Work, Power & Energy
769 Questions
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Q726
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 16
A pendulum of mass m, length l has angular amplitude $\theta$ . Tension in the string at extreme position is $T_{1}$ and at bottom is $T_{2}$ . Based on above information, answer the following questions.
If maximum tension which the string can bear is twice the weight of bob, then the maximum value of $\theta$ is
A.
$\cos^{-1}\left(\frac{1}{3}\right)$
B.
$\cos^{-1}\left(\frac{2}{3}\right)$
C.
$\cos^{-1}\left(\frac{3}{4}\right)$
D.
$\frac{\pi}{3}$
Q727
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 17
A small ball is given some velocity at point A towards right so that it moves on the semicircular track and does not leave contact up to the highest point B. After leaving the highest point B, it falls at the top of a building of height R and width $x(x \ll 2R)$ . (All the surfaces are frictionless).
Based on above information, answer the following questions.
The velocity given to the ball at point A so that it may hit the top of the building is
Based on above information, answer the following questions.
The velocity given to the ball at point A so that it may hit the top of the building is
A.
$\sqrt{gR}$
B.
$\sqrt{2gR}$
C.
$\sqrt{4gR}$
D.
$\sqrt{6gR}$
Q728
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 17
A small ball is given some velocity at point A towards right so that it moves on the semicircular track and does not leave contact up to the highest point B. After leaving the highest point B, it falls at the top of a building of height R and width $x(x \ll 2R)$ . (All the surfaces are frictionless).
Based on above information, answer the following questions.
If the collision of ball with the building is elastic, then the angle with the horizontal at which the ball will rebound from the top of the building is
Based on above information, answer the following questions.
If the collision of ball with the building is elastic, then the angle with the horizontal at which the ball will rebound from the top of the building is
A.
$30^{\circ}$
B.
$45^{\circ}$
C.
$60^{\circ}$
D.
$90^{\circ}$
Q729
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 17
A small ball is given some velocity at point A towards right so that it moves on the semicircular track and does not leave contact up to the highest point B. After leaving the highest point B, it falls at the top of a building of height R and width $x(x \ll 2R)$ . (All the surfaces are frictionless).
Based on above information, answer the following questions.
The horizontal distance of the ball from the foot of building where the ball strikes the horizontal ground is
Based on above information, answer the following questions.
The horizontal distance of the ball from the foot of building where the ball strikes the horizontal ground is
A.
$\sqrt{2}R$
B.
$(1+\sqrt{2})R$
C.
$2(1+\sqrt{2})R$
D.
12R
Q730
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 18
A particle of mass m is suspended vertically from a point O by an inextensible massless string of length L. A vertical line AB is at a distance $\frac{3\sqrt{3}}{8}L$ from O as shown in figure. The object is given a horizontal velocity $u=\sqrt{\frac{7gL}{2}}$ . At some point, its motion ceases to be circular and eventually the object passes through the line AB. At the instant of crossing AB, its velocity is horizontal. Based on above information, answer the following questions.
The magnitude of the change in its velocity as it reaches a position, where the string is horizontal
The magnitude of the change in its velocity as it reaches a position, where the string is horizontal
A.
$\sqrt{\frac{3gL}{2}}$
B.
$\sqrt{\frac{5gL}{2}}$
C.
$\sqrt{3gL}$
D.
$\sqrt{5gL}$
Q731
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 18
A particle of mass m is suspended vertically from a point O by an inextensible massless string of length L. A vertical line AB is at a distance $\frac{3\sqrt{3}}{8}L$ from O as shown in figure. The object is given a horizontal velocity $u=\sqrt{\frac{7gL}{2}}$ . At some point, its motion ceases to be circular and eventually the object passes through the line AB. At the instant of crossing AB, its velocity is horizontal. Based on above information, answer the following questions.
When the particle's motion ceases to be circular, the angle made by the string with the horizontal is
When the particle's motion ceases to be circular, the angle made by the string with the horizontal is
A.
$30^{\circ}$
B.
$60^{\circ}$
C.
$37^{\circ}$
D.
$\cos^{-1}\left(\frac{2}{3}\right)$
Q732
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 18
A particle of mass m is suspended vertically from a point O by an inextensible massless string of length L. A vertical line AB is at a distance $\frac{3\sqrt{3}}{8}L$ from O as shown in figure. The object is given a horizontal velocity $u=\sqrt{\frac{7gL}{2}}$ . At some point, its motion ceases to be circular and eventually the object passes through the line AB. At the instant of crossing AB, its velocity is horizontal. Based on above information, answer the following questions.
When the particle leaves the circular path its velocity vector is
When the particle leaves the circular path its velocity vector is
A.
$\sqrt{\frac{gL}{8}}\left(-\hat{i}+\sqrt{3}\hat{j}\right)$
B.
$\sqrt{\frac{gL}{2}}\left(-\hat{i}+\sqrt{3}\hat{j}\right)$
C.
$\sqrt{\frac{gL}{2}}\left(-\sqrt{3}\hat{i}+\hat{j}\right)$
D.
$\sqrt{\frac{gL}{8}}\left(-\sqrt{3}\hat{i}+\hat{j}\right)$
Q733
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 19
The Work Energy Theorem for a particle says, that the total work done on a particle equals the change in its kinetic energy. It means, to change the K.E. of a particle, have to apply some force on it and the force must do work on it. On a system of particles, the work done by external and internal forces on the system is equal to the change in K.E. of the system. Based on above information, answer the following questions.
When you lift a stone from the floor and put it on top of the table slowly, there is no change in its K.E. This implies.
A.
You have not done any work on the stone because had you done any work, then according to $W - E$ theorem the K.E. of the stone would have changed.
B.
Work done by the force of gravity on the stone is also zero as that has also not lead to any change in K.E. of stone.
C.
Gravity has done a positive work and you have done a negative work. Therefore, the total work done on the stone equals zero and hence change in K.E. equals zero
D.
The net work done on the stone is zero and hence its change in K.E. is zero
Q734
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 19
The Work Energy Theorem for a particle says, that the total work done on a particle equals the change in its kinetic energy. It means, to change the K.E. of a particle, have to apply some force on it and the force must do work on it. On a system of particles, the work done by external and internal forces on the system is equal to the change in K.E. of the system. Based on above information, answer the following questions.
Suppose a heavy box is kept on a smooth inclined plane and is pushed up by a force $F$ parallel to the plane.
A.
The work done by the force F as well as work done by the force of gravity both depend on how fast was the box moving at A and B as the box goes from A to B
B.
The work done by the force of gravity alone depends on how fast was the box moving at $A$ and $B$ as the box goes from $A$ to $B$
C.
The work done by the force F alone will depend on how fast was the box moving at A and B as the box goes from A to B
D.
None of them depends on how fast the box was moving at $A$ and $B$ as the box goes from $A$ to $B$
Q735
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 19
The Work Energy Theorem for a particle says, that the total work done on a particle equals the change in its kinetic energy. It means, to change the K.E. of a particle, have to apply some force on it and the force must do work on it. On a system of particles, the work done by external and internal forces on the system is equal to the change in K.E. of the system. Based on above information, answer the following questions.
A balls is given a speed $V_{0}$ on a rough horizontal surface. The ball travels through a distance $l$ on the surface and stops.
A.
The work done by $K$ -friction can be found by W.E. theorem as $-\mu mgl = 0 - \frac{1}{2} mv_0^2$ and hence work done by kinetic friction is positive
B.
The work done by the kinetic friction can never be positive. Therefore, work done is either $-\mu mgl$ or $-\frac{1}{2} mv_0^2$ , only expressed in two different ways
C.
In this case the work done by kinetic friction is negative and it can be expressed by either, $-\mu mgl$ or $-\frac{1}{2} mv_0^2$
D.
The work done by kinetic friction is negative and is $\mu mgl$ and not $-\frac{1}{2}mv_{0}^{2}$ because $\frac{1}{2}mv_{0}^{2}$ is the change in K.E. and that is a result of work done by all the forces like gravity and normal reaction as well
Q736
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 20
A block of mass m is released on a track shown in figure. The track is smooth from A to D and rough from point D onwards. Based on above information, answer the following questions.
The normal force $N_{B}$ exerted by the track just after crossing point B is
The normal force $N_{B}$ exerted by the track just after crossing point B is
A.
2mg
B.
3mg
C.
4mg
D.
5mg
Q737
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 20
A block of mass m is released on a track shown in figure. The track is smooth from A to D and rough from point D onwards. Based on above information, answer the following questions.
The normal force $N_{C}$ at $C$ is
The normal force $N_{C}$ at $C$ is
A.
$2mg$
B.
$3mg$
C.
$4mg$
D.
$7mg$
Q738
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 20
A block of mass m is released on a track shown in figure. The track is smooth from A to D and rough from point D onwards. Based on above information, answer the following questions.
The distance travelled along the incline beyond point D where the block stops is
The distance travelled along the incline beyond point D where the block stops is
A.
$\frac{2R}{\sqrt{3}\mu_{k}+1}$
B.
$\frac{4R}{\mu_{k}+1}$
C.
$\frac{4R}{\sqrt{3}\mu_{k}+1}$
D.
$\frac{4R}{\sqrt{3}\mu_{k}+3}$
Q739
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 21
A force $F = 50 \, \text{N}$ is applied at one end of a string, the other end of which is tied to a block of mass $10 \, \text{kg}$ . The block is free to move on a frictionless horizontal surface. Take initial instant as $\theta = 30^\circ$ and final instant as $\theta = 37^\circ$ . For the time between these two instants, answer the following questions? Take $\sin(37^\circ) = \frac{3}{5}$ .
Net work done by the force $F$ on the block is
Net work done by the force $F$ on the block is
A.
$\frac{20}{3}\mathrm{J}$
B.
$\frac{50}{3}\mathrm{J}$
C.
$\frac{100}{3}\mathrm{J}$
D.
$75\mathrm{J}$
Q740
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 21
A force $F = 50 \, \text{N}$ is applied at one end of a string, the other end of which is tied to a block of mass $10 \, \text{kg}$ . The block is free to move on a frictionless horizontal surface. Take initial instant as $\theta = 30^\circ$ and final instant as $\theta = 37^\circ$ . For the time between these two instants, answer the following questions? Take $\sin(37^\circ) = \frac{3}{5}$ .
What is the final velocity of the block if initially it was at rest?
What is the final velocity of the block if initially it was at rest?
A.
$\sqrt{5} \, ms^{-1}$
B.
$\sqrt{\frac{25}{3}} \, ms^{-1}$
C.
$\sqrt{\frac{20}{3}} \, ms^{-1}$
D.
$5 \, ms^{-1}$
Q741
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 21
A force $F = 50 \, \text{N}$ is applied at one end of a string, the other end of which is tied to a block of mass $10 \, \text{kg}$ . The block is free to move on a frictionless horizontal surface. Take initial instant as $\theta = 30^\circ$ and final instant as $\theta = 37^\circ$ . For the time between these two instants, answer the following questions? Take $\sin(37^\circ) = \frac{3}{5}$ .
Find the ratio of initial acceleration to final acceleration of the block.
Find the ratio of initial acceleration to final acceleration of the block.
A.
$\frac{3\sqrt{5}}{8}$
B.
$\frac{5\sqrt{3}}{8}$
C.
$\frac{8\sqrt{3}}{5}$
D.
$\frac{8\sqrt{5}}{3}$
Q742
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 22
A particle moves in x-y plane under the action of a path dependent force $F = y\hat{i} + x\hat{j}$ . The work done by the force as it moves in x-y plane can be evaluated by solving the line integral of force i.e. $\int\vec{F}\cdot d\vec{l}$ , where $d\vec{l}=dx\hat{i}+dy\hat{j}$ . The position coordinates x and y may vary according to some constraint determined by the path followed by the particle. Based on the information shared, answer the following questions.
When the particle moves along a straight-line path from origin to the point $(a, a)$ , the work done by the force on the particle is
A.
$a^2$
B.
$2a$
C.
$\frac{a^2}{2}$
D.
Zero
Q743
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 22
A particle moves in x-y plane under the action of a path dependent force $F = y\hat{i} + x\hat{j}$ . The work done by the force as it moves in x-y plane can be evaluated by solving the line integral of force i.e. $\int\vec{F}\cdot d\vec{l}$ , where $d\vec{l}=dx\hat{i}+dy\hat{j}$ . The position coordinates x and y may vary according to some constraint determined by the path followed by the particle. Based on the information shared, answer the following questions.
When the particle moves from $(0, 0)$ to $(a, 0)$ and then from $(a, 0)$ to $(a, a)$ in straight line paths, then work done by the force on the particle is
A.
$a^2$
B.
$2a$
C.
$\frac{a^2}{2}$
D.
Zero
Q744
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MSQ
Comprehension Passage: Comprehension - 22
A particle moves in x-y plane under the action of a path dependent force $F = y\hat{i} + x\hat{j}$ . The work done by the force as it moves in x-y plane can be evaluated by solving the line integral of force i.e. $\int\vec{F}\cdot d\vec{l}$ , where $d\vec{l}=dx\hat{i}+dy\hat{j}$ . The position coordinates x and y may vary according to some constraint determined by the path followed by the particle. Based on the information shared, answer the following questions.
The infinitesimal work done (dW) by the given force when displaced through $d\vec{l}$ is given by
$d W = \vec {F} \cdot d \vec {l} = y d x + x d y$
From this, it can be inferred that the work done by the force
A.
depends only on initial and final values of $x$ and $y$
B.
depends on initial and final values and on the path followed
C.
is zero for any values of $x$ and $y$ .
D.
is zero when object returns to original position after following any path.
Q745
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Match the following Column-I with Column-II
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (q) B → (s) C → (r) D → (r)
Explanation:
A → (q)
$\mathrm{B} \rightarrow (\mathrm{s})$
$\mathrm{C} \rightarrow (\mathrm{r})$
(D) → (r)
Since $W_{total} = \Delta K$ and $W_{C} = -\Delta U$
and $W_{\mathrm{ext}} = \Delta U + \Delta K = (U_f + K_f) - (U_i + K_i)$
$\mathrm{B} \rightarrow (\mathrm{s})$
$\mathrm{C} \rightarrow (\mathrm{r})$
(D) → (r)
Since $W_{total} = \Delta K$ and $W_{C} = -\Delta U$
and $W_{\mathrm{ext}} = \Delta U + \Delta K = (U_f + K_f) - (U_i + K_i)$
Q746
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A particle is suspended from a string of length $l$ and is given a velocity $u = 3\sqrt{gl}$ at the bottom. If $v_{B}, T_{B}$ and $v_{C}, T_{C}$ be the velocity, tension at points $B$ and $C$ , then match the following. Column-I with Column-II
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (s) B → (r) C → (p) D → (q)
Explanation:
A $\rightarrow$ (s)
$\mathrm{B} \rightarrow (\mathrm{r})$
$C \rightarrow (p)$
$\mathrm{D} \rightarrow (\mathrm{q})$
$v _ {B} ^ {2} = u _ {A} ^ {2} - 2 g h _ {A B} = (9 g \ell) - (2 g \ell) = 7 g \ell$
$\Rightarrow v _ {B} = \sqrt {7 g \ell}$
$\Rightarrow \frac {v _ {B} ^ {2}}{g ^ {\ell}} = 7$
Further, $T_{B}=\frac{mv_{B}^{2}}{\ell}=7mg$
$\Rightarrow \frac {T _ {B}}{2 m g} = \frac {7}{2}$
Again, $v_{C}^{2} = v_{A}^{2} - 2gh_{AC} = (9g\ell) - 2g(2\ell) = 5g\ell$
$\Rightarrow v _ {C} = \sqrt {5 g \ell}$
$\Rightarrow \frac {v _ {C} ^ {2}}{g ^ {\ell}} = 5$
Further, $T_{C} + mg = \frac{mv_{C}^{2}}{\ell}$
$\Rightarrow T _ {C} = 4 m g$
$\Rightarrow \frac {T _ {C}}{m g} = 4$
$\mathrm{B} \rightarrow (\mathrm{r})$
$C \rightarrow (p)$
$\mathrm{D} \rightarrow (\mathrm{q})$
$v _ {B} ^ {2} = u _ {A} ^ {2} - 2 g h _ {A B} = (9 g \ell) - (2 g \ell) = 7 g \ell$
$\Rightarrow v _ {B} = \sqrt {7 g \ell}$
$\Rightarrow \frac {v _ {B} ^ {2}}{g ^ {\ell}} = 7$
Further, $T_{B}=\frac{mv_{B}^{2}}{\ell}=7mg$
$\Rightarrow \frac {T _ {B}}{2 m g} = \frac {7}{2}$
Again, $v_{C}^{2} = v_{A}^{2} - 2gh_{AC} = (9g\ell) - 2g(2\ell) = 5g\ell$
$\Rightarrow v _ {C} = \sqrt {5 g \ell}$
$\Rightarrow \frac {v _ {C} ^ {2}}{g ^ {\ell}} = 5$
Further, $T_{C} + mg = \frac{mv_{C}^{2}}{\ell}$
$\Rightarrow T _ {C} = 4 m g$
$\Rightarrow \frac {T _ {C}}{m g} = 4$
Q747
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A block of mass 1 kg is stationary with respect to a rough wedge making an angle of $30^{\circ}$ with horizontal as shown in figure. Starting from rest with an acceleration of $2 \, ms^{-2}$ the work done on the block at t = 4 s. Column-I with Column-II
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (t) B → (p) C → (s) D → (q)
Explanation:
$\mathrm{A}\rightarrow (\mathfrak{t})$
$\mathrm{B} \rightarrow (\mathrm{p})$
$\mathrm{C} \rightarrow (\mathrm{s})$
$\mathrm{D} \rightarrow (\mathrm{q})$
$\mathrm{At} t = 4 \mathrm{s},$
$v = a t = 8 \mathrm{ms} ^ {- 1}$
and $s=\frac{1}{2}at^{2}=16m$
$K E = \frac {1}{2} m v ^ {2} = 3 2 \mathrm{J}$
From Work-Energy Theorem, we have
Work done by all the forces = $\Delta KE = 32$ J
Work done by gravity
$= - m g h = - (1) (1 0) (1 6) = - 1 6 0 \mathrm{J}$
Writing equation of motion, we have, $\sum F_y = ma$
$N \cos 3 0 ^ {\circ} + f \sin 3 0 ^ {\circ} - 1 0 = m a = 2$
$\Rightarrow \sqrt {3} \mathrm{N} + f = 2 4\tag{... (1}$
$\sum F _ {x} = 0$
$\Rightarrow \quad N \sin 3 0 ^ {\circ} = f \cos 3 0 ^ {\circ}$
$\Rightarrow N = \sqrt {3} f$
... (2)
Solving equations (1) and (2), we have
$f = 6 \mathrm{N}$
and $N = \frac{18}{\sqrt{3}} = 6\sqrt{3}\mathrm{N}$
So, $W_{N} = (N\cos \theta)(s) = (6\sqrt{3})\left(\frac{\sqrt{3}}{2}\right)(16) = 144 \mathrm{~J}$
and $W_{f} = (f\sin \theta)(s) = (6)\left(\frac{1}{2}\right)(16) = 48 \, \mathrm{J}$
$\mathrm{B} \rightarrow (\mathrm{p})$
$\mathrm{C} \rightarrow (\mathrm{s})$
$\mathrm{D} \rightarrow (\mathrm{q})$
$\mathrm{At} t = 4 \mathrm{s},$
$v = a t = 8 \mathrm{ms} ^ {- 1}$
and $s=\frac{1}{2}at^{2}=16m$
$K E = \frac {1}{2} m v ^ {2} = 3 2 \mathrm{J}$
From Work-Energy Theorem, we have
Work done by all the forces = $\Delta KE = 32$ J
Work done by gravity
$= - m g h = - (1) (1 0) (1 6) = - 1 6 0 \mathrm{J}$
Writing equation of motion, we have, $\sum F_y = ma$
$N \cos 3 0 ^ {\circ} + f \sin 3 0 ^ {\circ} - 1 0 = m a = 2$
$\Rightarrow \sqrt {3} \mathrm{N} + f = 2 4\tag{... (1}$
$\sum F _ {x} = 0$
$\Rightarrow \quad N \sin 3 0 ^ {\circ} = f \cos 3 0 ^ {\circ}$
$\Rightarrow N = \sqrt {3} f$
... (2)
Solving equations (1) and (2), we have
$f = 6 \mathrm{N}$
and $N = \frac{18}{\sqrt{3}} = 6\sqrt{3}\mathrm{N}$
So, $W_{N} = (N\cos \theta)(s) = (6\sqrt{3})\left(\frac{\sqrt{3}}{2}\right)(16) = 144 \mathrm{~J}$
and $W_{f} = (f\sin \theta)(s) = (6)\left(\frac{1}{2}\right)(16) = 48 \, \mathrm{J}$
Q748
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Acceleration versus x and potential energy versus x graph of a particle of mass 1 kg moving along x-axis are shown in figure. Assume that at x = 0, the velocity of the particle is $4 \, ms^{-1}$ , match the following at x = 8 m. Column-I with Column-II
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (r) B → (q) C → (p) D → (t)
Explanation:
$\mathrm{A}\rightarrow (\mathrm{r})$
$\mathrm{B} \rightarrow (\mathrm{q})$
$C \rightarrow (p)$
$\mathrm{D} \rightarrow (\bar {t})$
$v _ {f} - v _ {i} = \mathrm{areaunder} a - x \mathrm{graph} = 1 2 \mathrm{ms} ^ {- 1}$
$\Rightarrow v _ {f} = 1 2 + 4 = 1 6 \mathrm{ms} ^ {- 1}$
$\Rightarrow \Delta K E = \frac {1}{2} m \left(v _ {f} ^ {2} - v _ {i} ^ {2}\right) = 1 2 0 \mathrm{J}$
Work done by all the forces = $\Delta KE = 120$ J
$K _ {f} = \frac {1}{2} m v _ {f} ^ {2} = 1 2 8 \mathrm{J}$
Work done by conservative forces = $U_{i} - U_{f} = 240$ J
Work done by external force i.e. $W_{ext}$ is given by
$W _ {\text {ext}} = \left( \begin{array}{c} \text {Total} \\ \text {work} \\ \text {done} \end{array} \right) - \left( \begin{array}{c} \text {Work done by} \\ \text {conservative} \\ \text {forces} \end{array} \right)$
$\Rightarrow W _ {\mathrm{ext}} = - 1 1 2 \mathrm{J}$
$\mathrm{B} \rightarrow (\mathrm{q})$
$C \rightarrow (p)$
$\mathrm{D} \rightarrow (\bar {t})$
$v _ {f} - v _ {i} = \mathrm{areaunder} a - x \mathrm{graph} = 1 2 \mathrm{ms} ^ {- 1}$
$\Rightarrow v _ {f} = 1 2 + 4 = 1 6 \mathrm{ms} ^ {- 1}$
$\Rightarrow \Delta K E = \frac {1}{2} m \left(v _ {f} ^ {2} - v _ {i} ^ {2}\right) = 1 2 0 \mathrm{J}$
Work done by all the forces = $\Delta KE = 120$ J
$K _ {f} = \frac {1}{2} m v _ {f} ^ {2} = 1 2 8 \mathrm{J}$
Work done by conservative forces = $U_{i} - U_{f} = 240$ J
Work done by external force i.e. $W_{ext}$ is given by
$W _ {\text {ext}} = \left( \begin{array}{c} \text {Total} \\ \text {work} \\ \text {done} \end{array} \right) - \left( \begin{array}{c} \text {Work done by} \\ \text {conservative} \\ \text {forces} \end{array} \right)$
$\Rightarrow W _ {\mathrm{ext}} = - 1 1 2 \mathrm{J}$
Q749
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A pendulum is released from point A as shown in figure. At any instant, the net force on the bob is making an angle $\theta$ with the string. Then match the following.
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (p, q) B → (t) C → (r) D → (s)
Explanation:
A $\rightarrow$ (p, q) B $\rightarrow$ (t) C $\rightarrow$ (r) D $\rightarrow$ (s)
Angle between net force and the string can never be obtuse. It is $90^{\circ}$ at A, $0^{\circ}$ at B and acute in between.
Angle between net force and the string can never be obtuse. It is $90^{\circ}$ at A, $0^{\circ}$ at B and acute in between.
Q750
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Match the following.
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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Correct Answer: A → (p, q, r) B → (p, q, r) C → (q, r) D → (p, r)
Explanation:
A → (p, q, r)
B → (p, q, r)
C → (q, r)
D → (p, r)
Conceptual
B → (p, q, r)
C → (q, r)
D → (p, r)
Conceptual