energy and Momentum
MCQ
A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to
A.
$\frac{3}{2} D$
B.
$D$
C.
$\frac{7}{5} D$
D.
$\frac{5}{4} D$
JEE (Main) 2018
Q142
Errorless
Conservation of Energy and Momentum
MCQ
A particle is moving in a circular path of radius a under the action of an attractive potential $U = \frac{k}{2r^2}$ . Its total energy is
A.
$\frac{k}{2a^2}$
B.
Zero
C.
$-\frac{3}{2}\frac{k}{a^2}$
D.
$-\frac{k}{4a^2}$
JEE (Main) 2018
Q143
Errorless
Elastic and Inelastic Collision
MCQ
In a collinear collision, a particle with an initial speed $v_{0}$ strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is
A.
$\sqrt{2}v_{0}$
B.
$\frac{v_{0}}{2}$
C.
$\frac{v_{0}}{\sqrt{2}}$
D.
$\frac{v_{0}}{4}$
JEE (Main) 2018
Q144
Errorless
Elastic and Inelastic Collision
MCQ
It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is $p_d$ ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is $p_c$ . The values of $p_d$ and $p_c$ are respectively
A.
(.28, .89)
B.
(0, 0)
C.
(0, 1)
D.
(.89, .28)
JEE (Advanced) 2018
Q145
Errorless
JEE Section
MSQ
A particle of mass $m$ is initially at rest at the origin. It is subjected to a force and starts moving along the $x$ -axis. Its kinetic energy $K$ changes with time as $dK / dt = \gamma t$ , where $\gamma$ is a positive constant of appropriate dimensions. Which of the following statements is (are) true
A.
The force applied on the particle is constant
B.
The speed of the particle is proportional to time
C.
The distance of the particle from the origin increases linearly with time
D.
The force is conservative
2018
Q146
Advance
ARCHIVE: JEE MAIN
MCQ
A particle is moving in a circular path of radius $a$ under the action of an attractive potential $U = -\frac{k}{2r^2}$ . Its total energy is
A.
$-\frac{k}{4a^2}$
B.
$\frac{k}{2a^2}$
C.
Zero
D.
$-\frac{3}{2}\frac{k}{a^2}$
Online 2018
Q147
Advance
ARCHIVE: JEE MAIN
MCQ
Two particles of the same mass m are moving in circular orbits because of force, given by $F(r) = -\frac{16}{r} - r^{3}$ . The first particle is at a distance r = 1 and the second, at r = 4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to
A.
$10^{-1}$
B.
$6 \times 10^{-2}$
C.
$6 \times 10^{2}$
D.
$3 \times 10^{-3}$
Online 2018
Q148
Advance
ARCHIVE: JEE MAIN
MCQ
A body of mass m starts moving from rest along x-axis so that its velocity varies as $v = a\sqrt{s}$ where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first t seconds after the start of the motion is
A.
$\frac{1}{4}ma^{4}t^{2}$
B.
$4ma^{4}t^{2}$
C.
$\frac{1}{8}ma^{4}t^{2}$
D.
$8ma^{4}t^{2}$
JEE (Main) 2017
Q149
Errorless
Elastic and Inelastic Collision
MCQ
A particle $A$ of mass $m$ and initial velocity $\nu$ collides with a particle $B$ of mass $\frac{m}{2}$ which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths $\lambda_{A}$ to $\lambda_{B}$ after the collision is
A.
$\frac{\lambda_A}{\lambda_B} = \frac{1}{2}$
B.
$\frac{\lambda_A}{\lambda_B} = \frac{1}{3}$
C.
$\frac{\lambda_A}{\lambda_B} = 2$
D.
$\frac{\lambda_A}{\lambda_B} = \frac{2}{3}$
JEE (Advanced) 2017
Q150
Errorless
JEE Section
MSQ
A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at x=0, in a coordinate system fixed to the table. A point mass m is released from rest at the topmost point of the path as shown and its slides down. When the mass loses contact with the block, its position is x and the velocity is v. At the instant, which of the following options is/are correct
A.
The velocity of the point mass $m$ is: $v = \sqrt{\frac{2gR}{1 + \frac{m}{M}}}$
B.
The $x$ component of displacement of the center of mass of the block $M$ is: $-\frac{mR}{M + m}$
C.
The position of the point mass is: $x = -\sqrt{2}\frac{mR}{M + m}$
D.
The velocity of the block $M$ is: $V = -\frac{m}{M}\sqrt{2gR}$